Taking a Fresh Look at the Physics of Radiometers.
NoEinstein
me a curious present that was shaped somewhat like a light bulb. It
had four white squares mounted on tiny cross arms, 90 degrees apart.
The back of each square—facing in the same orientation around a
central axis—was painted black. When exposed to light, the squares
would spin on a long pin. When exposed to direct sunlight, those
squares would spin very fast. I don’t think I knew that the device is
called a Radiometer. It lasted about a year before being broken. In
important ways that present from my Father enabled me to understand
the nature of light, and to understand the mechanism of gravity. It
was a very, very important ‘toy’!
More recently, I chanced to see that Edmund Optics still sells those
light bulb-shaped devices, like I had enjoyed watching as a kid.
Since I wasn’t looking to purchase another one, I didn’t read what the
things are called. But after remembering my childhood present, and
not recollecting in which direction the things spin, I used my
knowledge of light to figure that out. Fortunately, I still had an
Edmund Optics catalogue; found the ad; and learned that the devices
are called Radiometers. Wikipedia showed an animated photo indicating
the spin was away from the black sides of the squares—exactly what I
had just figured out. Several other curious qualities of Radiometers
were explained. The rationale for those qualities was mostly wrong.
So, I wish to share with you how my New Science does a better job.
Because of the radial symmetry of Radiometers, both sides of the four
squares will receive identical amounts of radiant energy. The status
quo physics attributes the rotation direction—away from the black—to
the fact that the black sides become hotter; heat up the dilute argon
gas inside the bulb; and then the thermal motion of the gas imparts
more ‘push’ on the black sides. However, since thermal motion of the
Argon would be random, such should push on the white sides of the
squares, too…
Enter Albert Einstein, the moron. He ‘supposed’ that there is a
portion of the squares near the edges that can receive ‘thrust’ from
the argon gas that doesn’t have to ‘hit’ the white sides of the
squares. That supposition must have been considered to be brilliant,
because of the status… of the person postulating such. But Einstein—
as he was so wont to do—ignored Newton’s 3rd Law of Motion by having
there be a THRUST, without there being “an equal and opposite
reaction”! That would be akin to having Tom Sawyer, on a raft, BLOW
on the sail to make the raft move. For over a century, the latter has
been the accepted explanation for the rotation direction of
Radiometers, and such is patently WRONG!
I’ve just learned, and provisionally accept as true, that: Radiometers
won’t rotate at all in a perfect vacuum; rotate backwards when exposed
to cold; and rotate best with dilute argon gas. I’ll explain the
actual reasons for the observed qualities of Radiometers:
The radiant energy hitting the white and the black sides of the
squares is identical. The white sides, immediately, re emit light
that has a spectrum similar to the source. The black sides re emit
light of equal TOTAL energy, but in the infrared portion of the
spectrum. Photons have ZERO mass, and exert zero force upon
‘striking’ a reflecting surface. The thrust that causes Radiometers
to rotate away from the black is caused, mainly, by the higher ether
pressure on the black sides, very similar to the explanation given at
the first part of my post:
“There is no "pull" of gravity, only the PUSH of flowing ether!”
http://groups.google.com/group/sci.physics/browse_thread/thread/a8c26d2eb535ab8/efdbea7b0272072f?hl=en
The above necessitates that the ether pressure on the white sides be
less. Because of the very small mass of the four squares
(insufficient to overcome the device’s friction by gravity alone), it
is the argon atoms, within the glass bulb, moving toward the black
sides that accounts for most of the spinning. The argon isn’t moving
because of the heat, but because the argon is caught in the outward
flowing ether induced by the higher energy photons being emitted from
WHITE squares.
There is ether throughout the bulb. The sunlight coming in imparts
little or no thrust to the ether (toward the squares). But, because
of the concentration of ether within the matter of the squares, the re
emitted light carries some of that ether outward. The lower-energy
(per train) infrared light that’s emitted from the black sides carries
ether outward, too; and much more efficiently than from the white
squares, because of the greater number of photon trains (to have the
same total energy as the ‘white’ light). And because some of the
light reflected from the white squares carries ether toward the black
squares, the latter ether gets banked against the black sides of the
squares. Partially, that‘s because the heavy argon gas atoms being
propelled by the white light help to keep the ether pressure on the
black sides of the squares high. If the devices were totally
frictionless, the rotation would occur in the identical direction
without that gas being there.
The depleted ether from within the squares gets replenished by the
ether that freely flows through the glass bulb. Newton’s 3rd Law
isn’t violated in this process, because the ether carried outward
(beyond the glass bulb) is conserved and contributes to the pressure
that causes the squares to rotate in the light or heat.
— NoEinstein —
Timo Nieminen
>
> I’ve just learned, and provisionally accept as true, that: Radiometers
> won’t rotate at all in a perfect vacuum;
True enough, but misleading, since there is still a measurable force.
Only the friction of the bearings stops it from rotating.
> If the devices were totally
> frictionless, the rotation would occur in the identical direction
> without that gas being there.
This isn't true. The force on the vanes has been measured in vacuum,
and the force is in the opposite direction to the usual Crookes
radiometer thermal force. If not for friction, the radiometer in
vacuum would rotate "backwards".
See http://en.wikipedia.org/wiki/Nichols_radiometer
This reverse force due to radiation pressure was measured in 1901.
(Published in 1901, anyway. I think Nichols and Hull did their
measurement in 1901, but Lebedev did his in1899, but didn't publish in
a journal until 1901.)
> Photons have ZERO mass, and exert zero force upon
> ‘striking’ a reflecting surface.
Non-zero force. This has been measured. Microscopic objects can be
easily pushed around with this force. Macroscopic objects have been
levitated against gravity. It's more common to use the force due to
refraction (which is also non-zero), since then you don't cook the
object being pushed, but reflection works too. (Also absorption.)
Sam Wormley
> Photons have ZERO mass, and exert zero force upon
> ‘striking’ a reflecting surface.
Photons have zero REST mass, but have momentum of
p = hν/c = h/λ
A perfect reflector's momentum would increase by
2p = 2h/λ
for every photon. Ever heard of a solar sail?
Uncle Al
>
> In keeping with my early childhood interest in science, my Father gave
> me a curious present that was shaped somewhat like a light bulb.
The solution is on the Web, jackass, including pressure dependence of
rotation sense. Thrust accrues from the *edges* not the faces.
idiot
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz4.htm
Darwin123
> In keeping with my early childhood interest in science, my Father gave
> me a curious present that was shaped somewhat like a light bulb. It
> had four white squares mounted on tiny cross arms, 90 degrees apart.
> The back of each square—facing in the same orientation around a
> central axis—was painted black. When exposed to light, the squares
> would spin on a long pin. When exposed to direct sunlight, those
> squares would spin very fast.
about it.
The radiometer that was commercially sold to the public did not
work by radiation pressure. The radiometer was filled with air.
Because it was filled with air, the pressure caused by air currents
was stronger than the radiation pressure. Therefore, propellers moved
in the opposite direction of what would be predicted from radiation
pressure alone.
> I don’t think I knew that the device is
> called a Radiometer. It lasted about a year before being broken. In
> important ways that present from my Father enabled me to understand
> the nature of light, and to understand the mechanism of gravity.
to check. I remember that before they had the Internet, they had the
library.
> It
> was a very, very important ‘toy’!
>
> More recently, I chanced to see that Edmund Optics still sells those
> light bulb-shaped devices, like I had enjoyed watching as a kid.
> Since I wasn’t looking to purchase another one, I didn’t read what the
> things are called. But after remembering my childhood present, and
> not recollecting in which direction the things spin, I used my
> knowledge of light to figure that out. Fortunately, I still had an
> Edmund Optics catalogue; found the ad; and learned that the devices
> are called Radiometers. Wikipedia showed an animated photo indicating
> the spin was away from the black sides of the squares—exactly what I
> had just figured out.
the black sides. Reflection causes a greater momentum transfer than
absorption.
What happened is that the black side had a higher temperature due
to the absorption of light. The heat energy on the black side conducts
into the gas on the black side. The heat causes the gas to expand,
causing an increase in pressure. This increase in pressure is much
greater than the pressure due to radiation pressure. The increased
pressure on the black side causes the propeller to turn away from the
black sides.
> Several other curious qualities of Radiometers
> were explained. The rationale for those qualities was mostly wrong.
Radiation pressure can be measured if the bulb was evacuated. If
it were totally evacuated, the spin would occur toward the black
sides.
> So, I wish to share with you how my New Science does a better job.
>
> Because of the radial symmetry of Radiometers, both sides of the four
> squares will receive identical amounts of radiant energy.
The black side will absorb more light energy, causing the black
side to heat up more than the white side. The white side will absorb
more light momentum, causing more radiation pressure on the white side
than the black side.
> The status
> quo physics attributes the rotation direction—away from the black—to
> the fact that the black sides become hotter; heat up the dilute argon
> gas inside the bulb; and then the thermal motion of the gas imparts
> more ‘push’ on the black sides. However, since thermal motion of the
> Argon would be random, such should push on the white sides of the
> squares, too…
the heat energy will go into random directions of molecules, like you
said. However, the molecules on the black side are going to move on
the average faster than on the white side.
>
> Enter Albert Einstein, the moron. He ‘supposed’ that there is a
> portion of the squares near the edges that can receive ‘thrust’ from
> the argon gas that doesn’t have to ‘hit’ the white sides of the
> squares.
radiometer. Can you provide a citation? I thought the principle of the
radiometer, with air, was well known.
>That supposition must have been considered to be brilliant,
> because of the status… of the person postulating such. But Einstein—
> as he was so wont to do—ignored Newton’s 3rd Law of Motion by having
> there be a THRUST, without there being “an equal and opposite
> reaction”! That would be akin to having Tom Sawyer, on a raft, BLOW
> on the sail to make the raft move.
wondered how the wind blows despite the Third Law of Newton?
There is an equal and opposite reaction. A packet of air on the
black side is expanding and pushes against the black panel. The black
panel moves away. However, the packet of hot air moves in the opposite
direction. Hence, the force on the panel is equal and opposite the
force on the packet of air.
> For over a century, the latter has
> been the accepted explanation for the rotation direction of
> Radiometers, and such is patently WRONG!
You are wrong.
>
> I’ve just learned, and provisionally accept as true, that: Radiometers
> won’t rotate at all in a perfect vacuum;
That isn't precisely true. If the propeller is extremely well
lubricated, so the friction is negligible, it was turn in the opposite
direction. However, getting a junction to be nearly frictionless is
very hard. If you would have evacuated that toy radiometer, it would
probably have stood still. The friction is too high.
>rotate backwards when exposed
> to cold;
?
>and rotate best with dilute argon gas.
I'll bet. You need those currents in the gas. Under high pressure, the
viscosity of the gas is too high and its conductivity too high to
maintain a large temperature difference. So very low pressure gas is
best.
> I’ll explain the
> actual reasons for the observed qualities of Radiometers:
And the name of NoEinstein will be immortalized forever!-)
>
> The radiant energy hitting the white and the black sides of the
> squares is identical.
However, the radiant energy that is absorbed is different.
>The white sides, immediately, re emit light
> that has a spectrum similar to the source. The black sides re emit
> light of equal TOTAL energy, but in the infrared portion of the
> spectrum. Photons have ZERO mass, and exert zero force upon
> ‘striking’ a reflecting surface. The thrust that causes Radiometers
> to rotate away from the black is caused, mainly, by the higher ether
> pressure on the black sides, very similar to the explanation given at
> the first part of my post:
Notice that you explanation doesn't explain why there is no
motion, or opposite motion, in a vacuum. You were the one who said
that dilute Argon gas was necessary. What ahppened? You forgot?
>
> The above necessitates that the ether pressure on the white sides be
> less.
I was wondering when you got to that.
> Because of the very small mass of the four squares
> (insufficient to overcome the device’s friction by gravity alone), it
> is the argon atoms, within the glass bulb, moving toward the black
> sides that accounts for most of the spinning.
> The argon isn’t moving
> because of the heat,
BS
> but because the argon is caught in the outward
> flowing ether induced by the higher energy photons being emitted >from WHITE squares.
So why would high energy photons carry away ether? In fact, how can
there be both ether and photons? Light acts like a wave. It shows
interference. That is the entire basis of the ether theory. Waves
can't act like particles.
It looks to me like the ether is superfluous in this case. If the
Argon gas is so important, why not just say that the Argon gas is
moving?
>
> There is ether throughout the bulb. The sunlight coming in imparts
> little or no thrust to the ether (toward the squares).
This is correct. Make that little thrust. The sunlight definitely
produces a very small thrust. Almost negligible in the radiometer your
father gave you.
> But, because
> of the concentration of ether within the matter of the squares, the re
> emitted light carries some of that ether outward.
Have you been sniffing that ether again?
> The lower-energy
> (per train) infrared light that’s emitted from the black sides carries
> ether outward, too; and much more efficiently than from the white
> squares, because of the greater number of photon trains (to have the
> same total energy as the ‘white’ light).
Photon trains?
> And because some of the
> light reflected from the white squares carries ether toward the black
> squares, the latter ether gets banked against the black sides of the
> squares.
So the light curves around to hit the black sides? It sounds like
the photons are shaped like boomarangs.
> Partially, that‘s because the heavy argon gas atoms being
> propelled by the white light help to keep the ether pressure on the
> black sides of the squares high.
Yep. The photons are shaped like boomarangs. The Argon is causing
them to curve backward and hit the black side. That is my
interpretation of your model. Show me how I am wrong.
> If the devices were totally
> frictionless, the rotation would occur in the identical direction
> without that gas being there.
??????????????
>
> The depleted ether from within the squares gets replenished by the
> ether that freely flows through the glass bulb.
It is the Argon! Not the aether!
> Newton’s 3rd Law
> isn’t violated in this process, because the ether carried outward
> (beyond the glass bulb) is conserved and contributes to the pressure
> that causes the squares to rotate in the light or heat.
So the momentum is carried outside the light bulb? Show me where
this pressure going outside the light bulb is manifested!
I refuse to call you an idiot. That would be an insult to all true
idiots.
Phony! Bigot! Intentional ignoramus!
Androcles
"Darwin123" <drose...@yahoo.com> wrote in message
news:7962468c-d8e9-41de...@v18g2000vbc.googlegroups.com...
On May 30, 10:31 am, NoEinstein <noeinst...@bellsouth.net> wrote:
> In keeping with my early childhood interest in science, my Father gave
> me a curious present that was shaped somewhat like a light bulb. It
> had four white squares mounted on tiny cross arms, 90 degrees apart.
> central axis�was painted black. When exposed to light, the squares
> would spin on a long pin. When exposed to direct sunlight, those
> squares would spin very fast.
I remember that toy. There is something you apparently don't know
about it.
The radiometer that was commercially sold to the public did not
work by radiation pressure. The radiometer was filled with air.
Because it was filled with air, the pressure caused by air currents
was stronger than the radiation pressure. Therefore, propellers moved
in the opposite direction of what would be predicted from radiation
pressure alone.
> called a Radiometer. It lasted about a year before being broken. In
> important ways that present from my Father enabled me to understand
> the nature of light, and to understand the mechanism of gravity.
You should have gone to the library and looked up. It always pays
to check. I remember that before they had the Internet, they had the
library.
> It
>
> More recently, I chanced to see that Edmund Optics still sells those
> light bulb-shaped devices, like I had enjoyed watching as a kid.
> things are called. But after remembering my childhood present, and
> not recollecting in which direction the things spin, I used my
> knowledge of light to figure that out. Fortunately, I still had an
> Edmund Optics catalogue; found the ad; and learned that the devices
> are called Radiometers. Wikipedia showed an animated photo indicating
> the fact that the black sides become hotter; heat up the dilute argon
> gas inside the bulb; and then the thermal motion of the gas imparts
> Argon would be random, such should push on the white sides of the
No, because the white side is cooler than the black side. Most of
the heat energy will go into random directions of molecules, like you
said. However, the molecules on the black side are going to move on
the average faster than on the white side.
>
> portion of the squares near the edges that can receive �thrust� from
> the argon gas that doesn�t have to �hit� the white sides of the
> squares.
I didn't know Einstein made any comment on this type of
radiometer. Can you provide a citation? I thought the principle of the
radiometer, with air, was well known.
>That supposition must have been considered to be brilliant,
> as he was so wont to do�ignored Newton�s 3rd Law of Motion by having
> there be a THRUST, without there being �an equal and opposite
> reaction�! That would be akin to having Tom Sawyer, on a raft, BLOW
> on the sail to make the raft move.
More like the motion of wind at a cold front. Have you ever
wondered how the wind blows despite the Third Law of Newton?
There is an equal and opposite reaction. A packet of air on the
black side is expanding and pushes against the black panel. The black
panel moves away. However, the packet of hot air moves in the opposite
direction. Hence, the force on the panel is equal and opposite the
force on the packet of air.
> For over a century, the latter has
> been the accepted explanation for the rotation direction of
> Radiometers, and such is patently WRONG!
You are wrong.
>
> I�ve just learned, and provisionally accept as true, that: Radiometers
> won�t rotate at all in a perfect vacuum;
That isn't precisely true. If the propeller is extremely well
lubricated, so the friction is negligible, it was turn in the opposite
direction. However, getting a junction to be nearly frictionless is
very hard. If you would have evacuated that toy radiometer, it would
probably have stood still. The friction is too high.
>rotate backwards when exposed
> to cold;
?
>and rotate best with dilute argon gas.
I'll bet. You need those currents in the gas. Under high pressure, the
viscosity of the gas is too high and its conductivity too high to
maintain a large temperature difference. So very low pressure gas is
best.
> I�ll explain the
> actual reasons for the observed qualities of Radiometers:
And the name of NoEinstein will be immortalized forever!-)
>
> The radiant energy hitting the white and the black sides of the
> squares is identical.
However, the radiant energy that is absorbed is different.
>The white sides, immediately, re emit light
> that has a spectrum similar to the source. The black sides re emit
> light of equal TOTAL energy, but in the infrared portion of the
> spectrum. Photons have ZERO mass, and exert zero force upon
> �striking� a reflecting surface. The thrust that causes Radiometers
> to rotate away from the black is caused, mainly, by the higher ether
> pressure on the black sides, very similar to the explanation given at
> the first part of my post:
Notice that you explanation doesn't explain why there is no
motion, or opposite motion, in a vacuum. You were the one who said
that dilute Argon gas was necessary. What ahppened? You forgot?
>
> The above necessitates that the ether pressure on the white sides be
> less.
I was wondering when you got to that.
> Because of the very small mass of the four squares
> (insufficient to overcome the device�s friction by gravity alone), it
> is the argon atoms, within the glass bulb, moving toward the black
> sides that accounts for most of the spinning.
> The argon isn�t moving
> because of the heat,
BS
> but because the argon is caught in the outward
> flowing ether induced by the higher energy photons being emitted >from
> WHITE squares.
So why would high energy photons carry away ether? In fact, how can
there be both ether and photons? Light acts like a wave. It shows
interference. That is the entire basis of the ether theory. Waves
can't act like particles.
It looks to me like the ether is superfluous in this case. If the
Argon gas is so important, why not just say that the Argon gas is
moving?
>
> There is ether throughout the bulb. The sunlight coming in imparts
> little or no thrust to the ether (toward the squares).
This is correct. Make that little thrust. The sunlight definitely
produces a very small thrust. Almost negligible in the radiometer your
father gave you.
> But, because
> of the concentration of ether within the matter of the squares, the re
> emitted light carries some of that ether outward.
Have you been sniffing that ether again?
> The lower-energy
> (per train) infrared light that�s emitted from the black sides carries
> ether outward, too; and much more efficiently than from the white
> squares, because of the greater number of photon trains (to have the
> same total energy as the �white� light).
Photon trains?
> And because some of the
> light reflected from the white squares carries ether toward the black
> squares, the latter ether gets banked against the black sides of the
> squares.
So the light curves around to hit the black sides? It sounds like
the photons are shaped like boomarangs.
> Partially, that�s because the heavy argon gas atoms being
> propelled by the white light help to keep the ether pressure on the
> black sides of the squares high.
Yep. The photons are shaped like boomarangs. The Argon is causing
them to curve backward and hit the black side. That is my
interpretation of your model. Show me how I am wrong.
> If the devices were totally
> frictionless, the rotation would occur in the identical direction
> without that gas being there.
??????????????
>
> The depleted ether from within the squares gets replenished by the
> ether that freely flows through the glass bulb.
It is the Argon! Not the aether!
> Newton�s 3rd Law
> isn�t violated in this process, because the ether carried outward
> (beyond the glass bulb) is conserved and contributes to the pressure
> that causes the squares to rotate in the light or heat.
So the momentum is carried outside the light bulb? Show me where
this pressure going outside the light bulb is manifested!
I refuse to call you an idiot. That would be an insult to all true
idiots.
Phony! Bigot! Intentional ignoramus!
===========================================
How come you can be rational while dealing with idiots like "NoEinstein"
and yet are totally irrational over Einstein's crap?
Phony! Bigot! Intentional ignoramus!
alie...@gmail.com
(snip to the crash)
> There is ether throughout the bulb. The sunlight coming in imparts
> little or no thrust to the ether (toward the squares). But, because
> of the concentration of ether within the matter of the squares, the re
> emitted light carries some of that ether outward.
The light emitted from the squares carries ether with it, but the
sunlight coming into the bulb does *not*?
Inconsistent.
Mark L. Fergerson
NoEinstein
>
Dear Timo: I like that you have had a broad exposure to the world of
physics. My New Physics is different in that it is based almost
solely on analysis and on reason. To avoid being ‘corrupted’ by the
status quo, I relish the observations of valid experiments—while
always being open minded to the possibility of errors. I avoid
“automatically” accepting the explanations, by supposed authorities,
for the observed phenomena.
White and black squares are two competing “gravity” experiments
combined into one. In the Crookes Radiometer, the black squares
exhibit more repulsion from the light (or heat) source than the white
squares. Reverse rotation has been observed (by others) to occur if
the glass is made to be cooler than the vanes, themselves. You say…
“The force on the vanes has been measured in vacuum, and the force is
in the opposite direction to the usual Crookes radiometer thermal
force. If not for friction, the radiometer in vacuum would rotate
"backwards". That observation may or may not be a true analogy to the
Crookes. I don’t make it a point to shoehorn anyone’s “observations”
unless and until I know most of the particulars.
*** I invite you to reply with a concise PARAPHRASE of how that “in
vacuum” experiment was done. (Note: I do not read links to the words
of others.) The thermal qualities of the vacuum container must be
considered, as well as the thermal isolation of the white paint from
the black paint, if present. Since there was no rotation, how was…
“the force” measured?
Like I have said, conclusively, massless photons, alone, exert no
force on objects. What is actually happening to move small objects is
that photons create a gravity effect, as explained in:
There is no "pull" of gravity, only the PUSH of flowing ether!
The latter involves having the varying ether flow and density push the
object in proportion to the object’s cross section that is in the
photon stream. Dust particles adjacent to laser beams can be seen to
move in the direction of the beam. But that is due to the air
molecules, and the ether being moved, together. The dust is pushed by
the air gases and by the flowing ether, not by the photons.
That Wikipedia article on Radiometers mentioned that there is an
induced gas flow through porous ceramic plates that is toward the side
that is heated. [ Note: That is consistent with the ether flow
direction predicted by my New Science. ] The rather iffy porosity of
the ‘edges’ of the squares in the Crookes Radiometer has, for over a
century, been considered to be the primary source for the thrust. The
errant rationale has been: The edges of the black squares heat, and
then shoot-out, the argon atoms, causing the observed rotation. The
latter concocted ‘science‘, combined with Einstein’s heated gas
nonsense, supposedly accounts for 100% of the observed rotation of the
vanes.
Photons are concentrations of energy which, in high enough
concentrations, can burn through steel. Those photons don’t “force”
through the steel. You could say: They “energy” through the steel!
Timo, I’ve observed over the past month that you have, occasionally,
been adversarial regarding aspects of my New Science. To the extent
that you bring up valid points which I can explain to the many
readers, I welcome your comments. But I don’t seek to have a time
consuming one-on-one conversation with you just for your edification.
Though this reply is long, don’t take that to be an invitation that
you have been selected as the spokes-person for the status quo.
Because of my obvious huge contributions to science, you should ask
questions, not sit in judgment. You are welcomed to make your own
‘+new post(s)’ to pontificate your science if you differ with me.
Lastly, please TOP post, and limit yourself to about two paragraphs.
I really don’t need to hear what you think about every little thing
that I’ve ever said. No more… PDs are wanted, here. Thanks! —
NoEinstein —
>
> On May 31, 12:31 am, NoEinstein <noeinst...@bellsouth.net> wrote:
>
> > I’ve just learned, and provisionally accept as true, that: Radiometers
> > won’t rotate at all in a perfect vacuum;
>
> True enough, but misleading, since there is still a measurable force.
> Only the friction of the bearings stops it from rotating.
>
> > If the devices were totally
> > frictionless, the rotation would occur in the identical direction
> > without that gas being there.
>
> This isn't true. The force on the vanes has been measured in vacuum,
> and the force is in the opposite direction to the usual Crookes
> radiometer thermal force. If not for friction, the radiometer in
> vacuum would rotate "backwards".
>
> Seehttp://en.wikipedia.org/wiki/Nichols_radiometer
NoEinstein
>
Dear Sam: You are a former teacher, and a recognized proponent of the
Status Quo. By copying the equations of others, you are showing
yourself to be a pedant. Mostly, you don't understand a thing about
true science. If you would like to expound your "thorough
understanding" of the equations of others, please make your own '+new
post' entitled, say, "Why NoEinstein is wrong about 'this or
that'..." I'm sure you will find… FEW readers. — NoEinstein —
NoEinstein
>
Uncle Al, the Plumber, likes to snip crap. I encourage him to make
his own '+new post' and see how many people read what he has to say.
— NE —
>
> NoEinstein wrote:
>
> > In keeping with my early childhood interest in science, my Father gave
> > me a curious present that was shaped somewhat like a light bulb.
>
> [snipc rap]
>
> The solution is on the Web, jackass, including pressure dependence of
> rotation sense. Thrust accrues from the *edges* not the faces.
>
> idiot
>
> --
NoEinstein
>
Dear Darwin123: If verbiage determined science truths, you would
win. Apparently, you aren't a very good reader, because I don't
disagree with the oft-observed directions of rotation. I just state
(after more than a century) the CORRECT reasons for the rotations. I
invite you to give the names and links to ‘+new posts’ that you have
made. Some of mine are below. — NoEinstein —
Where Angels Fear to Fall
http://groups.google.com/group/sci.physics/browse_frm/thread/8152ef3e...
Last Nails in Einstein's Coffin
http://groups.google.com/group/sci.physics.relativity/browse_frm/thre...
Pop Quiz for Science Buffs!
http://groups.google.com/group/sci.physics/browse_frm/thread/43f6f316...
An Einstein Disproof for Dummies
http://groups.google.com/group/sci.physics/browse_thread/thread/f7a63...
Another look at Einstein
http://groups.google.com/group/sci.physics/browse_frm/thread/41670721...
Three Problems for Math and Science
http://groups.google.com/group/sci.physics/browse_thread/thread/bb07f30aab43c49c?hl=en
Matter from Thin Air
http://groups.google.com/group/sci.physics/browse_thread/thread/ee4fe3946dfc0c31/1f1872476bc6ca90?hl=en#1f1872476bc6ca90
Curing Einstein’s Disease
http://groups.google.com/group/sci.physics/browse_thread/thread/4ff9e866e0d87562/f5f848ad8aba67da?hl=en#f5f848ad8aba67da
Replicating NoEinstein’s Invalidation of M-M (at sci.math)
http://groups.google.com/group/sci.math/browse_thread/thread/d9f9852639d5d9e1/dcb2a1511b7b2603?hl=en&lnk=st&q=#dcb2a1511b7b2603
Cleaning Away Einstein’s Mishmash
http://groups.google.com/group/sci.physics/browse_thread/thread/5d847a9cb50de7f0/739aef0aee462d26?hl=en&lnk=st&q=#739aef0aee462d26
Dropping Einstein Like a Stone
http://groups.google.com/group/sci.physics/browse_thread/thread/989e16c59967db2b?hl=en#
Plotting the Curves of Coriolis, Einstein, and NoEinstein (is
Copyrighted.)
http://groups.google.com/group/sci.physics/browse_thread/thread/713f8a62f17f8274?hl=en#
Are Jews Destroying Objectivity in Science?
http://groups.google.com/group/sci.physics/browse_thread/thread/d4cbe8182fae7008/b93ba4268d0f33e0?hl=en&lnk=st&q=#b93ba4268d0f33e0
The Gravity of Masses Doesn’t Bend Light.
http://groups.google.com/group/sci.physics/browse_thread/thread/efb99ab95e498420/cd29d832240f404d?hl=en#cd29d832240f404d
KE = 1/2mv^2 is disproved in new falling object impact test.
http://groups.google.com/group/sci.physics/browse_thread/thread/51a85ff75de414c2?hl=en&q=
Light rays don’t travel on ballistic curves.
http://groups.google.com/group/sci.physics/browse_thread/thread/c3d7a4e9937ab73e/c7d941d2b2e80002?hl=en#c7d941d2b2e80002
A BLACK HOLE MYTH GETS BUSTED:
http://groups.google.com/group/sci.physics/browse_thread/thread/a170212ca4c36218?hl=en#
SR Ignored the Significance of the = Sign
http://groups.google.com/group/sci.physics/browse_thread/thread/562477d4848ea45a/92bccf5550412817?hl=en#92bccf5550412817
Eleaticus confirms that SR has been destroyed!
http://groups.google.com/group/sci.math/browse_thread/thread/c3cdedf38e749bfd/0451e93207ee475a?hl=en#0451e93207ee475a
NoEinstein Finds Yet Another Reason Why SR Bites-the-Dust!
http://groups.google.com/group/sci.physics/browse_thread/thread/a3a12d4d732435f2/737ef57bf0ed3849?hl=en#737ef57bf0ed3849
NoEinstein Gives the History & Rationale for Disproving Einstein
http://groups.google.com/group/sci.physics/browse_thread/thread/81046d3d070cffe4/f1d7fbe994f569f7?hl=en#f1d7fbe994f569f7
There is no "pull" of gravity, only the PUSH of flowing ether!
http://groups.google.com/group/sci.physics/browse_thread/thread/a8c26d2eb535ab8/efdbea7b0272072f?hl=en&
PD has questions about science. Can any of you help?
http://groups.google.com/group/sci.physics/browse_thread/thread/4a2edad1c5c0a4c1/2d0e50d773ced1ad?hl=en&
> ...
>
> read more »- Hide quoted text -
>
> - Show quoted text -
NoEinstein
>
Androcles: Argon has mass. Thus argon can be moved by, and can move
the ether. The 'problem' with Einstein's heat thrust on the argon is
that those vanes, in sweeping around so fast, will cause a more
perfect vacuum on the black sides. So there won't be any argon atoms
to be pushed faster by the heat of the black squares. Einstein
countered by saying that the heat traveled to the edges, where the
faster molecules of the black paint can... thrust the argon. But,
because there are so few argon atoms, many of those which have thrust
imparted (sic) would strike the opposing white squares, stopping the
rotation. Einstein's non-logic requires 100% of the argon to fly
outward enough to miss hitting the white squares. His theory would be
like a paddle-wheel machine, with the argon getting propelled over-and-
over. But anyone understanding fluid dynamics knows that won't
happen; the argon within the rotation radius of the squares would drag
things to a stop. — NoEinstein —
>
> "Darwin123" <drosen0...@yahoo.com> wrote in message
>
> news:7962468c-d8e9-41de...@v18g2000vbc.googlegroups.com...
> On May 30, 10:31 am, NoEinstein <noeinst...@bellsouth.net> wrote:> In keeping with my early childhood interest in science, my Father gave
> > me a curious present that was shaped somewhat like a light bulb. It
> > had four white squares mounted on tiny cross arms, 90 degrees apart.
> > central axis—was painted black. When exposed to light, the squares
> > would spin on a long pin. When exposed to direct sunlight, those
> > squares would spin very fast.
>
> I remember that toy. There is something you apparently don't know
> about it.
> The radiometer that was commercially sold to the public did not
> work by radiation pressure. The radiometer was filled with air.
> Because it was filled with air, the pressure caused by air currents
> was stronger than the radiation pressure. Therefore, propellers moved
> in the opposite direction of what would be predicted from radiation
> > called a Radiometer. It lasted about a year before being broken. In
> > important ways that present from my Father enabled me to understand
> > the nature of light, and to understand the mechanism of gravity.
>
> You should have gone to the library and looked up. It always pays
> to check. I remember that before they had the Internet, they had the
> library.> It
>
> > More recently, I chanced to see that Edmund Optics still sells those
> > light bulb-shaped devices, like I had enjoyed watching as a kid.
> > things are called. But after remembering my childhood present, and
> > not recollecting in which direction the things spin, I used my
> > knowledge of light to figure that out. Fortunately, I still had an
> > Edmund Optics catalogue; found the ad; and learned that the devices
> > are called Radiometers. Wikipedia showed an animated photo indicating
> > had just figured out.
>
> If it had been from radiation pressure, the spin would be toward
> the black sides. Reflection causes a greater momentum transfer than
> absorption.
> What happened is that the black side had a higher temperature due
> to the absorption of light. The heat energy on the black side conducts
> into the gas on the black side. The heat causes the gas to expand,
> causing an increase in pressure. This increase in pressure is much
> greater than the pressure due to radiation pressure. The increased
> pressure on the black side causes the propeller to turn away from the
> black sides.> Several other curious qualities of Radiometers
> > were explained. The rationale for those qualities was mostly wrong.
>
> No, you are wrong. The booklet is right.
> Radiation pressure can be measured if the bulb was evacuated. If
> it were totally evacuated, the spin would occur toward the black
> sides.> So, I wish to share with you how my New Science does a better job.
>
> > Because of the radial symmetry of Radiometers, both sides of the four
> > squares will receive identical amounts of radiant energy.
>
> Receive means what?
> The black side will absorb more light energy, causing the black
> side to heat up more than the white side. The white side will absorb
> more light momentum, causing more radiation pressure on the white side
> than the black side.> The status
> > the fact that the black sides become hotter; heat up the dilute argon
> > gas inside the bulb; and then the thermal motion of the gas imparts
> > Argon would be random, such should push on the white sides of the
>
> No, because the white side is cooler than the black side. Most of
> the heat energy will go into random directions of molecules, like you
> said. However, the molecules on the black side are going to move on
> the average faster than on the white side.
>
> > portion of the squares near the edges that can receive ‘thrust’ from
> > squares.
>
> I didn't know Einstein made any comment on this type of
> radiometer. Can you provide a citation? I thought the principle of the
> radiometer, with air, was well known.>That supposition must have been considered to be brilliant,
> > as he was so wont to do—ignored Newton’s 3rd Law of Motion by having
> > there be a THRUST, without there being “an equal and opposite
> > on the sail to make the raft move.
>
> More like the motion of wind at a cold front. Have you ever
> wondered how the wind blows despite the Third Law of Newton?
> There is an equal and opposite reaction. A packet of air on the
> black side is expanding and pushes against the black panel. The black
> panel moves away. However, the packet of hot air moves in the opposite
> direction. Hence, the force on the panel is equal and opposite the
> force on the packet of air.
>
> > For over a century, the latter has
> > been the accepted explanation for the rotation direction of
> > Radiometers, and such is patently WRONG!
> You are wrong.
>
> > won’t rotate at all in a perfect vacuum;
>
> That isn't precisely true. If the propeller is extremely well
> lubricated, so the friction is negligible, it was turn in the opposite
> direction. However, getting a junction to be nearly frictionless is
> very hard. If you would have evacuated that toy radiometer, it would
> probably have stood still. The friction is too high.>rotate backwards when exposed
> > to cold;
>
> ?
> >and rotate best with dilute argon gas.
> I'll bet. You need those currents in the gas. Under high pressure, the
> viscosity of the gas is too high and its conductivity too high to
> maintain a large temperature difference. So very low pressure gas is
> > actual reasons for the observed qualities of Radiometers:
>
> And the name of NoEinstein will be immortalized forever!-)
>
> > The radiant energy hitting the white and the black sides of the
> > squares is identical.
>
> However, the radiant energy that is absorbed is different.>The white sides, immediately, re emit light
> > that has a spectrum similar to the source. The black sides re emit
> > light of equal TOTAL energy, but in the infrared portion of the
> > spectrum. Photons have ZERO mass, and exert zero force upon
> > to rotate away from the black is caused, mainly, by the higher ether
> > pressure on the black sides, very similar to the explanation given at
> > the first part of my post:
>
> Notice that you explanation doesn't explain why there is no
> motion, or opposite motion, in a vacuum. You were the one who said
> that dilute Argon gas was necessary. What ahppened? You forgot?
>
> > The above necessitates that the ether pressure on the white sides be
> > less.
>
> I was wondering when you got to that.> Because of the very small mass of the four squares
> > is the argon atoms, within the glass bulb, moving toward the black
> > sides that accounts for most of the spinning.
> > because of the heat,
> BS
> > but because the argon is caught in the outward
> > flowing ether induced by the higher energy photons being emitted >from
> > WHITE squares.
>
> So why would high energy photons carry away ether? In fact, how can
> there be both ether and photons? Light acts like a wave. It shows
> interference. That is the entire basis of the ether theory. Waves
> can't act like particles.
> It looks to me like the ether is superfluous in this case. If the
> Argon gas is so important, why not just say that the Argon gas is
> moving?
>
> > There is ether throughout the bulb. The sunlight coming in imparts
> > little or no thrust to the ether (toward the squares).
>
> This is correct. Make that little thrust. The sunlight definitely
> produces a very small thrust. Almost negligible in the radiometer your
> father gave you.> But, because
> > of the concentration of ether within the matter of the squares, the re
> > emitted light carries some of that ether outward.
>
> Have you been sniffing that ether again?> The lower-energy
> > ether outward, too; and much more efficiently than from the white
> > squares, because of the greater number of photon trains (to have the
> Photon trains?
> > And because some of the
> > light reflected from the white squares carries ether toward the black
> > squares, the latter ether gets banked against the black sides of the
> > squares.
>
> So the light curves around to hit the black sides? It sounds like
> > propelled by the white light help to keep the ether pressure on the
> > black sides of the squares high.
>
> Yep. The photons are shaped like boomarangs. The Argon is causing
> them to curve backward and hit the black side. That is my
> interpretation of your model. Show me how I am wrong.> If the devices were totally
> > frictionless, the rotation would occur in the identical direction
> > without that gas being there.
> ??????????????
>
> > The depleted ether from within the squares gets replenished by the
> > ether that freely flows through the glass bulb.
>
> It is the Argon! Not the aether!
>
>
>
> > isn’t violated in this process, because the ether carried outward
NoEinstein
>
Dear Mark: Excellent point! The ether which pervades the Universe,
nurtures light on its way, and doesn't drag the light. If light were
dragged, we would all be dead and in the total darkness! James C.
Maxwell certainly wasn't being logical when he suggested to A. A.
Michelson that Michelson measure the drag on the velocity of the light
in M-M. There is none! Photons being emitted from within matter
carry some of the internal ether outward. Most of that falls off
before too long, Light from the Sun has, say, 95% of the ether
dropped off. But the reflected (re emitted) light from the vans
carries ‘fresh’ ether outward. The 'none' in, 'some' out is why the
Radiometer can function (friction being the limiting factor) by
varying ether pressure alone. Those argon atoms give a secondary mode
of pressure, as I've already explained. — NoEinstein —
Uncle Al
>
> On May 30, 10:33 pm, "n...@bid.nes" <alien8...@gmail.com> wrote:
> >
> Dear Mark: Excellent point! The ether which pervades the Universe,
HEY STOOOPID: ***OBSERVATION*** SAYS YOU ARE A PIECE OF SHIT TO 15
DECIMAL PLACES. Pull your thumb out of your ass and read what people
have *done* in the past five years. You are insurmountably
empirically stooopid.
http://arXiv.org/abs/0706.2031
Physics Today 57(7) 40 (2004)
http://physicstoday.org/vol-57/iss-7/p40.shtml
Phys. Rev. D8, pg 3321 (1973)
Phys. Rev. D9 pg 2489 (1974)
<http://cfa-www.harvard.edu/Walsworth/pdf/PT_Romalis0704.pdf>
No aether
http://arxiv.org/abs/0905.1929
<http://relativity.livingreviews.org/Articles/lrr-2005-5/index.html>
Phys. Rev. D 81 022003 (2010)
http://arxiv.org/abs/0801.0287
No Lorentz violation
idiot
--
Uncle Al
BURT
>
> The latter involves having the varying ether flow and density push the
> object in proportion to the object’s cross section that is in the
> photon stream. Dust particles adjacent to laser beams can be seen to
> move in the direction of the beam. But that is due to the air
> molecules, and the ether being moved, together. The dust is pushed by
> the air gases and by the flowing ether, not by the photons.
>
> That Wikipedia article on Radiometers mentioned that there is an
> induced gas flow through porous ceramic plates that is toward the side
> that is heated. [ Note: That is consistent with the ether flow
> direction predicted by my New Science. ] The rather iffy porosity of
> the ‘edges’ of the squares in the Crookes Radiometer has, for over a
> century, been considered to be the primary source for the thrust. The
> errant rationale has been: The edges of the black squares heat, and
> then shoot-out, the argon atoms, causing the observed rotation. The
> latter concocted ‘science‘, combined with Einstein’s heated gas
> nonsense, supposedly accounts for 100% of the observed rotation of the
> vanes.
>
> Photons are concentrations of energy which, in high enough
> concentrations, can burn through steel. Those photons don’t “force”
> through the steel. You could say: They “energy” through the steel!
Radio waves pass through steel.
Mitch Raemsch
> > object being pushed, but reflection works too. (Also absorption.)- Hide quoted text -
Timo Nieminen
> *** I invite you to reply with a concise PARAPHRASE of how that “in
> vacuum” experiment was done. (Note: I do not read links to the words
> of others.)
Why should I bother paraphrasing such things? _You_ don't bother answering
simple direct question that should take a genius like yourself only 10
minutes or so to answer. Does somebody of your intellect need others to
pre-digest your reading for you? Of course not! Since you're clearly
capable of reading, read:
P.N. Lebedev
Untersuchungen über die Druckkräfte des Lichtes
Annalen der Physik 6, 433 (1901)
E.F. Nichols and G.F. Hull
A preliminary communication on the pressure of heat and light radiation
Physical Review 13, 307 (1901)
> Like I have said, conclusively, massless photons, alone, exert no
> force on objects.
Simply wrong. Force due to electromagnetic waves is routinely observed.
Fundamentally, no different from the force that drive electric motors.
And since you're already whining about your two-paragraph attention span
being exceeded, I'll stop here.
Tim BandTech.com
I've recently done some research on the radiometer, and this math
analysis is definitely not what makes it tick. The sun is putting out
1300 watts per square meter. If all of this energy goes into
mechanical energy then we'd be making turbines rather than PV cells.
There is a 'radiation pressure' claim of one third the energy density
(loose language) supposedly from Maxwell. I haven't found that
computation yet.
The Nichols radiometer is a torsion balance, whereas Crook's spins
freely. Somehow Nichols claims to have provided one that works in
vacuum, and yet the argument should pass freely onto the Crook's
version. How much friction can there be in a well made pin bearing
versus a quartz fiber? The Nichol's data is weird near perfect vacuum.
An inversion takes place that he just graphs and doesn't care to
explaim. He claims that the atmospheric effects can be cancelled as
shown by the zeros in his graphs. Well, at these zeros there will be
no effect. I find the Nichols claims unconvincing.
In terms of the atmospheric effects a thermal differential causing
fluid flow at the edges is supposedly the cause, discovered by
Reynolds. I've not seen any analysis in terms of Bernoulli effect, but
if the air is rising at the black surface then the velocity of the gas
is greater there. This is the simplest explanation, but there are so
many explanations with very little proof.
I'm pretty sure the radiation pressure argument is false. Photons are
more like AC sources than DC sources. If the pressure is alternating
then to claim a propulsive acceleration is a misnomer. It would be
like putting a ping pong ball in front of a speaker, blasting out 100
watts and expecting the ping pong ball to accelerate away from the
speaker.
It's interesting how easy it is to become opinionated on this topic.
Without gravity the effects of the atmosphere with temperature
difference would be diminished so it would be neat to see this
experiment done in outer space, and at various atmospheric pressures.
I was researching this in a recent thread
http://groups.google.com/group/sci.math/browse_frm/thread/4835ecd7ca63f1f1#
There are some links in there, starting with
http://math.ucr.edu/home/baez/physics/General/LightMill/light-mill.html
The most blatant farce is in terms of conservation of energy. The
claim is that the light hitting a reflective surface will provide
twice the momentum; one kick when the light hits it and one kick again
from the light when it leaves. This concept offends the conservation
of energy. 1300 watts in with 1300 watts out leaves no acceleration
whatsoever for the perfect reflector. Consider taking a full length
mirror out into the sun and being boled over by three horsepower of
push. This area is loaded with misnomers and is a great subject for
it.
- Tim
Timo Nieminen
> On May 30, 2:58 pm, Sam Wormley <sworml...@gmail.com> wrote:
> > On 5/30/10 9:31 AM, NoEinstein wrote:
> >
> > > Photons have ZERO mass, and exert zero force upon
> > > ‘striking’ a reflecting surface.
> >
> > Photons have zero REST mass, but have momentum of
> >
> > p = hν/c = h/λ
> >
> > A perfect reflector's momentum would increase by
> >
> > 2p = 2h/λ
> >
> > for every photon. Ever heard of a solar sail?
>
> I've recently done some research on the radiometer, and this math
> analysis is definitely not what makes it tick. The sun is putting out
> 1300 watts per square meter. If all of this energy goes into
> mechanical energy then we'd be making turbines rather than PV cells.
Apart from some very early claims about the Crookes radiometer (such as
by Crookes), shown to be wrong a few years later, and their modern
repetition, who claims that radiation pressure makes the radiometer tick?
1300W per square metre means 9 micronewtons per square meter, on a
perfect reflector. That it would take a square kilometer to be able to use
solar radiation pressure to produce the same force as the weight of 1kg
pretty much says why we don't do this.
(A little more at the bottom of post.)
> There is a 'radiation pressure' claim of one third the energy density
> (loose language) supposedly from Maxwell. I haven't found that
> computation yet.
Try Maxwell's Treatise. Art 792, pp 391-392 in 1st edition.
> I'm pretty sure the radiation pressure argument is false. Photons are
> more like AC sources than DC sources. If the pressure is alternating
> then to claim a propulsive acceleration is a misnomer. It would be
> like putting a ping pong ball in front of a speaker, blasting out 100
> watts and expecting the ping pong ball to accelerate away from the
> speaker.
Why not? Just don't expect a large acceleration; I don't think that 100W
would be enough to levitate against gravity.
It's been done, google or youtube for "acoustic levitation".
> The most blatant farce is in terms of conservation of energy. The
> claim is that the light hitting a reflective surface will provide
> twice the momentum; one kick when the light hits it and one kick again
> from the light when it leaves. This concept offends the conservation
> of energy. 1300 watts in with 1300 watts out leaves no acceleration
> whatsoever for the perfect reflector.
If the reflector is moving, there will be a Doppler shift, and energy in
will be different from energy out. If the reflector is stationary, then
you could have, e.g., 1300W in and 1300W out. But no work would be done.
Even better, if the light is slowing the reflector down, the reflected
light has more energy than the incident light. Just as one would expect
from conservation of energy.
Back to the idea of using solar radiation to produce mechanical energy. If
you can move the 1 square kilometer reflector at 1km/s downwards, then
you'd get a power of 9kW, from a total of 1.9GW input solar power. This is
very poor efficiency compared to even poor solar cells.
--
Timo
Tim BandTech.com
> On Mon, 31 May 2010, Tim BandTech.com wrote:
> > On May 30, 2:58 pm, Sam Wormley <sworml...@gmail.com> wrote:
> > > On 5/30/10 9:31 AM, NoEinstein wrote:
> > > > Photons have ZERO mass, and exert zero force upon
> > > > ‘striking’ a reflecting surface.
> > > Photons have zero REST mass, but have momentum of
> > > p = hν/c = h/λ
> > > A perfect reflector's momentum would increase by
> > > 2p = 2h/λ
> > > for every photon. Ever heard of a solar sail?
> > I've recently done some research on the radiometer, and this math
> > analysis is definitely not what makes it tick. The sun is putting out
> > 1300 watts per square meter. If all of this energy goes into
> > mechanical energy then we'd be making turbines rather than PV cells.
> Apart from some very early claims about the Crookes radiometer (such as
> by Crookes), shown to be wrong a few years later, and their modern
> repetition, who claims that radiation pressure makes the radiometer tick?
The Nichols radiometer is claimed by some to work in a vacuum,
including Nichols himself.
Here is a link to his original work which I'm snipping from a few
weeks ago of that past thread:
Here is a quote from Nichols published work:
"At the close of the pressure and energy measurements when the
reflecting power of the silver faces of the vanes was compared with
that of the glass silver faces the reflection from the silver faces
was found very much higher than that for the glass faces backed by
silver. This result was the more surprising because the absorption of
the unsilvered vanes was found by measurement to be negligibly small.
This unexpected difference in reflecting power of the two faces of the
mirrors prevented the elimination of the gas action by the method
described from being as complete as had been hoped for. But by
choosing a gas pressure where the gas action after long exposure is
small the whole gas effect during the time of a ballistic exposure may
be so reduced as to be of little consequence in any case."
- http://books.google.com/books?id=8n8OAAAAIAAJ&pg=RA5-PA327&dq=torsion...
E.F. Nichols and G.F. Hull, The Pressure due to Radiation, The
Astrophysical Journal,Vol.17 No.5, p.315-351 (1903)
There is a standard claim that the Nichols radiometer is operating on
different principles than the Crookes radiometer, but this passage
exposes that gas effects were never eliminated, and furthermore, if
the Nichols radiometer truely operated in vacuum, then why shouldn't
the Crookes? This question seems to go unanswered, and is to me a sore
point. The stranger reversal in deflection from .05 mm Hg to .02 mm Hg
seems to go observed yet unanswered within the analysis. Nichol's
seems to have pushed for a specific result, and having gotten it, is
happy not to look back. Is this science?
The link is worth the read but above I've quoted a passage that my
skepticism picks upon. Periods were missing, and this text is coming
straight from Google's 'cut feature of their image reader; pretty damn
slick and righteuous that we've got access to this information. If all
journals would do this... I guess they will in time; that is, those
that do not wish to exclude amateurs from access.
To amplify the contradiction of the quote I pick out two portions:
"the mirrors prevented the elimination of the gas action by the
method described from being as complete as had been hoped for"
"the whole gas effect during the time of a ballistic exposure may
be so reduced as to be of little consequence in any case"
There is no data demonstrating these nulls, and the sharp inversion
from the 0.02 to 0.05 suggests some sort of nonlinearity at that null,
which will prevent stable experimentation.
Anyway, selection of that null should actually yield no results. The
dynamics exposed by the experiment go ignored in the analysis. I do
not feel strongly enough to declare a farce here, but I am leaning in
that direction.
(end of copy from recent prior thread)
>
> 1300W per square metre means 9 micronewtons per square meter, on a
> perfect reflector. That it would take a square kilometer to be able to use
> solar radiation pressure to produce the same force as the weight of 1kg
> pretty much says why we don't do this.
Yes, well, I'd like to understand how even this slight figure comes
about. It is easy to falsify the e=hv momentum claim (which is a big
enough statement) but even this slender acceleration is questionable
by the argument on conservation of energy.
>
> (A little more at the bottom of post.)
>
> > There is a 'radiation pressure' claim of one third the energy density
> > (loose language) supposedly from Maxwell. I haven't found that
> > computation yet.
>
> Try Maxwell's Treatise. Art 792, pp 391-392 in 1st edition.
Thank you Timo for such a clean reference. There is a square within
the expression that is rectifying the wave. If we were to attempt an
instantaneous measurement of acceleration then I think the result
would read more true. I only just started studying this and it does
rely on much of the previous content. Also I haven't made it to the
one third argument.
Why oh why would we rely upon a volume figure to assess the force on a
plate? The figuring does not extend generally either. For instance,
staying with Watts and Joules and seconds what if we consider a one
second interval of the solar power? Now rather than a 1mx1mx1m box we
have a box extending 3e8 meters toward the sun, and now the
computation gets trickier, for the assumption on this 1kJ box of
energy now has to worry about the proximity to the sun.
I am not familiar with modern physics using the electokinetic energy
versus electrostatic energy. What seems most familiar is
e = h v
where v is proportional to the frequency of a supposed photon. You
seem to have validated that photon momentum is not responsible for the
radiation pressure.
I'm honestly not clear on whether you are in support of a mechanical
pressure or not. This does not really matter, but because the
ballistic theory of light and black body principles rest nearby, then
there is some cause for interest beyond the toy light-mill's
functionality. I believe it is accurate to translate the 1300 W/m/m of
the sun directly into photon energy. If one photon
e = h v
is a bullet then how much force does it impart? Is Wormley correct?
Then why doesn't it translate? You see, we are knocking on the back of
the black door.
>
> > I'm pretty sure the radiation pressure argument is false. Photons are
> > more like AC sources than DC sources. If the pressure is alternating
> > then to claim a propulsive acceleration is a misnomer. It would be
> > like putting a ping pong ball in front of a speaker, blasting out 100
> > watts and expecting the ping pong ball to accelerate away from the
> > speaker.
>
> Why not? Just don't expect a large acceleration; I don't think that 100W
> would be enough to levitate against gravity.
>
> It's been done, google or youtube for "acoustic levitation".
This isn't quite what I had in mind. Particularly at the Earth's
distance from the sun we are not free to rely upon access to the sun
itself within our argument, though I will get near to this below.
>
> > The most blatant farce is in terms of conservation of energy. The
> > claim is that the light hitting a reflective surface will provide
> > twice the momentum; one kick when the light hits it and one kick again
> > from the light when it leaves. This concept offends the conservation
> > of energy. 1300 watts in with 1300 watts out leaves no acceleration
> > whatsoever for the perfect reflector.
>
> If the reflector is moving, there will be a Doppler shift, and energy in
> will be different from energy out. If the reflector is stationary, then
> you could have, e.g., 1300W in and 1300W out. But no work would be done.
Very good. We have agreement, particularly at the low velocities that
these instruments work at. As far as we can tell there is no reliance
upon red shifting of light, which could provide some work.
- Tim
>
> Even better, if the light is slowing the reflector down, the reflected
> light has more energy than the incident light. Just as one would expect
> from conservation of energy.
>
> Back to the idea of using solar radiation to produce mechanical energy. If
> you can move the 1 square kilometer reflector at 1km/s downwards, then
> you'd get a power of 9kW, from a total of 1.9GW input solar power. This is
> very poor efficiency compared to even poor solar cells.
I'm still trying to understand how to arrive at your 9kW figure, other
than copying other's work. It is a strange figure to me to arrive at
which somehow depends upon a volume, even though the surface is not a
volume. This makes me think (not joking either) of space creation of
the sun as a paradigm. This sort of thing would validate the
acceleration, provide the repulsive force desired by cosmologists,
invalidate dark matter, and so on. Linking into thermodynamics and
gravity, well, lets just consider gravity alone and thermodynamics
alone, but here they tieing into electromagnetism. That's awfully
speculative, but gravitational shadowing as a principle is nearby as
well, since light would be involved. Too much and too loose, so I am
sorry, but I'm not going to delete.
Thanks Timo for your strong response and that JCM reference.
>
> --
> Timo
Timo Nieminen
> On Jun 1, 12:24 am, Timo Nieminen <t...@physics.uq.edu.au> wrote:
> > On Mon, 31 May 2010, Tim BandTech.com wrote:
> > > On May 30, 2:58 pm, Sam Wormley <sworml...@gmail.com> wrote:
> > > > On 5/30/10 9:31 AM, NoEinstein wrote:
> > > > > Photons have ZERO mass, and exert zero force upon
> > > > > ‘striking’ a reflecting surface.
> > > > Photons have zero REST mass, but have momentum of
> > > > p = hν/c = h/λ
> > > > A perfect reflector's momentum would increase by
> > > > 2p = 2h/λ
> > > > for every photon. Ever heard of a solar sail?
> > > I've recently done some research on the radiometer, and this math
> > > analysis is definitely not what makes it tick. The sun is putting out
> > > 1300 watts per square meter. If all of this energy goes into
> > > mechanical energy then we'd be making turbines rather than PV cells.
> > Apart from some very early claims about the Crookes radiometer (such as
> > by Crookes), shown to be wrong a few years later, and their modern
> > repetition, who claims that radiation pressure makes the radiometer tick?
>
> The Nichols radiometer is claimed by some to work in a vacuum,
> including Nichols himself.
Nichols and Hull (and Lebedev very shortly before them) were operating
at the limits of what they could detect or measure, with the best
vacuums they could manage affordably. That their results aren't
terribly good shouldn't be a surprise. But it's been done again, and
it doesn't have to be done in vacuum. We discussed some of these
experiments in http://arxiv.org/abs/0710.0461 - the work by R. V.
Jones was excellent, his 1954 measurements might well be the best pre-
laser measurements of optical radiation pressure.
So, whatever your doubts about Nichols' experiment, it's been repeated
and amply confirmed many times (and at least some of these were by
people who didn't Nichols' results were really good enough). These
days, it's almost trivial.
(I will be brief for the rest, since it is late. Perhaps I can give
more detail or refs later.)
> > 1300W per square metre means 9 micronewtons per square meter, on a
> > perfect reflector. That it would take a square kilometer to be able to use
> > solar radiation pressure to produce the same force as the weight of 1kg
> > pretty much says why we don't do this.
>
> Yes, well, I'd like to understand how even this slight figure comes
> about. It is easy to falsify the e=hv momentum claim (which is a big
> enough statement) but even this slender acceleration is questionable
> by the argument on conservation of energy.
This is just the momentum of a photon p=h/lambda claim. But it isn't a
quantum result; quantum mechanics just inherits this from the
classical theory. The standard result in the classical theory is that
the momentum flux of a parallel beam of power P is P/c (for light in
free space or some medium with refractive index close enough to 1).
For light in some other medium, such as water, momentum flux = nP/c.
You can obtain this result via at least 3 different routes:
(a) the Lorentz force, (i) by directly calculating the force for a
specific case, or (ii) by obtaining the momentum flux of an
elecromagnetic wave along the lines of the derivation of Poynting's
theorem.
(b) the Lagrangian formulation of EM theory + Noether's theorem.
(c) the thermodynamics of moving energy around without moving mass
around.
Jackson's book, Classical electrodynamics, does (a)(ii) and (b). (c)
goes back to N. A. Umov, 1874 (I haven't seen this last in English).
> Why oh why would we rely upon a volume figure to assess the force on a
> plate? The figuring does not extend generally either. For instance,
> staying with Watts and Joules and seconds what if we consider a one
> second interval of the solar power? Now rather than a 1mx1mx1m box we
> have a box extending 3e8 meters toward the sun, and now the
> computation gets trickier, for the assumption on this 1kJ box of
> energy now has to worry about the proximity to the sun.
We don't have to rely on a volume to find the force on a plate. But
the density and flux of a quantity are related by the speed of
transport of the quantity, so once you know the density, you know the
flux.
> I am not familiar with modern physics using the electokinetic energy
> versus electrostatic energy. What seems most familiar is
> e = h v
> where v is proportional to the frequency of a supposed photon. You
> seem to have validated that photon momentum is not responsible for the
> radiation pressure.
Maxwell's "electrokinetic energy" isn't modern. Maxwell can be hard to
read, since he is archaic.
> I'm honestly not clear on whether you are in support of a mechanical
> pressure or not. This does not really matter, but because the
> ballistic theory of light and black body principles rest nearby, then
> there is some cause for interest beyond the toy light-mill's
> functionality. I believe it is accurate to translate the 1300 W/m/m of
> the sun directly into photon energy. If one photon
> e = h v
> is a bullet then how much force does it impart? Is Wormley correct?
> Then why doesn't it translate? You see, we are knocking on the back of
> the black door.
I don't think we're anywhere mysterious here. Inertia of energy (which
means that moving energy has momentum) goes back to Umov 1874 on
thermodynamic grounds, Einstein 1905 on more general relativistic
grounds, that electromagnetic energy has inertia goes back to Poynting
and Heaviside in 1884. Given the energy of a photon as E=hf, then you
have mometum p=h/lambda, since p=E/c.
And it all works for other kinds of waves, and angular momentum too.
But it used to be mysterious. Back in the early 1800s, it was argued
that light was a wave phenomenon rather than a corpuscular phenomenon
since radiation pressure was not observed - how could something
massless like a wave exert a force?
These days, with the idea that electromagnetic fields can exert
forces, and light being an electromagnetic wave, optical radiation
pressure is about as mysterious as an electric motor. Which is to say,
that it exists and works is clear, whatever fundamental mysteries of
"why" remain.
It's well-tested quantitatively. It isn't too hard to verify the
momentum of light to within 10% using optical tweezers (e.g., our
experiment in http://arxiv.org/abs/physics/0610087 . Yes, we say
better than 1%, but that's with a known particle for calibration, to
find the power at the focus of our trap. Without this, we'd have a
larger error, mainly in the estimate of this power. Still, no more
than 10%.)
> > > The most blatant farce is in terms of conservation of energy. The
> > > claim is that the light hitting a reflective surface will provide
> > > twice the momentum; one kick when the light hits it and one kick again
> > > from the light when it leaves. This concept offends the conservation
> > > of energy. 1300 watts in with 1300 watts out leaves no acceleration
> > > whatsoever for the perfect reflector.
>
> > If the reflector is moving, there will be a Doppler shift, and energy in
> > will be different from energy out. If the reflector is stationary, then
> > you could have, e.g., 1300W in and 1300W out. But no work would be done.
>
> Very good. We have agreement, particularly at the low velocities that
> these instruments work at. As far as we can tell there is no reliance
> upon red shifting of light, which could provide some work.
The force doesn't depend on redshift. But it doesn't take any
expenditure of energy to produce a force - just ask a fridge magnet or
a paperweight. Having the force do work does very much depend on
redshift, since the reflector has to move for the radiation pressure
to do work on it.
NoEinstein
>
Dear U. Al: You know that I never read links to the words of others.
Yet you keep replying with links. I must offend you, intellectually,
because you keep call 'me' STOOOPID. If 'low IQ’ is something you...
hate so much, go to any State Mental Hospital and cuss out the
unfortunate ones. While you are there, let the White Coats examine
YOU. Though you are a hopeless case, I'm sure they will be nice to
you, there. — NE —
>
> NoEinstein wrote:
>
> > On May 30, 10:33 pm, "n...@bid.nes" <alien8...@gmail.com> wrote:
>
> > Dear Mark: Excellent point! The ether which pervades the Universe,
>
> [snip crap]
>
> HEY STOOOPID: ***OBSERVATION*** SAYS YOU ARE A PIECE OF SHIT TO 15
> DECIMAL PLACES. Pull your thumb out of your ass and read what people
> have *done* in the past five years. You are insurmountably
> empirically stooopid.
>
> http://arXiv.org/abs/0706.2031
> Phys. Rev. D8, pg 3321 (1973)
> Phys. Rev. D9 pg 2489 (1974)
> <http://cfa-www.harvard.edu/Walsworth/pdf/PT_Romalis0704.pdf>
> No aether
>
> http://arxiv.org/abs/0905.1929
> <http://relativity.livingreviews.org/Articles/lrr-2005-5/index.html>
> No Lorentz violation
>
> idiot
>
> --
NoEinstein
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
No disagreement, there; provided the steel isn't too thick. — NE —
NoEinstein
>
Dear Timo: You are being adversarial. You belittle me by claiming
that giving answers is simple for 'me'. Yet, you can't find or figure
out simple things for yourself. I refuse to walk-you-through your…
not very determined science journeys. Pick one or two issues or
comments; top post a paragraph or two in your own words, and I will
reply as apt. Thanks. — NE —
NoEinstein
>
Dear Tim: Where have you been?! Your logical mind, justifies cloning
you... or putting you out-to-stud! Thanks for giving me a hint of how
the Nichols Radiometer works. If you will make a concise top post(s)
regarding science questions that relate to light or to gravity, I will
be most happy to reply. But know this: I am a generalist, not a
specialist. Don't try to make me into a mathematician. Those aren't
generalists, nor scientists!
NoEinstein
>
Dear Timo: If the (supposed) pressure exerted by 'solar radiation' is
correct, then, such can be used to determine what percentage of the
ether is still trapped between the photons. The ether that's present
between the mass-less photons is what 'moves' the vanes of a Nichols
Interferometer Fresh radiation—as from a highly heated Cavendish ball—
will be carrying outward a higher percentage of ether. The latter
could affect the relative gravity strength. This will be an
interesting area for study, by others, down the road. — NE —
Timo Nieminen
> On May 31, 8:44 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> Dear Timo: You are being adversarial.
You ask for information, I give you information, and you complain. OK,
you did ask for it to be paraphrased, but since you are a literate
adult, who claims to be a super-genius of exceedingly high ability, I
figure you can read it for yourself.
You ask, I provide, and then you complain. You are a disgusting excuse
for a human being. If you can't maintain at least a minimally
acceptable level of civility, then fuck off.
> You belittle me by claiming
> that giving answers is simple for 'me'.
_You're_ the one who claims over and over and over what a wonderfully
superior genoius you are, how your talents exceed those of any other
10 scientists, how you have well-developed mathematical abilities. All
I'm doing is assuming that your claims are true. Why would you lie
about such things? All that would happen is that people might assume
it is true, and expect an outcome matching your claims.
> Yet, you can't find or figure
> out simple things for yourself.
Given your stated assumptions, that gravity is linearly proportional
to "photon emission", it's a simple back-of-the-envelope calculation
to come up with a quantitative model based on this, just a few minutes
will do. But this model is wildly and obvious incorrect; it has no
serious relationship with observation or measurements of gravity.
Since you're such a super-genius, this is obviously obvious to you.
Since you keep claiming that you theory does indeed match reality,
then what you have bothered to reveal so far about your theory isn't
your whole theory, since the revealed part simply doesn't work. I
don't know what the unrevealed part of your theory is, since you
haven't revealed it.
And since you claim that the development of your theory is such a
stupendous and revolutionary achievement, you can't expect others to
figure out the unrevealed parts by themselves.
But coming up with a quantitative model from the revealed portion is
simple. Just as finding data for the Sirius system was simple (how
hard can it be to look up on Wikipedia?). And calculating the
fractions of the light from each that is intercepted by the other was
simple. Divining the unrevealed part of your theory isn't so simple.
Hence the questions.
> I refuse to walk-you-through your…
> not very determined science journeys.
It's your theory. I'm just interested in whether it's plausible. You
refuse, and keep refusing to answer questions about this. A safe
asumption therefore seems to be that your theory is wrong.
You've kept saying that more information is needed. I've given you all
of the information that you've said is required, and all you can do is
refuse to answer. Again, a safe asumption therefore seems to be that
your theory is wrong. Or that you're really stupid, a moron so stupid
as to claim to be a super-genius of monstrously great talent, with
well-developed mathematical ability, but completely incapable. In
which case, a safe assumption is that the theory of such a moron,
given the obvious conflicts it already has with reality, is wrong.
> Pick one or two issues or
> comments; top post a paragraph or two in your own words, and I will
> reply as apt. Thanks.
I was replying to _your requests_. I kept it to a couple of
paragraphs. I didn't top post, being an Old School believer in
netiquette. You replied rudely. So fuck off.
Timo Nieminen
> On Jun 1, 12:24 am, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> > 1300W per square metre means 9 micronewtons per square meter, on a
> > perfect reflector. That it would take a square kilometer to be able to use
> > solar radiation pressure to produce the same force as the weight of 1kg
> > pretty much says why we don't do this.
>
> Yes, well, I'd like to understand how even this slight figure comes
> about. It is easy to falsify the e=hv momentum claim (which is a big
> enough statement) but even this slender acceleration is questionable
> by the argument on conservation of energy.
Here is another derivation:
Consider a steady beam of monchromatic light incident on a stationary
reflector. Each optical cycle has energy E (i.e., the total energy of
a length of the beam equal to 1 wavelength). The incident power P is
thus P = Ef, where f is the optical frequency.The output power is the
same, since the reflected beam is identical, except for the reversed
direction. Since the reflector is stationary, no work is being done on
it.
Now change to a coordinate system where the reflector is in motion, at
a speed v away from the source of the incident beam. (For simplicity,
assume v << c.) The incident beam is now blue-shifted, and the
frequency of the incident beam in this coordinate system is now f(1+v/
c), and the power is P_in = Ef(1+v/c). Similarly, the reflected beam
now has power P_out = Ef(1-v/c).
Since energy is conserved, the difference in power must be the rate of
doing work on the reflector.
P_reflector = P_in - P_out = 2Efv/c = 2vP/c.
Since P_reflector = F_reflector * v, we have:
F_reflector = 2P/c,
which is the radiation pressure due to complete reflection at normal
incidence.
This is the simplest derivation of radiation pressure I know of.
--
Timo
NoEinstein
>
Dear Timo: That Nichols Interferometer which you claimed showed
thrust toward the first surface mirror (the white analogy in the
Crookes) doesn't show that at all. Those guys got so caught up in
"precision", that the basic design of their interferometer wasn't even
a scientific extension of the Crookes. Their most basic mistake was
to change away from having black squares—the essence of the Crookes.
They wrongly assumed that the glass-out side of the squares wouldn't
heat up, but rather insulate the silver from speeding up the argon
atoms.
The very simple reason that the Crookes required some air to operate
was because the induced GRAVITY on the squares (due to photon
exchange) wasn't strong enough to overcome the spindle friction. The
dilute argon functioned like 'bullets" fired toward the black by the
ether flowing out along with the re emitted photons from the white
sides of the vanes. It was the mass of the argon, compressing the
ether coming out of the black sides of the vanes that caused the vanes
to rotate (that along with the gravity PUSH).
Try as you may, I will not get involved in wordy discussion with you
on anything. So far you haven't shown that you can carry the ball…
one yard, alone. — NoEinstein —
spudnik
apparently Reynold and Maxwell showed that it stopped working
with too little gas ... and somehow goes the other way,
with even less gas, but it seems that the upshot is that
it is magnetohydrodynamical. anyway, on quantum grounds,
it must be that & not "momentum," since electromagnetism operates
on the electronic orbitals, the *angular* momentum of the atom; then,
waht ever happens, does happen, and that is that.
anyway, it's nice to have an experimentalist, instead
of all of us prestidigitators, like Shakespeare's monkeys:
"no; but the'yre producing *gobs* of Stephen King!"
anyway, I'd just like to say,
Nature abhors an index of refraction, equal to 1.0000..., although
Pascal thought that he'd dyscovered that.
> The force doesn't depend on redshift. But it doesn't take any
> expenditure of energy to produce a force - just ask a fridge magnet or
> a paperweight. Having the force do work does very much depend on
> redshift, since the reflector has to move for the radiation pressure
> to do work on it.
thusNso:
whoah, stinky feet that are burning!... seriously,
I doubt that there is one other guy or gal in Buckyland
that has gotten into Cliff's system, other than
the encyclopaedist for the Wolframites. anyway, what Bucky saith
about angular momentum is just crap, perhaps just
as problematic with homogenous (4D) spatial coordinates, and
taht is where the action is -- not with either
of your systems (til further noticingness .-)
there are many areas in which Bucky is not even wrong, but
he is definitely also not entirely incorrect (to wit,
tensegrity, as fundamental a priciple of Universe,
as may ever be).
well, Cliff at least has a thing that is formalized enough,
apparently, to encode it in Wolframiteware;
Tim's stuff, on the other hand, is just verbiage:
notice that he sort-of defines his additions
as mere concatenations of his polysignosis,
which would be no different than using 4 unary operators,
as with quaternions (but those are *not* homogenous;
they're "space+time" a la Minkowski), in a tuplet
with only positive entries, somewhat like quadrays (or,
exactly like that; who, knows?)
but, he clouds the idea by spacing his tuplets as "equations,"
so that the unary operators look like binary operators,
like plus and minus and times, and further clouds this
by using plus and minus and times *signs* -- I mean,
symbols, but he calls it, Polysigns Yahoo!TM ... and
that is just God-am annoying, maybe on purpose.
well, can any thing actually be done with such a thing?...
remains to be seen, because mister Polysignosis won't. so,
in the case of 4D homogenous coordiantion,
just do it, the old way!
> Still, I am not afraid to hold my feet close to the fire.
thusNso:
I meant, shouldn't force kids to read Shakespeare (or
any thing else) til they are 11 *years* old, or about 5th grading.
the proper hands-on study for pre-pubescence ought
to be spatial geometry, astronomy, music & numberthory,
a.k.a. *mathematica* or *quadrivium* --
not the God-am 3 Rs of the *trivium*, to impose life-
long impedimentia.
> It would seem bizarre, if Dudley did not think
> that geometry was part of mathematics.
thusNso:
all that it shows, since no violation of causality is known,
is that the pair of waves are correlated,
from the "splitting" to the absorption, and how could that be?
get rid of the "particle" ideal of Einstein's "photon,"
and most (or all) all of the quandary goes away.
(then, dump Minkowski's silly slogan
about phase-space-and-then-he-died .-)
> Quantum mechanics predicts that measuring the spin of
> one proton in an entangled pair will affect the state
> of the other proton.
--Stop BP's capNtrade rip-off, "hey,
let's just let a bunch of arbitrageurs/hackers
make as much money as they can, trading CO2 credits!"
http://wlym.com
Tim BandTech.com
> Jones was excellent, his 1954 measurements might well be the best pre-
> laser measurements of optical radiation pressure.
>
> So, whatever your doubts about Nichols' experiment, it's been repeated
> and amply confirmed many times (and at least some of these were by
> people who didn't Nichols' results were really good enough). These
> days, it's almost trivial.
I've reviewed your paper "Momentum of an electromagnetic wave in
dielectric media".
I wonder to what degree the name 'energy-momentum tensor' is a strict
meaning?
If we accept that the electric field is sinusoidal and so has zeros
then all of the expressions that you consider contain instaneous zeros
within their representation. This suggests that conservation of
'energy-momentum' does not exist, but then, these are not claims on
just momentum are they? As you say we should consider a monochromatic
and coherent source to simplify matters. This is fairly analogous to
Maxwell plane wave isn't it?
With the experiment of Ashkin and Dziedzic did they ever bother to
drive the beam up out of the water into the air and observe a reversal
of their effect? Did they ever bother to consider that the thermal
heating of the water could be providing their effect?
In the experiment of Jones and Richard you say
"Jones sought
to verify the prediction that when a mirror was immersed
in a dielectric medium, the radiation pressure exerted on
the mirror would be proportional to the refractive index
of the medium."
yet this goes in exact contradiction to the claimed Reynolds effect,
where immersion in a fluid is subject again to thermodynamic effects.
As you say above here to just relax, because these experiments have
been done again, and that the weakness of Nichols work does not exist
in perpetutity, well, I believe that I have presented evidence here
that these experimenters are the ones who relaxed on top of the work
of Nichols.
Lastly, I would argue from a very simplistic perspective that if
conservation of energy and momentum is upheld that these figures in
vacuum should match these figures in solid media.
Otherwise where did the energy or momentum go?
I admit that my own belief is that the momentum is actually zero in
terms of a steady unidirectional push on a plate of material, but in
that light does propagate in a given direction we must admit the
energy itself is directed, but when for instance we absorb all of that
energy as in a black ideal absorber, that it has nothing to do with
mechanical momentum. The claims of propulsion doubling from a perfect
reflector only help to boost my position. I guess to pose the simplest
question that I can on this is to ask whether the energy of a photon
e = h f
is the entire energy of the photon? What have you concluded about
Wormley's claim on the momentum figure here? Isn't this figure the
classical blackbody figure of momentum? How did the discrepancy then
to a very slight momentum figure arise and how do we split that
smaller energy off of this photon energy without corrupting physics?
I'm sorry but the effect in a perfect vacuum has not been clearly
demonstrated. As you've accepted the flaws of Nichols work then the
question of whether the radiation pressure is observable in vacuum
still exists. The experiments that you've exposed in your paper must
not be the modern experiments that you speak of.
>
> It's well-tested quantitatively. It isn't too hard to verify the
> momentum of light to within 10% using optical tweezers (e.g., our
> experiment inhttp://arxiv.org/abs/physics/0610087. Yes, we say
> better than 1%, but that's with a known particle for calibration, to
> find the power at the focus of our trap. Without this, we'd have a
> larger error, mainly in the estimate of this power. Still, no more
> than 10%.)
>
> > > > The most blatant farce is in terms of conservation of energy. The
> > > > claim is that the light hitting a reflective surface will provide
> > > > twice the momentum; one kick when the light hits it and one kick again
> > > > from the light when it leaves. This concept offends the conservation
> > > > of energy. 1300 watts in with 1300 watts out leaves no acceleration
> > > > whatsoever for the perfect reflector.
>
> > > If the reflector is moving, there will be a Doppler shift, and energy in
> > > will be different from energy out. If the reflector is stationary, then
> > > you could have, e.g., 1300W in and 1300W out. But no work would be done.
>
> > Very good. We have agreement, particularly at the low velocities that
> > these instruments work at. As far as we can tell there is no reliance
> > upon red shifting of light, which could provide some work.
>
> The force doesn't depend on redshift. But it doesn't take any
> expenditure of energy to produce a force - just ask a fridge magnet or
> a paperweight. Having the force do work does very much depend on
> redshift, since the reflector has to move for the radiation pressure
> to do work on it.
Awww, come on.... The claim is that the force is doing work, and the
existence of any redshift in doing that work has been completely
ignored in any of these theories or experiments. Without doing some
work there will be no indication as of a vane which rotates on a
tensioned fiber. We know we're at extremely low velocity in this
experiment, so any redshift observed would be a highly impressive
mechanism.
I don't mean to get too ornery, but you are sweeping over these points
as if there is no fundamental problem. This is somewhat the attitude
of modern physics. It is true that proffessional physicists must
produce work in order to be proffessional, and this I do not mean as
an insult to you, but to explain the overall attitude of the
community. I really appreciate your input and it is very detailed, but
particularly on the claim that modern experiments have gone beyond the
shadow of doubt, well, we will first need something working in vacuum
which observes the supposed radiation pressure. From the theoretical
side there is a reliance upon a third of the energy density for this
pressure that we still have not arrived at, though you've gotten very
close in Maxwell's treatise. Here is the strongest link that I have
found in support of challenging the existing theory:
http://www.neumann-alpha.org/lightpressure.pdf
This site is a bit out there, but then, so too was Nichols. Any who
will rest on top of his work, and that is how science is done, are on
a shaky ladder.
- Tim
Tim BandTech.com
> On May 31, 8:57 pm, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> Dear Tim: Where have you been?! Your logical mind, justifies cloning
> you... or putting you out-to-stud! Thanks for giving me a hint of how
> the Nichols Radiometer works. If you will make a concise top post(s)
> regarding science questions that relate to light or to gravity, I will
> be most happy to reply. But know this: I am a generalist, not a
> specialist. Don't try to make me into a mathematician. Those aren't
> generalists, nor scientists!
>
Geeze, what a compliment. So rare on usenet. I share somwhat your
concept of a generalist and have posed that question in the past of
where these generalists are? It is pretty easy to grow skeptical of
the course of science given the system's structure of censors on a
quantity of journals whose access is protected by a price tag that few
can afford, and whose accumulation is overwhelming. Still, there is a
tension between freedom to construct and correctness. I will argue for
the freedom to construct, and, to accept all that is printed in
journals is perfect is a farcical stance. It is more apt to state that
what is printed in journals fits the censors' agendas. Well, hooray
for arxiv.org! What about right here on usenet? Why shouldn't
scientists be free to disagree with each other in the open? What
better means of falsification could there be? In terms of seeking the
truth I suspect that this medium could be more effective than the
journal system, which is a protected form. Why shouldn't the
conversations be open? Probably because in a climate of falsification
everyone will appear to be failures. Still, shouldn't we seek out
those falsifications as a means to drive toward some fundamental
truth? Just imagine the level of slander that would go on with the
behemoths who struggle for millions of dollars for their pet projects.
We are humans practicing science as our only alternative, other than
resignation to reality on her own terms. Is the over exuberance of
wall street's elite possible in the scientific community? As an
outsider looking in, and seeing that fundamentals do still remain to
be constructed by humans, then the idea that we are overlooking
something fairly simple is an affirmation of the faith of science;
that answers do exist. The problems are open. Those who attempt to
shut others down are practicing dead science. Falsification of one
idea will lead to another idea, and that is how we work; the current
position should not take the attitude that it has taken, but we are
still caught nearby. The clean answers may exist, but these are not
necessarily consistent with the current position. Here is an agile
medium for propagation of ideas and the corrective falsifications that
are needed, with no worry of a control freak censor stepping in.
- Tim
spudnik
it's not just a "visualization programme" from teh Wolframites!
> > > http://math.ucr.edu/home/baez/physics/General/LightMill/light-mill.html
Dear Editor;
The staff report on plastic bags, given when SM considered a ban,
before, refused to list the actual fraction of a penny, paid for them
by bulk users like grocers & farmers at markets. Any rational EIR
would show that, at a fraction of a gram of "fossilized fuel (TM)" per
bag, a)
they require far less energy & materiel than a paper bag, and b)
that recycling them is impractical, beyond reusing the clean ones for
carrying & garbage.
As I stated at that meeting, perhaps coastal communities *should* ban
them -- except at farmers' markets -- because they are such efficient
examples of "tensional integrity," that they can clog stormdrains by
catching all sorts of leaves, twigs & paper. But, a statewide ban is
just too much of an environmental & economic burden.
--Stop British Petroleum's capNtrade rip-off;
tell your legislators, a tiny tax on carbon could achieve the result,
instead of "let the arbitrageurs/hedgies/daytrippers make
as much money as they can on CO2 credits!"
http://wlym.com
Timo Nieminen
> On Jun 1, 9:40 am, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> > So, whatever your doubts about Nichols' experiment, it's been repeated
> > and amply confirmed many times (and at least some of these were by
> > people who didn't Nichols' results were really good enough). These
> > days, it's almost trivial.
>
> I've reviewed your paper "Momentum of an electromagnetic wave in
> dielectric media".
> I wonder to what degree the name 'energy-momentum tensor' is a strict
> meaning?
> If we accept that the electric field is sinusoidal and so has zeros
> then all of the expressions that you consider contain instaneous zeros
> within their representation. This suggests that conservation of
> 'energy-momentum' does not exist, but then, these are not claims on
> just momentum are they? As you say we should consider a monochromatic
> and coherent source to simplify matters. This is fairly analogous to
> Maxwell plane wave isn't it?
>
> With the experiment of Ashkin and Dziedzic did they ever bother to
> drive the beam up out of the water into the air and observe a reversal
> of their effect? Did they ever bother to consider that the thermal
> heating of the water could be providing their effect?
They did it with the beam going from air to water, and from water to
air (in which case the water surface still bulges outwards). Yes, they
did consider thermal effects.
> In the experiment of Jones and Richard you say
> "Jones sought
> to verify the prediction that when a mirror was immersed
> in a dielectric medium, the radiation pressure exerted on
> the mirror would be proportional to the refractive index
> of the medium."
> yet this goes in exact contradiction to the claimed Reynolds effect,
> where immersion in a fluid is subject again to thermodynamic effects.
The two general difficulties in doing experiments of this kind are
that (a) the forces are small, and (b) thermal effects. Jones (in both
the '54 and the '78 experiments) did a reasonably good job of dealing
with the thermal effects (don't have a copy here with me to check the
details, but he does discuss this).
> As you say above here to just relax, because these experiments have
> been done again, and that the weakness of Nichols work does not exist
> in perpetutity, well, I believe that I have presented evidence here
> that these experimenters are the ones who relaxed on top of the work
> of Nichols.
No, you haven't presented any _evidence_ of this here. Read the
relevant papers, and make sure you read the later ones. The pioneer
experiments aren't always the best.
One difficulty with thermal effects is that, to first order, they'll
be proportional to the power, just like the radiation forces. What to
do? Firstly, immersion in a liquid reduces thermal effects as compared
with a gas - the thermal conductivity of the liquid is much higher, so
temperatures don't rise as much. Doing this with very small objects (a
few microns in size, as in optical tweezers) helps too, since the
distances over which heat needs to be conducted is very small.
Next, it would be quite a coincidence for the thermal force to give a
force equal to (i.e., within a reasonably small experimental error)
that expected due to radiation pressure. But it could happen. So you
try it with liquids of different refractive indices. If the expected
dependence on refractive index is there, it's a very, very stong
indication that the force is due to radiation momentum, not thermal
forces. This is what Jones did. This is also confirmed by lots of
optical tweezers experiments. In particular, in some of these, it's
known that the thermal forces would move the object in the wrong
direction.
One of the best experiments is:
http://apl.aip.org/applab/v87/i22/p221109_s1
where they measured the optical force on a microsphere, as the
wavelength is changes (using a tunable laser). They see the Mie
resonances and whispering gallery resonances, as expected from
radiation momentum, and not from thermal forces.
The other thing to try, changing materials so that the optical forces
should be same but the thermal forces different, I haven't seen
explicitly tried. That said, thermal forces can be seen in optical
tweezers in some circumstances, especially with surfaces present
(e.g., due to bubbles). We've seen the effects of convection. But a
clear and simple experiment to look at the size of thermal forces
could be good. (Maybe it's been done?)
But the absolutely conclusive experiments showing that there are force
that aren't due to thermal effects are various atom trapping
experiments (including Bose-Einstein condenstate (BEC) experiments).
> Lastly, I would argue from a very simplistic perspective that if
> conservation of energy and momentum is upheld that these figures in
> vacuum should match these figures in solid media.
> Otherwise where did the energy or momentum go?
As a beam enters a non-absorbing and non-reflecting (use an anti-
reflection coating if you wish) medium, the power can't change - where
would or could the energy go? But the momentum can change. A force is
a transfer of momentum - all you need to change the momentum of the
beam is a force on the surface.
> I admit that my own belief is that the momentum is actually zero in
> terms of a steady unidirectional push on a plate of material, but in
> that light does propagate in a given direction we must admit the
> energy itself is directed, but when for instance we absorb all of that
> energy as in a black ideal absorber, that it has nothing to do with
> mechanical momentum.
And yet, a force acts on the absorber.
> The claims of propulsion doubling from a perfect
> reflector only help to boost my position.
Why? Force due to reflection is observed. The amount is such that the
force on a perfect reflector would be double that on a perfect
absorber.
> I guess to pose the simplest
> question that I can on this is to ask whether the energy of a photon
> e = h f
> is the entire energy of the photon?
The results on force and momentum are entirely compatible with this.
The classical experiments don't say anything about each photon, but
the various atom trapping experiments do. I dont' know this literature
well, but perhaps a good starting point would be the Chu/Phillips/
Cohen-Tannoudhi Nobel speeches/papers.
> What have you concluded about
> Wormley's claim on the momentum figure here? Isn't this figure the
> classical blackbody figure of momentum? How did the discrepancy then
> to a very slight momentum figure arise and how do we split that
> smaller energy off of this photon energy without corrupting physics?
??? What do you mean "split that smaller energy off"?
Momentum = energy/c for a photon, momentum flux = power/c for a
parallel beam. This is a "small" momentum compared to the energy (but
be careful when comparing - these are in different units!). How is
this a problem?
> I'm sorry but the effect in a perfect vacuum has not been clearly
> demonstrated. As you've accepted the flaws of Nichols work then the
> question of whether the radiation pressure is observable in vacuum
> still exists. The experiments that you've exposed in your paper must
> not be the modern experiments that you speak of.
Perfect vacuum, no. Very good vacuums, yes, especially with atom
trapping. I've seen classical experiments done in vacuum (can't recall
how good), where absorbing particles were blasted by short pulses of
light, to measure their radiation pressure cross-sections.
The bulk of the modern experiments aren't experiments on radiation
momentum; they just use it in the experiment. But the experiment by
Calos Lenz Cesar's group (http://apl.aip.org/applab/v87/i22/p221109_s1
as cited above) is an excellent modern experiment.
> > > > If the reflector is moving, there will be a Doppler shift, and energy in
> > > > will be different from energy out. If the reflector is stationary, then
> > > > you could have, e.g., 1300W in and 1300W out. But no work would be done.
>
> > > Very good. We have agreement, particularly at the low velocities that
> > > these instruments work at. As far as we can tell there is no reliance
> > > upon red shifting of light, which could provide some work.
>
> > The force doesn't depend on redshift. But it doesn't take any
> > expenditure of energy to produce a force - just ask a fridge magnet or
> > a paperweight. Having the force do work does very much depend on
> > redshift, since the reflector has to move for the radiation pressure
> > to do work on it.
>
> Awww, come on.... The claim is that the force is doing work, and the
> existence of any redshift in doing that work has been completely
> ignored in any of these theories or experiments.
This is just wrong. It's true that a lot of the theory doesn't take
any redshift into account, but that's because they're calculating the
force on a stationary object, when there's no redshift. (more below)
> Without doing some
> work there will be no indication as of a vane which rotates on a
> tensioned fiber. We know we're at extremely low velocity in this
> experiment, so any redshift observed would be a highly impressive
> mechanism.
At very low speeds, you don't need to deal with redshift to find
forces in classical experiments. The force exists at v=0 (when there
is no redshift and no work), and for small v, the force is almost
exactly the same. The difference between the v=0 force and the small
but non-zero v force is much, much, smaller than can be detected in
most experiments.
It does matter in atom trapping and cooling, and is essential for the
proper functioning.
> I don't mean to get too ornery, but you are sweeping over these points
> as if there is no fundamental problem.
In this case, there isn't any fundamental problem. The known theory
gives a very good quantitative match for the observed results.
> we will first need something working in vacuum
> which observes the supposed radiation pressure.
Basically, the atom-trapping experiments are as close to "perfect
vacuum" experiments as will likely be available.
> From the theoretical
> side there is a reliance upon a third of the energy density for this
> pressure that we still have not arrived at, though you've gotten very
> close in Maxwell's treatise.
No, there isn't any such reliance; as I already said, there are many
pathways (Lorentz force, Noether's theorem, thermodynamics, the simple
derivation in my other post). (The 1/3 turns up for omnidirectional
radiation, like blackbody radiation. For a directed beam or a plane
wave, p = P/c.)
> Here is the strongest link that I have
> found in support of challenging the existing theory:
> http://www.neumann-alpha.org/lightpressure.pdf
> This site is a bit out there, but then, so too was Nichols. Any who
> will rest on top of his work, and that is how science is done, are on
> a shaky ladder.
Will look when I have time.
--
Timo
NoEinstein
>
Dear Tim: I agree with you, totally! Your reply should be required
reading for the stone brains who... censor True Science from the
journals. I was impressed by your assessment, earlier, that the
Nichols Radiometer 'might not be analogous’ to the Crookes, which it
purported to be an extension of. After reading the entire
Astrophysical Journal article, I was moved to make a new post, today,
entitled: A proposed Gravity-Propelled Swing Experiment. Read it and
let me hear what you think. There are simple experiments which can
finally get us out of the Dark Ages of Einstein! — NoEinstein —
NoEinstein
>
Dear Timo: Too much verbiage hurts your science content. Most will
read two concise paragraphs. You've got those... in you, but you
don't let others see them, because you enjoy... conversation too
much. Someone once said: "Nothing constructive was ever done while
talking." True science requires thought and action, not talk. —
NoEinstein —
>
> read more »- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
Timo Nieminen
> On Jun 2, 6:09 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> Dear Timo: Too much verbiage hurts your science content.
So why do you post so many long posts? Over and over?
> Most will
> read two concise paragraphs.
I've given you replies of two concise paragraphs. All you did was
refuse to reply. And rudely at that.
If you didn't want the information, why did you ask?
> Someone once said: "Nothing constructive was ever done while
> talking." True science requires thought and action, not talk. —
Very true. And since there is no action (and apparently no thought)
from you, I can safely assume that you're not doing science, whatever
you might fantasise.
Tim BandTech.com
> > I'm sorry but the effect in a perfect vacuum has not been clearly
> > demonstrated. As you've accepted the flaws of Nichols work then the
> > question of whether the radiation pressure is observable in vacuum
> > still exists. The experiments that you've exposed in your paper must
> > not be the modern experiments that you speak of.
>
> Perfect vacuum, no. Very good vacuums, yes, especially with atom
> trapping. I've seen classical experiments done in vacuum (can't recall
> how good), where absorbing particles were blasted by short pulses of
> light, to measure their radiation pressure cross-sections.
>
> The bulk of the modern experiments aren't experiments on radiation
> momentum; they just use it in the experiment. But the experiment by
> Calos Lenz Cesar's group (http://apl.aip.org/applab/v87/i22/p221109_s1
> as cited above) is an excellent modern experiment.
There are a number of weak points in the post that I am responding to,
but I want to focus on one for now. I'd like to come back to some of
the others later, but here is one crux that still is unanswered. The
link
http://www.mcallister.com/vacuum2.html#params
exposes that modern vacuum apparatus is capable of 10E-11 torrs(mm of
mercury).
One experiment exposing the Nichols or Crooks radiometers operating
cleanly at this gas pressure would be quite a piece of evidence.
Nichols got down to .02 torrs, and had an anomaly between .05 and .02.
As you identified, there is a confluence between radiation absorption
as temperature and a figure of radiation pressure. Specific dependence
upon atmospheric effects is already understood. The word 'radiometer'
is intended as a device to measure the amount of radiation. We would
like a 'radiation pressure meter' which will not generate any
behaviors due to thermal absorption from the 'radiation energy' which
I believe you will agree is encompassed in
e = hv
as a sum of photon energies. Because the 'radiation energy' is so
substantial at 1300 W/m/m for sunlight then eliminating the effect of
this large energy which could easily power quite some motive work is a
substantial problem for the verification of the claimed pittance that
is 'radiation pressure'. I see no work so far that proves the
existence of radiation pressure since all of the experiments that I
have seen do rely upon some gas effects. The only claimed exception
that I've read so far is Nichols, and his logic is falsifiable. None
of the experiments you've studied consider a solid dielectric placed
in a vacuum of 10E-11 torrs, which would be a fine experiment with a
Nichols torsion fiber style experiment.
Here is a link I just came accross very similar to your own work
though more concerned with semiconductors:
http://www.colin-baxter.com/academic/research/downloads/prl063802.pdf
I do not have access to pay for links like the aip.org link you site
about optical tweezers.
Here is another link in support of the challenge to radiation
pressure's validity:
"It is therefore much more likely that in a given case the apparent
'radiation pressure' is caused either by thermal surface effects or
electrons which are released from the surface by the radiation."
- http://www.physicsmyths.org.uk/#radpress
This would really be something if Nichols anomaly was repeatable and
turned out to be a mild photoelectric effect.
It was after all an inversion. A freed electron would exit the
silvered side; not the insulated glass side. The photovoltaic cell is
the rectifier that converts AC light to DC. You've ignored the
instantaneous momentum argument I made haven't you? This AC/DC
problem, well, now I'm going too far again. One thing at a time, eh?
- Tim
Timo Nieminen
>
> As you identified, there is a confluence between radiation absorption
> as temperature and a figure of radiation pressure.
Not quite. For a _given experiment_, the radiation pressure and the
temperature increase are proportional. This is because the absorbed
power, and hence the temperature increase, and the radiation pressure
are both proportional to the radiation power.
Across experiments, this isn't so. As already said, R. V. Jones, and
pretty much all optical tweezers experiments. That the forces in
optical tweezers on weakly depend on the absorption (over orders of
magnitude of change in the absorption and heating) while strongly
depending on the refractive index amply demonstrates that the effect
is not thermal.
What shows the weak dependence on absorption? Trapping the same object
at different wavelengths, with different absorption at the different
wavelengths, and looking at the effect on the forces; this has been
done to look at thermal damage to the trapped object, especially
living trapped objects. Trapping at 1064nm in water vs heavy water,
which have very different absorption, gives very different temperature
rises. We've trapped objects that absorb about 10x as much as water,
and therefore get significantly hotter (we can see the effects of
convection at high power), and objects that absorb significantly less
than the surrounding water. The temperature makes little difference.
> Specific dependence
> upon atmospheric effects is already understood. The word 'radiometer'
> is intended as a device to measure the amount of radiation. We would
> like a 'radiation pressure meter' which will not generate any
> behaviors due to thermal absorption from the 'radiation energy' which
> I believe you will agree is encompassed in
> e = hv
> as a sum of photon energies. Because the 'radiation energy' is so
> substantial at 1300 W/m/m for sunlight then eliminating the effect of
> this large energy which could easily power quite some motive work is a
> substantial problem for the verification of the claimed pittance that
> is 'radiation pressure'.
This is why the most reliable experiments use reflection/refraction by
dielectric surfaces to change the momentum of the light. But one can
see radiation pressure due to absorption.
Some freely-available references for you:
Trapping polymer molecules, with a discussion of thermal effects:
http://arxiv.org/abs/physics/0702044
Forces and torques due to absorption:
http://arxiv.org/abs/physics/0310003
and more details:
http://arxiv.org/abs/physics/0310022
For comparison, torques not due to absorption:
http://arxiv.org/abs/physics/0308113
Note that the same theory works for both cases, absorption and no
absorption.
> I see no work so far that proves the
> existence of radiation pressure since all of the experiments that I
> have seen do rely upon some gas effects. The only claimed exception
> that I've read so far is Nichols, and his logic is falsifiable. None
> of the experiments you've studied consider a solid dielectric placed
> in a vacuum of 10E-11 torrs, which would be a fine experiment with a
> Nichols torsion fiber style experiment.
Re: "all of the experiments that I have seen do rely upon some gas
effects." Most optical tweezers experiments are done in liquid (the
ones done in air do have significant "gas effects"), Jones's
experiments were in liquid. The BEC experiments require very good
vacuums. True, they don't use a solid dielectric, unless you consider
an atom to be a solid dielectric. What difference does that make?
Radiation pressure is radiation pressure whether it acts on an atom or
a group of atoms.
> Here is another link in support of the challenge to radiation
> pressure's validity:
> "It is therefore much more likely that in a given case the apparent
> 'radiation pressure' is caused either by thermal surface effects or
> electrons which are released from the surface by the radiation."
> -http://www.physicsmyths.org.uk/#radpress
That is a _really_ bad webpage! Just consider its claim: "Even if one
assumes a momentum, a radiation pressure force could only be caused by
a momentum change dp/dt, but this is not possible because the speed of
light c has to be constant" (1) Direction of motion matters when it
comes to momentum, (2) refractive index.
> You've ignored the
> instantaneous momentum argument I made haven't you? This AC/DC
> problem, well, now I'm going too far again. One thing at a time, eh?
Your comment: "I'm pretty sure the radiation pressure argument is
false. Photons are
more like AC sources than DC sources. If the pressure is alternating
then to claim a propulsive acceleration is a misnomer. It would be
like putting a ping pong ball in front of a speaker, blasting out 100
watts and expecting the ping pong ball to accelerate away from the
speaker."?
Acoustic radiation pressure is observed. The ping pong ball _will_
accelerate. How does this not address this argument?
In the optical experiments, the time-averaged force is what is
observed. The predicted time-averaged force, which agrees with the
observations, is the average of the predicted instantaneous force, so
what reason is there to think that the predicted instantaneous force
is wrong?
So where lies the problem?
What did you think of my simple derivation of radiation pressure? What
do you disagree with in it?
Consider a steady beam of monchromatic light incident on a stationary
reflector. Each optical cycle has energy E (i.e., the total energy of
a length of the beam equal to 1 wavelength). The incident power P is
thus P = Ef, where f is the optical frequency.The output power is the
same, since the reflected beam is identical, except for the reversed
direction. Since the reflector is stationary, no work is being done on
it.
Now change to a coordinate system where the reflector is in motion, at
a speed v away from the source of the incident beam. (For simplicity,
assume v << c.) The incident beam is now blue-shifted, and the
frequency of the incident beam in this coordinate system is now f(1+v/
c), and the power is P_in = Ef(1+v/c). Similarly, the reflected beam
now has power P_out = Ef(1-v/c).
Since energy is conserved, the difference in power must be the rate of
doing work on the reflector.
P_reflector = P_in - P_out = 2Efv/c = 2vP/c.
Since P_reflector = F_reflector * v, we have:
F_reflector = 2P/c,
which is the radiation pressure due to complete reflection at normal
incidence.
And, again, one can predict all of this just from the Lorentz force.
So, fundamentally, electromagnetic radiation pressure is just what one
would expect, given that electric motors etc work. If there is no
radiation pressure due to electromagnetic waves, then the Lorentz
force is bunk. Thus, some choices: (1) light is not an electromagnetic
wave (but radiation forces and torques have been done at RF), (2) the
Lorentz force is bunk, and electric motors work by some other mystery
mechanism, (3) the forces which have been measured at optical
frequencies and RF, and which, when thermal effects are made
observably negligible - that is, their effect is shown to be small -
agree with prediction based on the Lorentz force or equivalent, are
really radiation pressure. Seriously, I think (3) is the best-
supported of these choices.
--
Timo
spudnik
"photonics" is a huge field, but you don't have to use equations
with momentum to describe lightwaves; the photoelectrical effect
is electronic. how do photons interact with atoms,
excepting electromagnetically via electron orbitals?
thusNso:
current Sci ... no, it's in Nude Scientist, issue-before-last cover-
article, almost like it's New Science, covering the main three
orbital variations of Milankovitch. now, it is just not true that
humans cannot (or aren't) affecting the climate;
the tiny differences of insolation form the orbitals probably
are not "forcing" anything, just tweaking them.
any real theory of glaciation has to account for "glass house gasses"
as the primary forcers, and how the biota & tectonic systems interact
with that.
> There is not a single article anywhere describing what the Earth's
> orbit is doing as it moves along its annual circumference so this is
> http://daphne.palomar.edu/jthorngren/tutorial.ht
> Commentariolis
thusNso:
yes, but special relativity assumes general relativity
in the "twin paradox," because acceleration is required
to get the home-leaving twin, relativistical (I mean, Duh .-)
> Special Relativity considers that relativistic effects such as time
> dilation and length contraction are perspective effects that occur
> when an observer observes an object moving relative to himself.
thusNso:
anyway, Einstien's **** is not really dysprovable, if
it is merely a matter of odd interpretations (viz, *photon*
means "particle" ipso facto "herr Albert thought, So.")
thusNso:
always the "doubling" of CO2 is used as an outcome in the GCMs,
when it is clear that there would be change of the whole phase
of the weather, before that was reached (if you are familiar
with studies of the Quaternary Period, Shackleton et al e.g.).
Dear Editor;
The staff report on plastic bags, given when SM considered a ban,
before, refused to list the actual fraction of a penny, paid for them
by bulk users like grocers & farmers at markets. A rational EIR'd
show that, at a fraction of a gram of "fossilized fuel (TM)" per bag,
a)
they require far less energy & materiel than a paper bag, and b)
that recycling them is impractical & unsanitary,
beyond reusing the clean ones for carrying & garbage. (Alas,
the fundy Greenies say that the bags are not biodegradeable,
but everyday observation shows, they just don't last so long.)
As I stated at that meeting, perhaps coastal communities *should* ban
them -- except at farmers' markets -- because they are such efficient
examples of "tensional integrity," that they can clog stormdrains by
catching all sorts of leaves, twigs & paper. But, a statewide ban is
just too much of an environmental & economic burden.
--Stop BP's and Waxman's capNtrade arbitrageur rip-off!
http://wlym.com
Tim BandTech.com
Yes, and this is very similar to the arguments that I have provided.
The conservation of momentum is a strict principle, one that you have
already cast aside in your argument about changing momenta as media
change in dielectric quality. You already accepted once the farce of
the reflector as a doubling agent of radition pressure effects, and
then go back to supporting it. What do you have to say about
conservation of momentum?
>
> > You've ignored the
> > instantaneous momentum argument I made haven't you? This AC/DC
> > problem, well, now I'm going too far again. One thing at a time, eh?
>
> Your comment: "I'm pretty sure the radiation pressure argument is
> false. Photons are
> more like AC sources than DC sources. If the pressure is alternating
> then to claim a propulsive acceleration is a misnomer. It would be
> like putting a ping pong ball in front of a speaker, blasting out 100
> watts and expecting the ping pong ball to accelerate away from the
> speaker."?
>
> Acoustic radiation pressure is observed. The ping pong ball _will_
> accelerate. How does this not address this argument?
>
> In the optical experiments, the time-averaged force is what is
> observed. The predicted time-averaged force, which agrees with the
> observations, is the average of the predicted instantaneous force, so
> what reason is there to think that the predicted instantaneous force
> is wrong?
Well, for one, there is an observation that a black surface is drawn
toward the light rather than being driven away from the light, but I'm
not that strong on this as an argument to hinge upon since the
information is not well presented.
I can accept that in a plane wave of acoustic pressure that the ping
pong ball will accelerate in an AC fashion, that is to say a
sinusoidal acceleration, yielding no DC acceleration, which is what
the radiation pressure claim is, and thus that it might propel
spacecraft with no need of thrusters, which I am currently in
disbelief of. Even if you were correct about a ping pong ball
accelerating in an acoustic field, it would not be wise to hinge this
discussion on that sideline of thought. You have not provided any
evidence beyond a ping pong ball with a bolt fastened to it sitting in
the magnetic field of a loudspeaker, which I actually provided as you
suggested to youtube this effect. Please provide a more apt link if
you wish. I did not go on to look at every video on youtube and
instead was so displeased with this first one that I plopped it down,
but you do seem to be serious so please do give us a good link. When
the ball is in the proximity of the speaker cone we will have more
dynamics to worry about than in a plane field, so again I say this may
not be a wise point to hinge any discussion of electromagnetic
radiation pressure.
Within your own analysis of radiation pressure you use cross products
of the E field. We should accept that the E field is sinusoidal, and
so contains instantaneous zeros. These zeros are points in time at
which your effect has gone to zero. This was somewhat the argument
that I was making, which you've somewhat demolished by throwing in the
ping pong ball.
Really I am trying to discorrupt this thread, as my last post went
toward one simple argument, and by not addressing that argument you
are providing a form of evidence. You have backed off of any radiation
pressure effect in vacuum and instead support it in fluid mediums
only. This is evidence.
>
> So where lies the problem?
>
> What did you think of my simple derivation of radiation pressure? What
> do you disagree with in it?
If this is the expressions with velocity and red shifting then I have
commented in line on that. Summarily the radiation pressure effect
should be apparent at zero velocity and so there is a fault in that
attempt.
>
> Consider a steady beam of monchromatic light incident on a stationary
> reflector. Each optical cycle has energy E (i.e., the total energy of
> a length of the beam equal to 1 wavelength). The incident power P is
> thus P = Ef, where f is the optical frequency.The output power is the
> same, since the reflected beam is identical, except for the reversed
> direction. Since the reflector is stationary, no work is being done on
> it.
OK. You've somewhat legitimated my sinusoid argument elsewhere here,
but yes, I am following this construction here.
>
> Now change to a coordinate system where the reflector is in motion, at
> a speed v away from the source of the incident beam. (For simplicity,
> assume v << c.) The incident beam is now blue-shifted, and the
> frequency of the incident beam in this coordinate system is now f(1+v/
> c), and the power is P_in = Ef(1+v/c). Similarly, the reflected beam
> now has power P_out = Ef(1-v/c).
In the frame of the reflector the light will appear redder, since the
reflector is moving away from the source by v, so I disagree with your
blue-shifted claim of the incident beam.
As this light will be reflected and returned toward its source I
believe that an observer in the source frame will observe light that
is red shifted relative to the source off of that receding reflector.
Will the light be doubly red shifted to this observer over an observer
in the reflector's frame? I do fear that we could suffer some
breakdowns here as to whether or not the reflector is perfect, well,
if it does not return all the energy that it received then it is not a
perfect reflector. As long as we are studying breakdowns we may also
admit that the photon takes an interval of zero seconds to do all of
this that we are discussing.
We are more engaged at this part in relativistic cosmology than in
radiation pressure.
>
> Since energy is conserved, the difference in power must be the rate of
> doing work on the reflector.
Well, this is a different topic isn't it? We are supposed to be able
to start with a Nichols radiometer in stasis; no acceleration or
velocity relative to the lab bench. I do find the possibility of red
shifting light a more plausible way to extract energy, but this is not
at all the claim of radiation pressure. We need to get an acceleration
at zero velocity in order to accept radiation pressure.
>
> P_reflector = P_in - P_out = 2Efv/c = 2vP/c.
>
> Since P_reflector = F_reflector * v, we have:
>
> F_reflector = 2P/c,
>
> which is the radiation pressure due to complete reflection at normal
> incidence.
>
> And, again, one can predict all of this just from the Lorentz force.
> So, fundamentally, electromagnetic radiation pressure is just what one
> would expect, given that electric motors etc work. If there is no
> radiation pressure due to electromagnetic waves, then the Lorentz
> force is bunk. Thus, some choices: (1) light is not an electromagnetic
> wave (but radiation forces and torques have been done at RF), (2) the
> Lorentz force is bunk, and electric motors work by some other mystery
> mechanism, (3) the forces which have been measured at optical
> frequencies and RF, and which, when thermal effects are made
> observably negligible - that is, their effect is shown to be small -
> agree with prediction based on the Lorentz force or equivalent, are
> really radiation pressure. Seriously, I think (3) is the best-
> supported of these choices.
>
> --
> Timo
I really appreciate your effort here Timo, and I do not want to
detract from or inhibit your own thinking. Still, wouldn't you think
that the Nichols radiometer could act as a radiation pressure meter if
it were put in a strong vacuum? Why rely upon fluid media when vacuums
of 10E-11 torr exist? You are quite a dodger on this simplistic
thinking. I feel very hesitant heading for relativity theory when the
claims are supposedly satisfiable on a lab bench with a laser source
and vane in vacuum, all right there on the bench. Yes, it is
interesting to consider the red shift of the light off of the moving
reflector especially in terms of energy conservation, but it is not
the same problem.
- Tim
NoEinstein
>
Dear Timo: I expected you to ask about "the length" of my comments.
There is a difference between what you write and what I write: First,
the '+new post' is mine, not yours. I'm aware that there are
thousands of people reading what I say. I'm also aware that few of
those bother to go to the original post to read about what we are
discussing. Because I know my subject, I willingly explain things
from square one, with minor differences in the subtleties. I wish to
be as clear as possible, so that the largest number of people will
agree with what I say. Second, if you (knee-jerk) question any part of
my New Science, you force me to reply aggressively so that the many
readers will know that I know from whence I come.
When skimming over your replies, I sense that the latter 2/3rds is
copied, as though you are BUYING credibility with verbiage. If you
wish for my replies to be shorter, stop making unjustified put-downs
of any of my science truths. If I feel you are a friend rather than
an adversary, I'll be more likely to reply shorter. That said, I only
have so many hours in a day. I'm about "at" my talking limit, now.
Don't reply to me... just for recreation. I only have time for work!
— NoEinstein —
NoEinstein
>
Dear Tim: You have the right sort of analytical mind! I've made a
new post about my proposed Gravity Swing experiment. That should
answer most of your questions. If you have access to funding, that
experiment would be well suited to your intellect. — NoEinstein —
NoEinstein
>
Dear Spudnik: Please leave the name of the person to whom you are
replying at the top. Or just say: "Dear Timo, or etc." And try to
shorten your replies. "Truth in science is INVERSELY proportional to
the amount of words needed to explain things! — NoEinstein —
NoEinstein
>
Dear Tim (and Timo): Photons are massless ENERGY. Such don't impart a
kick to the mirror when such arrive, nor do they impart a
"thrust" (force), 1/2 phase later, when they are re emitted. My
proposed Gravity Swing experiment (Please search and find.) should
prove those points, conclusively. — NoEinstein —
>
> read more »- Hide quoted text -
Timo Nieminen
! I haven't seen you directly arguing such complete nonsense. That web
page was saying that the direction of motion doesn't affect momentum,
that if an object changes direction and goes, e.g., in the exact
opposite direction, there is no change in momentum. Before you claim
that the argument given there is very similar to yours, you should
understand just how completely defective that argument is.
> The conservation of momentum is a strict principle, one that you have
> already cast aside in your argument about changing momenta as media
> change in dielectric quality
?? Conservation of momentum is why, if the momentum flux of the beam
changes, there must be a matching force on the surface, or, if there
is a force on the surface of the liquid (as we observe
experimentally), then the momentum flux of the beam must change?
Why do you say this "casts aside" conservation of momentum? It's
directly based on the conservation of momentum.
> You already accepted once the farce of
> the reflector as a doubling agent of radition pressure effects, and
> then go back to supporting it. What do you have to say about
> conservation of momentum?
You don't think the momentum of an object that reverses direction
changes? You don't think that a force (i.e., a change in momentum as
per Newton) is needed to change the direction of an object?
If one instead, at least provisionally, accepts Newton's laws of
motion as correct, if an object with momentum p reverses direction,
while maintaining the same speed, the final momentum is -p, for a
change in momentum of -2p, with an average force of -2p/t where t is
the time over which the force that causes the change in direction is
applied. If instead, the object is "absorbed", has a sticky collision,
then the change in momentum is -p, only half of the "reflection"
change in momentum. What's so mysterious about radiation pressure on a
reflector being double?
> > > You've ignored the
> > > instantaneous momentum argument I made haven't you? This AC/DC
> > > problem, well, now I'm going too far again. One thing at a time, eh?
>
> > Your comment: "I'm pretty sure the radiation pressure argument is
> > false. Photons are
> > more like AC sources than DC sources. If the pressure is alternating
> > then to claim a propulsive acceleration is a misnomer. It would be
> > like putting a ping pong ball in front of a speaker, blasting out 100
> > watts and expecting the ping pong ball to accelerate away from the
> > speaker."?
>
> > Acoustic radiation pressure is observed. The ping pong ball _will_
> > accelerate. How does this not address this argument?
>
> > In the optical experiments, the time-averaged force is what is
> > observed. The predicted time-averaged force, which agrees with the
> > observations, is the average of the predicted instantaneous force, so
> > what reason is there to think that the predicted instantaneous force
> > is wrong?
>
> Well, for one, there is an observation that a black surface is drawn
> toward the light rather than being driven away from the light, but I'm
> not that strong on this as an argument to hinge upon since the
> information is not well presented.
This (the classic Crookes radiometer) is _not_ a radiation force
experiment. In Crookes, the radiation force is negligible; in a
typical instrument, it isn't enough to turn the vanes against
friction. The pressure needs to be in the correct range for the device
to work - it depends on non-thermal equilibrium low-pressure gas flow.
To much gas, and it doesn't work, too little gas, and it doesn't work
either.
> I can accept that in a plane wave of acoustic pressure that the ping
> pong ball will accelerate in an AC fashion, that is to say a
> sinusoidal acceleration, yielding no DC acceleration, which is what
> the radiation pressure claim is, and thus that it might propel
> spacecraft with no need of thrusters, which I am currently in
> disbelief of. Even if you were correct about a ping pong ball
> accelerating in an acoustic field, it would not be wise to hinge this
> discussion on that sideline of thought. You have not provided any
> evidence beyond a ping pong ball with a bolt fastened to it sitting in
> the magnetic field of a loudspeaker, which I actually provided as you
> suggested to youtube this effect.
Do a better search (and don't search for ping pong balls, just search
for acoustic levitation). For example, search on Google Scholar for
acoustic levitation.
Since you appear to be arguing (above) in direct contradiction to
Newton's laws of motion, and are supporting a web page that certainly
does so, will a link here really help? If you don't accept Newton's
laws of motion, what of modern science will you accept? That the above-
suggested Google Scholar search throws up a screenful of links where
acoustic levitation is being put to practical use might convince some,
but they're products of modern science.
I don't just mean this as rhetoric! You're supporting your argument
with a page spouting complete anti-Newtonian nonsense, and rejecting
the acoustic radiation pressure just because you didn't like the first
youtube video you found (in which, if the force was magnetic due to
the bolt, it would attract it towards the speaker magnet).
http://www.youtube.com/watch?v=jpvbqdXJ1lM
http://www.youtube.com/watch?v=9iQF61Ez000
> When
> the ball is in the proximity of the speaker cone we will have more
> dynamics to worry about than in a plane field, so again I say this may
> not be a wise point to hinge any discussion of electromagnetic
> radiation pressure.
So what if there are more dymanics? If there is a time-averaged force,
there is acoustic radiation pressure. The electromagnetic experiments
we do don't use a plane wave.
> Within your own analysis of radiation pressure you use cross products
> of the E field. We should accept that the E field is sinusoidal, and
> so contains instantaneous zeros. These zeros are points in time at
> which your effect has gone to zero.
Yes, and? For a plane polarised beam, you can think of it as a
succession of blobs of energy, 2 per wavelength, moving at the wave
speed. It's the blobs of energy that have the momentum (since they
have the energy). As the direction of each blob is reversed as the
beam is reflected (or each blob is absorbed, if you prefer to consider
radiation pressure by absorption), there is a force. Between blobs,
nothing is happening, so why should there be a force? No energy is
changing direction then, so the momentum is changing, so a non-zero
force isn't expected, since that would violate conservation of
momentum.
> This was somewhat the argument
> that I was making, which you've somewhat demolished by throwing in the
> ping pong ball.
>
> Really I am trying to discorrupt this thread, as my last post went
> toward one simple argument, and by not addressing that argument you
> are providing a form of evidence.
Sorry, I didn't see what the AC, instantaneous zero thing has to do
with it. If you intended the point to be that the time-average of a
sinusoidal force is zero, and that therefore radiation pressure can't
exist, the example of radiation pressure in action - examples of
acoustic radiation pressure, for example - show that the force isn't
purely sinusoidal. (Sorry, I don't know a simple direct theoretical
argument. If the object scatters any of the incident light/sound, the
force isn't purely sinusoidal. The circuit analogy is simple enough. A
lossless non-radiating antenna looks like a capacitor to the circuit
driving it. P=VI gives exactly sinusoidal power, average power is
zero. If there is any Ohmic resistive loss, then it looks like a
capacitor + a resistor. Then, P=VI no longer gives zero average
("power factor"). If the antenna radiates energy, then it has non-zero
radiation resistance, which looks, to the driving circuit, identical
to Ohmic resistance. As far as the circuit is concerned, loss of
energy is loss of energy, and the same non-zero time-average of P=VI
results with a radiating lossless antenna.)
If you mean something else, please explain.
> > So where lies the problem?
>
> > What did you think of my simple derivation of radiation pressure? What
> > do you disagree with in it?
>
> If this is the expressions with velocity and red shifting then I have
> commented in line on that. Summarily the radiation pressure effect
> should be apparent at zero velocity and so there is a fault in that
> attempt.
OK, you misunderstood the example. I was not clear enough. Exactly the
same physical system is being analysed in two (or more, if you prefer)
reference frames. All we do is look at the source and reflector in a
frame when both are stationary, and look at the work done on the
reflector (based on conservation of energy and redshift/blueshift of
the incident and reflected light), and then re-do for a reference
frame where the reflector is moving (and the source is moving with
velocity indentical to the reflector).
It does say something - a lot, even - about force at v=0, but let us
first see if you agree with this analysis of work, with this
clarification about no relative motion between source and reflector.
(Else there isn't any point in discussing the consequences of this
analysis, is there?) (Most of your comments deleted for brevity, the
relevant one kept.)
> > Consider a steady beam of monchromatic light incident on a stationary
> > reflector. Each optical cycle has energy E (i.e., the total energy of
> > a length of the beam equal to 1 wavelength). The incident power P is
> > thus P = Ef, where f is the optical frequency.The output power is the
> > same, since the reflected beam is identical, except for the reversed
> > direction. Since the reflector is stationary, no work is being done on
> > it.
>
> > Now change to a coordinate system where the reflector is in motion, at
> > a speed v away from the source of the incident beam. (For simplicity,
> > assume v << c.) The incident beam is now blue-shifted, and the
> > frequency of the incident beam in this coordinate system is now f(1+v/
> > c), and the power is P_in = Ef(1+v/c). Similarly, the reflected beam
> > now has power P_out = Ef(1-v/c).
>
> In the frame of the reflector the light will appear redder, since the
> reflector is moving away from the source by v, so I disagree with your
> blue-shifted claim of the incident beam.
We only changed the coordinate system; we didn't introduce any
relative motion between the source and the reflector. All we do is
choose a new coordinate system with origin moving in the direction
from reflector to source.
In the frame of the reflector, everything should (must!) remain
identical. The question is, what does it look like in _another_
reference frame.
> > P_reflector = P_in - P_out = 2Efv/c = 2vP/c.
>
> > Since P_reflector = F_reflector * v, we have:
>
> > F_reflector = 2P/c,
>
> > which is the radiation pressure due to complete reflection at normal
> > incidence.
>
> > And, again, one can predict all of this just from the Lorentz force.
> > So, fundamentally, electromagnetic radiation pressure is just what one
> > would expect, given that electric motors etc work. If there is no
> > radiation pressure due to electromagnetic waves, then the Lorentz
> > force is bunk. Thus, some choices: (1) light is not an electromagnetic
> > wave (but radiation forces and torques have been done at RF), (2) the
> > Lorentz force is bunk, and electric motors work by some other mystery
> > mechanism, (3) the forces which have been measured at optical
> > frequencies and RF, and which, when thermal effects are made
> > observably negligible - that is, their effect is shown to be small -
> > agree with prediction based on the Lorentz force or equivalent, are
> > really radiation pressure. Seriously, I think (3) is the best-
> > supported of these choices.
> I really appreciate your effort here Timo, and I do not want to
> detract from or inhibit your own thinking. Still, wouldn't you think
> that the Nichols radiometer could act as a radiation pressure meter if
> it were put in a strong vacuum? Why rely upon fluid media when vacuums
> of 10E-11 torr exist? You are quite a dodger on this simplistic
> thinking.
It might be a nice radiation pressure meter, but we don't need it as a
radiation pressure meter. The experiments in fluid media are
sufficient, and the many, many atom trapping and cooling experiments
are sufficient vacuum experiments (and are done in such ultra-high
vacuums).
I really don't see a need to do this experiment. I don't know of any
non-atomic scale experiments confirming Newton's laws of motion in
ultra-high vacuum, and I don't see any need for that to be done
either.
What will it show? It still isn't a perfect vacuum, so how is a
macroscopic experiment done in such vacuum better than one done in an
easier to achiever vacuum? Isn't all that is needed a vacuum good
enough so that radiometer forces (i.e., the thermal force that drive a
Crookes radiometer) are negligibly small? Are the laser-cooling of
mirrors experiments good enough? (I don't know these experiments well,
so I don't know how good their vacuums are.)
For atomic-scale experiments, the effect of pressure (a collision with
another atom!) on an atom is rather catastrophic. Pressure = collision
rate = rate at which atoms are lost from the trap. That they're
trapped until collision is an excellent high vacuum experiment. What
else is needed? There was a recent experiment in which a macroscopic
(in the sense that it's many many atoms, so bulk matter, but it was
microscopic in size, as in you want a microscope to look at it) object
was trapped and tracked between individual collisions in a low-
pressure gas, effectively an equivalent experiment, although not in
high vacuum (this is to be published in Science, and is still on
Science Express, their online prepublication service, at
http://www.sciencemag.org/cgi/content/abstract/science.1189403 , not
free though, apart from abstract).
If I had a spare ultra-high vacuum set-up available, with no serious
science going in it, and a student who wanted to (or was wanted to)
learn about UHV systems, it would be a good project. Otherwise, it's a
big investment in time (and money if you don't have the pumps already,
for what gain? It doesn't do any new science, it doesn't produce a
practically useful device, so what is the gain?
(Don't underestimate the time, effort, and money needed for ultra-high
vacuum work!)
> I feel very hesitant heading for relativity theory when the
> claims are supposedly satisfiable on a lab bench with a laser source
> and vane in vacuum, all right there on the bench. Yes, it is
> interesting to consider the red shift of the light off of the moving
> reflector especially in terms of energy conservation, but it is not
> the same problem.
As above, it is the same problem (or is meant to be). v<<c, so
Galileian relativity is entirely sufficient (I hope that you have no
argument with Galileian relativity!).
--
Timo
NoEinstein
>
Dear Timo: You had said, earlier: "This isn't true. The force on the
vanes has been measured in vacuum, and the force is in the opposite
direction to the usual Crookes radiometer thermal force. If not for
my New Science has the direction of rotation identical to the Crookes
(black squares trailing), your statement seemed to be saying that my
New Science is wrong—which it of course, isn't. The wrongly assumed
'forward' rotation is for the white squares to trail. Friction can
STOP or prevent a rotation, but never change its direction. If you
are wishing to change the subject to "momentum", you are way over your
head. I wrote the book on momentum and KE. PD has fought on those
subjects for three years, and has lost (to me). The world doesn't
need any more PDs! — NoEinstein —
Timo Nieminen
>
> Dear Timo: You had said, earlier: "This isn't true. The force on the
> vanes has been measured in vacuum, and the force is in the opposite
> direction to the usual Crookes radiometer thermal force. If not for
> friction, the radiometer in vacuum would rotate "backwards"." Since
> my New Science has the direction of rotation identical to the Crookes
> (black squares trailing), your statement seemed to be saying that my
> New Science is wrong—which it of course, isn't. The wrongly assumed
> 'forward' rotation is for the white squares to trail. Friction can
> STOP or prevent a rotation, but never change its direction. If you
> are wishing to change the subject to "momentum", you are way over your
> head. I wrote the book on momentum and KE. PD has fought on those
> subjects for three years, and has lost (to me). The world doesn't
> need any more PDs! — NoEinstein —
Job 8:2
NoEinstein
>
Dear Timo: Have I turned you... religious? "How long wilt (I) speak
these things? And how long will the words of (my) mouth be like a
strong wind?" Answers: I'll keep speaking the truth until the
largely dull 'scientific' community wises up. Since my New Science
fits the observations of the entire Universe, I can speak, with
authority, that no "detail" of my science will change the whole of my
discoveries. Note: Throughout the Universe, "the wind" is most often
ether flow caused by pressure differentials. The maximum pressure is
always closest to where the mass(es) is most concentrated. Photon
emission (including infrared) depletes part of the ether INSIDE
matter. That causes a vacuum that keeps drawing in new ether from
outside. Matter that receives photons from other matter isn't as
'deficient' in ether on the facing sides. So, the net 'replacement'
energy will flow in on the opposing sides. It is the DRAG of the
flowing ether on the matter on the opposing sides of the masses that
PUSHES the masses together. I. e.: the moon and the Earth are
orbitally bound in this way. That’s how gravity works! — NoEinstein
—
Tim BandTech.com
> On Jun 4, 9:59 pm, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
> > On Jun 3, 5:59 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
> > > Consider a steady beam of monchromatic light incident on a stationary
> > > reflector. Each optical cycle has energy E (i.e., the total energy of
> > > a length of the beam equal to 1 wavelength). The incident power P is
> > > thus P = Ef, where f is the optical frequency.The output power is the
> > > same, since the reflected beam is identical, except for the reversed
> > > direction. Since the reflector is stationary, no work is being done on
> > > it.
>
> > > Now change to a coordinate system where the reflector is in motion, at
> > > a speed v away from the source of the incident beam. (For simplicity,
> > > assume v << c.) The incident beam is now blue-shifted, and the
> > > frequency of the incident beam in this coordinate system is now f(1+v/
> > > c), and the power is P_in = Ef(1+v/c). Similarly, the reflected beam
> > > now has power P_out = Ef(1-v/c).
>
> > In the frame of the reflector the light will appear redder, since the
> > reflector is moving away from the source by v, so I disagree with your
> > blue-shifted claim of the incident beam.
>
> We only changed the coordinate system; we didn't introduce any
> relative motion between the source and the reflector. All we do is
> choose a new coordinate system with origin moving in the direction
> from reflector to source.
>
> In the frame of the reflector, everything should (must!) remain
> identical. The question is, what does it look like in _another_
> reference frame.
You've contradicted yourself Timo. In your previous post you
state(above):
"Now change to a coordinate system where the reflector is in
motion, at
a speed v away from the source of the incident beam. (For
simplicity,
assume v << c.) The incident beam is now blue-shifted..."
which is clearly in error, and now to double cover your error you've
misinterpreted your own post:
"We only changed the coordinate system; we didn't introduce any
relative motion between the source and the reflector."
This is a twist on the old dodge tactic that usenet suffers from. Come
Timo, let's please admit our mistakes in order to seek the truth. The
self contradiction above I can accept as overlooked by you, but we are
tendrilling away from the core argument within this context, as I've
already argued. Therefor I isolate this information here in order to
see if we can come to some agreement.
Without credibility what is the value of discussion? I will pursue the
momentum conservation argument on the reflective surface in another
post later, upon which you argue that my own credibility is suffering.
Here please, lets at least clean up and move onward. Redshifting of
light has nothing to do with the claimed existence of radiation
pressure. Still, it is nearby so I will not dismiss this topic
completely, and the definition of a 'perfect reflector' does come
under scrutiny under these conditions, which could be beneficial in
the provision of a new theory.
There are fundamental problems with humans doing science and the
assumption that the scientist makes of getting beyond his human
behavior cannot be granted unconditionally. This is cause to consider
all problems as open problems, for assumptions that we make and
observations that we fail to make can go invisible. This will be
proven in time, and already has been proven many times. The ability to
practice false belief systems is beyond no man, including myself.
Falsification is a worthy method, which is a skeptical approach, but
the ideal outcome is a replacement theory. It is not quite enough to
declare a belief system wrong.
- Tim
Tim BandTech.com
Hi NoEinstein. I honestly haven't carefully considered your argument,
but I do believe that it will be consistent with gravitational
shadowing, and perhaps a means of describing some of what modern jibe
calls 'dark matter'.
This is a bit of a scary subject, for it would bring huge interest
from astrologers who concern themselves with the dynamics of such
situations. Will we be forced to admit under this theory that when the
moon passes between the sun and the earth that some of the
gravitational force has disappeared? This is what I mean by
gravitational shadowing. It is an interesting concept, especially when
tied into electromagnetic radiation. I don't feel it is a complete
theory, and feel spread pretty thin trying to consider it within the
radiation pressure claim. Still, they are nearby to each other.
There are so many possible variations on the Crooks and Nichols
radiometers. There is not necessarily any need to expose more than one
vane of the Crooks device to light in order to experiment, so that the
black/silver conundrum could be deleted. Further I have not found any
analysis on a spinning bearing versus a quartz fiber's friction, but I
presume that a fine enough pin style bearing on a hard jewel concavity
could have a very slight amount of friction relative to the torque of
a fairly distant vane, not to mention more advanced options such as a
magnetic bearing. I've attempted some research on this but found
nothing yet. Any links on this bearing friction analysis are welcome.
I have made some simple bearing out of copper arms resting on a nail
and it is such a nice simple thing that a child can do it. Even wood
on wood bearings are nicely behaved, and can make pretty windvanes, a
few of which I have around my garden.
- Tim
Timo Nieminen
As I said (and you cut), I was not clear. The key phrase is "now
change to a coordinate system". How can changing to a different
coordinate system change the relative speed of the reflector and
source. Of course, the reflector is still moving in some direction,
and the direction of motion is away from the source. The source is
also moving, and the direction is towards the reflector. As I said
already, this wasn't clear enough.
* Change of coordinate system, no change in relative motion. *
> The
> self contradiction above I can accept as overlooked by you, but we are
> tendrilling away from the core argument within this context, as I've
> already argued. Therefor I isolate this information here in order to
> see if we can come to some agreement.
> Without credibility what is the value of discussion? I will pursue the
> momentum conservation argument on the reflective surface in another
> post later, upon which you argue that my own credibility is suffering.
> Here please, lets at least clean up and move onward. Redshifting of
> light has nothing to do with the claimed existence of radiation
> pressure. Still, it is nearby so I will not dismiss this topic
> completely, and the definition of a 'perfect reflector' does come
> under scrutiny under these conditions, which could be beneficial in
> the provision of a new theory.
If light carries energy, redshift/blueshift and conservation of energy
tells you everything you need to know about radiation pressure.
Why avoid discussing it? Is there anything wrong with what is below?
Consider a steady beam of monchromatic light incident on a stationary
reflector. Each optical cycle has energy E (i.e., the total energy of
a length of the beam equal to 1 wavelength). The incident power P is
thus P = Ef, where f is the optical frequency.The output power is the
same, since the reflected beam is identical, except for the reversed
direction. Since the reflector is stationary, no work is being done on
it.
Now change to a coordinate system where the reflector and source are
in mutual motion, at a speed v, in a direction along a line joining
the
source to the reflector. (For simplicity,
assume v << c.) The incident beam is now blue-shifted, and the
frequency of the incident beam in this coordinate system is now
f(1+v/c), and the power is P_in = Ef(1+v/c). Similarly, the reflected
beam now has power P_out = Ef(1-v/c).
Since energy is conserved, the difference in power must be the rate of
doing work on the reflector. Since work is being done on the
reflector, there must be a force acting on the reflector
due to the reflection.
Tim BandTech.com
To support my claim above I quote from your own words
"If the reflector is moving, there will be a Doppler shift,
and energy in will be different from energy out.
If the reflector is stationary, then you could have, e.g.,
1300W in and 1300W out. But no work would be done."
- http://groups.google.com/group/sci.physics/msg/f064666482b7c3b5
I don't mean to be crass but I do mean to use this medium with
accountability, which is one of its strong points. I would like you to
find such falsifications in my own writing.
The radiation pressure effect is claimed to be operant even with no
initial relative velocity between the source and reflector. Further,
as far as I can tell none of the existing work on radiation pressure
relies upon any red shift argument as the balance of energy. Simply
put, if the perfect reflector reflects all of the beams light back
toward the source then it cannot have absorbed any energy and hence it
will not accelerate.
Next, if we do take the accepted photon momentum derived from
e = h f
then we would witness the entire 1300 watts of power on a square meter
of mirror, which should do quite some work according to this accepted
momentum equation, particularly for those fond of a doubling effect.
Clearly this is not the source of radiation pressure. Even if we go to
a black body which absorbs the energy, we must admit that there is no
effect; The energy is converted into heat, to be reradiated, and if we
assume that this reradiation is isotropic then we need not bother with
any directed effect from it within the already corrupted concept of
photon momentum. All the while you stand by the photon momentum
equation, even while it fails to provide any momentum. This argument
on momentum does not necessarily have anything to do with the claimed
existence of radiation pressure.
The heating effect is nearby to the redshift concept, but again, we
are not necessarily talking about radiation pressure here. This is an
altogether different effect as far as I can tell. Wouldn't we have to
get into thermodyamics more and Planck's energy distribution to
discuss this effect? You see, it is a bit crazy how many sidelines
there are, and so long as we wish to discuss the radiometer as a means
of measuring radiation pressure, well, we are not really discussing a
radiometer any more. These fine points you seem not to have
acknowledged though I've stated them several times. You have pretty
much accepted (I believe) that there will be no effect in the best
vacuum labs available. Isn't this enough to admit that you do not
believe that radiation pressure exists?
As I see it the problem of human as scientist is embedded within this
discussion. Upon an 'accepted' theory being posited, for one to break
away from that norm places one as a potential quack. And yet, without
this freedom of thought, the quality of science will suffer. Such a
person has relinquished their freedom to falsify, and in doing so has
guaranteed their success within a system where mimicry is a merit. I
respect your rights, particularly here on this free medium, but I
would point out to you that as a partially accurate sounding board to
existing theory you are in some regards helping to disprove that
theory.
As you argue for the radiation pressure on a perfect reflector in
outer space why would you deny that it can be provided in a lab at
10E-11 torrs with a 10kW laser? Should this be enough to turn the
spindle of a Crooks style radiometer, or is the friction too great?
Really, I must admit that I am growing cynical here, if for no other
reason than the fact that your response to this question will be
miserable to read.
You are a strong poster Timo and I don't mean you any harm. I can't
have this discussion without you. I find it poor that people who claim
to be scientific will not apply themselves at this level. I ask that I
be falsified, and you are the only one who steps forward. I am still
open to such falsification, and it could turn on some minor point or
discrepancy that we are overlooking. Still, could it be the other way
around too? One must give ones self enough credit to declare something
false if one is to give one's self enough credit to declare something
to be true. Otherwise mimicry ensues, and this is a transparent human
factor inculcated within the schooling. I give you as much credit as I
can but attempt to hold you accountable. Thanks, Timo, for hanging in
there. I half take back my prediction on your response.
- Tim
NoEinstein
>
Dear Tim: A Sun-Moon alignment would reduce the much greater "sun"
gravity. But there is a built-in inertia to there being instantaneous
changes in Earth's orbit. Here's why: The mass of the Earth is huge
compared to the solar energy portion which could be received during a
solar eclipse. Gravity forces are due to the ether density near the
Earth. That reserve of ether would cause orbit changes to occur
slowly. Though I haven't thought out the particulars, it's likely
that the solar energy of the Sun on the exact back-side of the moon
would momentarily increase the Earth-moon gravity so that there would
be no... fly-away Earth. And there is the possibility that if Earth
did move outward, that Earth would get back in the correct orbit once
the full Sun-Earth gravity returned.
Interestingly, I've figured that "shading" could be a means of
diverting asteroids or comets on collision courses with the Earth.
Explode an aluminum foil bomb(s) along the object's path to shade from
the sun, and the object will move out on its tangent, and miss the
Earth.
An exceedingly low friction bearing is to stick two double-edge razor
blades into a soft cork, parallel. Place a sewing needle through two
large soda straw, perpendicular. You can use the device as a beam
balance that is sensitive enough to weigh a hair. — NoEinstein —
> - Tim- Hide quoted text -
BURT
>
> - Show quoted text -
Sun Moon depending on which side; the moon close to the sun or on the
other side will both add to gravity and cancel it.
Note that a circular orbit has no strength of gravity. This means
angles of enregy flow through space determine a partial strength of
gravity.
Only partial strength applies in all orbits. For light which is
parabola path the space flow strength of gravity is always at maximum.
Nobody in physics can explain where the strength of gravity goes in a
circular orbit. But it is proof of partial gravity at angles
approaching it. And that is all there is to it.
Mitch Raemsch
Timo Nieminen
>
> To support my claim above I quote from your own words
>
> "If the reflector is moving, there will be a Doppler shift,
> and energy in will be different from energy out.
> If the reflector is stationary, then you could have, e.g.,
> 1300W in and 1300W out. But no work would be done."
>
>
> I don't mean to be crass but I do mean to use this medium with
> accountability, which is one of its strong points. I would like you to
> find such falsifications in my own writing.
>
> The radiation pressure effect is claimed to be operant even with no
> initial relative velocity between the source and reflector. Further,
> as far as I can tell none of the existing work on radiation pressure
> relies upon any red shift argument as the balance of energy. Simply
> put, if the perfect reflector reflects all of the beams light back
> toward the source then it cannot have absorbed any energy and hence it
> will not accelerate.
Yes, it will work with no relative velocity between source and
reflector. Don't handwave! Say specifically where my redshift/
blueshift derivation is wrong. If it isn't wrong, it's right.
> Next, if we do take the accepted photon momentum derived from
> e = h f
> then we would witness the entire 1300 watts of power on a square meter
> of mirror, which should do quite some work according to this accepted
> momentum equation, particularly for those fond of a doubling effect.
Why? 1300W gives you 4 micronewtons (9 micronewtons if reflected).
This gives you "quite some work"? That's a tiny force, expecially
considering it's acting on something 1m^2 in area.
> You have pretty
> much accepted (I believe) that there will be no effect in the best
> vacuum labs available. Isn't this enough to admit that you do not
> believe that radiation pressure exists?
Absolutely not. Go back and read what I wrote!
> As you argue for the radiation pressure on a perfect reflector in
> outer space why would you deny that it can be provided in a lab at
> 10E-11 torrs with a 10kW laser? Should this be enough to turn the
> spindle of a Crooks style radiometer, or is the friction too great?
> Really, I must admit that I am growing cynical here, if for no other
> reason than the fact that your response to this question will be
> miserable to read.
I don't deny it at all that it would work in a lab at ultra-high
vacuum. I said that it does work in the lab at UHV, as seen by the
various atom trapping/cooling and BEC experiments.
Would it be enough to turn a Crookes radiometer? Depends on the power.
Levitate the rotor magnetically, avoid friction. Nichols and Hull
avoided friction.
> You are a strong poster Timo and I don't mean you any harm. I can't
> have this discussion without you. I find it poor that people who claim
> to be scientific will not apply themselves at this level. I ask that I
> be falsified, and you are the only one who steps forward. I am still
> open to such falsification,
Are you? I say that UHV experiments have been done, on atoms. I say
that radiation pressure forces have been observed on macroscopic
objects, between collisions with atoms in the low-pressure gas, so
equivalent to UHV in principle. I say that radiation pressure forces
would be observable in an UHV experiment. Your response: "why would
you deny that it can be provided in a lab at 10E-11 torrs with a 10kW
laser?" Apart from the fact that I didn't mention a 10kW laser, nor
did you until now, what you are claiming I said is the opposite of
what I said. So, is there any point discussing the various experiments
if you don't read what I write?
Don't avoid the redshift/blueshift derivation then. It's simple, and
follows only from the energy of light and conservation of energy. You
don't need to assume that light carries momentum; this follows as a
consequence.
NoEinstein
>
Dear Tim: My reading speed can't take in all that you are saying.
But I don't disagree. The 'missing link' of your Black Body rationale
is that the white light coming in has more energy per 'ray'. To have
the same infrared radiation, there have to be a higher number of
rays. It's likely that it's the number of rays that account for the
amount of ether ejected along with the photons. In the Crookes, light
reflecting from the white squares, and the ether ejected in the same
direction, push some of the argon atoms toward the adjacent (opposed)
black squares. It is the mass of those argon atoms which causes the
vanes to rotate, not heat "rocketing" from the porous edges of the
vanes! — NoEinstein —
BURT
Hey NoEinsein
Light bulbs are flooding quadrillions of light waves into a room every
second. But what happens to them when the light is turned off? They
don't leave the room. But if they go into matter are they absorbed or
unabsorbed as heat?
An atom at a matterial surface after absorbing light has the
probability for emmiting light either outward of itself or inward
inside the atom toward the nucleus or sideways to the atom Emmision
goes spherical 360 degrees and is stochastic appearently. If there is
order to emmision I have yet to see it.
This means that atoms can keep emmiting light deeper into the surface.
And all of this is probability.
Mitch Raemsch
NoEinstein
>
Dear Burt: Photons don't get "lost" inside of reflecting surfaces.
The photon energy will be expelled 1/2 wavelength later, even if there
are hundreds of internal, atom-to-atom exchanges. Light of any kind
is radiant energy which can warm a room. Black (or dark) surfaces are
slower to release the radiant energy, and might even release some of
the heat by conduction or convection. In the present discussion on
Radiometers, the convection is small due to the vacuum in the bulb.
And the conduction is small due to the isolated design of the four
vanes. — NoEinstein —
Tim BandTech.com
Hey, great snip here Timo! Thanks for clipping out your own
contradiction. The usenet dodge is in effect. I will attempt to do
better as you will see below. The one thing for sure about this medium
is that mistakes can be made, and this is entirely acceptable. This
allows us to exercise mistake detection, which is a skill that needs
more development. It's going to be the only way out of the pile-up
that is modern physics. Beyond this, credibility exists more for those
who admit their mistakes than it does for the usenet dodger. Well, I
should admit we each are actually doing pretty well at holding our
feet to the fire.
> > To support my claim above I quote from your own words
>
> > "If the reflector is moving, there will be a Doppler shift,
> > and energy in will be different from energy out.
> > If the reflector is stationary, then you could have, e.g.,
> > 1300W in and 1300W out. But no work would be done."
>
> > -http://groups.google.com/group/sci.physics/msg/f064666482b7c3b5
>
> > I don't mean to be crass but I do mean to use this medium with
> > accountability, which is one of its strong points. I would like you to
> > find such falsifications in my own writing.
>
> > The radiation pressure effect is claimed to be operant even with no
> > initial relative velocity between the source and reflector. Further,
> > as far as I can tell none of the existing work on radiation pressure
> > relies upon any red shift argument as the balance of energy. Simply
> > put, if the perfect reflector reflects all of the beams light back
> > toward the source then it cannot have absorbed any energy and hence it
> > will not accelerate.
>
> Yes, it will work with no relative velocity between source and
> reflector. Don't handwave! Say specifically where my redshift/
> blueshift derivation is wrong. If it isn't wrong, it's right.
This accusation of handwaving is false. I find this an interesting
form of denial. I was recently criticized for constructing a straw man
argument, which was itself a straw man argument. Such iterative
discrepancies are fascinating and perhaps are operant within the human
mind, and would help to explain our poor observational abilities. What
does it mean to observe? Well, the gedanken observer merely marks down
some strict number, but for the human the observant have already
entered the theoretical realm. Perhaps we are suffering from this
ambiguation and should resolve it. Here is a fine point on detachment
from philosophy within the modern physics position, and a minor
falsification of Einstein's construction technique. These subtleties
coupled to human nature are like a substrate that we inhabit, with no
direct access to the substrate.
Well, I have gone back to check what I remember pretty well that you
wrote, and you state that it will not be observed in perfect vacuum:
"Perfect vacuum, no. Very good vacuums, yes, especially with atom
trapping. I've seen classical experiments done in vacuum (can't recall
how good), where absorbing particles were blasted by short pulses of
light, to measure their radiation pressure cross-sections."
in response to my statement:
"I'm sorry but the effect in a perfect vacuum has not been clearly
demonstrated. As you've accepted the flaws of Nichols work then
the
question of whether the radiation pressure is observable in vacuum
still exists. The experiments that you've exposed in your paper
must
not be the modern experiments that you speak of."
I stand corrected and apologize for the misinformation. I suppose that
what you meant was that we cannot generate a perfect vacuum. Now, the
only thing we need to do is turn up an experiment on solids rather
than fluids or particles that takes into consideration the strength of
the vacuum, no different than Nichols himself did, but beyond the .02
torr mark; preferably heading all the way to the limit of the modern
equipment, and observe the characteristic curve of diminishment of the
gas effect, as Nichols did.
I have been avoiding your references to optical tweezers and light
traps, partially because I have only limited familiarity with those.
Anyway, we are in agreement that radiation pressure should be
measurable at 10E-11 torrs using a solid plate device.
>
> Don't avoid the redshift/blueshift derivation then. It's simple, and
> follows only from the energy of light and conservation of energy. You
> don't need to assume that light carries momentum; this follows as a
> consequence.
No. I will continue to attempt these as seperate tendrils of this
thread. The redshift concept stands alone away from the photon
momentum, and this too stands alone away from the radiation pressure.
These three concepts may be related, but it is their isolation that we
are after. Again, there is no need of any velocity other than the
velocity of the light under the initial conditions of a target in
stasis with a source, as at a laboratory light table, with a laser
pinned to the table shining into an evacuated compartment pinned to
the table with either a Crooks device or a Nichols device inside
initially at zero velocity and under the Nichols case settling to zero
velocity via a torsion fiber. The only parameter here are the power of
the light, which supposedly is related to radiation pressure, the
strength of the vacuum, and perhaps the amount of friction in the
measuring device. Well, clearly there are more parameters, but the
isolation of radiation pressure seems possible under this situation,
with the exception of the reradiated energy of the detector, and
possibly freed electrons as one of the links in support of my position
mentions. When we start thinking of those electrons then the tweezers
and traps start to make more sense I think, but I cannot argue on
these devices without doing some research.
If this experiment has been done why can't we find it? If it has not
been done then why not?
If Nichols work were falsifiable then we should see a next gen, no
different than Nichols followed Crooks.
Ahah! I just found one!
http://iopscience.iop.org/0959-5309/45/2/315
Reads pretty much like some of this discussion, but in 1933.
He calls Nichols and Hull's work a 'paper dagger'.
He gets down to 10E-6 torr.
I can't access the whole paper, but the intro reads easy.
If I was a real hanger on I would try to consider the free electron
problem. For instance, did he try using some static E fields to vary
the experiment? I will try to go through the theory again. The
doubling claim still troubles me.
I guess if you want to argue the doubling effect some more I can try,
but I'd rather go over to the black body stuff and try to understand
just how literally the photon momentum is taken. If it's all in e=hf
then it is a lie, because the 1300 watts per square meter of sun would
be observed as a mechanical force. This is likely a matter of
interpretation. As I approach this as an open problem, then it seems
blatantly apparent that what we just claimed as an isolational
experiment on radiation pressure can actually be taken as a claim to
have isolated photon momentum as much more miniscule than the e=hf
derivation. This then could be twisted into photon mass, and as it is
such a slight figure, then all the better.
The radiation pressure is not unlike electron spin, in that it is a
small effect that is difficult to notice and for me, difficult to
understand. Incidentally, the p=u/3 that I see in other sources is not
the equation you rely upon. The gyrations possible within this realm
are ridiculously complex. Electromagnetism itself is a kaleidoscope of
behaviors and we already are working under the conflicted wave/
particle duality. This is a great topic. Thanks Timo for your time on
this thread.
- Tim
NoEinstein
>
Dear Tim: Since I replied to this same reply of yours, earlier, I've
realized that the Moon is being held in orbit about the Sun; and being
held in orbit about the Earth, by the radiations exchanged among those
three. Lunar eclipses don't send the moon flying out on its tangent,
probably because the ether pressure "reserve" is greater than the
ether pressure lost due to being in the shadow. The shadow must be
maintained long enough so the ether pressure reserve gets used up.
The following might be considered science 'fiction', now, but one day
mankind may be able to maneuver, say, the planet Mars completely out
of the solar system. This will require having an artificial
satellite, with a huge reserve of FUSION power, to provide the "solar"
energy to maintain life on the revolving planet. The destination
would be a star young enough to sustain human life for a few billion
years. And... it might be possible, by controlling solar flares and
sunspots, to turn the Sun into a propulsion system to carry the entire
solar system close enough to another 'younger' star, to facilitate the
relocation. Forgive my "imagined" science. Real physics would have
to apply there, too! — NoEinstein —
> - Tim- Hide quoted text -
Timo Nieminen
>
> Well, I have gone back to check what I remember pretty well that you
> wrote, and you state that it will not be observed in perfect vacuum:
> "Perfect vacuum, no. Very good vacuums, yes, especially with atom
> trapping. I've seen classical experiments done in vacuum (can't recall
> how good), where absorbing particles were blasted by short pulses of
> light, to measure their radiation pressure cross-sections."
>
> in response to my statement:
> "I'm sorry but the effect in a perfect vacuum has not been clearly
No, I said the experiment hasn't been *done* in a perfect vacuum.
Because we can't achieve a perfect vacuum in the lab. Technically, you
were correct.
We also haven't done an experimental test of Newton's 1st law, because
we don't have a friction free, viscous drag free, force free
laboratory. This doesn't mean that I think that Newton's 1st law
doesn't work in perfect vacuum.
It's an important point. How do we know when we can extrapolate a good
vacuum result (or poor vacuum, or atomospheric result) to perfect
vacuum? It's important to know (by measuring) the effect of what is
there, e.g., damping due to viscous drag, drag due to convective flow,
the radiometer force, the effect of the gas on the temperature
difference between the two sides of the vane, etc.
> I have been avoiding your references to optical tweezers and light
> traps, partially because I have only limited familiarity with those.
They're the main modern application of radiation pressure. *Thousands*
of experiments have been done. Early experiments to show that the
method works, and to explore the theoretical principles, experiments
to use the force e.g. for restraining or moving live cells,
experiments to quantitatively measure forces between biomolecules.
> Anyway, we are in agreement that radiation pressure should be
> measurable at 10E-11 torrs using a solid plate device.
In principle. Whether the force is too small to measure in practice is
the question, and this depends on the particular setup.
> ... and
> possibly freed electrons as one of the links in support of my position
> mentions. When we start thinking of those electrons then the tweezers
> and traps start to make more sense I think, but I cannot argue on
> these devices without doing some research.
Freed electrons are not responsible for the force in general. The atom
trapping experiments are quite conclusive on this (since freeing an
electron would result in ionisation, which would result in the ion not
being trapped). (Perhaps in some specific experiment? In the modern
laser experiments, there isn't enough photon energy for (significant)
photoejection of electrons. Given 1/2 of the photon energy used to
eject the electron, the rest becoming its KE, electron ejection would
give about 400 times as much force per photon. If this was significant
in the various tweezers experiments, it would be observed, as it would
thoroughly overcome the trapping force. It would also give an
interesting wavelength dependence.)
> If this experiment has been done why can't we find it? If it has not
> been done then why not?
> If Nichols work were falsifiable then we should see a next gen, no
> different than Nichols followed Crooks.
>
> Ahah! I just found one!
> http://iopscience.iop.org/0959-5309/45/2/315
> Reads pretty much like some of this discussion, but in 1933.
> He calls Nichols and Hull's work a 'paper dagger'.
> He gets down to 10E-6 torr.
> I can't access the whole paper, but the intro reads easy.
So, a later and better repeat of Nichols and Hull. N&H (or the
widespread quoting of their result as "definitive") are criticised
since the discrepancy between theory and experiment was 10% (due to
the effect of the gas). Hull had a paper a few years after N&%, 1905
iirc, on the elimination of gas action, or similar title. This might
give some idea of what N&H thought at the time. Bell (and Green) had
another paper the next year, with Hull as 1st author. (I briefly read
the paper you link above some years ago, but not the others).
> The
> doubling claim still troubles me.
>
> I guess if you want to argue the doubling effect some more I can try,
The doubling is no more, no less, than you have with similar doubling
in the reflection of ping pong balls, billiard balls, etc. The impulse
needed to reverse the momentum of an object is double that needed to
stop the momentum of an object (straight from Newton 2). Acting in the
same time, the reflection force must be double the stopping force.
> but I'd rather go over to the black body stuff and try to understand
> just how literally the photon momentum is taken. If it's all in e=hf
> then it is a lie, because the 1300 watts per square meter of sun would
> be observed as a mechanical force.
Why would it be observed? 4 (or 9) micronewtons per square meter?
Compare this with the gravitational force on a practical plate of the
same area. (Perhaps also compare this with forces due to Brownian
motion?)
> This is likely a matter of
> interpretation. As I approach this as an open problem, then it seems
> blatantly apparent that what we just claimed as an isolational
> experiment on radiation pressure can actually be taken as a claim to
> have isolated photon momentum as much more miniscule than the e=hf
> derivation. This then could be twisted into photon mass, and as it is
> such a slight figure, then all the better.
You keep saying that this is problem, but I don't see why. Why is it a
problem? E=hf gives a very, very, small force.
E = hf, energy of photon, momentum = E/c = E/(f*lambda) = h/lambda,
the standard QM result. h is small; this isn't much momentum.
This is small enough so it isn't casually observable. Observed by
Nichols and Hull (to about 10%, according to Bell and Green), and
others since. Observed in special circumstances before and since
(e.g., as contributor to comet tails).
> The radiation pressure is not unlike electron spin, in that it is a
> small effect that is difficult to notice and for me, difficult to
> understand. Incidentally, the p=u/3 that I see in other sources is not
> the equation you rely upon.
The 1/3 is for omni-directional radiation, such as blackbody
radiation. What is done experimentally usually uses a beam of some
kind. For a plane wave, or a parallel (i.e., collimated) beam, p=P/c
(or p=nP/c=P/v in a medium). For a focussed beam, it is a little less
(and this difference is seen, and is essential to 3D trapping in
optical tweezers).
And of course, u=P/c.
> and for me, difficult to
> understand.
You're not the first. Until Maxwell's calculation of electromagnetic
radiation pressure, it was generally assuming that there was no
radiation pressure, that no wave would exert a non-zero time-averaged
force (since it has no mass). In the 1800s, the lack of observabed
wave pressure was used as an argument for the wave theory of light and
against corpuscular theories. One of the last works of the late Thomas
Gold was a short note on solar sails and why they don't work. You are
not alone.
Maxwell's result (and Poynting's and Heaviside's) was specifically
electromagnetic, so N. A. Umov's general result (not given the
attention it deserves in the West) in 1874 was also a significant
advance. Then Einstein in 1905 with E=mc^2.
NoEinstein
>
Dear Tim: In my invalidation of the M-M experiment, I chose to look
at light in S L O W motion. I realized that the easiest way to
understand what was happening with the light was to consider just a
SINGLE photon racing along each of the two light courses. That is so
much simpler than trying to say, as Timo does, that the light red
shifts or blue shifts. Einstein declared that light never blue
shifts, because that would require the photons to exceed 'c'. Using
simple 9th grade algebra, I have proved, conclusively, that light
velocity is: V = 'c' plus or minus v (or the velocity of the light
source). Each time Timo talks about... a reflector, he could just as
well be talking about the mirrors in the M-M experiment. Below are
the links which explain these things. Also, I have determined that
there is always a slight "Friction of Reflection" which red-shifts all
reflected light. Some of the energy imparted to the reflector
converts to heat that isn't instantaneously re emitted with the
reflected light. That necessitates that reflected light will be red
shifted, otherwise the Law of the Conservation of Energy is violated.
— NoEinstein —
>
> - Show quoted text -- Hide quoted text -
NoEinstein
>
Dear Burt: Where did you get the notion that circular orbits have no
gravity? If that were so, then, how are those telecommunications
satellites held in orbit? I've got gravity nailed as: Flowing ether,
replenished by photon exchange. Nothing that you've ever said changes
those facts. — NE —
NoEinstein
>
Dear Timo: When you've said that the energy in (to the reflector)
doesn't = the energy out, isn't THAT a violation of the Law of the
Conservation of Energy? — NE —
Tim BandTech.com
No. This is not a clean analogy. In order for a receiver to return a
billiard ball to a sender at the same energy the receiver must not
accelerate. No work can be done on the receiver, and this is why I
claim this portion of this subject farcical. If momentum is imparted
upon the receiver, then in order to return the ball with the same
energy to the sender the receiver will have to expend energy. There is
no free lunch, though I'm getting awfully close to it.
>
> > but I'd rather go over to the black body stuff and try to understand
> > just how literally the photon momentum is taken. If it's all in e=hf
> > then it is a lie, because the 1300 watts per square meter of sun would
> > be observed as a mechanical force.
>
> Why would it be observed? 4 (or 9) micronewtons per square meter?
> Compare this with the gravitational force on a practical plate of the
> same area. (Perhaps also compare this with forces due to Brownian
> motion?)
>
> > This is likely a matter of
> > interpretation. As I approach this as an open problem, then it seems
> > blatantly apparent that what we just claimed as an isolational
> > experiment on radiation pressure can actually be taken as a claim to
> > have isolated photon momentum as much more miniscule than the e=hf
> > derivation. This then could be twisted into photon mass, and as it is
> > such a slight figure, then all the better.
>
> You keep saying that this is problem, but I don't see why. Why is it a
> problem? E=hf gives a very, very, small force.
No. 1300 watts on a square meter is nearly two horsepower.
This is not a trivial amount of power. This is the energy in e=hf that
I am discussing. In one second of time we have 1.3 kiloJoules of
energy on a square meter, if all that energy is absorbed. Obviously
this is not momentum, so any blackbody interpretation which treats it
as such is a fraud.
>
> E = hf, energy of photon, momentum = E/c = E/(f*lambda) = h/lambda,
> the standard QM result. h is small; this isn't much momentum.
It's 1300 watts of power per square meter from the sun at the earth's
surface.
h doesn't matter. It's a large quantity of photons.
"A person having a mass of 100 kilograms who climbs
a 3 meter high ladder in 5 seconds is
doing work at a rate of about 600 watts."
- http://en.wikipedia.org/wiki/Watt
And so we should be getting quite a push when we hold a mirror up to
the sun under the standard photon momentum argument. It is clearly a
misinterpretation and disambiguating this failure is not happening on
this thread.
>
While you bring it up I had better just run my own simple thinking on
photon momentum against this e=mcc. A photon has energy
e = h f
where f is the frequency, not worrying about any 2pi factor.
The velocity of the photon is
v = c
and so using
e = m c c
we have
h f = m c c
and so the photon's equivalent mechanical momentum
p = m v
can be expressed as
p = m c = e / c = h f / c.
This also leads to a mass expression for the photon, or at least an
equivalent mass for the photon. This expression takes what I believe
to be the entire energy of the photon and expresses a mechanical
momentum. Thus the 1300 watts of solar power should provide quite some
motive force if the translation is accurate. Thus far we do not see
any such mechanical force, and instead see the ability to absorb a
large amount of this as heat or with less efficiency to turn it into
eletricity with special diodes, called solar cells. The last two
phenomena are acceptable from AC principles, whereas the mechanical
momentum claim is not. All that we can hope for in an AC situation is
to shake the matter about quite alot, which it traditionally thought
of as heat. Well, this is a loose, but consistent logic with observed
systems. Photon momentum is wrongly addressed as a mechanical vector
of a static variety, and would be better expressed in an AC form. Yes,
it propagates in a direction (though often from omnidirectional
sources), but this does not inherently imply a force vector ala
Newton. This vector marks a passage of energy that is obviously not at
all to do with physical momentum. Therefor the equations in use are
too loosely presented, as a high school student will easily get a
photon mass out of the product relationships. We are doing little
better than this sort of derivation; we merely abide by 'photons don't
have mass' ruling, and that ruling is all that keeps us from going
over the edge on that side. We have instead barreled over the edge on
the other side. This is the current state. The creature called a
lemming comes to mind, and we are perhaps little better than them at
some base level. In fact, we provably are designing our own cliff and
it's collapse on a very grand scale. To presume that physicists are
somehow better than this overlooks a fundamental problem of human
existence. Likewise for mathematicians. We come from a blind state of
a blank slate and without the mimicry we would have nothing, but with
it we are likewise stuck.
- Tim
Tim BandTech.com
> On Jun 5, 3:25 pm, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> Dear Tim: Since I replied to this same reply of yours, earlier, I've
> realized that the Moon is being held in orbit about the Sun; and being
> held in orbit about the Earth, by the radiations exchanged among those
> three. Lunar eclipses don't send the moon flying out on its tangent,
> probably because the ether pressure "reserve" is greater than the
> ether pressure lost due to being in the shadow. The shadow must be
> maintained long enough so the ether pressure reserve gets used up.
>
> The following might be considered science 'fiction', now, but one day
> mankind may be able to maneuver, say, the planet Mars completely out
> of the solar system. This will require having an artificial
> satellite, with a huge reserve of FUSION power, to provide the "solar"
> energy to maintain life on the revolving planet. The destination
> would be a star young enough to sustain human life for a few billion
> years. And... it might be possible, by controlling solar flares and
> sunspots, to turn the Sun into a propulsion system to carry the entire
> solar system close enough to another 'younger' star, to facilitate the
> relocation. Forgive my "imagined" science. Real physics would have
> to apply there, too! — NoEinstein —
Yowser !
The sun propulsion thing sounds fascinating.
- Tim
Timo Nieminen
Not "not accelerate", but "not move". Work = F*d, power = F*v.
So consider a stationary reflector, of billiard balls or photons.
Reflected balls/photons have the same energy as the incident ones, so from
conservation of energy, no work is being done on the reflector. The
momentum is still reversed, so from the conservation of momentum, there
must be force. Power = F*v means that these aren't incompatible.
Then look at this in a coordinate system where the reflector is moving
(but not accelerating).
> and this is why I
> claim this portion of this subject farcical.
Is Newtonian mechanics farcical? Is the conservation of momentum farcical?
> If momentum is imparted
> upon the receiver, then in order to return the ball with the same
> energy to the sender the receiver will have to expend energy. There is
> no free lunch, though I'm getting awfully close to it.
If you want, do the analysis for perfectly elastic reflection of balls,
say of mass m1, from a plate of mass m2. Use conservation of momentum and
energy. Compare with "absorption" of completely sticky balls.
If you don't believe the prediction of what happens from Newtonian
mechanics and conservation of energy and momentum, then I don't think
there's much point in continuing this.
The reflection/"absorption" of balls isn't exactly the same as
reflection/absorption of light, but the basic principles are the same.
For the balls case, don't start with the plate at rest, but with the
centre of mass of ball+plate at rest. Then look at what these results
become when the plate is initially at rest (and moving at the end).
> > > but I'd rather go over to the black body stuff and try to understand
> > > just how literally the photon momentum is taken. If it's all in e=hf
> > > then it is a lie, because the 1300 watts per square meter of sun would
> > > be observed as a mechanical force.
> >
> > Why would it be observed? 4 (or 9) micronewtons per square meter?
> > Compare this with the gravitational force on a practical plate of the
> > same area. (Perhaps also compare this with forces due to Brownian
> > motion?)
> >
> > > This is likely a matter of
> > > interpretation. As I approach this as an open problem, then it seems
> > > blatantly apparent that what we just claimed as an isolational
> > > experiment on radiation pressure can actually be taken as a claim to
> > > have isolated photon momentum as much more miniscule than the e=hf
> > > derivation. This then could be twisted into photon mass, and as it is
> > > such a slight figure, then all the better.
> >
> > You keep saying that this is problem, but I don't see why. Why is it a
> > problem? E=hf gives a very, very, small force.
>
> No. 1300 watts on a square meter is nearly two horsepower.
> This is not a trivial amount of power. This is the energy in e=hf that
> I am discussing. In one second of time we have 1.3 kiloJoules of
> energy on a square meter, if all that energy is absorbed. Obviously
> this is not momentum, so any blackbody interpretation which treats it
> as such is a fraud.
Yes, it isn't momentum. And it isn't force. It's power (and, over 1
second, an amount of energy).
> > E = hf, energy of photon, momentum = E/c = E/(f*lambda) = h/lambda,
> > the standard QM result. h is small; this isn't much momentum.
>
> It's 1300 watts of power per square meter from the sun at the earth's
> surface.
> h doesn't matter. It's a large quantity of photons.
> "A person having a mass of 100 kilograms who climbs
> a 3 meter high ladder in 5 seconds is
> doing work at a rate of about 600 watts."
> - http://en.wikipedia.org/wiki/Watt
>
> And so we should be getting quite a push when we hold a mirror up to
> the sun under the standard photon momentum argument. It is clearly a
> misinterpretation and disambiguating this failure is not happening on
> this thread.
If h doesn't matter, why did you introduce h?
Do it without h then. P = 1300W, so F=P/c = 9 micronewtons. I wouldn't
call this "quite a push".
1300W incident on a reflecting object _doesn't_ mean that 1300W of work
will be done. Or even the same on a perfect absorber. Again, do the
calculation for reflection/"absorption" of balls.
To repeat yet again, the predicted force is 9 micronewtons per square
metre. Why do you think this should be easy to observe?
... which is the same as p = f/lambda as already stated earlier.
> This also leads to a mass expression for the photon, or at least an
> equivalent mass for the photon.
For some definitions of "mass", this is mass, what is sometimes called
"relativistic mass". For the usual modern technical definition of mass
(which is rest mass), it isn't mass.
It does tell you the
> This expression takes what I believe
> to be the entire energy of the photon and expresses a mechanical
> momentum. Thus the 1300 watts of solar power should provide quite some
> motive force if the translation is accurate. Thus far we do not see
> any such mechanical force,
Do the actual calculation, and you will see that the force is small. Note
the 1/c in your expression; c is a large number.
> and instead see the ability to absorb a
> large amount of this as heat or with less efficiency to turn it into
> eletricity with special diodes, called solar cells.
Just as we would see with the collision of lightweight balls sticking to a
massive object.
Do the calculation, using Newtonian mechanics, for a stream of Newtonian
particles moving at c, with kinetic energy (for each particle) equal to hf
for, say, 500nm light, and total KE per second in the stream of 1300W.
Do this for reflection or "absorption" of the particles. Don't just say
that 1300W must give a huge force, do the calculation instead.
> The last two
> phenomena are acceptable from AC principles, whereas the mechanical
> momentum claim is not.
Compare with the result of the suggested Newtonian calculation above. Is
it still not acceptable?
--
Timo
Tim BandTech.com
Light is not the same as massive objects colliding. I've merely
pointed out where the analogy breaks down. In effect again, there will
be no perfect reflector, except in the mechanical case of a grounded
plate which is not free to accelerate. I'm not breaking any physical
laws here, and if I were you should be able to find the flaw in my
wording and falsify directly rather than building more argument to
sift through. I've been through this one already with another guy
about passing a ball back and forth in space. It is not at all the
same problem, since light will go at speed c, and only speed c,
staying within that existing theory. Yes, now we can gyrate onto the
redshift concept, but it just isn't there for the practical problem.
Taking photon momentum literally, then we should consider what 1300
watts is in terms of mechanical systems. I did this in the quote of my
last post from the wiki.
> > -http://en.wikipedia.org/wiki/Watt
>
> > And so we should be getting quite a push when we hold a mirror up to
> > the sun under the standard photon momentum argument. It is clearly a
> > misinterpretation and disambiguating this failure is not happening on
> > this thread.
>
> If h doesn't matter, why did you introduce h?
>
> Do it without h then. P = 1300W, so F=P/c = 9 micronewtons. I wouldn't
> call this "quite a push".
>
> 1300W incident on a reflecting object _doesn't_ mean that 1300W of work
> will be done. Or even the same on a perfect absorber. Again, do the
> calculation for reflection/"absorption" of balls.
>
> To repeat yet again, the predicted force is 9 micronewtons per square
> metre. Why do you think this should be easy to observe?
We've moved to photon momentum here, and not radiation pressure. I
guess we're near to a stalemate on this, because of the quantity of
repetitions that we are going onto. I encourage you to falsify me as I
have falsified you, and remind you of your own admission:
"If the reflector is moving, there will be a Doppler shift, and
energy in will be different from energy out. If the reflector
is stationary, then you could have, e.g., 1300W in and 1300W out.
But no work would be done."
- http://groups.google.com/group/sci.physics/msg/f064666482b7c3b5
This is essentially the same thing that I am communicating here.
OK. I just did set it up in a spreadsheet. Each photon at 500nm has
energy of
5.03E-019 Joules
The total number of photons in one second that will stream is
2.58E+021 photons
The momentum per photon is
1.32E-027 kg m / s
The momentum for one seconds worth is
3.42E-006 kg m / s
I agree that the momentum is small, but this does nothing to diminish
the power argument and the net energy of this momentum when absorbed.
The momentum figure is operant at the speed of light and damn well
better be small or we'll all be vaporized. It's tough enough already
being out in the sun all day. Rather than getting pushed over by the
sunlight we're getting heated by it instead. There is no photon
momentum in a literal sense.
So long as the momentum is derived from the power then there is
nothing wrong with my own logic. The power is still there, even in
this slight figure, because momentum is not the same as energy or
power. It's still 1300 watts, and a one second pulse of this is quite
some energy, though I was just looking at lasers and most can only
pulse this kind of power for shorter durations. Energy takes many
forms, and we see that they can be converted, but this does not mean
that they actually are converted, though we can still write the
equation. Somehow the physics culture has come to take the photon
momentum as fundamental but instead it is a trick, and we may as well
give them mass from another trick by the same trickster.
Is my logic a trick?
No, I do not think so. If it is then I do wish you would expose it as
such via falsification rather than via more construction. We've
already constructed the things that we need. There is no need for any
more.
>
> Do this for reflection or "absorption" of the particles. Don't just say
> that 1300W must give a huge force, do the calculation instead.
Look, 1300 watts is a very sound figure. It means nothing more or
nothing less. If anything it is just a matter of how long you have the
1300 watts for, and if we presume to consider one second, well, then
we have an energy figure. The fact is that the interpretation of
photon momentum places all of the energy of the photon into momentum,
and there is a lack of truth in this, yet it has come to be a
foundational concept for much thinking. It is easily disproven, and I
have done so here. I am still open to direct falsification of my
statements. The oddities of conjoining this with radiation pressure,
and the numerous interpretations possible are baffling.
- Tim
Timo Nieminen
> On Jun 6, 9:02 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
[big cut for brevity, the core seems to be:]
[E=hf]
OK, this will do for a start. This _isn't_ the suggested calculation,
but it _is_ a start. Note that E = hf = 3.98e-19J, so you underestimate
photon flux by 20%, and this the momentum by 20%. With the right E, this
brings you back to to the already-mentioned 4 or 9 micronewtons.
> The momentum figure is operant at the speed of light and damn well
> better be small or we'll all be vaporized. It's tough enough already
> being out in the sun all day. Rather than getting pushed over by the
> sunlight we're getting heated by it instead.
Why would 4 or 9 (or 3 or 7) micronewtons push you over? Do you expect to
feel this force?
> being out in the sun all day. Rather than getting pushed over by the
> sunlight we're getting heated by it instead. There is no photon
> momentum in a literal sense.
There's a reason why I suggested the calculation:
> > Do the calculation, using Newtonian mechanics, for a stream of Newtonian
> > particles moving at c, with kinetic energy (for each particle) equal to hf
> > for, say, 500nm light, and total KE per second in the stream of 1300W.
Try this! Don't do this for real photons, do this for Newtonian particles
with the same kinetic energy as the photons.
> So long as the momentum is derived from the power then there is
> nothing wrong with my own logic. The power is still there, even in
> this slight figure, because momentum is not the same as energy or
> power. It's still 1300 watts, and a one second pulse of this is quite
> some energy, though I was just looking at lasers and most can only
> pulse this kind of power for shorter durations. Energy takes many
> forms, and we see that they can be converted, but this does not mean
> that they actually are converted, though we can still write the
> equation. Somehow the physics culture has come to take the photon
> momentum as fundamental but instead it is a trick, and we may as well
> give them mass from another trick by the same trickster.
>
> Is my logic a trick?
> No, I do not think so. If it is then I do wish you would expose it as
> such via falsification rather than via more construction. We've
> already constructed the things that we need. There is no need for any
> more.
Power isn't force, energy isn't force, work isn't force.
As far as I can tell, your argument here is solely that 1300W is a lot of
power, and if light has momentum, 1300W of light must have lots of
momentum, and since we're not pushed around (as far as we can notice) by
sunlight, therefore light doesn't have momentum.
You did the calculation above, and you can see that the force would only
be micronewtons. Isn't this a direct falsification of any "1300W must
correspond to a large force" argument?
If you don't like to do a calculation for photons, or light, try it for
classical Newtonian particles of the same energy and speed. (You should
get a force double that of light, 9 micronewtons for "absorption", and 17
micronewtons for reflection. That's for KE coming in at 1300W.)
> > Do this for reflection or "absorption" of the particles. Don't just say
> > that 1300W must give a huge force, do the calculation instead.
>
> Look, 1300 watts is a very sound figure. It means nothing more or
> nothing less. If anything it is just a matter of how long you have the
> 1300 watts for, and if we presume to consider one second, well, then
> we have an energy figure. The fact is that the interpretation of
> photon momentum places all of the energy of the photon into momentum,
> and there is a lack of truth in this, yet it has come to be a
> foundational concept for much thinking. It is easily disproven, and I
> have done so here.
First, what does "the interpretation of photon momentum places all of the
energy of the photon into momentum" mean? The standard accepted physics
theories of radiation pressure, whether by classical light or photons,
don't say this. In particular, who says that energy is momentum, or that
momentum is energy?
Second, you calculated yourself that the momentum of 1300J worth of
photons is 3.4e-6 kg.m/s (OK, 20% too low, but let us not worry about a
little error.) Is this 3.4e-6 kg.m/s "all of" the energy of that 1300J of
photons?
Which did you disprove? If the former, it's already accepted that energy
and momentum are different things; this doesn't say anything useful about
radiation pressure. If the latter, how did you disprove it?
> I am still open to direct falsification of my
> statements. The oddities of conjoining this with radiation pressure,
> and the numerous interpretations possible are baffling.
--
Timo
Sue...
You are trying to work with half a tweezers. My spell chequer
won't let me type the singular form. Probably because the
notion of half a tweezers is absurd. :-))
<<Optical tweezers can be used for three-dimensional manipulation
of transparent particles around 1–100 ñm in diameter, or other
small particles which behave as reactive dipoles.
The single-beam gradient optical trap consists of a single
laser beam, tightly focused to create a very strong field
gradient both radially and axially, which acts on polarisable
particles to cause a dipole force. Polarisable particles are
attracted to the strongest part of the field, at the beam focus,
due to the gradient force. A scattering force results from
momentum transfer to the particle when light is scattered by it.
Under the right conditions the gradient force can balance the
scattering and gravity forces, to trap particles three-dimensionally
in the laser beam. If the laser beam is not tightly focused, the
axial component of the gradient force will be weak, and only
radial trapping will be possible [1, 4].>>
"Optical trapping of absorbing particles"
Authors: H. Rubinsztein-Dunlop, T. A. Nieminen, M. E. J. Friese, N. R.
Heckenberg
http://arxiv.org/abs/physics/0310022
See also:
http://en.wikipedia.org/wiki/Van_der_Waals_force
http://en.wikipedia.org/wiki/London_dispersion_force
http://en.wikipedia.org/wiki/Induced_gravity
Sue...
Tim BandTech.com
> On Jun 6, 7:49 pm, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > On Jun 6, 4:44 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> > > On Jun 6, 11:05 pm, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
Thanks Sue.
I suggested the electric forces were at work in the trap a few posts
ago, and here you have Timo stating it in prior work.
As we discuss AC and DC principles, we are probably going to agree
that the electron is a constant charge, and so in effect is a DC
concept. On top of this the idea of stopping an electon does not exist
as far as I can tell. On top of this when we allow for electron spin
then we have gone beyond Maxwell, and even Einstein, and the raw
charge which Maxwell managed his equations on does not exist. In
effect this is the ultimate Maxwellian behavior, but denied charge as
fundamental to magnetism as in a displacement current model. Beyond
this thermodynamics remains open in my book. I'm not sure if you are
insinuating something more specific with the tweezer (no spell checker
here other than myself). I do appreciate the electrical nature of the
balance, though I can't say that I understand the existing theory. I
need to understand the atomic level better, so I appreciate your
persistent linkage into VdW and so forth.
- Tim
Tim BandTech.com
I had a typo on the constant c. Now I am in agreement. Nice to see you
are serious.
>
> > The momentum figure is operant at the speed of light and damn well
> > better be small or we'll all be vaporized. It's tough enough already
> > being out in the sun all day. Rather than getting pushed over by the
> > sunlight we're getting heated by it instead.
>
> Why would 4 or 9 (or 3 or 7) micronewtons push you over? Do you expect to
> feel this force?
Let's see now, I've got a momentum of 4.3E-6 kg m / s
but you've got a figure of 4E-6 kg m / s / s .
We have a units mismatch on momentum versus newtons, but most
importantly each of these figures includes the dimension of kilograms,
and these kilograms are travelling at the speed of light, and
furthermore since we used the relativistic
e = m c c
to get here there is no reason not to use it to get back a usual
amount of work, otherwise we are discussing an impossibility, right?
The math simply inverts, and we wind back up at the 1300 watt figure,
which is expressible in units through
1 J = 1 N m = 1 W s
and so I must ask you what happened to all of the energy that we
started the calculation with? Where is it? We had 1.3 kJ at the
beginning of the calculation and now over one second you are claiming
to have just something tiny. Well, multiply by c and you'll discover
that the energy is all there in that momentum figure. When does the
usage of relativity end? only after the thing travelling at the speed
of light is absorbed; not before. Isn't this coherent?
This really suggests that what is regarded as radiation pressure might
better be interpreted as photon momentum, whereas this figure that we
are computing above is a fraud.
>
> > being out in the sun all day. Rather than getting pushed over by the
> > sunlight we're getting heated by it instead. There is no photon
> > momentum in a literal sense.
>
> There's a reason why I suggested the calculation:
>
> > > Do the calculation, using Newtonian mechanics, for a stream of Newtonian
> > > particles moving at c, with kinetic energy (for each particle) equal to hf
> > > for, say, 500nm light, and total KE per second in the stream of 1300W.
>
> Try this! Don't do this for real photons, do this for Newtonian particles
> with the same kinetic energy as the photons.
I understand that the
K = m v v / 2
does not hold within relativity, and that there is another term, but
on top of this we will not get any sense at c. Why move the chase over
here? We already are using
E = m c c
to get momentum, and that is bad enough. To mix the two even more does
not seem appropriate. Already we are considering a Newtonian momentum
figure on something that is claimed to break Newtonian mechanics, but
then you don't care to perform the reversal do you? Again Timo, there
should be a more direct falsification. You ask for yet more
construction here and fail to falsify my own argument.
Your argument is claiming somewhat that the energy has disappeared. It
has not, and the math which has generated the argument holds, and its
reversal should be considered, since otherwise the conservation of
energy will be destroyed.
Look Timo, this is hardly the right place to be asking this question.
You are the one who set up the computations above, and we've
considered a one second pulse of laser light. This is 1.3kJ that we've
converted to momentum through e=hv and e=mcc. That you criticize this
figuring is to say that you are critical of existing theory.
>
> Second, you calculated yourself that the momentum of 1300J worth of
> photons is 3.4e-6 kg.m/s (OK, 20% too low, but let us not worry about a
> little error.) Is this 3.4e-6 kg.m/s "all of" the energy of that 1300J of
> photons?
Yes, at the speed c. You seem to think that the math is a one-way
framework. If this is the case then this is perhaps how we come to
disagreement. Why don't you answer this question yourself. Clearly we
fed the equations all of this energy. If we go back through them we
will get 1300 watts on the other side.
>
> Which did you disprove? If the former, it's already accepted that energy
> and momentum are different things; this doesn't say anything useful about
> radiation pressure. If the latter, how did you disprove it?
I'm feeling more secure that the term 'radiation pressure' is a
misnomer.
We've just derived it as photon momentum, so what is the difference?
One is distributed over a specific area whereas the other is a
conglomeration. This is nearly a quip on the photon itself. Mostly I'm
very interested in this topic, and not interested in studying it via a
mimic's approach. You are great to spend so much time on this. I'm
working on a new method, but I don't have anything substantial to
share yet. There are a number of places that I see frailties in
existing theory, and herabouts they are tied together.
- Tim
Sue...
> On Jun 7, 3:37 am, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
>
> > See also:http://en.wikipedia.org/wiki/Van_der_Waals_forcehttp://en.wikipedia.o...
>
> > Sue...
>
> Thanks Sue.
>
> I suggested the electric forces were at work in the trap a few posts
> ago, and here you have Timo stating it in prior work.
They are always at work but for the light-mill I
assumed convection currents explained it simply.
If not, long range induction forces should not
not be overlooked. Particularly for better
vacuums, solar sails, or even the fundamental
mechanism of inertia.
> As we discuss AC and DC principles, we are probably going to agree
> that the electron is a constant charge, and so in effect is a DC
> concept. On top of this the idea of stopping an electon does not exist
> as far as I can tell. On top of this when we allow for electron spin
> then we have gone beyond Maxwell, and even Einstein, and the raw
> charge which Maxwell managed his equations on does not exist.
No doubt, Maxwell and Einstein overlooked something
that W.Weber and F.London picked up on. (VdW)
> In
> effect this is the ultimate Maxwellian behavior, but denied charge as
> fundamental to magnetism as in a displacement current model. Beyond
> this thermodynamics remains open in my book. I'm not sure if you are
> insinuating something more specific with the tweezer (no spell checker
> here other than myself).
That is funny. So I am clapping with one hand.
> I do appreciate the electrical nature of the
> balance, though I can't say that I understand the existing theory. I
> need to understand the atomic level better, so I appreciate your
> persistent linkage into VdW and so forth.
Persistent? Thank you. :-)
It is a quality learned from my mother who did not
realise that she was not Jewish, Cass Elliot was
not attractive and cream cheese is not fat-free
even on a bagel.
Sue...
NoEinstein
>
I had intended to include the following links that relate to the time
of travel of photons around light courses defined by mirrors. — NE —
Where Angels Fear to Fall
http://groups.google.com/group/sci.physics/browse_frm/thread/8152ef3e...
Replicating NoEinstein’s Invalidation of M-M (at sci.math)
http://groups.google.com/group/sci.math/browse_thread/thread/d9f9852639d5d9e1/dcb2a1511b7b2603?hl=en&lnk=st&q=#dcb2a1511b7b2603
NoEinstein
>
Dear Tim: Once I can get further... off of square two with my New
Science, there are "infinite" avenues for experiments, research, and
just plain fun! I know that there are other rational thinkers, like
you, amid the thousands reading my posts. Come on, readers (the
rational ones), let your presence be known! — NoEinstein —
Timo Nieminen
Momentum per second has the units of force. Force times time is
impulse, same units as momentum. Per Newton 2, force is the rate of
transfer of momentum.
> but most
> importantly each of these figures includes the dimension of kilograms,
> and these kilograms are travelling at the speed of light, and
> furthermore since we used the relativistic
> e = m c c
> to get here there is no reason not to use it to get back a usual
> amount of work, otherwise we are discussing an impossibility, right?
> The math simply inverts, and we wind back up at the 1300 watt figure,
> which is expressible in units through
> 1 J = 1 N m = 1 W s
> and so I must ask you what happened to all of the energy that we
> started the calculation with? Where is it?
We didn't do anything with the energy; it's still there.
> We had 1.3 kJ at the
> beginning of the calculation and now over one second you are claiming
> to have just something tiny. Well, multiply by c and you'll discover
> that the energy is all there in that momentum figure.
Momentum _isn't_ energy, energy _isn't_ momentum.
> When does the
> usage of relativity end? only after the thing travelling at the speed
> of light is absorbed; not before. Isn't this coherent?
When does the usage of relativity begin? You get the same kind of
result using Newtonian mechanics and tiny particles.
> This really suggests that what is regarded as radiation pressure might
> better be interpreted as photon momentum, whereas this figure that we
> are computing above is a fraud.
>
>
>
> > > being out in the sun all day. Rather than getting pushed over by the
> > > sunlight we're getting heated by it instead. There is no photon
> > > momentum in a literal sense.
>
> > There's a reason why I suggested the calculation:
>
> > > > Do the calculation, using Newtonian mechanics, for a stream of Newtonian
> > > > particles moving at c, with kinetic energy (for each particle) equal to hf
> > > > for, say, 500nm light, and total KE per second in the stream of 1300W.
>
> > Try this! Don't do this for real photons, do this for Newtonian particles
> > with the same kinetic energy as the photons.
>
> I understand that the
> K = m v v / 2
> does not hold within relativity, and that there is another term, but
> on top of this we will not get any sense at c. Why move the chase over
> here?
Your main complaint with the result above is that you have difficulty
reconciling the large energy with the small momentum, the large power
with the small force.
Do the calculation for Newtonian particles, and you get the same type
of qualitative result (even similar numbers, just different by a
factor of 2). Do you disbelieve this result?
Note well, the _same kind_ of result. If you photon momentum
calculation makes you think it's all rubbish, will getting the same
kind of result with a Newtonian calculation make you think that
Newtonian mechanics is all rubbish?
Yes, of course (1/2)mv^2 is known to be incorrect at v=c. If this
bothers you, do the calculation for v=c/2, where Newtonian mechanics
is a reasonable (even if no longer very accurate) approximation.
> We already are using
> E = m c c
> to get momentum, and that is bad enough. To mix the two even more does
> not seem appropriate. Already we are considering a Newtonian momentum
> figure on something that is claimed to break Newtonian mechanics, but
> then you don't care to perform the reversal do you? Again Timo, there
> should be a more direct falsification.
> You ask for yet more
> construction here and fail to falsify my own argument.
It's been falsified many times already. Once should be enough.
Where is the logic in your claim? Your only basis for it above is that
a large energy results in a small momentum, so must be wrong. But
momentum and energy aren't the same thing! Qualitatively, one gets the
same kind of result from Newtonian mechanics; it shouldn't be
mysterious (unfamiliar, and offensive to our common sense is OK, since
this is outside our common everyday experience of mechanics).
The falsifications so far:
(1) Radiation pressure is observed experimentally, for RF, light, and
acoustic waves at least.
(2) The result of high energy/small momentum is exactly as expected
from either relativistic or Newtonian mechanics. That energy and
momentum are different things is also clear in both mechanics. Both
are also observed to give a good description of reality (Newtonian
mechanics is it's regime of applicability).
(3) Conservation of energy and momentum gives a prediction of
radiation pressure.
(4) Direct calculation of the forces gives electromagnetic radiation
pressure, using classical electromagnetic theory (which is also
observed to give a good description of reality). In particular, the
same physics that is used to (successfully) describe electric motors
etc. results in a prediction of radiation pressure.
The results of (3) and (4) are the same, and the same as (2) using
relativistic mechanics of zero rest-mass photons. The theoretical
models agree with experiment (with 10% for Nichols and Hull, 4% for
Bell and Green, 0.1% for R. V. Jones, within 5% for our own absolute
measurements).
If you choose to outright reject relativistic and Newtonian mechanics,
the conservation of energy and momentum, and electric motors, and
ignore a diverse range of thousands of experiments, what can be said?
(Other than you _not_ being open to falsification!)
> Your argument is claiming somewhat that the energy has disappeared. It
> has not, and the math which has generated the argument holds, and its
> reversal should be considered, since otherwise the conservation of
> energy will be destroyed.
Of course the energy hasn't disappeared! Momentum isn't energy, energy
isn't momentum! They're not the same thing!
Seriously, it's time to do the Newtonian calculation (at v=c, or v=c/2
if you prefer).
Why claim that I criticise this? It's correct, and I said so (apart
from your calculation error from typo).
> > Second, you calculated yourself that the momentum of 1300J worth of
> > photons is 3.4e-6 kg.m/s (OK, 20% too low, but let us not worry about a
> > little error.) Is this 3.4e-6 kg.m/s "all of" the energy of that 1300J of
> > photons?
>
> Yes, at the speed c. You seem to think that the math is a one-way
> framework. If this is the case then this is perhaps how we come to
> disagreement. Why don't you answer this question yourself. Clearly we
> fed the equations all of this energy. If we go back through them we
> will get 1300 watts on the other side.
Yes, do it the other way. The energy is still all there. We haven't
magicced the energy into becoming momentum; momentum and energy are
two different things.
In Newtonian mechanics, do you get rid of kinetic energy (1/2)mv^2 by
calculating the momentum mv? You get a different number, with
different units. One can be small, and the other large. Do you think
this spells trouble for conservation of energy?
> > Which did you disprove? If the former, it's already accepted that energy
> > and momentum are different things; this doesn't say anything useful about
> > radiation pressure. If the latter, how did you disprove it?
>
> I'm feeling more secure that the term 'radiation pressure' is a
> misnomer.
> We've just derived it as photon momentum, so what is the difference?
Force is rate of transfer of momentum. Newton 2 links force and
momentum together. If light carries momentum, it must be able to exert
force. If light exerts force, it must carry momentum.
This would have been the clear and evident result from the redshift/
blueshift thing by now (hopefully), but you've avoided this so far.
From conservation of energy, the work-energy theorem, and Galileian
invariance, out comes radiation force and momentum.
NoEinstein
>
Dear Timo: "Momentum", first and foremost, requires that there be a
moving MASS. Photons are MASSLESS, and thus are without momentum.
So, there is nothing to be... "reversed". — NoEinstein —
>
> > And so we should be getting quite a push when we hold a mirror up to
> > the sun under the standard photon momentum argument. It is clearly a
> > misinterpretation and disambiguating this failure is not happening on
> > this thread.
>
> If h doesn't matter, why did you introduce h?
>
> Do it without h then. P = 1300W, so F=P/c = 9 micronewtons. I wouldn't
> call this "quite a push".
>
> 1300W incident on a reflecting object _doesn't_ mean that 1300W of work
> will be done. Or even the same on a perfect absorber. Again, do the
> calculation for reflection/"absorption" of balls.
>
> To repeat yet again, the predicted force is 9 micronewtons per square
> metre. Why do you think this should be easy to observe?
>
>
>
> > > This is small enough so it isn't casually observable. Observed by
> > > Nichols and Hull (to about 10%, according to Bell and Green), and
> > > others since. Observed in special circumstances before and since
> > > (e.g., as contributor to comet tails).
>
> > > > The radiation pressure is not unlike electron spin, in that it is a
> > > > small effect that is difficult to notice and for
>
>
> read more »- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
Sam Wormley
> I had intended to include the following links that relate to the time
> of travel of photons around light courses defined by mirrors. — NE —
From the quantum mechanical perspective,
1. photons are emitted (by charged particles)
2. photons propagate at c
3. photons are absorbed (by charged particles)
Photon momentum
p = hν/c = h/λ
Photon Energy
E = hν
NoEinstein
>
Dear Tim: Light always EMITS at speed 'c' (which I rename IVL, for
the Intrinsic Velocity of Light). But the velocity of the traveling
light is influenced by the speed of the moving source—plus or minus
'v''. All blue shifted light is EXCEEDING velocity 'c'. Timo should
explain WHY he is concerned with "momentum" in photons? The Crookes
Radiometer rotates black sides trailing, because of the greater
concentration of ether near the black sides. And that ether pressure
is being augmented by the pressure/impacts of argon atoms hurled out
by the light (and ether) from the white side of the vanes. Nothing
about Einstein's speed limit on light is valid. That doesn't require
"faith", it only requires understanding. — NoEinstein —
NoEinstein
>
Dear Timo: About the time I decided to disprove Einstein (by
invalidation of the M-M experiment and rubber rulers), I read an early
book by Stephen Hawking. In it Hawking explained that the first step
in formulating any theory is to express the ideas, verbally. My
entire New Science has done that so well, and, I think, conclusively,
that no "equations" are needed. Equations are for quantifying the
variables. Much of what tries to pass for physics and cosmology,
today, is to formulate ridiculous (or not intuitive) notions about the
relationships of the variables, hoping to... "confirm" that the theory
is correct. I know my that New Science is correct without needing to
know any of the variables!
For example: Since gravity requires that there be photon emission OUT,
in order to have forces of gravity, IN; and since Black Holes have no
photon emissions out; then, Black Holes can't have gravity! The star
distribution data for the center of Andromeda PROVES that fact! No
black holes means there are no singularities, AND no Big Bang (nor Big
Crunch). So, it isn't necessary to quantify the forces in the
Universe to "aid" finding the missing mass to hold everything
together. There isn’t any missing mass! That’s why there's no
practical reason to know the thermal correction to Newton's Law of
Universal (sic) Gravitation. One only needs to know that temperature
affects gravity.
In your replies to Tim, you keep stating simple equations. My feeling
is that you don't yet understand the VERBAL expressions of the theory,
while you try to hide that fact by seeming "learned" with those
equations. Think really clearly, Timo, and equations won't be needed
in any conversation about basic science. — NoEinstein —
> Timo- Hide quoted text -
Sam Wormley
> About the time I decided to disprove Einstein (by
> invalidation of the M-M experiment and rubber rulers)
There are USENET Posters that understand relativity
theory and many that don't, including you.
Relativity theory is self consistent and has no contradictions.
In fact, there has yet to be an observation that contradicts a
prediction of relativity.
Are There Any Good Books on Relativity Theory?
http://math.ucr.edu/home/baez/physics/Administrivia/rel_booklist.html
Physics FAQ: What is the experimental basis of special relativity?
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
Sue...
[...]
> Beyond this thermodynamics remains open in my book.
This is a bit disturbing at this point in the thread.
Reynolds explained Crooks radiometer with
~thermodynamics~ .
Your discussion with Timo seems to be about
Nichols radiometer. The distinction was
made earlier in the thread but you will be
talking past one another if you are not
about the same device and effect.
http://en.wikipedia.org/wiki/Nichols_radiometer
http://en.wikipedia.org/wiki/Crookes_radiometer
Apologies if I am covering old ground but
it is a long thread and I got here late.
Sue...
>
> > > - Tim
>
>
Sue...
> On Jun 6, 9:02 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> Dear Timo: "Momentum", first and foremost, requires that there be a
> moving MASS. Photons are MASSLESS, and thus are without momentum.
> So, there is nothing to be... "reversed". — NoEinstein —
See equation 2.2
"Radiation Pressure
and Momentum Transfer in Dielectrics:
The Photon Drag Effect"
Rodney Loudon, Stephen M. Barnett, C. Baxter
http://www.colin-baxter.com/academic/research/downloads/prl063802.pdf
Sue...
Tim BandTech.com
> On Jun 7, 7:41 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
> > Beyond this thermodynamics remains open in my book.
>
> This is a bit disturbing at this point in the thread.
> Reynolds explained Crooks radiometer with
> ~thermodynamics~ .
>
> Your discussion with Timo seems to be about
> Nichols radiometer. The distinction was
> made earlier in the thread but you will be
> talking past one another if you are not
> about the same device and effect.
Thanks for the attempt at resolution, but we are on to other aspects
now.
He seems to think that when you attribute all of a photons energy to
momentum that somehow the energy is independent of that momentum. I'm
not going to buy that, but if you can falsify either of us then I
think that input is very welcome. We're really not beyond anything
more than
e = h f , e = m c c ,
and such simple product relationships.
> http://en.wikipedia.org/wiki/Nichols_radiometer
> http://en.wikipedia.org/wiki/Crookes_radiometer
>
> Apologies if I am covering old ground but
> it is a long thread and I got here late.
Hey Sue, no problem. I guess one of the key points is that the
radiometer itself is not quite what most of the discussion is about.
Isolation of radiation pressure from the radiation is more like it.
What I now understand and had overlooked for much of the thread is
that the radiation pressure is merely the photon momentum, as is
overlooked at
http://en.wikipedia.org/wiki/Radiation_pressure
and likely elsewhere. Nichols work is here:
http://books.google.com/books?id=8n8OAAAAIAAJ&pg=RA5-PA329#v=onepage&q&f=false
which is actually linked to in that wiki you just gave.
Some if his argumentation is quite poor imo. There is a 1933 paper by
a woman Bell that I do not have access to which claims to resolve the
study down to 10E-6 torr. I posted that link a few days ago here.
- Tim
Timo Nieminen
> He seems to think that when you attribute all of a photons energy to
> momentum that somehow the energy is independent of that momentum.
No! Absolutely not!
ENERGY IS NOT MOMENTUM, AND MOMENTUM IS NOT ENERGY!
THIS IS ELEMENTARY NEWTONIAN PHYSICS, THAT THE TWO ARE DIFFERENT!
TO SAY SOMETHING HAS BOTH KINETIC ENERGY AND MOMENTUM IS NOT THE SAME AS
"ATTRIBUTING THE ENERGY TO MOMENTUM"!
And, although energy and momentum are different, they're not independent
either! Stop a classical particle, and it has zero momentum and zero KE.
--
Timo
BURT
> On May 31, 8:11 pm, BURT <macromi...@yahoo.com> wrote:
>
>
>
>
>
> > On May 31, 3:31 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
>
> > > On May 30, 2:48 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> > > Dear Timo: I like that you have had a broad exposure to the world of
> > > physics. My New Physics is different in that it is based almost
> > > solely on analysis and on reason. To avoid being ‘corrupted’ by the
> > > status quo, I relish the observations of valid experiments—while
> > > always being open minded to the possibility of errors. I avoid
> > > “automatically” accepting the explanations, by supposed authorities,
> > > for the observed phenomena.
>
> > > White and black squares are two competing “gravity” experiments
> > > combined into one. In the Crookes Radiometer, the black squares
> > > exhibit more repulsion from the light (or heat) source than the white
> > > squares. Reverse rotation has been observed (by others) to occur if
> > > the glass is made to be cooler than the vanes, themselves. You say…
> > > “The force on the vanes has been measured in vacuum, and the force is
> > > in the opposite direction to the usual Crookes radiometer thermal
> > > force. If not for friction, the radiometer in vacuum would rotate
> > > Crookes. I don’t make it a point to shoehorn anyone’s “observations”
> > > unless and until I know most of the particulars.
>
> > > *** I invite you to reply with a concise PARAPHRASE of how that “in
> > > vacuum” experiment was done. (Note: I do not read links to the words
> > > of others.) The thermal qualities of the vacuum container must be
> > > considered, as well as the thermal isolation of the white paint from
> > > the black paint, if present. Since there was no rotation, how was…
> > > “the force” measured?
>
> > > Like I have said, conclusively, massless photons, alone, exert no
> > > force on objects. What is actually happening to move small objects is
> > > that photons create a gravity effect, as explained in:
>
> > > There is no "pull" of gravity, only the PUSH of flowing ether!http://groups.google.com/group/sci.physics/browse_thread/thread/a8c26....
>
> > > The latter involves having the varying ether flow and density push the
> > > object in proportion to the object’s cross section that is in the
> > > photon stream. Dust particles adjacent to laser beams can be seen to
> > > move in the direction of the beam. But that is due to the air
> > > molecules, and the ether being moved, together. The dust is pushed by
> > > the air gases and by the flowing ether, not by the photons.
>
> > > That Wikipedia article on Radiometers mentioned that there is an
> > > induced gas flow through porous ceramic plates that is toward the side
> > > that is heated. [ Note: That is consistent with the ether flow
> > > direction predicted by my New Science. ] The rather iffy porosity of
> > > the ‘edges’ of the squares in the Crookes Radiometer has, for over a
> > > century, been considered to be the primary source for the thrust. The
> > > errant rationale has been: The edges of the black squares heat, and
> > > then shoot-out, the argon atoms, causing the observed rotation. The
> > > latter concocted ‘science‘, combined with Einstein’s heated gas
> > > nonsense, supposedly accounts for 100% of the observed rotation of the
> > > vanes.
>
> > > Photons are concentrations of energy which, in high enough
> > > concentrations, can burn through steel. Those photons don’t “force”
> > > through the steel. You could say: They “energy” through the steel!
>
> > Radio waves pass through steel.
>
> > Mitch Raemsch
>
> > > Timo, I’ve observed over the past month that you have, occasionally,
> > > been adversarial regarding aspects of my New Science. To the extent
> > > that you bring up valid points which I can explain to the many
> > > readers, I welcome your comments. But I don’t seek to have a time
> > > consuming one-on-one conversation with you just for your edification.
> > > Though this reply is long, don’t take that to be an invitation that
> > > you have been selected as the spokes-person for the status quo.
> > > Because of my obvious huge contributions to science, you should ask
> > > questions, not sit in judgment. You are welcomed to make your own
> > > ‘+new post(s)’ to pontificate your science if you differ with me.
> > > Lastly, please TOP post, and limit yourself to about two paragraphs.
> > > I really don’t need to hear what you think about every little thing
> > > that I’ve ever said. No more… PDs are wanted, here. Thanks! —
> > > NoEinstein —
>
> > > > On May 31, 12:31 am, NoEinstein <noeinst...@bellsouth.net> wrote:
>
> > > > > I’ve just learned, and provisionally accept as true, that: Radiometers
> > > > > won’t rotate at all in a perfect vacuum;
>
> > > > True enough, but misleading, since there is still a measurable force.
> > > > Only the friction of the bearings stops it from rotating.
>
> > > > > If the devices were totally
> > > > > frictionless, the rotation would occur in the identical direction
> > > > > without that gas being there.
>
> > > > This isn't true. The force on the vanes has been measured in vacuum,
> > > > and the force is in the opposite direction to the usual Crookes
> > > > radiometer thermal force. If not for friction, the radiometer in
> > > > vacuum would rotate "backwards".
>
>
> > > > This reverse force due to radiation pressure was measured in 1901.
> > > > (Published in 1901, anyway. I think Nichols and Hull did their
> > > > measurement in 1901, but Lebedev did his in1899, but didn't publish in
> > > > a journal until 1901.)
>
> > > > > Photons have ZERO mass, and exert zero force upon
> > > > > ‘striking’ a reflecting surface.
>
> > > > Non-zero force. This has been measured. Microscopic objects can be
> > > > easily pushed around with this force. Macroscopic objects have been
> > > > levitated against gravity. It's more common to use the force due to
> > > > refraction (which is also non-zero), since then you don't cook the
> > > > object being pushed, but reflection works too. (Also absorption.)- Hide quoted text -
>
> > > - Show quoted text -- Hide quoted text -
>
> > - Show quoted text -- Hide quoted text -
>
> > - Show quoted text -
>
>
> - Show quoted text -
What is it about the steel that allows for different absorption for
thin and then thick?
What is the chance order of getting absorbed? Why would more steel
atoms provide better absorption?
If it is pure steel why would steel atoms absorb differently for
thiness or thickness. I have a problem that absorption is less
probable. How can it be made into probability?
I want to lnow what blocks radio signals.
MItch Raemsch
Sue...
> On Jun 7, 5:55 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
> > On Jun 7, 7:41 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
> > > Beyond this thermodynamics remains open in my book.
>
> > This is a bit disturbing at this point in the thread.
> > Reynolds explained Crooks radiometer with
> > ~thermodynamics~ .
>
> > Your discussion with Timo seems to be about
> > Nichols radiometer. The distinction was
> > made earlier in the thread but you will be
> > talking past one another if you are not
> > about the same device and effect.
>
> Thanks for the attempt at resolution, but we are on to other aspects
> now.
> He seems to think that when you attribute all of a photons energy to
> momentum that somehow the energy is independent of that momentum. I'm
> not going to buy that, but if you can falsify either of us then I
> think that input is very welcome.
I'll take your side because radiation~suction
fits an induction gravity mechanism better.
(Timo can conscript a few students if he
thinks we are ganging up on him.)
> We're really not beyond anything
> more than
e = h f , e = m c c ,
> and such simple product relationships.
I don't see how e = hf applies where there
may be no atomic absorption.
<<The requirements of energy and momentum
conservation generally forbid the absorption
of photons by free carriers, and the process can
only take place by interband transitions or with
the assistance of phonon absorption or emission. >>
http://www.colin-baxter.com/academic/research/downloads/prl063802.pdf
>
http://en.wikipedia.org/wiki/Nichols_radiometer
http://en.wikipedia.org/wiki/Crookes_radiometer
>
> > Apologies if I am covering old ground but
> > it is a long thread and I got here late.
>
> Hey Sue, no problem. I guess one of the key points is that the
> radiometer itself is not quite what most of the discussion is about.
> Isolation of radiation pressure from the radiation is more like it.
> What I now understand and had overlooked for much of the thread is
> that the radiation pressure is merely the photon momentum, as is
> overlooked at
> http://en.wikipedia.org/wiki/Radiation_pressure
> and likely elsewhere. Nichols work is here:
> http://books.google.com/books?id=8n8OAAAAIAAJ&pg=RA5-PA329#v=onepage&...
<<Theory
It may be shown by electromagnetic theory, by quantum
theory, or by thermodynamics, making no assumptions as
to the nature of the radiation, that the pressure against
a surface exposed in a space traversed by radiation
uniformly in all directions is equal to one third of the
total radiant energy per unit volume within that space.>>
http://en.wikipedia.org/wiki/Radiation_pressure
Hmmm... 1/3 is a pretty nice number and the statement
is very inverse square-ish. Is it too late to
switch to Timo's team? I am a sore looser.
I see the traversed volume here:
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html
But all of it, half of it or 1/3 of it is not
doing anything for me unless we put some gas in
it and give it a temperature.
Yeah... That's the steam.
Eggs explode in my microwave oven because
of radiation pressure.
> which is actually linked to in that wiki you just gave.
> Some if his argumentation is quite poor imo. There is a 1933 paper by
> a woman Bell that I do not have access to which claims to resolve the
> study down to 10E-6 torr. I posted that link a few days ago here.
In fairness, I should read about Timo's light-bullets
a bit closer before we declare victory.
Photons (Phonons?) can be a pretty good model translating
angular momentum in a dielectric. I am becoming
sceptical however because acoustic radiation pressure
is lumped in, apparently as the same effect.
If it is just molecules in the traversed volume
jiggling more, induction gravity should be unscathed
and it really shouldn't matter how you describe the
heating process. light bullets, flaming arrows, or
Ella Fitzgerald on Memorex.
Sue...
>
> - Tim
Tim BandTech.com
Newton probably had a deep appreciation for light. He could tell there
was energy there as the sun warmed things. Still, how much energy was
probably quite an open puzzle. It would be possible to put a lower
bound on the energy by observing the heating, but was this all of the
energy? No, says modern theory, there is in addition to energy of the
photon itself a radiation pressure capable of providing a minor
propulsion to matter in addition to its radiant energy. Is this an
accurate portrayal of the modern theory? This I will argue is a
twisted interpretation, especially based on the mathematics that you
requested, and which until I did it had no idea that radiation
pressure actually is photon momentum, which is the entire theoretical
photon energy, and hence my claim of twistedness.
>
>
>
> > This really suggests that what is regarded as radiation pressure might
> > better be interpreted as photon momentum, whereas this figure that we
> > are computing above is a fraud.
>
> > > > being out in the sun all day. Rather than getting pushed over by the
> > > > sunlight we're getting heated by it instead. There is no photon
> > > > momentum in a literal sense.
>
> > > There's a reason why I suggested the calculation:
>
> > > > > Do the calculation, using Newtonian mechanics, for a stream of Newtonian
> > > > > particles moving at c, with kinetic energy (for each particle) equal to hf
> > > > > for, say, 500nm light, and total KE per second in the stream of 1300W.
>
> > > Try this! Don't do this for real photons, do this for Newtonian particles
> > > with the same kinetic energy as the photons.
>
> > I understand that the
> > K = m v v / 2
> > does not hold within relativity, and that there is another term, but
> > on top of this we will not get any sense at c. Why move the chase over
> > here?
>
> Your main complaint with the result above is that you have difficulty
> reconciling the large energy with the small momentum, the large power
> with the small force.
No, not at all. As I just stated in the 'twisted' paragraph it is that
all of these figures are the same energy. To represent them as
distinct things is breaking down under our careful analysis, which
from my side was to attempt to isolate them.
>
> Do the calculation for Newtonian particles, and you get the same type
> of qualitative result (even similar numbers, just different by a
> factor of 2). Do you disbelieve this result?
Alright, I guess I'll have to entertain you here.
K = m v v / 2 : kinetic energy
e(photon) = K : all of photon's energy kinetic
v(photon) = c : velocity of photon
p(photon) = m v = 2 K / c = 2 h f / c .
I guess this is what you are talking about.
I have no idea why you are pressing for this.
>
> Note well, the _same kind_ of result. If you photon momentum
> calculation makes you think it's all rubbish, will getting the same
> kind of result with a Newtonian calculation make you think that
> Newtonian mechanics is all rubbish?
The e = h f equation is not fundamental is it? It is derived. If we
were to simply suppose that each photon carried half the energy of
this claim then we would merely need to double the quantity of photons
to have an equivalent theory... up to a point. I'm not feeling solid
about this argument but you are asking me to think about it, so there
you have it. Newton did plenty with light, but I don't really know
about any Newtonian photon, or energy computation on light. Here is
some solid Newtonian thinking:
"RULE 1
We are to admit no more causes of natural things than such as are
both true and sufficient to explain their appearances.
To this purpose the philosophers say that Nature does nothing in
vain,
and more is in vain when less will serve; for Nature is pleased
with simplicity,
and affects not the pomp of superfluous causes."
- Isaac Newton, Principia Mathematica;
System Of The World; Rules Of Reasoning In Philosophy;
translation by Andrew Motte
>
> Yes, of course (1/2)mv^2 is known to be incorrect at v=c. If this
> bothers you, do the calculation for v=c/2, where Newtonian mechanics
> is a reasonable (even if no longer very accurate) approximation.
I'm not following any larger argument on this information. I see that
you claim an equivalence between light and massive objects, but then
that there are discrepancies too.
>
> > We already are using
> > E = m c c
> > to get momentum, and that is bad enough. To mix the two even more does
> > not seem appropriate. Already we are considering a Newtonian momentum
> > figure on something that is claimed to break Newtonian mechanics, but
> > then you don't care to perform the reversal do you? Again Timo, there
> > should be a more direct falsification.
> > You ask for yet more
> > construction here and fail to falsify my own argument.
>
> It's been falsified many times already. Once should be enough.
>
> Where is the logic in your claim? Your only basis for it above is that
> a large energy results in a small momentum, so must be wrong. But
> momentum and energy aren't the same thing! Qualitatively, one gets the
> same kind of result from Newtonian mechanics; it shouldn't be
> mysterious (unfamiliar, and offensive to our common sense is OK, since
> this is outside our common everyday experience of mechanics).
>
> The falsifications so far:
>
> (1) Radiation pressure is observed experimentally, for RF, light, and
> acoustic waves at least.
Actually we've just recently exposed that radiation pressure is photon
momentum in the e=hf, e=mcc sense.
>
> (2) The result of high energy/small momentum is exactly as expected
> from either relativistic or Newtonian mechanics. That energy and
> momentum are different things is also clear in both mechanics. Both
> are also observed to give a good description of reality (Newtonian
> mechanics is it's regime of applicability).
Yes, I'll go along with this, except to expose that by not reversing
the equations we are getting a figure that is supposedly
experimentally verifiable (the 4.6E-6 figure on 1370 Watts), and so in
some regards my earlier question about when the relativistic
conversions end might be important.
>
> (3) Conservation of energy and momentum gives a prediction of
> radiation pressure.
Nah. Photon momentum is radiation pressure. Radiation pressure is a
misnomer.
>
> (4) Direct calculation of the forces gives electromagnetic radiation
> pressure, using classical electromagnetic theory (which is also
> observed to give a good description of reality). In particular, the
> same physics that is used to (successfully) describe electric motors
> etc. results in a prediction of radiation pressure.
I'd like to understand this but don't yet. Back to Maxwell's Treatise
another day...
>
> The results of (3) and (4) are the same, and the same as (2) using
> relativistic mechanics of zero rest-mass photons. The theoretical
> models agree with experiment (with 10% for Nichols and Hull, 4% for
> Bell and Green, 0.1% for R. V. Jones, within 5% for our own absolute
> measurements).
>
> If you choose to outright reject relativistic and Newtonian mechanics,
> the conservation of energy and momentum, and electric motors, and
> ignore a diverse range of thousands of experiments, what can be said?
> (Other than you _not_ being open to falsification!)
>
> > Your argument is claiming somewhat that the energy has disappeared. It
> > has not, and the math which has generated the argument holds, and its
> > reversal should be considered, since otherwise the conservation of
> > energy will be destroyed.
>
> Of course the energy hasn't disappeared! Momentum isn't energy, energy
> isn't momentum! They're not the same thing!
When the momentum is absorbed is not the energy likewise absorbed?
>
> Seriously, it's time to do the Newtonian calculation (at v=c, or v=c/2
> if you prefer).
I think I did this above, but I'm not quite sure what you are getting
at.
So lead me on some more up there and I'll try to be a good dog.
Well, to me this is becoming the most interesting part of the
analysis.
On the one hand the math we used to generate the photon momentum takes
into account all of the photon's energy, yet upon absorbing the photon
we do not see this come out as mechanical energy. This math actually
does match experiment as you substantiate elsewhere, and so we should
accept this one way paradigm, but this is not a completed theory then,
for no discussion like this has been undertaken.
To reverse the math we simply admit that when a photon hits a cold
black surface that it's energy is transferred into that surface.
Knowing the momentum of the photon we simple backtrack the math and
get a small energy figure, and based on our earlier figures we will
find 1300 watts of energy was absorbed. Because we put all of this
energy in as momentum it follows that within the photon this became
kinetic energy, and that the surface therefor receives 1300 watts of
mechanical work. This is where you should falsify I believe.
I was just reading a wiki page on momentum and they claim that static
electric fields have momentum. Do you balk at this? At this point I
have no idea what to really believe. This is all a wake up call for
me. I am so confused as to why you would deny any connection between
momentum and energy, and it is clear that to uphold existing theory we
cannot admit the 1300 watts of mechanical work, otherwise we'll really
have high power solar mechanics going on, yet this is what the math
would do. There must be a glitch imo.
>
> In Newtonian mechanics, do you get rid of kinetic energy (1/2)mv^2 by
> calculating the momentum mv? You get a different number, with
> different units. One can be small, and the other large. Do you think
> this spells trouble for conservation of energy?
No. You're speaking with a lack of concession for the fact that when
the momentum rises so does the kinetic energy. Instead of admitting
this you say:
"You get a different number, with different units. One can be
small,
and the other large."
Now if you added the concession that I recommend then you refutation
here withers.
> > > Which did you disprove? If the former, it's already accepted that energy
> > > and momentum are different things; this doesn't say anything useful about
> > > radiation pressure. If the latter, how did you disprove it?
>
> > I'm feeling more secure that the term 'radiation pressure' is a
> > misnomer.
> > We've just derived it as photon momentum, so what is the difference?
>
> Force is rate of transfer of momentum. Newton 2 links force and
> momentum together. If light carries momentum, it must be able to exert
> force. If light exerts force, it must carry momentum.
>
> This would have been the clear and evident result from the redshift/
> blueshift thing by now (hopefully), but you've avoided this so far.
> From conservation of energy, the work-energy theorem, and Galileian
> invariance, out comes radiation force and momentum.> One is distributed over a specific area whereas the other is a
Well Timo, one thing at a time, perhaps two. I am sorry I have been so
rude, but I will be happy to take up the blueshift argument sometime
soon. After enough cycles of repetitious banter we should not engage
in more of the same, unless there is some avenue to change it up a
little. Well, I have suggested such an avenue, but I don't honestly
have a mechanism yet. To fully go here I think we'll need to break
e=hf into two parts; one rotational and one directed. Still, I am not
up to this yet, and obviously it won't go far with you. Maybe we could
drop this for now and go to the blue shift argument, which I can pick
up right where you left off. Still, if you have something new to add
on to this excessively repetitious and long post then please do, but
really if we're both just repeating ourselves we should call a
stalemate; temporarily at least. I do understand that momentum and
kinetic energy are not identical, but that they are related. Yes, they
are different, but not so terribly different. And yet experiment
agrees with you, but the theory does not; not when we attempt to
extract the 1300 watts as mechanical energy. In a few decades we may
have a nanotechnology that will do this. First for coherent light, and
then if we are lucky for the noisy sunlight. Even a 50% efficient
mechanical or electrical figure would be fine.
- Tim
Timo Nieminen
> On Jun 7, 3:55 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
[a very quick point, will return later]
> > Of course the energy hasn't disappeared! Momentum isn't energy, energy
> > isn't momentum! They're not the same thing!
>
> When the momentum is absorbed is not the energy likewise absorbed?
No! (I think "absorbed" is the wrong word here.)
Bounce a ball off a wall. KE_in = KE_out. No loss of KE. Change in
momentum = 2 * momentum_in.
ka...@nventure.com
>
>
> Relativity theory is self consistent and has no contradictions.
> In fact, there has yet to be an observation that contradicts a
> prediction of relativity.
>
>
not
without its problems.
And your second sentence is wrong.
The Eotvos experiment by R. H. Dicke at Princeton demonstrated
that mass is not relativistic.
Furthermore, there is the twin paradox.
Now don't give me all the BS that the relativistists use to try to
resolve this paradox.
Just tell me what twin is younger than the other when they meet
after the traveling twin has completed his/her trip and both are at
rest in what Einstein called the 'stationary system' on page 40 of
"The Principle of Relativity" copyrighted by Lorentz, Einstein,
Minkowski, and Weyl.
Moreover, there is the paradox of rapidly rotating discs (like a
Frisbee) moving near the speed of light, wherein the physical form
must constantly change due to the tensor mechanics. And these
physcal changes must occur without generating heat, or violate
the Laws of Thermodynamics.
D.Y.Kadoshima
ka...@nventure.com
>
> From the quantum mechanical perspective,
>
> 1. photons are emitted (by charged particles)
> 2. photons propagate at c
> 3. photons are absorbed (by charged particles)
>
> Photon momentum
> p = hν/c = h/λ
>
> Photon Energy
> E = hν
I would like to point out an inconsistency in your post
about the photon momentum
Momentum is a mathematically calculated dynamic property
of an uniformly moving mass, i.e., momentum is mass times
velocity by definition.
The photon is commonly given as a 'massless' quality, i.e.
a photon has no mass.
Thus any idea of photon momentum is an oxymoron.
Please do not think that I'm picking on you. I just choose to
respond to errors or inconsistencies your posts, and choose
decline to respond to the pure BS of others
D.Y. Kadoshima
Sue...
Is that ~consistent~ with the wikipedia page?
http://en.wikipedia.org/wiki/Radiation_pressure#Theory
If absorbed, the traversed volume is illuminated
once.
If reflected, the traversed volume is illuminated
twice.
Sue...
Timo Nieminen
Please, be more clear.
Is it OK for something (for example, a wave, a classical particle, or
a photon) to have a lot of energy and only a little momentum?
Is it OK for something (for example, a wave, a classical particle, or
a photon) to have a little energy and a lot of momentum?
Yes, for a given object, if the momentum is higher, the (kinetic)
energy is higher. Also, if the (kinetic) energy is higher, the
momentum is higher. This still doesn't mean that energy and momentum
are identical.
[moved]
> > The energy is still all there. We haven't
> > magicced the energy into becoming momentum; momentum and energy are
> > two different things.
>
> Well, to me this is becoming the most interesting part of the
> analysis.
> On the one hand the math we used to generate the photon momentum takes
> into account all of the photon's energy, yet upon absorbing the photon
> we do not see this come out as mechanical energy. This math actually
> does match experiment as you substantiate elsewhere, and so we should
> accept this one way paradigm, but this is not a completed theory then,
> for no discussion like this has been undertaken.
>
> To reverse the math we simply admit that when a photon hits a cold
> black surface that it's energy is transferred into that surface.
> Knowing the momentum of the photon we simple backtrack the math and
> get a small energy figure, and based on our earlier figures we will
> find 1300 watts of energy was absorbed. Because we put all of this
> energy in as momentum it follows that within the photon this became
> kinetic energy, and that the surface therefor receives 1300 watts of
> mechanical work. This is where you should falsify I believe.
[moved]
> I am so confused as to why you would deny any connection between
> momentum and energy, and it is clear that to uphold existing theory we
> cannot admit the 1300 watts of mechanical work, otherwise we'll really
> have high power solar mechanics going on, yet this is what the math
> would do. There must be a glitch imo.
[Big cut for brevity; here is the core issue for the moment.]
We can transfer - whether using classical particles, classical waves,
or quantum particles - momentum without transferring energy. You can
transfer energy without converting it all to work; that is, you can
transfer energy without it becoming KE. This works largely in the same
way in all three cases.
Here is where it is worth sitting down and looking at these a little
systematically. (It can also be worthwhile putting some numbers in and
doing actual calculations; feel free to try, using a 500nm photon, a
classical particle of the same KE and speed (v=c), and a wave of power
1300W.)
Let us assume Newtonian mechanics, the energy of a photon E=hf,
momentum of a photon E=h/lambda. Let the aim be to (a) look at the
similarities between the 3 cases (classical particle, quantum
particle, classical wave), and (b) determine the momentum of a wave,
and consequent radiation pressure on reflection and absorption.
Galileian transformation between moving coordinate systems (so not
using special relativity).
I don't know your background knowledge, so let me speak about
momentum, force, and impulse first. Consider Newton's 2nd law: the
force applied to an object is the rate of change of momentum of the
object. With Newton's 3rd law, this means that force is the rate of
change of momentum. So, a force is a rate of transfer of momentum, and
is convenient when we're transferring momentum continuously. Not so
convenient to consider effectively instantaneous impacts. For this,
the impulse is better. Impulse is commonly defined as force * time
(for a constant force), but it's better to think of impulse as the
change in momentum, and force as the rate of change of momentum. For a
constant force, then we have force = impulse/time. If we have an
impulse of I per particle, and N particles per second, then F=I*N.
Similarly, if the transfer of energy per particle is E, the power is
E*N.
1a. Reflection by a stationary reflector.
(i) Consider a particle of mass m and speed v being reflected with no
loss of KE from a stationary plate (for the refletor to remain
stationary, there will need to be another force acting to hold it in
place). There is no change in energy, and no work is done on the
plate. But the momentum of the particle is reversed. This change in
the momentum, equal to double the particle momentum, will be the
impulse applied to the plate.
Thus,
KE = E = (1/2) mv^2,
p = mv = 2E/v,
I = 2mv = 4E/v.
With N particles per second, F = 4EN/v.
No work is done, the energy transferred per particle is zero, and the
power is zero.
(ii) A photon. As above, but we have:
E = hf,
p=h/lambda = E/c (similar to the energy/momentum relationship for a
particle, but without the factor of 2).
I = 2*E/c,
F = 2EN/c.
(iii) For a classical wave of power P, we have no work being done. If
we don't know the energy/momentum relation for a classical wave, we
don't know anything more, yet.
1b. Reflection by a moving reflector. Consider the above, as seen from
a moving coordinate system, moving at speed v_cs such that the
incident particle is faster or the incident photon or wave is
blueshifted.
(i) The particle speed is now v_2 = v + v_cs. The mass is unchanged.
The reflected particle is now at speed v_3 = v - v_cs.
E_in = (1/2)m v_2^2 = (1/2)m (v + v_cs)^2 = (1/2)m (v^2 + v_cs^2 +
2*v*v_cs)
E_out = (1/2)m v_3^2 = (1/2)m (v - v_cs)^2 = (1/2)m (v^2 + v_cs^2 -
2*v*v_cs)
E_in - E_out = 2m*v*v_cs.
By conservation of energy, this energy is transferred to the
reflector. In 1a(i), no energy was transferred to the reflector, and
therefore there was no heating of the reflector. Looking at this from
a different (moving) coordinate system doesn't change this; there is
still no heating. Therefore, the transferred energy must be the work
done on the reflector by the particle. (The restraining force stopping
the reflector from accelerating does work of the same magnitude, but
negative, so the total work on the reflector is zero, as expected,
since the KE of the reflector doesn't change.)
W = 2m*v*v_cs
P = 2m*v*v_cs * N
Since P = Fv,
F = P/v = W*N/v_cs = 2mvN = 4EN/v, where E is the original energy in
1a(i); this is exactly the same force as before. This is as expected,
since the change of coordinate system won't affect the force.
The change is momentum is still the same as before (the v_cs parts of
the momentum in and momentum out cancel), so from conservation of
momentum, we again have F = 4EN/v.
(ii) Photon.
E_in = hf(1+v_cs/c),
E_out = hf(1-v_cs/c),
E_in - E_out = 2hf v_cs/c. As for the particle case, there must still
be no heating.
P = 2hf N v_cs/c = 2EN v_cs/c
F = P/v_cs = 2EN/c, as before. Again, the change in coordinate system
can't have changed this, so this is as expected.
1/lambda changes by the same (1+ v_cs/c) factor due to the change in
coordinate system, the v_cs/c part cancels when we find the change in
momentum, so again,
I = 2*E/c,
F = 2EN/c.
(iii) Classical wave of (original) speed v_w and original power P0.
The power is proportional to the frequency. So,
P_in = P0 (1+v_cs/v_w)
P_out = P0 (1-v_cs/v_w)
P_in - P_out = 2P0 v_cs/v_w.
In 1a(iii), there was no absorption, and therefore no heating. The
change in coordinate system doesn't change that, so there is still no
heating. Thus, this is the rate of doing work on the reflector.
F = P/v = 2P0 v_cs/v_w = 2P0/v_w.
From conservation of momentum, the momentum flux of the original wave
(in the coordinate system where the reflector was at rest) is
p_flux = P0/v_w.
OK, that's enough for the moment. Note that the photon and wave case
are identical, apart from the stream of N photons per second coming in
discrete chunks, and the classical wave being continuous. Note that
the classical particle result is the same except for a factor of 2.
Further note that, in all 6 cases, there was no _absorption_ of
energy. That is, no energy went towards heating the reflector.
Let me know what you think.
Next will be: 2a. Absorption by a stationary absorber, and 2b.
Absorption by a moving absorber. If you want, you can try this one for
yourself. In all 6 cases, there will be heating; changing to the
moving coordinate system won't change the rate of heating. The change
in momentum of the particles/photons/wave will be approximately half.
For v_cs that is small compared to v, c, and v_w, the work done on the
absorber will be much small than the energy that heats the absorber.
There is a surprising result here! For the stationary absorber, the
change in momentum is exactly half that for the reflector. So, the
force is force is exactly half of the reflector force. The change in
coordinate system to the moving one can't change this force, but the
change in momentum in the moving coordinate system is only
approximately half, not exactly half (it's a little bit more than
1/2). If the extra momentum or momentum flux doesn't give us extra
force, what happens to this extra momentum. Answer: the absorbed
energy, now thermal energy, must have momentum when its moving!
OK, a few more quick points to reply to:
> > (3) Conservation of energy and momentum gives a prediction of
> > radiation pressure.
>
> Nah. Photon momentum is radiation pressure. Radiation pressure is a
> misnomer.
Not a misnomer. Newton 2: force = rate of change of momentum. If
photons don't have momentum, they can't produce a force. If photons
can produce a force, they must carry momentum. (Electromagnetic)
radiation pressure is just the effect of changing the momentum of
photons (or a classical EM wave), by reflection, refraction, or
absorption.
> I was just reading a wiki page on momentum and they claim that static
> electric fields have momentum. Do you balk at this? At this point I
> have no idea what to really believe. This is all a wake up call for
> me.
I don't trust it, the statement that static EM fields have momentum.
What we get out of the conservation laws is the Maxwell stress tensor
(or the 4x4 energy-momentum tensor, if we're doing a full relativistic
treatment; the Maxwell stress tensor is a 3x3 chunk of this). What is
meaningful then is either the divergence of the stress tensor (which
tells you the rate of change of momentum density), or its integral
over a closed surface (which tells you the rate of change of momentum
in the enclosed volume). It doesn't directly tell you the momentum
density.
There is similar funny business with the Poynting vector for certain
static fields. But again, it's the divergence of the Poynting vector,
or its integral, that are physically meaningful.
It's possible to calculate what the momentum density of a static field
should be. Start with no field, which means momentum density of zero,
and calculate the divergence of the stress tensor as the field builds
up to a final value. The integral of this over time should be the
momentum density. At first glance, this gives zero, which suggests
that a static field doesn't have momentum. I haven't checked this
properly.
--
Timo
Timo Nieminen
> On Jun 8, 2:30 am, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> > On Mon, 7 Jun 2010, Tim BandTech.com wrote:
> > > On Jun 7, 3:55 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> > [a very quick point, will return later]
>
> > > > Of course the energy hasn't disappeared! Momentum isn't energy, energy
> > > > isn't momentum! They're not the same thing!
>
> > > When the momentum is absorbed is not the energy likewise absorbed?
>
> > No! (I think "absorbed" is the wrong word here.)
>
> > Bounce a ball off a wall. KE_in = KE_out. No loss of KE. Change in
> > momentum = 2 * momentum_in.
>
>
> If absorbed, the traversed volume is illuminated
> once.
>
> If reflected, the traversed volume is illuminated
> twice.
Sure it's consistent. Energy density with reflection is double the
energy density with absorption, pressure is likewise double. (Not 1/3
since not omnidirectional.)
NoEinstein
>
Dear Timo: The 'velocity' in the momentum equation F = mv is actually
the top half of v / 32.174 feet/sec.—which is a simple unitless
fraction. Momentum is: An increase (or decrease) in the apparent mass
of an object due to its velocity. The 'units' of momentum is pounds.
Power is: Any force which is available to be used continuously, but
which may be used for any length of time. The 'units' of power is
pounds. A pound-second, would be a one pound force utilized for one
second. A KWH is 1,000 watts of power, or the force equivalent,
utilized for one hour. — NoEinstein —
>
> read more »- Hide quoted text -
NoEinstein
not. 2. Photons propogate at 'c' plus or minus the velocity of the
source. 3. Photons are absorbed by all masses whether charged or
not, and such exchange will cause a gravitational attraction between
the masses.
Photons are energy, and thus have no momentum. And, lastly, photon
energy is identical throughout the Universe regardless of the
velocity. Higher energy light merely has the photons spaced closer
together in the trains. — NE —
NoEinstein
>
Dear Sam: You are this credential-less armchair, blow-hard who tries
to elevate your standing by attacking the person (me) who has made a
greater contribution to the understanding of science than all other
physicists combined. You, like PD, are in the bottom 2.5% of
incurable status quo junkies. Good luck in trying to climb up the
hill that I am King of. — NoEinstein —
NoEinstein
while not being conclusive about the basic science behind the design
of the detector. Thermal mechanics isn't involved in turning the
Crookes Radiometer. My gravity swing experiment is a much more
sensible design. — NoEinstein —
A Proposed Gravity-Propelled Swing Experiment.
http://groups.google.com/group/sci.physics/browse_thread/thread/3052e7f7b228a800/aef3ee7dc59b6e2f?hl=en&q=gravity+swing
>
> On Jun 7, 7:41 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> [...]
>
> > Beyond this thermodynamics remains open in my book.
>
> This is a bit disturbing at this point in the thread.
> Reynolds explained Crooks radiometer with
> ~thermodynamics~ .
>
> Your discussion with Timo seems to be about
> Nichols radiometer. The distinction was
> made earlier in the thread but you will be
> talking past one another if you are not
> about the same device and effect.
>
> http://en.wikipedia.org/wiki/Nichols_radiometerhttp://en.wikipedia.org/wiki/Crookes_radiometer
>
> Apologies if I am covering old ground but
> it is a long thread and I got here late.
>
> Sue...
>
>
>
>
>
> > > > - Tim- Hide quoted text -
NoEinstein
>
Dear Sue: As I say to everyone, here, I don't read links to the words
of others (unless you wrote the link). I wish to have the discussions
be in black and white. Please give a simple paraphrase of your link,
and I will reply. Intense photon emissions can move some of the ether
that is in the path. And moving ether can exert forces on masses.
But that pressure is NEVER from the photons. Photons are energy,
only. — NE —
Sue...
OK... I thought we were on the same ~page~ with that.
> (Not 1/3 since not omnidirectional.)
I am interpreting the 1/3 to be an allowance for
the fact that the pyramidal volumes are
only confined at the ends. Areas
of 1/9 are easly seen in this figure.
You'll have to imagine the lines
for 1/3.
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html
Risking accusation of practising numerology
without a licence, the factor 1/3 appears
prominently here:
http://en.wikipedia.org/wiki/Boltzmann_constant
Sue...