Solar sailing DOESN"T break laws of physics'
Geoffrey A. Landis
the laws of physics.
In an article in New Scientist recently, maverick astronomer Thomas
Gold cast doubt about solar sails:
"Thomas Gold from Cornell University in New York says the proponents
of solar sailing have forgotten about thermodynamics, the branch of
physics governing heat transfer. Solar sails are designed to be
perfect mirrors, meaning that they reflect all the photons that strike
them. Gold argues that when photons are reflected by a perfect mirror,
they do not suffer a drop in temperature. "
Unfortunately, Gold has apparently forgotten to account for a
well-known physical effect: the Doppler shift.
It's worth saying that the photon pressure on a spacecraft is not
theoretical; its effect on spacecraft is measurable, and it has been
observed and measured to great precision routinely in space. Photon
pressure-- the solar sail effect-- has already been used for an
operational space mission; it was for spacecraft attitude control on
the Pioneer Venus-Mercury mission.
The Crookes radiometer does not operate on photon pressure, and the
explanation for how it operates has been known for over a century.
The energy transfer to a solar sail can be accounted for from the
Doppler shift of reflected photons; even when the reflectivity is
100%, a photon looses energy when reflecting from a moving sail. This
effect exactly corresponds to the energy increase of the sail. No
sophisticated physics is needed to analyze this effect, it is a
problem suitable for a homework assignment for a college
undergraduate.
When the sail is moving, then the reflected photons are Doppler
shifted, and leave the sail with lower energy than they arrived. This
loss of energy exactly equals the energy imparted to the sail, a fact
which can be trivially verified by using Newton's laws, the Doppler
formula, and the Einstein equation for photon momentum p=E/c
If the sail is not moving, there is no Doppler shift. However, note
that since energy is proportional to momentum squared, the derivative
of energy with respect to momentum is zero for a non-moving sail.
Thus, when the sail is stationary, it can reflect photons with perfect
efficiency and still gain momentum at no energy cost.
For completeness, note that if the sail is moving *toward* the light
source, then the phtons are Doppler shifted to *higher* energy by the
reflection. This implies that the sail must lose energy-- which is
correct; when the sail moves toward the light source, it slows down.
--
Geoffrey A. Landis
http://www.sff.net/people/geoffrey.landis
Gregory L. Hansen
Geoffrey A. Landis <geoffre...@sff.net> wrote:
>Despite a recent article in New Scientist, a solar sail does not break
>the laws of physics.
>
>In an article in New Scientist recently, maverick astronomer Thomas
>Gold cast doubt about solar sails:
>"Thomas Gold from Cornell University in New York says the proponents
>of solar sailing have forgotten about thermodynamics, the branch of
>physics governing heat transfer. Solar sails are designed to be
>perfect mirrors, meaning that they reflect all the photons that strike
>them. Gold argues that when photons are reflected by a perfect mirror,
>they do not suffer a drop in temperature. "
http://www.newscientist.com/news/news.jsp?id=ns99993895
Dang, I don't know what to say. Gold is a putz, and I thought New
Scientist was better than that. I didn't realize a solar sail was a heat
engine.
--
"Is that plutonium on your gums?"
"Shut up and kiss me!"
-- Marge and Homer Simpson
Uncle Al
>
> Despite a recent article in New Scientist, a solar sail does not break
> the laws of physics.
Actually, it does as proposed. The sail will come into thermal
equilibrium with the radiation field and emit photons from its other
side, counter-thrusting. Solar wind will mostly stick rather than
bounce, incrementally increasing sail mass. The claimed efficencies
will be, surprise!, *much* higher than anything obtained. More
studies will be needed. Damn! It didn't scale linearly like it was
supposed to.
If the sail is a dielectric mirror, I don't know how it could be
fabricated to perform given engineering, cost, and weight constraints
including dense folding for launch. (This includes 3M multimicrolayer
plastic supermirrors). Dielectric sails will accumulate patches of
static charge from the solar wind. Judging from meter-scale
satellites, a square kilometer of arcing HV capacitor plate will be...
memorable. Hard UV will disintegrate organics.
If the sail is metallic, there will be god's own Hell of field
interactions. Space is rich with large scale electromagnetic this and
that. If you think solar flares are tough on power lines (EMP pulse),
imagine what they will do to a square kilometer of conductor. The
thing might buck like a bronco. Mechanical control lines must be
maximally thin or their mass is punitive. I don't care if you use
Kevlar - a klick of thin will stretch like taffy and suffer a low
speed of tensile conductance. There will be no real time control.
Uncle Al says, "Baikonur sleighride!"
Lastly, the payload or at least the control pod must be on the same
side as the sun since the solar sail can only bulge away from the
light. Any attempt at a building a rigid framework will negate the
payload. If the focus washes across the pod, hasta la vista baby.
Non-imaging caustics will also incinerate the pod.
I get 10^5 cm/km and 10^10 cm^2/km^2. At 10 mg/cm^2 the solar sail
alone weighs 100 metric tonnes. The thinnest capacitor aluminum foil
is 0.0015 inch for an areal density of 10.3 mg/cm^2. 0.0005"
aluminized Mylar is commercial, but it doesn't like getting warm or
irradiated and it is *fragile.* Hey... aluminum metal doesn't
tolerate heat either. 5-micron aluminized Kapton is about the best of
all worlds - unless you have to pay for a km^2 of the stuff as a
throwaway. Use of Parylene-C ultrathin membrane as in ornithopters
would bust even NASA's budget.
If you push optimistic numbers,
http://solarsails.jpl.nasa.gov/introduction/design-construction.html
you get a bowl of moldy farina.
Oh yeah... even chemically tough Kapton in orbit gets chewed -
especially its reflectance,
http://setas-www.larc.nasa.gov/esem/final_report_append_b.html
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Ed Keane III
Geoffrey A. Landis <geoffre...@sff.net> wrote in message
news:1bfb1890.03070...@posting.google.com...
>
> When the sail is moving, then the reflected photons are Doppler
> shifted, and leave the sail with lower energy than they arrived. This
> loss of energy exactly equals the energy imparted to the sail, a fact
> which can be trivially verified by using Newton's laws, the Doppler
> formula, and the Einstein equation for photon momentum p=E/c
>
Doppler shift is caused by relative velocity and has nothing
to do with acceleration or deceleration. Any energy imparted
to the motion of the sail will cause a change in velocity and
will be indicated by a change in Doppler shift.
> If the sail is not moving, there is no Doppler shift. However, note
> that since energy is proportional to momentum squared, the derivative
> of energy with respect to momentum is zero for a non-moving sail.
> Thus, when the sail is stationary, it can reflect photons with perfect
> efficiency and still gain momentum at no energy cost.
>
This is exactly what Gold argues. Except he points out, correctly,
that if there is no energy lost from the photons it violates the laws
of thermodynamics, or more simply conservation of energy, by
getting something for nothing if the sail gains momentum. Remember
that the sail is floating freely. It is stationary but elastic. It will not
be
stationary after the light hits it and the light is reflected after it hits
it.
-Ed
Gregory L. Hansen
A recoil energy loss isn't a Doppler shift, but it is a shift nonetheless.
Geoffrey A. Landis
> "Geoffrey A. Landis" wrote:
>> Despite a recent article in New Scientist, a solar sail does not
>> break the laws of physics.
> [snip]
>
> Actually, it does as proposed. The sail will come into thermal
> equilibrium with the radiation field and emit photons from its other
> side, counter-thrusting.
Yes, but if reflectivity is R, the amount of photons absorbed is A=(1-R).
These are then radiated from both sides. The re-radiation contributes
no force. For a sail perpendicular to the sun, therefore, the force (
assuming thermal equilibrium) is
F = I (2R+A)/c
accounting for the fact that the reflected photons contribute momentum
once on arrival and once on departure, for a factor of 2, and the
absorbed ones contribute momentum only on arrival, not on departure.
>...
Steve Harris
"Gregory L. Hansen" <glha...@steel.ucs.indiana.edu> wrote
in message news:be1s68$56o$1...@hood.uits.indiana.edu...
Well, of course it is if the light energy is being converted
to kinetic or gravitational potential energy (which it
generally is, else what's the point of the sail?) is
thermalized first. What it will be for a SOLAR sail (vs,
say, a laser or microwave powered sail). But Gold is still a
putz. Temperature of the sail material is irrelevant to the
basic process of photon reflection/scatter from a moving
object/charge.
Steve Harris
Doppler effect, but the Compton effect. Electrons are
perfect mirrors and have no temperature. But thermodynamics
is upheld anyway when they scatter photons.
"Geoffrey A. Landis" <geoffre...@sff.net> wrote in
message
news:1bfb1890.03070...@posting.google.com...
MSu1049321
incinerate it. Yes?
me...@cars3.uchicago.edu
>Despite a recent article in New Scientist, a solar sail does not break
>the laws of physics.
>
>In an article in New Scientist recently, maverick astronomer Thomas
>Gold cast doubt about solar sails:
"Maverick" is not the term I would use. In fact can't think of any
polite term I would use.
>"Thomas Gold from Cornell University in New York says the proponents
>of solar sailing have forgotten about thermodynamics, the branch of
>physics governing heat transfer. Solar sails are designed to be
>perfect mirrors, meaning that they reflect all the photons that strike
>them. Gold argues that when photons are reflected by a perfect mirror,
>they do not suffer a drop in temperature. "
>
>Unfortunately, Gold has apparently forgotten to account for a
>well-known physical effect: the Doppler shift.
>
Yep.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
me...@cars3.uchicago.edu
>"Geoffrey A. Landis" wrote:
>>
>> Despite a recent article in New Scientist, a solar sail does not break
>> the laws of physics.
>[snip]
>
>Actually, it does as proposed. The sail will come into thermal
>equilibrium with the radiation field and emit photons from its other
>side, counter-thrusting.
Nope. It radiates from both sides, thus its own emission results in
no momentum change (thus no "counter-thrust). As long as it is
illuminated unisotropically, it'll accelerate. Mind you, this does
not mean that it is practical for any serious purpose, but that's
another matter.
me...@cars3.uchicago.edu
>
>Geoffrey A. Landis <geoffre...@sff.net> wrote in message
>news:1bfb1890.03070...@posting.google.com...
>>
>> When the sail is moving, then the reflected photons are Doppler
>> shifted, and leave the sail with lower energy than they arrived. This
>> loss of energy exactly equals the energy imparted to the sail, a fact
>> which can be trivially verified by using Newton's laws, the Doppler
>> formula, and the Einstein equation for photon momentum p=E/c
>>
>
>Doppler shift is caused by relative velocity and has nothing
>to do with acceleration or deceleration. Any energy imparted
>to the motion of the sail will cause a change in velocity and
>will be indicated by a change in Doppler shift.
>
>> If the sail is not moving, there is no Doppler shift. However, note
>> that since energy is proportional to momentum squared, the derivative
>> of energy with respect to momentum is zero for a non-moving sail.
>> Thus, when the sail is stationary, it can reflect photons with perfect
>> efficiency and still gain momentum at no energy cost.
>>
>
>This is exactly what Gold argues. Except he points out, correctly,
>that if there is no energy lost from the photons it violates the laws
>of thermodynamics, or more simply conservation of energy, by
>getting something for nothing if the sail gains momentum.
But there is energy lost from the photons. Do the math. That's where
the Doppler shift matters.
Henry Spencer
Ed Keane III <ke...@westelcom.com> wrote:
>> When the sail is moving, then the reflected photons are Doppler
>
>Doppler shift is caused by relative velocity and has nothing
>to do with acceleration or deceleration.
Acceleration or deceleration produces relative velocity. As soon as it
does, you get Doppler shift.
Doppler shift changes the wavelength of photons, and therefore their
energy... so just where *does* that extra or missing energy come from
or go to? Answer: from/to the mirror that's reflecting them.
If the mirror is approaching the light source, the light is blue-shifted,
increasing in energy by decelerating the mirror. If the mirror is going
away from the light source, the light is red-shifted, giving up some of
its energy to accelerate the mirror.
>...if there is no energy lost from the photons it violates the laws
>of thermodynamics, or more simply conservation of energy...
Right. Conservation of energy would *also* be violated if light could
be Doppler-shifted without an exchange of energy with the reflector.
--
MOST launched 1015 EDT 30 June, separated 1046, | Henry Spencer
first ground-station pass 1651, all nominal! | he...@spsystems.net
Henry Spencer
Uncle Al <Uncl...@hate.spam.net> wrote:
>> Despite a recent article in New Scientist, a solar sail does not break
>> the laws of physics.
>
>equilibrium with the radiation field and emit photons from its other
>side, counter-thrusting.
If the sail material has the same radiative properties on both sides,
there is no net counter-thrust at all -- the sail will emit equally from
*both* sides. In any case, for a good reflector, this is quite a small
effect compared to the thrust of the reflected light.
Guys, this stuff is not new. The basic physics have been understood quite
well for decades. Do a bit of research before sounding off, please. See,
for example, Appendix A of Jerome Wright's "Space Sailing".
>Solar wind will mostly stick rather than
>bounce, incrementally increasing sail mass.
Solar-wind atoms will stick *momentarily*, and then wander off again.
There may be a very slight initial increase in sail mass, but it will
reach equilibrium quickly. And in any case, the solar wind carries
several orders of magnitude less momentum than solar light pressure,
so its effects are slight.
>The claimed efficencies
>will be, surprise!, *much* higher than anything obtained. More
>studies will be needed. Damn! It didn't scale linearly like it was
>supposed to.
Sure it does. This stuff is well understood and is routinely allowed for
in precision tracking of deep-space missions. Moreover, unbalanced light
pressure is the biggest source of attitude disturbance for GSO comsats.
There is very little that isn't known about the subject by now. The major
unknowns of solar sails are the engineering hassles of deploying and
controlling large areas of very lightweight material.
>If the sail is metallic, there will be god's own Hell of field
>interactions.
Hardly. Almost all current sail designs are simply aluminized plastic...
quite like materials routinely used in the construction of spacecraft,
including some quite large ones.
>Space is rich with large scale electromagnetic this and that.
Very feeble this and that.
>If you think solar flares are tough on power lines...
Solar flares, by themselves, wouldn't do anything at all to power lines.
The trouble comes not from the flares themselves (more precisely, from
the associated particle events), but from the way they punch and pummel
Earth's magnetosphere. Outside the magnetosphere, nothing happens.
There have been deep-space spacecraft with very long wire antennas, for
low-frequency radio astronomy. They've had no trouble.
>Lastly, the payload or at least the control pod must be on the same
>side as the sun since the solar sail can only bulge away from the
>light. Any attempt at a building a rigid framework will negate the
>payload. If the focus washes across the pod, hasta la vista baby.
>Non-imaging caustics will also incinerate the pod.
You seem to be under the impression that solar sails are like parachutes.
They're not; they can't be, because light doesn't act like air. It won't
keep a flexible canopy inflated. Sails have to be *rigid*, either from
centrifugal force or from structural members. The structural mass is
annoying but not prohibitive. Again, please do some research before
sounding off; this problem has been explored in considerable depth and
reasonable solutions do exist.
>I get 10^5 cm/km and 10^10 cm^2/km^2. At 10 mg/cm^2 the solar sail
>alone weighs 100 metric tonnes.
10mg/cm^2 is 100g/m^2, which may sound light to you, but is a load of lead
bricks by solar-sail standards. JPL's late-1970s design for a solar-sail
Halley-rendezvous mission had a total mass (film, reflector, backside
emitter, joints) of 3.2-3.5g/m^2. They thought 1g/m^2 was a reasonable
near-term lower limit. Highly advanced sail materials could reach
0.1g/m^2 or perhaps a bit less, although not soon.
>The thinnest capacitor aluminum foil
>is 0.0015 inch for an areal density of 10.3 mg/cm^2.
Massive, horrible, hideous. That's 38um thick. The JPL Halley design was
2um of Kapton, topped by 100nm of aluminum, with 12.5nm of chromium on the
back. 2-3um Kapton is commercially available.
>0.0005"
>aluminized Mylar is commercial, but it doesn't like getting warm or
>irradiated and it is *fragile.*
Sails will be fragile. But they won't be exposed to much stress.
>Hey... aluminum metal doesn't tolerate heat either.
It doesn't get particularly hot.
>...Use of Parylene-C ultrathin membrane as in ornithopters
>would bust even NASA's budget.
Hardly. Parylene N was considered for the Halley mission, but its space
compatibility wasn't certain then (not sure if it is now), and it wasn't
available in large sheets.
>Oh yeah... even chemically tough Kapton in orbit gets chewed -
>especially its reflectance,
> http://setas-www.larc.nasa.gov/esem/final_report_append_b.html
Bare Kapton is quite vulnerable in low Earth orbit. Sails cannot operate
in low Earth orbit anyway, because of air drag. Atomic-oxygen erosion is
insignificant long before drag falls low enough for practical sail
operation.
Allen Thomson
> Dang, I don't know what to say. Gold is a putz,
Gold is, ah, interesting... Check out his theory on the origin of
petroleum.
> and I thought New Scientist was better than that.
Life teaches us bitter lessons.
> I didn't realize a solar sail was a heat engine.
It *is* a heat engine, properly analyzed. Others in this thread
have done that; Gold did not. In a sense, what a solar sail does
is thermodynamically not that much diffrent than what a piston in
your car's engine does. Although, thermodynamics aside, Newtonian
mechanics does just fine for understanding how a solar sail works.
Dale Trynor
Uncle Al wrote:
> "Geoffrey A. Landis" wrote:
> >
> > Despite a recent article in New Scientist, a solar sail does not break
> > the laws of physics.
> [snip]
>
> Actually, it does as proposed. The sail will come into thermal
> equilibrium with the radiation field and emit photons from its other
> side, counter-thrusting.
[snip]
Dale Trynor wrote:
I was going to point out that you needed to think about this a bit more.
That's unnecessary now but I am surprised you missed it. I am also very
surprised that an expert like Gold would make such a mistake and then get it
published. Makes me feel better about my own mistakes however I am just an
amateur.
I do remember reading that the solar force is something like 5 pounds per
square mile if that's of much use. It could really add up to a lot of
stresses if one were to have a 1000 mile wide mirror without using some
means to deal with that. Thickness would of course have to increase by a lot
if it was unsupported and no tricks used.
I did some posts recently on the idea that you could cover an entire planet
to 15 psi with a very thin film if each inch of area had a 15 pound weight
attached. In principle ones thickness would never need to be more than that
required for the 1 square inch of area no mater how much area was covered.
Mitchell
> the Doppler shift matters.
That's just not true. I'm a senior year physics undergrad at Cornell,
and I was in a group of 6 people who got to sit down and talk with
Gold about his "perfect mirror and violations of conservation of
energy/momentum" paper before it was published. Among the arguments we
attempted to refute is theory was the idea of Doppler shift. He
correctly countered that a Doppler shift does not affect the total
energy, only the power.
Another post mentioned that if a Doppler shift could occur without
interaction with the sail, there would be nowhere for the lost energy
to go. Well, a moving galaxy puts out Doppler shifted light, and since
each star radiates basically isotropically, there's no acceleration
from that. So where's that energy going when we view the galaxy? The
answer is that it's still all there, it's just coming in at a
different rate. Doppler shift only changes power, not total energy.
Mitchell
Geoffrey A. Landis
> Doppler shift is caused by relative velocity and has nothing
> to do with acceleration or deceleration. Any energy imparted
> to the motion of the sail will cause a change in velocity and
> will be indicated by a change in Doppler shift.
[Geoffrey A. Landis <geoffre...@sff.net> had written:]
> > If the sail is not moving, there is no Doppler shift. However, note
> > that since energy is proportional to momentum squared, the derivative
> > of energy with respect to momentum is zero for a non-moving sail.
> > Thus, when the sail is stationary, it can reflect photons with perfect
> > efficiency and still gain momentum at no energy cost.
>
> This is exactly what Gold argues.
Not quite. You understand this first point? If the sail is
stationary, then a reflection of a photon results in zero change in
energy (to first order), and thus the sail can gain momentum at no
energy cost?
> Except he points out, correctly,
> that if there is no energy lost from the photons it violates the laws
> of thermodynamics, or more simply conservation of energy, by
> getting something for nothing if the sail gains momentum.
Careful here; therodynamics says nothing about momentum.
Thermodynamics talks about energy; don't confuse energy and momentum
here. The sail gains *momentum*, but doesn't gain *energy*. For E
energy, p momentum, E= p^2/2m, and dE/dp is thus dE/dp= p/m. When p
is zero, dE/dP is also zero: you can gain momentum at NO cost in
energy.
>Remember that the sail is floating freely. It is stationary but
elastic. It
> will not be stationary after the light hits it and the light is reflected
> after it hits it.
Right. That's the second-order analysis. If you really want to do
the second order analysis, you have to take into account the sail
motion, which means that the photon is Doppler shifted, and thus loses
energy. The lost energy is exactly equal to the energy gained by the
sail.
The sail velocity change on absorbing a photon of energy hf is delta-V
= delta-p over m, where delta-P is hf/c (here I use hf for the photon
energy, to avoid confusing photon and sail energy)
The Doppler shift is proportional to velocity, v/c, so the change in
Doppler shift in frequency f is inversely proportional to the mass of
the sail. Since E=hf, the change in energy is linear with the Doppler
shift.
The energy of the sail is E=p^2/2m
So, calculating the energy gained by the sail and comparing it to the
energy lost by Doppler shift, the net result is that the energy change
of the sail divided by the incident photon momentum is exactly equal
to the energy lost by Doppler shift.
Henry Spencer
Mitchell <Quaz...@aol.com> wrote:
>He correctly countered that a Doppler shift does not affect the total
>energy, only the power.
You'll need to explain this in more detail, because it's not convincing
the way you've told it. Light is reflected from a perfectly-reflecting
moving solar sail; say it's moving away from the Sun for purposes of
argument. The light comes back red-shifted, having lost energy. Same
number of photons, less energy per photon. Where did the energy go?
>Another post mentioned that if a Doppler shift could occur without
>interaction with the sail, there would be nowhere for the lost energy
>to go. Well, a moving galaxy puts out Doppler shifted light...
Yes, but it's an emitter, not a reflector. As measured by an observer
motionless with respect to us, that light *started out* Doppler shifted;
at no point did its energy *change*. It's the change in energy as a
result of perfect reflection that requires explanation.
>...Doppler shift only changes power, not total energy.
However, changing the power carried by a specific beam of photons -- it
goes out from the Sun carrying X watts, it comes back after reflection
carrying X-Y watts -- still requires explanation. The missing power has
to go somewhere. Where?
Steve Harris
> Yeah, I was just thinking that a few hours after I'd sent
out the message.
> You could imagine the solar sail craft as a piston, the
Sun at so many
> milions of degrees, and the "sink" at about 3 kelvin. Not
a bad Carnot
> efficiency, actually.
COMMENT
Yeah. Though sunlight is at an effective 5800 K or so. It's
the same 5800 vs 2.7 K gradient that drives all
photosynthetic life on Earth.
Yeppers you got it: you can think of radiation in a cavity
as exerting pressure like a gas. If the cavity has a piston,
and the photon gas is sunlight, voila: a Carnot engine and
solar sail. Except the adiabatic volumetric exponential
coefficient is 4/3 for EM radiation, rather than gas gamma
number, which tends to be 5/3 or 7/5 (depending on gas).
Actual sails aren't that efficient, of course, since light
doesn't get to reflect of non-piston walls for another pass.
SBH
Dr John Stockton
news:sci.space.policy, Henry Spencer <he...@spsystems.net> posted at
Thu, 3 Jul 2003 21:48:17 :-
> The JPL Halley design was
>2um of Kapton, topped by 100nm of aluminum, with 12.5nm of chromium on the
>back. 2-3um Kapton is commercially available.
Such thicknesses have the property that solar radiation pressure
approximately balances solar gravity, at any distance from the Sun.
With exact balance (not forgetting payload), a sailing vessel would have
a constant-velocity path, and would reach Root2 AU in 1/2pi years, if
started with Earth's orbital velocity from an altitude where Earth's
gravity becomes unimportant.
--
© John Stockton, Surrey, UK. ?@merlyn.demon.co.uk Turnpike v4.00 MIME. ©
Web <URL:http://www.merlyn.demon.co.uk/> - FAQqish topics, acronyms & links;
some Astro stuff via astro.htm, gravity0.htm; quotes.htm; pascal.htm; &c, &c.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.
Steve Harris
"Geoffrey A. Landis" <geoffre...@sff.net> wrote in
message
news:1bfb1890.03070...@posting.google.com...
> > This is exactly what Gold argues.
>
> Not quite. You understand this first point? If the sail
is
> stationary, then a reflection of a photon results in zero
change in
> energy (to first order), and thus the sail can gain
momentum at no
> energy cost?
Come on. You're like the guy who says: when I heat up an
object, its weight doesn't change (to first order).
Therefore the equivalence of mass and energy is violated.
Duh.
There is no energy cost to move the stationary sail *to
first order.* Carnot's law is broken to exactly the degree
that you simplify the problem with approximation. But don't
confuse your approximation with violation of physical law.
>
> > Except he points out, correctly,
> > that if there is no energy lost from the photons it
violates the laws
> > of thermodynamics, or more simply conservation of
energy, by
> > getting something for nothing if the sail gains
momentum.
>
> Careful here; therodynamics says nothing about momentum.
> Thermodynamics talks about energy; don't confuse energy
and momentum
> here. The sail gains *momentum*, but doesn't gain
*energy*. For E
> energy, p momentum, E= p^2/2m, and dE/dp is thus dE/dp=
p/m. When p
> is zero, dE/dP is also zero: you can gain momentum at NO
cost in
> energy.
No, you can't. E^2 = (mc^2)+(pc)^2
Gregory L. Hansen
>> the Doppler shift matters.
>
>That's just not true. I'm a senior year physics undergrad at Cornell,
>and I was in a group of 6 people who got to sit down and talk with
>Gold about his "perfect mirror and violations of conservation of
>energy/momentum" paper before it was published. Among the arguments we
>attempted to refute is theory was the idea of Doppler shift. He
>correctly countered that a Doppler shift does not affect the total
>energy, only the power.
If you're a senior year physics undergrad at Cornell, then I'll bet you
can calculate a head-on collision between two billiard balls. Assume a
moving ball hits a stationary ball, and determine the energy that the
initially moving ball has when it bounces back.
Now repeat the calculation for a photon hitting a stationary billiard
ball. You'll obviously need to use different relations for the energy and
momentum of the photon, but you needn't go relativistic for that one.
I'll bet Gold could do the same calculation, if he'd thought to.
It's probably not proper to call a recoil effect a Doppler shift, but
there is certainly a recoil effect, and energy is lost from the reflected
photons.
Henry Spencer
Dr John Stockton <re...@merlyn.demon.co.uk> wrote:
>> The JPL Halley design was
>>2um of Kapton, topped by 100nm of aluminum, with 12.5nm of chromium on the
>>back. 2-3um Kapton is commercially available.
>
>Such thicknesses have the property that solar radiation pressure
>approximately balances solar gravity, at any distance from the Sun.
Not quite, alas. For that you need about 0.75g/m^2 -- about 1/4 of the
mass JPL was looking at, and somewhat better than the best they expected
for near-term advanced materials. And then, as John points out, it has to
go somewhat lighter yet because there is payload and other overhead to be
accommodated.
Near-term advanced materials with low-overhead designs do offer potential
for at least achieving thrust of a significant fraction of the Sun's
gravity, which is significant because it makes interplanetary trajectories
much more efficient than with lower thrusts.
Christopher M. Jones
> In article <VjUJUHB80aB$Ew...@merlyn.demon.co.uk>,
> Dr John Stockton <re...@merlyn.demon.co.uk> wrote:
> >Such thicknesses have the property that solar radiation pressure
> >approximately balances solar gravity, at any distance from the Sun.
>
> Not quite, alas. For that you need about 0.75g/m^2 -- about 1/4 of the
> mass JPL was looking at, and somewhat better than the best they expected
> for near-term advanced materials. And then, as John points out, it has to
> go somewhat lighter yet because there is payload and other overhead to be
> accommodated.
>
> Near-term advanced materials with low-overhead designs do offer potential
> for at least achieving thrust of a significant fraction of the Sun's
> gravity, which is significant because it makes interplanetary trajectories
> much more efficient than with lower thrusts.
Just to add a little clarification for the folks that aren't
quite up on the details of solar sails, any thickness is
sufficient for operation at some level, the thicknesses
Henry and John are talking about are only those which would
make for excellent solar sails. The trick is that you
aim your thrust ahead or behind the direction of travel
along the orbit, that way you don't have to fight the Sun's
gravity, you work with it. With any solar sail at all it's
possible to change to any orbit or to leave the Solar System
entirely, it's just a matter of how long it would take, with
a very poor solar sail it may take longer than the age of
the Solar System, with the sorts of sails they can build
these days (see above) it would take longer than using
modern chemical rockets, but not by all that much.
David M. Palmer
<sbha...@ix.RETICULATEDOBJECTcom.com> wrote:
> Come on. You're like the guy who says: when I heat up an
> object, its weight doesn't change (to first order).
> Therefore the equivalence of mass and energy is violated.
> Duh.
>
> There is no energy cost to move the stationary sail *to
> first order.* Carnot's law is broken to exactly the degree
> that you simplify the problem with approximation. But don't
> confuse your approximation with violation of physical law.
There is no energy cost to move the stationary sail to first order.
There is no change in photon energy to first order
There is an energy cost to move the stationary sail to second order.
There is an equal change in photon energy to second order.
There is no breaking of Carnot's law to any order.
--
David M. Palmer dmpa...@email.com (formerly @clark.net, @ematic.com)
Dr John Stockton
news:sci.space.policy, Dale Trynor <da...@nbnet.nb.ca> posted at Fri, 4
Jul 2003 09:54:45 :-
>I do remember reading that the solar force is something like 5 pounds per
>square mile if that's of much use.
I make it, for total reflection, about 9.6E-7 kgF/m^2, or nearly one
kilogram per square kilometre; I leave conversion to archaic units as an
exercise for the reader.
Gregory L. Hansen
And yet, there is an energy cost to move a stationary sail, and there is a
change in photon energy. And no matter what thermodynamics arguments are
brought up, light pressure has been used to orient satellites, there's
nothing controversial about it. Looks like someone is going to have to
figure out where their analysis went wrong.
Edward Green
> In article <1bfb1890.03070...@posting.google.com>,
> Geoffrey A. Landis <geoffre...@sff.net> wrote:
> >Despite a recent article in New Scientist, a solar sail does not break
> >the laws of physics.
> >
> >Gold cast doubt about solar sails:
> >of solar sailing have forgotten about thermodynamics, the branch of
> >physics governing heat transfer. Solar sails are designed to be
> >perfect mirrors, meaning that they reflect all the photons that strike
> >them. Gold argues that when photons are reflected by a perfect mirror,
> >they do not suffer a drop in temperature. "
>
>
> Dang, I don't know what to say. Gold is a putz, and I thought New
> Scientist was better than that. I didn't realize a solar sail was a heat
> engine.
Sure it is. In the limiting case of working between, say, the sun's
outgoing radiation at 5800K, and 3K cosmic background radiation, I'd
say we just _might_ be able to get a heat engine to work. ;-)
me...@cars3.uchicago.edu
>> But there is energy lost from the photons. Do the math. That's where
>> the Doppler shift matters.
>
>That's just not true. I'm a senior year physics undergrad at Cornell,
>and I was in a group of 6 people who got to sit down and talk with
>Gold about his "perfect mirror and violations of conservation of
>energy/momentum" paper before it was published. Among the arguments we
>attempted to refute is theory was the idea of Doppler shift. He
>correctly countered that a Doppler shift does not affect the total
>energy, only the power.
>
Consider an elastic collision of two particles. For convenience and
simplicity we'll make it a 1D problem. Two particles coming one
towards the other, colliding and flying away. Now, lets consider it
relative to three different reference frames.
1) In the center of mass frame, both particles approach (with the
same momentum), collide and recede, still with the same momentum and
same velocity (for each one) as before the collision (remember, the
collision is elastic). Total energy is preserved and the energy of
each of the particles individually is preserved as well.
2) In the initial reference frame of A, we see prior to the collision
A stationary and B approaching, after the collision both are moving.
Total energy still preserved but A gained some energy and B lost some.
3) In the (initial) frame of B, the situation is just reversed.
Total energy still preserved but as a result of the collision A lost
some energy and B gained some.
In all (infinity of) other reference frames you'll get other,
intermediate results. Mind you, we're not talking different processes.
We're talking about *same process* as viewed from different reference
frames. The question of who gains and who loses energy in a collision
is not "inherent" to the process but a matter of accounting, depending on
the reference frame.
So, if you take an elastic collision of a photon with the mirror, in
their CM reference frame, indeed, nobody gains or loses energy, the
photon bounces back with the same energy with which it hit. But,
you're not observing from the CM reference frame but from "our" frame,
stationed on the Sun. In this frame, the photon bounces back with
different energy. And the energy difference can be represented
through the Doppler shift between the CM reference frame and our.
Note, you've to apply the shift twice, first transforming from our
frame to the CM, then back.
As I said, do the math. Will do you good.
me...@cars3.uchicago.edu
here.
me...@cars3.uchicago.edu
figuring what went wrong is a matter for psychiatry, not physics.
Henry Spencer
>kilogram per square kilometre; I leave conversion to archaic units as an
>exercise for the reader.
Uh, "kgF" *is* an archaic unit. In modern units it is 9.126e-6 N/m^2.
Henry Spencer
Henry Spencer <he...@spsystems.net> wrote:
>>Such thicknesses have the property that solar radiation pressure
>>approximately balances solar gravity, at any distance from the Sun.
>
>Not quite, alas. For that you need about 0.75g/m^2 -- about 1/4 of the
>mass JPL was looking at, and somewhat better than the best they expected
>for near-term advanced materials.
Wups, my mistake -- multiply that number by 2, because I couldn't remember
the numeric value of solar light pressure, and had to look it up, and
forgot to check whether I was looking at the number for an absorber or a
reflector.
That puts balance possibly within reach of a near-term advanced sail,
given a low-overhead design.
Christopher M. Jones
> JRS: In article <3F0587AA...@nbnet.nb.ca>, seen in
> news:sci.space.policy, Dale Trynor <da...@nbnet.nb.ca> posted at Fri, 4
> Jul 2003 09:54:45 :-
>
> >I do remember reading that the solar force is something like 5 pounds per
> >square mile if that's of much use.
>
> I make it, for total reflection, about 9.6E-7 kgF/m^2, or nearly one
> kilogram per square kilometre; I leave conversion to archaic units as an
> exercise for the reader.
Both of these determinations are wrong, without the
stipulation that they are for a distance from the Sun
of 1 AU. The "solar force" scales as the inverse
square of distance from the Sun. Since Solar Sails
are meant for sailin' it doesn't make a whole lotta
sense to force the assumption of 1 AU distance into
the measurement of "solar force", as actually *using*
a Solar Sail will change that, hopefully by a lot.
Dr John Stockton
>In article <VjUJUHB80aB$Ew...@merlyn.demon.co.uk>,
>Dr John Stockton <re...@merlyn.demon.co.uk> wrote:
>>> The JPL Halley design was
>>>2um of Kapton, topped by 100nm of aluminum, with 12.5nm of chromium on the
>>>back. 2-3um Kapton is commercially available.
>>
>>Such thicknesses have the property that solar radiation pressure
>>approximately balances solar gravity, at any distance from the Sun.
>
>Not quite, alas. For that you need about 0.75g/m^2 -- about 1/4 of the
>mass JPL was looking at, and somewhat better than the best they expected
>for near-term advanced materials.
I'll accept a factor of two for my "such ... approximately" (I was not
paying much attention to the density of the material); also up to 25% in
data and arithmetic.
However, I make it not 0.75 g/m^2 but 1.6 g/m^2 for the areal density of
a 100% reflective sail that balances with solar gravity.
Now 4 MT/s of photons over 4 pi * (150 Gm)^2 times speed of 3E8 m/s
gives a pressure about 5 microNewtons per square metre, which is 0.5
mgF/m^2, confirming a result obtained by starting with 1.4kW/m^2.
The solar gravity field here, since it is matched by w^2r, is
approximately (2 pi/3155000)^2 * 150E9 = 6E-3 m/s^2 = 6E-3 N/kg.
F = ma, so m = F/a, so the result is 5E-6/6E-3 = 0.8E-3 kg/m^2.
Thus for a reflective sail, which is twice as good, about 1.6 g/m^2 can
be supported. I think your figure omits reflection, the remaining
discrepancy being insignificant.
Steve Harris
"David M. Palmer" <dmpa...@email.com> wrote in message
news:040720031618552800%dmpa...@email.com...
If there's no change in photon energy to first order, then
obviously that's a breaking of Carnot's law to first order,
since Carnot requires an decrease in photon temperature
(photon energy) for work to be extracted. Carnot requires
two thermal baths of different temperatures for kinetic
energy to be gained.
But it's okay, because in any inertial frame where work is
being done on the sail, you see two populations of photons
(those coming and those leaving), and these two DO have two
different temperatures. That's it.
SBH
David M. Palmer
<sbha...@ix.RETICULATEDOBJECTcom.com> wrote:
> "David M. Palmer" <dmpa...@email.com> wrote in message
> news:040720031618552800%dmpa...@email.com...
> > There is no breaking of Carnot's law to any order.
> If there's no change in photon energy to first order, then
> obviously that's a breaking of Carnot's law to first order,
> since Carnot requires an decrease in photon temperature
> (photon energy) for work to be extracted. Carnot requires
> two thermal baths of different temperatures for kinetic
> energy to be gained.
(In the sail's frame of reference)
To first order, there is no change in photon temperature and no work
extracted, so Carnot's law is not broken to first order.
To second order, there is a change in photon temperature and work is
extracted, and Carnot's law is not broken to second order.
It is only if you mix first and second order that you think it breaks
Carnot's law.
(We are in agreement that light sails work and that Gold is wrong.)
Dr John Stockton
news:sci.space.policy, Christopher M. Jones <spic...@dualboot.net>
posted at Sat, 5 Jul 2003 08:25:53 :-
They are for 1 AU, of course; but I think we can assume the inverse
square law to be known in these two newsgroups.
But the areal density at which gravity matches light pressure is
independent of distance, just depending on W, the average solar power
per unit mass. It seems to be, but ICBW,
p = W c / 4 pi G
For other stars, W will differ.
me...@cars3.uchicago.edu
Two points:
1) Photon energy is *not* "photon temperature". Photon (or any
particle) does not have a temperature). Temperature is an ensemble
property.
2) When a stationary body (i.e. one with p = 0) receives an momentum
increment dp, its energy increases by (dp)^2/(2*m) (that's classical,
I'll let you work out the relativistic equivalent itself), thus the
energy increment is *second order in dp*. The first order increment
is zero. The energy loss of whatever caused the energy increment is
the same, so it is also second order. No laws are violated.
> Carnot requires
>two thermal baths of different temperatures for kinetic
>energy to be gained.
>
Nope, Carnot requires the above when you're converting thermal energy
to kinetic energy. That's all. It has nothing to do with energy
transfer in two-body collisions.
John Ordover
sailing would be to move a terrestrial sailboat by hitting the sail
with a whole lot of high-speed ping-pong balls.
Light is massless but not energy-less, natch, so it winds up having a
similar effect on objects to the ping-pong balls. Just as some of the
momementum of the ping-pong balls would be transfered to the sails,
some of the energy of the light is transfered to the sail. Just as
the ping-pong balls would therefore lose some energy, the photons in
the light lose some energy, but light does that all the time.
So what's the problem?
Alex Terrell
> In article <HHIIM...@spsystems.net>,
> Henry Spencer <he...@spsystems.net> wrote:
> >>Such thicknesses have the property that solar radiation pressure
> >>approximately balances solar gravity, at any distance from the Sun.
> >
> >Not quite, alas. For that you need about 0.75g/m^2 -- about 1/4 of the
> >mass JPL was looking at, and somewhat better than the best they expected
> >for near-term advanced materials.
>
> Wups, my mistake -- multiply that number by 2, because I couldn't remember
> the numeric value of solar light pressure, and had to look it up, and
> forgot to check whether I was looking at the number for an absorber or a
> reflector.
>
> That puts balance possibly within reach of a near-term advanced sail,
> given a low-overhead design.
Does that include when you've pitted the material with subwavelength
holes, thereby reducing (halving?) the mass whilst maintaining the
refelctivity?
Has anyone done any theory on making the sustrate from carbon
nanotubes? If these were interlaced at a sub light wavelength? It
might even make sense to forget the reflector (if it more than doubles
the mass).
Alex Terrell
to be one of many experiments that appear to break a fundamental law
of Physics (Carnot's rule).
(I read about another where light was made to trevel faster than
light, but no information went faster than light).
The interesting bit is always finding out how the laws of physics are
conserved, and Geoffrey's done just that.
May I ask a more basic question?
If photons have mass, and if they travel at c, how come they don't
have infinite mass? Since they don't have infinite mass, they can have
no mass, and therefore no momentum.
I know this is another one where at first glance it appears a law of
physics is broken.
geoffre...@sff.net (Geoffrey A. Landis) wrote in message news:<1bfb1890.03070...@posting.google.com>...
> Despite a recent article in New Scientist, a solar sail does not break
> the laws of physics.
>
> In an article in New Scientist recently, maverick astronomer Thomas
> Gold cast doubt about solar sails:
> "Thomas Gold from Cornell University in New York says the proponents
> of solar sailing have forgotten about thermodynamics, the branch of
> physics governing heat transfer. Solar sails are designed to be
> perfect mirrors, meaning that they reflect all the photons that strike
> them. Gold argues that when photons are reflected by a perfect mirror,
> they do not suffer a drop in temperature. "
>
> Unfortunately, Gold has apparently forgotten to account for a
> well-known physical effect: the Doppler shift.
>
> It's worth saying that the photon pressure on a spacecraft is not
> theoretical; its effect on spacecraft is measurable, and it has been
> observed and measured to great precision routinely in space. Photon
> pressure-- the solar sail effect-- has already been used for an
> operational space mission; it was for spacecraft attitude control on
> the Pioneer Venus-Mercury mission.
>
> The Crookes radiometer does not operate on photon pressure, and the
> explanation for how it operates has been known for over a century.
>
> The energy transfer to a solar sail can be accounted for from the
> Doppler shift of reflected photons; even when the reflectivity is
> 100%, a photon looses energy when reflecting from a moving sail. This
> effect exactly corresponds to the energy increase of the sail. No
> sophisticated physics is needed to analyze this effect, it is a
> problem suitable for a homework assignment for a college
> undergraduate.
>
> When the sail is moving, then the reflected photons are Doppler
> shifted, and leave the sail with lower energy than they arrived. This
> loss of energy exactly equals the energy imparted to the sail, a fact
> which can be trivially verified by using Newton's laws, the Doppler
> formula, and the Einstein equation for photon momentum p=E/c
>
> If the sail is not moving, there is no Doppler shift. However, note
> that since energy is proportional to momentum squared, the derivative
> of energy with respect to momentum is zero for a non-moving sail.
> Thus, when the sail is stationary, it can reflect photons with perfect
> efficiency and still gain momentum at no energy cost.
>
> For completeness, note that if the sail is moving *toward* the light
> source, then the phtons are Doppler shifted to *higher* energy by the
> reflection. This implies that the sail must lose energy-- which is
> correct; when the sail moves toward the light source, it slows down.
Gregory L. Hansen
Alex Terrell <alext...@yahoo.com> wrote:
>The effect of solar radiation pressure has been observed. This seems
>to be one of many experiments that appear to break a fundamental law
>of Physics (Carnot's rule).
No, they don't. People are analyzing it funny, in a way they'd never do
if they were looking at a piston in a cylinder pushed by gas under
pressure.
>
>(I read about another where light was made to trevel faster than
>light, but no information went faster than light).
The leading edge of the pulse triggered an excited medium to reshape the
pulse, pushing the peak forward. The leading edge still got there first,
and that contained the information.
>If photons have mass, and if they travel at c, how come they don't
>have infinite mass? Since they don't have infinite mass, they can have
>no mass, and therefore no momentum.
You were doing fine until that last part. How do you conclude that no
mass means no momentum?
Alex Terrell
> >If photons have mass, and if they travel at c, how come they don't
> >have infinite mass? Since they don't have infinite mass, they can have
> >no mass, and therefore no momentum.
>
> You were doing fine until that last part. How do you conclude that no
> mass means no momentum?
Please don't misintepret me. I do "believe" in solar sails. I'm just
trying to figure out how it works, because, in the normal world:
Momentum = mass * velocity
If mass = 0 and velocity = 3E8, then momentum = 0
Please tell me what I'm missing?
Rand Simberg
alext...@yahoo.com (Alex Terrell) made the phosphor on my monitor
glow in such a way as to indicate that:
>> >If photons have mass, and if they travel at c, how come they don't
>> >have infinite mass? Since they don't have infinite mass, they can have
>> >no mass, and therefore no momentum.
>>
>> You were doing fine until that last part. How do you conclude that no
>> mass means no momentum?
>
>Please don't misintepret me. I do "believe" in solar sails. I'm just
>trying to figure out how it works, because, in the normal world:
You mean the world in which light doesn't live?
--
simberg.interglobal.org * 310 372-7963 (CA) 307 739-1296 (Jackson Hole)
interglobal space lines * 307 733-1715 (Fax) http://www.interglobal.org
"Extraordinary launch vehicles require extraordinary markets..."
Swap the first . and @ and throw out the ".trash" to email me.
Here's my email address for autospammers: postm...@fbi.gov
Henry Spencer
Alex Terrell <alext...@yahoo.com> wrote:
>> >Not quite, alas. For that you need about 0.75g/m^2 -- about 1/4 of the
>> >mass JPL was looking at, and somewhat better than the best they expected
>> >for near-term advanced materials.
>> That puts balance possibly within reach of a near-term advanced sail,
>> given a low-overhead design.
>
>Does that include when you've pitted the material with subwavelength
>holes, thereby reducing (halving?) the mass whilst maintaining the
>refelctivity?
The JPL guys noted that option, but did not consider it to fit under the
"near-term" heading, because techniques for bulk manufacture of perforated
ultra-thin material are conjectural at best right now. Such materials
conceivably could go as low as 0.1g/m^2... eventually. Almost certainly
they would have to be made in space, with the sail assembled as a rigid
object -- deploying sails from small packages is tricky enough even with
more robust sail materials.
>Has anyone done any theory on making the sustrate from carbon
>nanotubes? If these were interlaced at a sub light wavelength? It
>might even make sense to forget the reflector (if it more than doubles
>the mass).
Nanotubes are not particularly reflective, as far as I know, and that is
the sine qua non of a good sail material. The way you get ultra-light
sail materials is not by leaving the reflector off, but by leaving
everything else off. :-)
That said, it has been suggested that you might be able to make an
intermediate-class sail material -- better than metallized plastic film,
easier to make and handle than ultra-thin metal -- by essentially draping
a very thin reflector over an open-weave cloth substrate.
Geoffrey A. Landis
> "Geoffrey A. Landis" <geoffre...@sff.net> wrote:
> > Careful here; therodynamics says nothing about momentum.
> > Thermodynamics talks about energy; don't confuse energy
> >and momentum here. The sail gains *momentum*, but doesn't gain
> > *energy*. For E energy, p momentum, E= p^2/2m, and dE/dp is thus dE/dp=
> > p/m. When p is zero, dE/dP is also zero: you can gain momentum at NO
> >cost in energy.
>
> No, you can't. E^2 = (mc^2)+(pc)^2
for reference, the correct equation is actually
E^2 = (mc^2)^2+(pc)^2
where m is rest mass.
For photons, this reduces to
E = pc
For objects with non-zero rest mass, you can expand
E = SQRT(Eo^2+(pc)^2)
(where Eo is rest energy mc^2)
For the nonrelativistic case (and do recall that we had been talking
about the case of a stationary sail, which is by definition
nonrelativistic), the leading term of the expansion is
E = p^2/2m
which is just Newtonian physics.
For the case where the photon energy is comparable to the rest energy
mc^2 of the sail, you may have to do a fully relativistic calculation.
If you do, however, you STILL discover that energy and momentum are
conserved, and the sail works as predicted.
--
Geoffrey A. Landis
http://www.sff.net/people/geoffrey.landis
Alan Anderson
> > >If photons have mass, and if they travel at c, how come they don't
> > >have infinite mass? Since they don't have infinite mass, they can have
> > >no mass, and therefore no momentum.
> >
> > You were doing fine until that last part. How do you conclude that no
> > mass means no momentum?
>
> Please don't misintepret me. I do "believe" in solar sails. I'm just
> trying to figure out how it works, because, in the normal world:
>
> Momentum = mass * velocity
>
> If mass = 0 and velocity = 3E8, then momentum = 0
>
> Please tell me what I'm missing?
You're missing the difference between "the normal world" and "Newtonian
physics", that's all. The relativistic formula for momentum has an extra
factor which increases as velocity approaches c, becoming infinite when
v=c, making that formula inapplicable to photons. What is zero times
infinity?
The true value of a photon's momentum "in the normal world" is
proportional to its frequency.
Allen Thomson
It's probably better/easier to think of the sail as flat and rigid,
like a sheet of plywood. But that said, the ping-pong ball analogy
is an excellent one and captures most of what's going on with solar
sails in the reflective case (the balls bounce off). A useful addition
is to consider the case in which the balls are sticky and stay on the
sail, corresponding to absorption (assume that the balls are extremely
light so the added mass is negligible). Real solar sails are a
combination of reflective and absorptive -- some balls bounce, some
stick.
The difference is important, because reflection allows the sail to
tack against the photon wind by canting itself, whereas the
absorptive case only produces a thrust in the direction of the wind.
A sail that can tack can add or subtract orbital angular momentum wrt
the sun and spiral between planets; one that can't tack can't do that.
George William Herbert
>> >If photons have mass, and if they travel at c, how come they don't
>> >have infinite mass? Since they don't have infinite mass, they can have
>> >no mass, and therefore no momentum.
>>
>> You were doing fine until that last part. How do you conclude that no
>> mass means no momentum?
>
>Please don't misintepret me. I do "believe" in solar sails. I'm just
>trying to figure out how it works, because, in the normal world:
>
>Momentum = mass * velocity
>
>If mass = 0 and velocity = 3E8, then momentum = 0
>
>Please tell me what I'm missing?
Photons have momentum inversely proportional to
their wavelength, independent of their rest
mass (which is zero).
P = h / lambda
This is also equal to their energy divided by the
speed of light:
P = E / c
See for example pp 1109 in Halliday and Resnick 2nd ed.
-george william herbert
gher...@retro.com
Gregory L. Hansen
Momentum isn't always mass * velocity. We find out what the real world is
like by running experiments, and light has momentum but no detectable
mass. It has energy, though, and mass itself is related to rest energy by
m=E/c^2.
Henry Spencer
Alex Terrell <alext...@yahoo.com> wrote:
>Momentum = mass * velocity
>If mass = 0 and velocity = 3E8, then momentum = 0
>Please tell me what I'm missing?
In a relativistic world, things moving at the speed of light follow
different rules.
Momentum is not m*v; rather, it is m*v/sqrt(1-v^2/c^2), which reduces to
approximately m*v when v<<c.
As is easy to see, for m=0 and v=c, momentum is 0/0, i.e. indeterminate.
Which is not surprising, since photons of different energies carry
different amounts of momentum. Photon momentum is, in fact, E/c, where
E is the energy.
Alex Terrell
> On 6 Jul 2003 10:44:05 -0700, in a place far, far away,
> alext...@yahoo.com (Alex Terrell) made the phosphor on my monitor
> glow in such a way as to indicate that:
>
> >> >If photons have mass, and if they travel at c, how come they don't
> >> >have infinite mass? Since they don't have infinite mass, they can have
> >> >no mass, and therefore no momentum.
> >>
> >> You were doing fine until that last part. How do you conclude that no
> >> mass means no momentum?
> >
> >Please don't misintepret me. I do "believe" in solar sails. I'm just
> >trying to figure out how it works, because, in the normal world:
>
> You mean the world in which light doesn't live?
No - the one in which Jonah Lomu lives
Alex Terrell
Thanks - what is the formula?
It seems that 0 (mass of photon) times infinity (1/(1-v^2/c^2)) equals
a small amount.
I can turn around Geoffrey's original explanation, and see from it
that a photon must have momentum, in order to ensure that both
momentum and energy are conserved.
me...@cars3.uchicago.edu
>The effect of solar radiation pressure has been observed. This seems
>to be one of many experiments that appear to break a fundamental law
>of Physics (Carnot's rule).
>
>(I read about another where light was made to trevel faster than
>light, but no information went faster than light).
>
>The interesting bit is always finding out how the laws of physics are
>conserved, and Geoffrey's done just that.
>
>May I ask a more basic question?
>
>If photons have mass,
They don't
and if they travel at c, how come they don't
>have infinite mass? Since they don't have infinite mass, they can have
>no mass, and therefore no momentum.
>
Nope, wrong. Mass and momentum are separate things.
me...@cars3.uchicago.edu
missed by many is that even in classical physics momentum *is not*
defined as mv (yes, there is quite a lot of physics beyond high school
physics). Momentum is defined as a gradient of the Lagrangian (yes, I
know this doesn't mean much to whoever didn't study it but, as I said,
there is lots of physics beyond high school physics). In the
particular case of a classical massive particle, this *evaluates* to
mv but that's a result, not a definition. For other entities you get
different result. Thus, even within classical physics electromagnetic
waves carry momentum even though they're massless.
Edward Green
> If there's no change in photon energy to first order, then
> obviously that's a breaking of Carnot's law to first order,
> since Carnot requires an decrease in photon temperature
> (photon energy) for work to be extracted.
True, I snipped context and all, but I love the way "to first order"
sometimes takes on a meaning all of its own -- as if it meant
something without even specifying "with respect to X".
> Carnot requires
> two thermal baths of different temperatures for kinetic
> energy to be gained.
Which are present when a solar sail is working.
> But it's okay, because in any inertial frame where work is
> being done on the sail, you see two populations of photons
> (those coming and those leaving), and these two DO have two
> different temperatures. That's it.
Ok. Start with a heat bath at temperature T. Extract a wee bit of
energy from it to do some useful work. Wind up with a heat bath at
temperature T - dt. Viola! Two different temperature heat baths! Ok
by thermo! :-)
No ... if that were the out, the second law would be empty. The two
different temperature heat baths were there at the start of the
problem, _before_ we introduced the sail. The are separated in
momentum space rather than physical space.
Of course the photons don't have to be charaterized by temperatures at
all ... the sail could work with pure monochromatic radiation. Yet
even if we had a complete Stefan-Boltzman distribution of frequencies
characteristic of TK, but with the momentum vectors all pointing in
one direction, that would _not_ correspond to envelopment in a heat
bath at temperture T. If diferent regions of momentum space are
populated by photons distributed differently in energy, whether or not
according to some temperatures, that's enough to get work out of the
system. And that's why a solar sail illuminated by the distant sun on
one side and exposed to the icy cold inky blackness space on the
other, can work.
So saying, the great Shaman snapped closed his dearhide covered book
of tales, and shooed the little braves off to bed to dream -- to dream
of the illiminatable black crow winging forever in blackness on sooty
wings while the coyote of the abyss howled his ceaseless song of
unnerving beauty and infinite sadness and the rosy sun-squaw is tucked
in her bed, waiting for a dawn that may never come.
And darkness descended on the world. Amen.
Laurel Amberdine
When you're not so busy you need to write a book, Mati. :) Something
like "All the Physics you Learned Wrong" or something.
After all this practice you could probably write it half asleep with your
eyes closed.
-Laurel
Laurel Amberdine
> "Steve Harris" <sbha...@ix.RETICULATEDOBJECTcom.com> wrote in message news:<be6vq2$jfb$1...@slb0.atl.mindspring.net>...
>
>> If there's no change in photon energy to first order, then
>> obviously that's a breaking of Carnot's law to first order,
>> since Carnot requires an decrease in photon temperature
>> (photon energy) for work to be extracted.
>
> True, I snipped context and all, but I love the way "to first order"
> sometimes takes on a meaning all of its own -- as if it meant
> something without even specifying "with respect to X".
Ignorance time: people are saying "to first (second, zeroth) order" etc,
quite often lately. What does it mean, anyway?
-Laurel
Sander Vesik
There isn't one.
--
Sander
+++ Out of cheese error +++
me...@cars3.uchicago.edu
>On Sun, 06 Jul 2003 22:35:55 GMT, me...@cars3.uchicago.edu <me...@cars3.uchicago.edu> wrote:
...
>>>
>> Before you even start getting to relativistic physics, the point
>> missed by many is that even in classical physics momentum *is not*
>> defined as mv (yes, there is quite a lot of physics beyond high school
>> physics). Momentum is defined as a gradient of the Lagrangian (yes, I
>> know this doesn't mean much to whoever didn't study it but, as I said,
>> there is lots of physics beyond high school physics). In the
>> particular case of a classical massive particle, this *evaluates* to
>> mv but that's a result, not a definition. For other entities you get
>> different result. Thus, even within classical physics electromagnetic
>> waves carry momentum even though they're massless.
>
>When you're not so busy you need to write a book, Mati. :) Something
>like "All the Physics you Learned Wrong" or something.
>
>After all this practice you could probably write it half asleep with your
>eyes closed.
>
Probably so, this ng is good practice for such purpose:-)
Christopher M. Jones
> Please don't misintepret me. I do "believe" in solar sails. I'm just
> trying to figure out how it works, because, in the normal world:
>
> Momentum = mass * velocity
>
> If mass = 0 and velocity = 3E8, then momentum = 0
>
> Please tell me what I'm missing?
This is one of the cases where you have to remember that
"mass" is really "rest-mass" but in this case rest-mass
is the wrong mass, so you want something more like the
old "relativistic mass".
In other words, you're missing that e=mc^2, and that
since photons have energy they also have mass (and
momentum).
Gregory L. Hansen
If you have some function f(x), perhaps an unknown that you're trying to
solve equations of motion to find, it's often possible to approximate it
and wind up with a polynomial expansion. So you might get
f(x) ~= a + bx + cx^2 + ...
First order would be linear in x, second order would be quadratic in x,
etc.
In this particular discussion, I don't think "first order" really means
anything, it's just something that people are saying.
me...@cars3.uchicago.edu
delta_p (momentum transfer).
Gregory L. Hansen
me...@cars3.uchicago.edu
relevant here at all. There appear to be two points of confusion in
the discussion. One is equating photons energy with temperature
(which it is not), the other is puzzlement over how can the mirror
gain anergy if the photons don't lose any (the answer is, yes, they
do). The "first order" business is primarily relevant to the second.
Of course, if somebody wants to calculate the entropy change of the
photon gas, then the "first order" business applies there as well.
Henry Spencer
>quite often lately. What does it mean, anyway?
While those terms have precise meanings, they are often applied somewhat
more loosely and generally. Speaking loosely and generally...
"To first order" means "considering only the biggest and most obvious
effects, neglecting details which don't change things very much".
"To second order" means "including the most significant of the details,
but neglecting really small ones which have rather smaller effects".
And so forth.
And by analogy, "to zeroth order" means "making drastic simplifications to
get an answer which will be somewhere in the right ballpark".
For example, if you wanted to know how much my little nephew weighed...
To zeroth order, you might try to pick him up (cautiously :-)).
To first order, you'd put him on a bathroom scale.
To second order, you'd have him strip to his undershorts, put him on a
doctor's scale, and tell him to stand still.
To third order, you'd need a research-grade scale, you'd have him strip
completely, you'd offer him an advance copy of the next Harry Potter book
if he stood very, very still, and you'd pay attention to issues like when
he last ate, whether his bladder is full or empty, etc.
Laurel Amberdine
> In article <bec7n8$3a6kb$4...@ID-45790.news.dfncis.de>,
> Laurel Amberdine <cir...@mtco.com> wrote:
<snip>
>>Ignorance time: people are saying "to first (second, zeroth) order" etc,
>>quite often lately. What does it mean, anyway?
>
> If you have some function f(x), perhaps an unknown that you're trying to
> solve equations of motion to find, it's often possible to approximate it
> and wind up with a polynomial expansion. So you might get
>
> f(x) ~= a + bx + cx^2 + ...
>
> First order would be linear in x, second order would be quadratic in x,
> etc.
Okay, thanks! It sounded familiar but I couldn't recall enough to piece a
meaning together.
> In this particular discussion, I don't think "first order" really means
> anything, it's just something that people are saying.
I don't know if the phrase is being used technically now or not, but it
does seem to have nearly become slang.
-Laurel
Laurel Amberdine
> On Mon, 7 Jul 2003 18:45:12 +0000 (UTC), Gregory L. Hansen <glha...@steel.ucs.indiana.edu> wrote:
>> In article <bec7n8$3a6kb$4...@ID-45790.news.dfncis.de>,
>> Laurel Amberdine <cir...@mtco.com> wrote:
I so like responding to myself.
> <snip>
>>>Ignorance time: people are saying "to first (second, zeroth) order" etc,
>>>quite often lately. What does it mean, anyway?
>>
>> If you have some function f(x), perhaps an unknown that you're trying to
>> solve equations of motion to find, it's often possible to approximate it
>> and wind up with a polynomial expansion. So you might get
>>
>> f(x) ~= a + bx + cx^2 + ...
>>
>> First order would be linear in x, second order would be quadratic in x,
>> etc.
>
> Okay, thanks! It sounded familiar but I couldn't recall enough to piece a
> meaning together.
And now I have come across "second order" and "third order" (etc)
determinants, and I don't know if (or how) that relates to the same
phrasing above. (I do know what it means in this context. Kinda obvious.)
That isn't really a question. I'm just babbling as a break from really
excessive quantities of multiplying, adding, and subtracting. Zeroes. I
just wish there were more zeroes...!
-Laurel
Steve Harris
<me...@cars3.uchicago.edu> wrote in message
news:jcPNa.77$Y4.2...@news.uchicago.edu...
> In article <be6vq2$jfb$1...@slb0.atl.mindspring.net>, "Steve
Harris" > >If there's no change in photon energy to first
order, then
> >obviously that's a breaking of Carnot's law to first
order,
> >since Carnot requires an decrease in photon temperature
> >(photon energy) for work to be extracted.
>
>
> 1) Photon energy is *not* "photon temperature". Photon
(or any
> particle) does not have a temperature). Temperature is an
ensemble
> property.
Yes, yes, but the point is you can have a thermalized bath
of photons, and if you do, then it has to follow the same
thermodynamic laws as any heat source. to wit, if you turn
some of the photon energy of such a collection into free
energy (charging a battery on the rocket or pushing the
rocket faster), then you have to pay some entropy price for
destroying that much heat. In simple reflective sails that
means creating another bunch of (reflected) photons that are
effectively thermalized at a lower temperature (heat going
into a lower temp reservoir). Or else you have to heat up
some sail material which was previously at a lower temp than
the illuminating radiation, and dump your entropy that way.
If you have a monochromatic source of photons like laser or
a microwave beam, then (so far as I can tell) nothing
prevents you from converting such a beam *entirely* into
free energy (charging a battery on board the rocket), and
destroying the EM radiation completely, so that no photons
are left at all, and sail heating is minimized (perhaps
doesn't occur at all). All the energy could be extracted
from the beam and used to charge a battery or make chemical
fuel, or something. Which could then be used indirectly for
propulsion, subject only to a rocket's ability to extract
fuel energy and turn it into rocket kinetic energy (not a
thermodynamic problem, but certainly a practical one which I
supposed must be related to the thermodynamic one).
For reasons that are obscure to me, there are additional
limitations for using the energy in photons that come in a
blackbody spectrum. They act like heat, and only some or
their energy is available for conversion into free energy at
the target, and thence into rocket kinetic energy.
Gold, I'm sure, specified a "heat" driven sail (thermalized
light from the sun), because thermodynamics limits in that
case what fraction of light energy can be extracted from the
heat beam as a first step. But I agree it does just serve to
complicate the problem, since the major shortcoming in using
EM radiation to power a rocket, is in converting all that
energy into something useful for propulsion. If you just
reflect it, you waste most of your energy, and that has
nothing to do with thermodynamics.
SBH
me...@cars3.uchicago.edu
And, indeed, you will. If you take a, say, blackbody radiation
impinging on you sail, the reflected beam will have blackbody spectrum
corresponding to slightly lower temperature. Doppler shifted
blackbody spectrum is blackbody spectrum. No problem.
The confusion present arises from thinking that the reflected photons
are at same energy as the incoming ones. They're not.
Or else you have to heat up
>some sail material which was previously at a lower temp than
>the illuminating radiation, and dump your entropy that way.
>
>If you have a monochromatic source of photons like laser or
>a microwave beam, then (so far as I can tell) nothing
>prevents you from converting such a beam *entirely* into
>free energy (charging a battery on board the rocket) and
>destroying the EM radiation completely, so that no photons
>are left at all, and sail heating is minimized (perhaps
>doesn't occur at all). All the energy could be extracted
>from the beam and used to charge a battery or make chemical
>fuel, or something.
How do you plan to use the energy to make chemical fuel? To use it to
create protons and electrons, then assemble these into atoms etc?:-)
I don't think so. You can much better carry the fuel with you, to
begin with. But that defeats the idea of the light sail which is to
get around the inherent reaction mass limitations.
>
>For reasons that are obscure to me, there are additional
>limitations for using the energy in photons that come in a
>blackbody spectrum. They act like heat, and only some or
>their energy is available for conversion into free energy at
>the target, and thence into rocket kinetic energy.
>
The limits are the same as in any other process. Entropy cannot
decrease.
Gregory L. Hansen
It is actually slang, in the right groups. Hang around with physicists
during lunch time and eventually someone will say something like "To first
order, it rained during the entire vacation", or "A three sigma apple is
still a good apple" (when comparing to e.g. peaches).
Shows like Star Trek seem a little artificial sometimes because none of
that sort of thing creeps in when engineers talk to engineers. And also
because they spout off with all the technical words on the job, when
scientists in real life might ask for "the magic wrench", or suggest doing
"the hokey pokey" on a detector.
But I suppose if they try to get those sorts of language issues in there,
it would seem more artificial than otherwise.
Edward Green
> In article <becmsa$3p7e2$2...@ID-45790.news.dfncis.de>,
> Laurel Amberdine <cir...@mtco.com> wrote:
> >On Mon, 7 Jul 2003 18:45:12 +0000 (UTC), Gregory L. Hansen
> ><glha...@steel.ucs.indiana.edu> wrote:
>
> >> In this particular discussion, I don't think "first order" really means
> >> anything, it's just something that people are saying.
> >
> >I don't know if the phrase is being used technically now or not, but it
> >does seem to have nearly become slang.
>
> It is actually slang, in the right groups. Hang around with physicists
> during lunch time and eventually someone will say something like "To first
> order, it rained during the entire vacation", or "A three sigma apple is
> still a good apple" (when comparing to e.g. peaches).
Hah. Similar to hacker humor: "For sufficiently large values of
zero".
Except ... gasp ... I can hear your line above not even being used
ironically.
Apropros the present problem: like interesting, man: colon: I suspect,
to first order, that when a wee tiny body recoils elastically off a
great massie body, and the great massie body is originally at rest,
that energy loss of the wee tiny body is, to first order in dp, zero.
My suspicions are also aroused to the possibility that once the great
massie body is put in your state of motion in your original state of
reference, that this is no longer true.
Which is, however, as Mati suggests, a complete red herring here;
though there are so many on the plate I think we bought them in tomato
sauce. It does sound like a nice little oolieish answer that the
second law of thermo is violated only to second order in dp, or some
such nonsense, though it is in fact nonsense and, whatsmore,
irrelevant nonsense. Is that an oxymoron?
Speaking of slang and argot and jargon* and all that, this thread
illustrates another habit of language. While I am familiar with the
second law of thermo, and Carnot cycles, and the approximate relation
of these concepts, I have never made the aquaintance of "Carnot's
Rule", or whatever the precise form being brutish about here.
Now maybe my thermo education is incredibly inadequate or non-standard
or shows my heritage as a failed p-chemist rather than a failed
physicist, but it seems to me just _possible_ that other threadies had
no greater familiarity with this rule than I did -- though probably
understanding the second law perfectly adequately. Yet here we are,
present company (inconsistently) excluded, all brandying about
"Carnot's rule" as if it is something we have known and loved from
childhood, and never spoke about violating the second law of thermo in
any other way that "violating Carnot's rule".
I don't think it's so.
The habit of language illustrated is that if somebody uses a
catch-phrase which we are not _quite_ familiar with, but feel we ought
to be, we are apt to immediately adopt it to show that we too are
totally hep, cool and jargonationally hep with it, dadeo.
There was an amazing phenomena in NYC a few years back where sidewalk
scaffolding -- you know, that stuff erected under construction sites
to protect you from being hit on the head by the smaller debris --
suddenly became known as "sidewalk bridges". Nobody knows who started
this word virus, yet suddenly all discussion of sidewalk scaffolding,
themselves enjoying a temporary newsworthyness for equally obscure
reasons, had to refer to them by the new shiboleth, or reveal the
writer to be horribly, horribly technically inept and inadequate and
to have been sired out of wedlock in Spanish Harlem on a warm summer
night and grown up in the street!
If you know what I mean.
I believe the concept intended was the, now abundantly wrong,
assertion that solar sails "violated the second law of
thermodynamics". Repeat after me: "Solar sails violate the second law
of thermodynamics" -- except they don't.
> Shows like Star Trek seem a little artificial sometimes because none of
> that sort of thing creeps in when engineers talk to engineers.
I feel like an attempt was made once or twice, but maybe I err.
> And also
> because they spout off with all the technical words on the job, when
> scientists in real life might ask for "the magic wrench", or suggest doing
> "the hokey pokey" on a detector.
That's good!
Spock and Data may have been given the role of using technical
language in ordinary conversation; but you are right -- I never heard
the reverse phenomenon emulated, or even emulsified, though the
writers doubtless received substantial emollient to effect it.
> But I suppose if they try to get those sorts of language issues in there,
> it would seem more artificial than otherwise.
And most people, not even being around engineers and scientists ever,
just wouldn't get it.
*also shiboleths and sacred kine
Gregory L. Hansen
Edward Green <null...@aol.com> wrote:
>glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
>news:<bed4vf$oso$1...@hood.uits.indiana.edu>...
>> In article <becmsa$3p7e2$2...@ID-45790.news.dfncis.de>,
>> Laurel Amberdine <cir...@mtco.com> wrote:
>Speaking of slang and argot and jargon* and all that, this thread
>illustrates another habit of language. While I am familiar with the
>second law of thermo, and Carnot cycles, and the approximate relation
>of these concepts, I have never made the aquaintance of "Carnot's
>Rule", or whatever the precise form being brutish about here.
>
>Now maybe my thermo education is incredibly inadequate or non-standard
>or shows my heritage as a failed p-chemist rather than a failed
>physicist, but it seems to me just _possible_ that other threadies had
>no greater familiarity with this rule than I did -- though probably
>understanding the second law perfectly adequately. Yet here we are,
>present company (inconsistently) excluded, all brandying about
>"Carnot's rule" as if it is something we have known and loved from
>childhood, and never spoke about violating the second law of thermo in
>any other way that "violating Carnot's rule".
>
>I don't think it's so.
>
>The habit of language illustrated is that if somebody uses a
>catch-phrase which we are not _quite_ familiar with, but feel we ought
>to be, we are apt to immediately adopt it to show that we too are
>totally hep, cool and jargonationally hep with it, dadeo.
Interesting. I assumed Carnot's Rule was something like that the
efficiency of a heat engine was the Carnot efficiency or less, but I don't
think it was ever really defined. So I Googled "Carnot's Rule", and the
only hits on it link to Gold and solar sails. And no hits on a Web of
Science search. Or in the McGraw-Hill Dictionary of Scientific and
Technical Terms.
But that does suggest a debating tactic that would probably have been old
news to Gorgias. Invent almost correct terms that your opponent is
embarassed to question. "But the Green's polynomials don't converge...",
and watch the discussion turn to the important of convergence rather than
"What the hell is a Green's polynomial?"
Gregory L. Hansen
>But that does suggest a debating tactic that would probably have been old
>news to Gorgias. Invent almost correct terms that your opponent is
And did I mention making casual historical references that your opponent
feels shamed not to know?
Geoffrey A. Landis
[Geoffrey Landis had written]:
>> > There is no energy cost to move the stationary sail to
>> > first order.
> If there's no change in photon energy to first order, then
> obviously that's a breaking of Carnot's law to first order,
> since Carnot requires an decrease in photon temperature
> (photon energy) for work to be extracted.
For a stationary sail, there's no work extracted, so there is no
breaking of Carnot's law.
Let's phrase it slightly differently.
power (which is work per unit time) is force times velocity.
For a stationary sail, velocity is zero, so power is zero. No work is
extracted.
> Carnot requires two thermal baths of different temperatures for
> kinetic energy to be gained.
Right. And Carnot says nothing whatsoever about requirements for
momentum to be extracted.
> But it's okay, because in any inertial frame where work is
> being done on the sail, you see two populations of photons
> (those coming and those leaving), and these two DO have two
> different temperatures. That's it.
Correct, and since the reflected photon ensemble has a blackbody
spectrum (for a non-dichroic sail), that analysis works and gives the
right result. It was only the specific case of a stationary sail where
the analysis has to take into account the case of no change in photon
energy.
--
Geoffrey A. Landis
http://www.sff.net/people.landis
geoff
> In article <d81e59c9.0307...@posting.google.com>, alext...@yahoo.com (Alex Terrell) writes:
> >The effect of solar radiation pressure has been observed. This seems
> >to be one of many experiments that appear to break a fundamental law
> >of Physics (Carnot's rule).
> >
> Nothing is broken here.
>
> me...@cars.uchicago.edu | chances are he is doing just the same"
OK, none of us doubt solar sails work now?
But this is an interesting problem to consider. Firstly, we would not
expect to understand photons by analogy with classical particles, as
in some of the snooker ball arguments given above. Most importantly,
photon velocity is always the same when measured in any frame of
reference but this would not be the case with classical particles.
Thus any consideration of the motions involved must be related to a
frame of reference because special relativity leads us to expect
observers in motion with respect to each other to make different
measurements.
Surprisingly, we also need to be careful about frames of reference
when we make classical considerations. For example kinetic energy and
momentum are not unique properties of the sail and depend upon the
frame of reference of the measurement. Also, when two classical
particles collide the momentum transfer between them is unique but the
kinetic energy transfer depends upon the frame of reference. This is
because momentum transfer is equal to force times time but energy
transfer is force times distance. The time is unique (in classical
systems), but the distance covered during the collision depends upon
the observer's frame of reference.
The upshot of all this is that observers in different places will see
something different. An observer on the sail will see the sail as
stationary, with zero kinetic energy at all times. Light from the sun
will approach at velocity c and wavelength l and be reflected with the
same velocity and wavelength. She will see no energy transfer as
proposed by Gold.
An observer on the Sun would see photons reflected from the sail that
were doppler shifted because of the sail's velocity in his frame of
reference. He sees a different photon velocity relative to the sail
before and after reflection because the photon velocity must always be
c in his frame of reference. The energy lost or gained by the photons
would match the kinetic energy change of the sail, again in his frame
of reference.
Strictly speaking it may be invalid to apply special relativity to the
sail because it is accelerating, although with a large velocity and
very small acceleration this assumption may be OK. What is certain
here is that all observers will experience the same laws of physics in
their frame of reference and make the same measurement of the velocity
of light at all times. This appears to lead to the conclusions above.
Laurel Amberdine
> In article <becmsa$3p7e2$2...@ID-45790.news.dfncis.de>,
> Laurel Amberdine <cir...@mtco.com> wrote:
>>On Mon, 7 Jul 2003 18:45:12 +0000 (UTC), Gregory L. Hansen
>><glha...@steel.ucs.indiana.edu> wrote:
>
>>> In this particular discussion, I don't think "first order" really means
>>> anything, it's just something that people are saying.
>>
>>I don't know if the phrase is being used technically now or not, but it
>>does seem to have nearly become slang.
>
> It is actually slang, in the right groups. Hang around with physicists
> during lunch time and eventually someone will say something like "To first
> order, it rained during the entire vacation", or "A three sigma apple is
> still a good apple" (when comparing to e.g. peaches).
Uh huh. And you guys wonder why no one else sits with you at lunch. :D
I am, of course, only kidding. I think it's cute, and would happily hang around
and listen, but there aren't any physicists around where I am. :(
-Laurel
Steve Harris
<me...@cars3.uchicago.edu> wrote in message
news:IXnOa.94$Y4.3...@news.uchicago.edu...Harris" > >For reasons that are obscure to me, there are
additional
> >limitations for using the energy in photons that come in
a
> >blackbody spectrum. They act like heat, and only some or
> >their energy is available for conversion into free energy
at
> >the target, and thence into rocket kinetic energy.
> >
> The limits are the same as in any other process. Entropy
cannot
> decrease.
Yes, but I'm curious about the mechanism for why entropy
cannot decrease. Something about quantum states expanding in
phase-space, but can you put it into easier terms? Why (for
example) can you extract all the energy in a monochromatic
beam and use it to charge a battery (or make new atoms of
matter and antimatter if you must), but there are limits on
what fraction of a beam of blackbody radiation you can use.
Don't just give me an entropy argument (you were ragging on
me for "explaining" wing forces on winds by momentum
conservation, remember?).
What's going on *mechanistically* that you can in theory
completely convert monochromatic EMR to any other kind of
energy you like, but not the mix of wavelengths of EMR that
comes off a black body?
SBH
me...@cars3.uchicago.edu
>In <be6vq2$jfb$1...@slb0.atl.mindspring.net> Steve Harris wrote:
>
>[Geoffrey Landis had written]:
>>> > There is no energy cost to move the stationary sail to
>>> > first order.
>
>> If there's no change in photon energy to first order, then
>> obviously that's a breaking of Carnot's law to first order,
>> since Carnot requires an decrease in photon temperature
>> (photon energy) for work to be extracted.
>
>For a stationary sail, there's no work extracted, so there is no
>breaking of Carnot's law.
>
>Let's phrase it slightly differently.
>power (which is work per unit time) is force times velocity.
>
>For a stationary sail, velocity is zero, so power is zero. No work is
>extracted.
>
>> Carnot requires two thermal baths of different temperatures for
>> kinetic energy to be gained.
>
>Right. And Carnot says nothing whatsoever about requirements for
>momentum to be extracted.
>
is the source of some of the confusion present.
No, the second law (as Ed rightly pointed out, people here say
"Carnot" when what they really mean is the second law of
thermodynamics) does ***not*** say that you need two thermal baths of
different temperatures for kinetic energy to be gained.
Consider an ensamble of particles (and it doesn't matter in the least
whether they're photons or anything else). The energy of the ensamble
can be represented as a sum of two parts:
1) The kinetic energy of the CM motion (i.e. the energy associated
with the total momentum of the ensamble).
2) The "internal" energy, i.e. the energy of the ensamble in its CM
frame.
Now, the first part is freely transferable to anything the ensamble
interacts with, subject only to conservation of energy and momentum
laws. The second law is not involved at all. It is only when you
want to extract some of the second part (the internal energy) that the
second law comes into play.
me...@cars3.uchicago.edu
>me...@cars3.uchicago.edu wrote in message news:<IV0Oa.83$Y4.2...@news.uchicago.edu>...
>> In article <d81e59c9.0307...@posting.google.com>, alext...@yahoo.com (Alex Terrell) writes:
>> >The effect of solar radiation pressure has been observed. This seems
>> >to be one of many experiments that appear to break a fundamental law
>> >of Physics (Carnot's rule).
>> >
>> Nothing is broken here.
>>
>
>
>But this is an interesting problem to consider. Firstly, we would not
>expect to understand photons by analogy with classical particles, as
>in some of the snooker ball arguments given above.
Just a comment here: There is no need to consider photons at all.
Just EM waves.
> Most importantly,
>photon velocity is always the same when measured in any frame of
>reference but this would not be the case with classical particles.
>
>Thus any consideration of the motions involved must be related to a
>frame of reference because special relativity leads us to expect
>observers in motion with respect to each other to make different
>measurements.
Not only special relativity, since ...
>
>Surprisingly, we also need to be careful about frames of reference
>when we make classical considerations.
... yep.
>For example kinetic energy and
>momentum are not unique properties of the sail and depend upon the
>frame of reference of the measurement. Also, when two classical
>particles collide the momentum transfer between them is unique but the
>kinetic energy transfer depends upon the frame of reference. This is
>because momentum transfer is equal to force times time but energy
>transfer is force times distance. The time is unique (in classical
>systems), but the distance covered during the collision depends upon
>the observer's frame of reference.
>
Yes. And this appears to confuse some of the participants.
>The upshot of all this is that observers in different places will see
>something different. An observer on the sail will see the sail as
>stationary, with zero kinetic energy at all times. Light from the sun
>will approach at velocity c and wavelength l and be reflected with the
>same velocity and wavelength. She will see no energy transfer as
>proposed by Gold.
>
>An observer on the Sun would see photons reflected from the sail that
>were doppler shifted because of the sail's velocity in his frame of
>reference. He sees a different photon velocity relative to the sail
>before and after reflection because the photon velocity must always be
>c in his frame of reference. The energy lost or gained by the photons
>would match the kinetic energy change of the sail, again in his frame
>of reference.
>
>Strictly speaking it may be invalid to apply special relativity to the
>sail because it is accelerating, although with a large velocity and
>very small acceleration this assumption may be OK.
Oh, there is absolutely no problem applying SR to situations involving
accelerations, this is routinely done. You just refer everything, at
any given moment, to the inertial frame momentarily comoving with your
system.
> What is certain
>here is that all observers will experience the same laws of physics in
>their frame of reference and make the same measurement of the velocity
>of light at all times. This appears to lead to the conclusions above.
That's true but even this goes beyond the issues involved as there is
nothing special about light in this context. The story remains the
same if you replace light with a stream of massive particles.
me...@cars3.uchicago.edu
>
><me...@cars3.uchicago.edu> wrote in message
>additional
>> >limitations for using the energy in photons that come in
>a
>> >blackbody spectrum. They act like heat, and only some or
>> >their energy is available for conversion into free energy
>at
>> >the target, and thence into rocket kinetic energy.
>> >
>> The limits are the same as in any other process. Entropy
>cannot
>> decrease.
>
>
>cannot decrease.
No "mechanism", just probabilities. And strictly speaking, it is not
"cannot decrease", just "is extremely unlikely to decrease".
> Something about quantum states expanding in
>phase-space, but can you put it into easier terms?
What you need here is a primer to statistical mechanics. That goes a
tad beyond a casual 1-2 paragraph note, some ot he concepts are quite
subtle. But no, you don't need to invoke QM, entropy was defined within
the classical framework to begin with (though it generalizes beautifully
to QM).
Entropy is (up to a multiplicative constant of no inherent
significance) the log of the number of microstates compatible with a
given macrostate. Now, given different macrostates realizable within
some given constraints (say, available energy), the probability of any
such macrostate is proportional to the number of microstates
compatible with it. So, higher entropy state is more probable.
Another take on this (for which you've to thank Shannon) is that
entropy of a system is proportional to the amount of missing
information about the system (i.e. of the information that would have
to be provided to characterize the system uniquely). Thus a fully
characterized system has zero entropy (no information missing).
> Why (for
>example) can you extract all the energy in a monochromatic
>beam and use it to charge a battery (or make new atoms of
>matter and antimatter if you must), but there are limits on
>what fraction of a beam of blackbody radiation you can use.
For the same reason for which you can convert a perfect optical image
to a perfect digital recording of such, and vise versa, but you cannot
convert a blurred and grainy image into a perfect one. There is
missing information.
>Don't just give me an entropy argument (you were ragging on
>me for "explaining" wing forces on winds by momentum
>conservation, remember?).
yep:-)
>
>What's going on *mechanistically* that you can in theory
>completely convert monochromatic EMR to any other kind of
>energy you like, but not the mix of wavelengths of EMR that
>comes off a black body?
>
It is not "machanistically". It is a matter of missing information.
You can shuffle an ordered deck, but it is highly improbable to get a
suffled deck back to an ordered one by continued shuffling.
For more than this, you'll have to consult textbooks.
Steve Harris
<me...@cars3.uchicago.edu> wrote in message
news:6dDOa.107$Y4.3...@news.uchicago.edu...
> In article <20030708101...@newsread.grc.nasa.gov>,
> Consider an ensamble of particles (and it doesn't matter
in the least
> whether they're photons or anything else). The energy of
the ensamble
> can be represented as a sum of two parts:
>
> 1) The kinetic energy of the CM motion (i.e. the energy
associated
> with the total momentum of the ensamble).
>
> 2) The "internal" energy, i.e. the energy of the ensamble
in its CM
> frame.
>
> Now, the first part is freely transferable to anything the
ensamble
> interacts with, subject only to conservation of energy and
momentum
> laws. The second law is not involved at all. It is only
when you
> want to extract some of the second part (the internal
energy) that the
> second law comes into play.
Well, here's the source of come of my confusion. You tell me
that I can't extract the energy of a beam of photons that is
coming from a pinhole in a blackbody and has been focused by
a parabolic reflector (I'll call this a heat beam), because
some of the information is missing WRT a beam of the same
power that has been generated in the same fashion, which is
non-coherent but monochromatic. But if all the photons are
moving in the same direction in the "heat beam," why *can't*
we convert all of their energy to free energy?
It seems to me that a very important thing about loss of
free energy in heat energy is the loss of *directionality*
of the kinetic energy of the particles, and I'm thinking
this must also translate in some analogous fashion to heat
beams as well, when you dispense with material particles
entirely. If you focus heat EM into a beam, you've partly
directionalized it. And yet nature doesn't allow you to take
a confined gas of some temperature and pass it though a
pinhole and bounce if off a reflector and thereby somehow
gain more than the thermodynamically available amount of
energy out of it. Er--- or does it? You're letting the gas
expand into a larger volume as you "focus" it, and you pay
the entropy cost that way. Perhaps if you're allowed to use
volume liberally in this fashion you can extract all the
heat energy in a hot gas that you like (get as close as you
like to all of it out as free energy). Yes? And the same
for a thermalized photon gas?
SBH
me...@cars3.uchicago.edu
>
><me...@cars3.uchicago.edu> wrote in message
>news:6dDOa.107$Y4.3...@news.uchicago.edu...
>> In article <20030708101...@newsread.grc.nasa.gov>,
>
>
>in the least
>> whether they're photons or anything else). The energy of
>the ensamble
>> can be represented as a sum of two parts:
>>
>> 1) The kinetic energy of the CM motion (i.e. the energy
>associated
>> with the total momentum of the ensamble).
>>
>> 2) The "internal" energy, i.e. the energy of the ensamble
>in its CM
>> frame.
>>
>> Now, the first part is freely transferable to anything the
>ensamble
>> interacts with, subject only to conservation of energy and
>momentum
>> laws. The second law is not involved at all. It is only
>when you
>> want to extract some of the second part (the internal
>energy) that the
>> second law comes into play.
>
>
>
>that I can't extract the energy of a beam of photons that is
>coming from a pinhole in a blackbody and has been focused by
>a parabolic reflector (I'll call this a heat beam), because
>some of the information is missing WRT a beam of the same
>power that has been generated in the same fashion, which is
>non-coherent but monochromatic. But if all the photons are
>moving in the same direction in the "heat beam," why *can't*
>we convert all of their energy to free energy?
Oh, you can absorb them, converting *all* their energy to heat. No
problem. It is only when you try to convert all of it into
macroscopic motion that you run into themodynamic limits. And, by the
way, same is true for your "monochromatic but non-coherent" beam
(though, strictly speaking, if it is non coherent it cannot be quite
monochromatic, either). You're still missing phase information and
you cannot fully convert the beam into useful power.
>
>It seems to me that a very important thing about loss of
>free energy in heat energy is the loss of *directionality*
>of the kinetic energy of the particles, and I'm thinking
>this must also translate in some analogous fashion to heat
>beams as well, when you dispense with material particles
>entirely. If you focus heat EM into a beam, you've partly
>directionalized it. And yet nature doesn't allow you to take
>a confined gas of some temperature and pass it though a
>pinhole and bounce if off a reflector and thereby somehow
>gain more than the thermodynamically available amount of
>energy out of it. Er--- or does it? You're letting the gas
>expand into a larger volume as you "focus" it, and you pay
>the entropy cost that way. Perhaps if you're allowed to use
>volume liberally in this fashion you can extract all the
>heat energy in a hot gas that you like (get as close as you
>like to all of it out as free energy). Yes?
You just described the principle behind a rocket nozzle. That's
exactly what it does.
> And the same for a thermalized photon gas?
Can't think of an exact analogy, off hand, but in principle I see no
reason why not. Well, come to think of it, I can think about an
analogy, though it is a bit artificial.
Edward Green
> In article <2a0cceff.0307...@posting.google.com>,
> Edward Green <null...@aol.com> wrote:
> >glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
> >While I am familiar with the
> >second law of thermo, and Carnot cycles, and the approximate relation
> >of these concepts, I have never made the aquaintance of "Carnot's
> >Rule", or whatever the precise form being brutish about here.
> >
> >Now maybe my thermo education is incredibly inadequate or non-standard
> >or shows my heritage as a failed p-chemist rather than a failed
> >physicist, but it seems to me just _possible_ that other threadies had
> >no greater familiarity with this rule than I did -- though probably
> >understanding the second law perfectly adequately. Yet here we are,
> >present company (inconsistently) excluded, all brandying about
> >"Carnot's rule" as if it is something we have known and loved from
> >childhood, and never spoke about violating the second law of thermo in
> >any other way that "violating Carnot's rule".
> >
> >I don't think it's so.
> >
> >The habit of language illustrated is that if somebody uses a
> >catch-phrase which we are not _quite_ familiar with, but feel we ought
> >to be, we are apt to immediately adopt it to show that we too are
> >totally hep, cool and jargonationally hep with it, dadeo.
>
> Interesting. I assumed Carnot's Rule was something like that the
> efficiency of a heat engine was the Carnot efficiency or less, but I don't
> think it was ever really defined. So I Googled "Carnot's Rule", and the
> only hits on it link to Gold and solar sails. And no hits on a Web of
> Science search. Or in the McGraw-Hill Dictionary of Scientific and
> Technical Terms.
Beautiful! _I_ assumed it was at least a known, if obscure, concept,
which we were picking up as jargon for "the second law", whereas you
went the extra mile and found out that if it's obscure, it's _really_
obscure! That's too good.
Anyway, we both know why at gatherings it's only the most senior
emeritus emertus who will dare ask questions like "Er, excuse me: what
is 'Carnot's Rule'?". It's not even necessarily that they are braver
than everybody else: it's just that, given sufficient experience, they
_know_ if they haven't heard of something that probably nobody else in
the room has either -- so they can ask the question without fear of
looking foolish.
> But that does suggest a debating tactic that would probably have been old
> news to Gorgias.
Who's Gorgias? (He asked with a certain embarassment :-).
> Invent almost correct terms that your opponent is
> embarassed to question. "But the Green's polynomials don't converge...",
> and watch the discussion turn to the important of convergence rather than
> "What the hell is a Green's polynomial?"
Thanks, Gregory! You've really made my day. LOL, etc.
Edward Green
> In article <beeki3$8c0$1...@hood.uits.indiana.edu>,
> Gregory L. Hansen <glha...@steel.ucs.indiana.edu> wrote:
> >In article <2a0cceff.0307...@posting.google.com>,
> >Edward Green <null...@aol.com> wrote:
>
> >But that does suggest a debating tactic that would probably have been old
> >news to Gorgias. Invent almost correct terms that your opponent is
>
> And did I mention making casual historical references that your opponent
> feels shamed not to know?
You bastard. ;-)
Gregory L. Hansen
Edward Green <null...@aol.com> wrote:
>glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
>news:<beeklh$8c0$2...@hood.uits.indiana.edu>...
>> In article <beeki3$8c0$1...@hood.uits.indiana.edu>,
>> Gregory L. Hansen <glha...@steel.ucs.indiana.edu> wrote:
>> >In article <2a0cceff.0307...@posting.google.com>,
>> >Edward Green <null...@aol.com> wrote:
>>
>> >But that does suggest a debating tactic that would probably have been old
>> >news to Gorgias. Invent almost correct terms that your opponent is
>>
>> And did I mention making casual historical references that your opponent
>> feels shamed not to know?
>
>You bastard. ;-)
Heh! Gorgias was an ancient Greek sophist that peddled debating skills
for money. His thing was that it didn't matter whether you're right or
wrong, he'd teach you how to win the argument. I'd remembered there was a
name associated with that, but it took a bit of Googling before I found
the reference to casually drop. Couldn't have happened in real-time
conversation.
Edward Green
Besides, he didn't specify 3 sigma which way.
Apropros of nothing, I am totally targetted by the new Chrysler ad
campaign. I totally wanted one _before_ they mentioned casually it
was clocked at 150 mph on the autobahn -- just based on its looks. Of
course, don't try this at home guys. Right. Now of course, my
Maseratti does 185, they took away my license and now I can't drive,
and etc. But at $35,000 "fully equipped", this is a fast car for
pocket change... well, within reach of Joe Average. AND the added
zinger "be one of the first 9000 to call to get one in 2003. And, it
looks cool. I must be in the target audience, and they hit about 5
sigma on my buying buttons: ACT NOW! GOES REALLY FAST! LOOKS REALLY
COOL! GREAT VALUE!! (Yeah, I'm enough of a Jewish grandmother to think
"great value" adds to a product's cache ;-).
Sigh ... now, just where did my Joe Average disposable income go ...
Mary Shafer
<sbha...@ix.RETICULATEDOBJECTcom.com> wrote:
>
> "Geoffrey A. Landis" <geoffre...@sff.net> wrote in
> message
> news:1bfb1890.03070...@posting.google.com...
>
> > > This is exactly what Gold argues.
> >
> > Not quite. You understand this first point? If the sail
> is
> > stationary, then a reflection of a photon results in zero
> change in
> > energy (to first order), and thus the sail can gain
> momentum at no
> > energy cost?
>
> Come on. You're like the guy who says: when I heat up an
> object, its weight doesn't change (to first order).
> Therefore the equivalence of mass and energy is violated.
> Duh.
>
> There is no energy cost to move the stationary sail *to
> first order.* Carnot's law is broken to exactly the degree
> that you simplify the problem with approximation. But don't
> confuse your approximation with violation of physical law.
What's Carnot's Law?
I looked in both my thermodynamics books, Halliday & Resnick, and
Feynman, but couldn't find anything about it. I found a bunch about
the Carnot cycle, naturally, but no Law. I know Carnot is long dead,
so he can't have invented it in the last thirty years, either.
This is annoying, because I can't look up anything about airplanes or
aeronautics because all those books are packed away, which irritates
me, and what I can look up, about physics, I can't find.
Mary
--
Mary Shafer Retired aerospace research engineer
mil...@qnet.com
"Turn to kill, not to engage." LCDR Willie Driscoll, USN
Mark
Carnot's Law == The 2nd law of thermodynamics
a short mention here:
http://www.benwiens.com/energyFAQ.html#energyFAQ.25
Mark
Keith F. Lynch
> But if all the photons are moving in the same direction in the "heat
> beam," why *can't* we convert all of their energy to free energy?
You can, but only if you have an infinite heat sink at absolute zero,
and are willing to take literally forever. (Radiating heat at low
temperatures is slow. Radiating heat at absolute zero is infinitely
slow.)
If you have monchromatic light, you can have each photon strike an
easily ionized target and cause one electron to be emitted for each
photon. Since the photons all have the same energy, then so will the
electrons. The electrons can be made to go "uphill" by an amount
equal to their voltage, to charge a capacitor.
But if the photons all have different energies, then so will the
electrons. Each electron will have a different voltage (momentum).
You will have replaced your "hot" photons with "hot" electrons.
If you try to use those hot electrons to charge your capacitor, some
will have too much energy. The excess energy will be wasted heating
the capacitor plate, which will lose that heat either by radiating
electrons or photons. Other electros will have too little energy.
They'll fall back to the emitter, and heat *it* without doing anything
useful whatsoever.
A similar argument can be made for any other kind of photon collector.
> It seems to me that a very important thing about loss of free energy
> in heat energy is the loss of *directionality* of the kinetic energy
> of the particles, ...
If the thermalized photons are coming from all directions equally then
the whole environment is equally hot, and you can get no useful work
whatsoever. Well, maybe a little, but only until your absorber heats
up to the same temperature as the background. After that, each part
of it will be just as likely to emit photons as to absorb them, so
your engine is just as likely to run backwards as forwards, regardless
of the details of its construction. Everything will be at maximum
entropy for that temperature.
(Well, actually, you could have thermalized photons are come from all
directions equally, but *not* at the equilibrium temperature of the
environment. A cloudy day on earth is an example. Solar panels still
work pretty well even if the light is not at all directional.)
> Perhaps if you're allowed to use volume liberally in this fashion
> you can extract all the heat energy in a hot gas that you like
> (get as close as you like to all of it out as free energy). Yes?
Yes. The coldest things in nature -- colder than the 3 K background
-- are espanding gas clouds from supernovas, ironically enough.
But this cooling sure takes a lot of space and time.
> And the same for a thermalized photon gas?
Yes, given sufficient space and time. You may want to wait for the
cosmic background to cool off some first.
Exercise for the student: What's the optimal radius to build a Dyson
sphere around the sun to capture as much energy as possible? Would it
be useful to build several concentric spheres, each using the waste
heat from the next one inwards?
--
Keith F. Lynch - k...@keithlynch.net - http://keithlynch.net/
I always welcome replies to my e-mail, postings, and web pages, but
unsolicited bulk e-mail (spam) is not acceptable. Please do not send me
HTML, "rich text," or attachments, as all such email is discarded unread.
Rand Simberg
he...@spsystems.net (Henry Spencer) made the phosphor on my monitor
glow in such a way as to indicate that:
>In article <bec7n8$3a6kb$4...@ID-45790.news.dfncis.de>,
>Laurel Amberdine <cir...@mtco.com> wrote:
>>Ignorance time: people are saying "to first (second, zeroth) order" etc,
>>quite often lately. What does it mean, anyway?
>
>While those terms have precise meanings, they are often applied somewhat
>more loosely and generally. Speaking loosely and generally...
>
>"To first order" means "considering only the biggest and most obvious
>effects, neglecting details which don't change things very much".
>
>"To second order" means "including the most significant of the details,
>but neglecting really small ones which have rather smaller effects".
>
>And so forth.
>
>And by analogy, "to zeroth order" means "making drastic simplifications to
>get an answer which will be somewhere in the right ballpark".
This is a useful discussion in the context of space policy as well.
Most discussion in Congress is about second and third order items,
when in fact the problem is at the level of first and zeroth order, or
as Tom Rogers says, "page 1."
--
simberg.interglobal.org * 310 372-7963 (CA) 307 739-1296 (Jackson Hole)
interglobal space lines * 307 733-1715 (Fax) http://www.interglobal.org
"Extraordinary launch vehicles require extraordinary markets..."
Swap the first . and @ and throw out the ".trash" to email me.
Here's my email address for autospammers: postm...@fbi.gov
Edward Green
> On Mon, 7 Jul 2003 19:24:20 GMT, in a place far, far away,
> he...@spsystems.net (Henry Spencer) made the phosphor on my monitor
> glow in such a way as to indicate that:
>
> >In article <bec7n8$3a6kb$4...@ID-45790.news.dfncis.de>,
> >Laurel Amberdine <cir...@mtco.com> wrote:
> >>Ignorance time: people are saying "to first (second, zeroth) order" etc,
> >>quite often lately. What does it mean, anyway?
> >
> >While those terms have precise meanings, they are often applied somewhat
> >more loosely and generally. Speaking loosely and generally...
> >
> >"To first order" means "considering only the biggest and most obvious
> >effects, neglecting details which don't change things very much".
> >
> >"To second order" means "including the most significant of the details,
> >but neglecting really small ones which have rather smaller effects".
> >
> >And so forth.
> >
> >And by analogy, "to zeroth order" means "making drastic simplifications to
> >get an answer which will be somewhere in the right ballpark".
>
> This is a useful discussion in the context of space policy as well.
> Most discussion in Congress is about second and third order items,
> when in fact the problem is at the level of first and zeroth order, or
> as Tom Rogers says, "page 1."
Yes ... and that doesn't seem to apply exclusively to space policy.
Most involved discussions, particularly those involving experts,
probably tend to devolve to 2nd and 3rd order effects, while the 1st
and, especially, the zero order effects and assumptions remain
unexamined. I suppose the contrary idea gives us "zero basing".
Of course to make any progress, one eventually must settle on some
assumptions at each order ... but not, one hopes, unexamined and
immutable assumptions, which seems to be the norm.