If
Dr ***
how would we measure the time dilation and length contraction that we were
being subject to ?
--
Dr *** Vacuum Energy zero state = 4/3 pi duration + length ^3 perhaps
http://home.freeuk.com/paulps/
Maybe updates. The spuds, beans and onions are coming up nicely. Ooh
ah.{:-)
Spoonfed
this regardless of what reference frame you watched it from, the center
would be stationary--only in some, the milky way galaxy would be in the
center, uncontracted, and aging at full speed, and in others, the milky
way galaxy would be at the edge, contracted, and aging slowly.
*** rD
"Spoonfed" <jonatha...@spoonfedrelativity.com> wrote in message
news:1117321308.9...@f14g2000cwb.googlegroups.com...
Interesting, but missed the point of how an observer in a velocity
contracted frame might measure their contraction.
--
*** rD
Spoonfed
It is only moving objects which appear contracted.
Consider this.
A giant stick is flying past you, marked off with distances.
The distances on that stick are now length contracted.
You see two objects attached to the stick--one moving away and one
moving toward you.
Due to the doppler effect or simple calculation of where the object is
compared to where the image must appear, the object moving toward you
appears to be moving superluminally. It's actual "current" position of
the object is MUCH closer than the image.
On the other hand, the object moving away can be no further away than
twice as far as the image.
Suddenly you accelerate, matching pace with the stick. You are
accelerating TOWARD the object that is moving away.
The stick is no longer length contracted.
The distances to both objects has suddenly increased.
You will see the same light both before and after the acceleration.
However, the wave front has changed shape. The sphere of light coming
from the object you've moved toward has GROWN (making it appear to be
further away which it actually is.), while the sphere of light from the
object you've accelerated away from has SHRUNK, making it appear to be
closer to you (which it actually isn't).
Some clue of this shrinking/growing effect is shown in
http://www.spoonfedrelativity.com/files/newYears2.swf
http://www.spoonfedrelativity.com/files/acceleration4.swf
and
http://www.spoonfedrelativity.com/files/ninephotons.swf
Anyway, I think I've got off topic here, as usual, but the main point
is that you should always feel that your own frame is uncontracted.
*** rD
Thank you but I don't subscribe to the view that simple observational
limitations and mathematical perspective are of any interest but academic
and I am more interested in the *actual* contraction and dilation of matter
under acceleration, deceleration, velocity near c and gravitational
variation.
--
*** rD Vacuum Energy = 4 pi duration+length ^3 perhaps
Spoonfed
All right. There is no actual contraction and dilation of matter at
velocity near c. There is actual contraction for accelerated bodies,
which is very clear from watching people's faces under hi G. Also
there is an actual time dilation of an object in a gravitational field.
The clock on top goes faster than the clock on the bottom, and on an
accelerated body, the clock in front goes faster than the clock in back.
Spoonfed
matter
> under acceleration, deceleration, velocity near c and gravitational
> variation.
Oh and deceleration is the same as acceleration.
See:
http://www.spoonfedrelativity.com/files/myGravity-contractionvelocity.swf
*** rD
Not so sure of this last explain why the clock in front going at the same
velocity as the one in front should show a different rate ?
the clock in front goes faster than the clock in back.
|
--
Spoonfed
> | All right. There is no actual contraction and dilation of matter at
> | velocity near c. There is actual contraction for accelerated bodies,
> | which is very clear from watching people's faces under hi G. Also
> | there is an actual time dilation of an object in a gravitational field.
> | The clock on top goes faster than the clock on the bottom, and on an
> | accelerated body,
>
> Not so sure of this last explain why the clock in front going at the same
> velocity as the one in front should show a different rate ?
>
> the clock in front goes faster than the clock in back.
> |
>
> --
> *** rD Vacuum Energy = 4 pi duration+length ^3 perhaps
> http://home.freeuk.com/paulps/
> Maybe updates. The spuds, beans and onions are coming up nicely. Ooh
> ah.{:-)
If a solid object is accelerated, the internal forces (chemical
bonding, nuclear bonding, etc.) will cause the object to maintain its
shape, as long as the acceleration forces do not overpower the internal
forces.
If it maintains it's shape in its own reference frame, then as it
accelerates it will appear to contract (or as it decelerates, it will
appear to uncontract) in any inertial frame. This contraction and
decontraction occurs over a period of time so it has its own velocity
which can be added to or subtracted from the velocity of whatever
portion of the object is receiving the force of acceleration.
Thus the front and back of the object are moving at different
velocities, and thus the clocks are going at different speeds.
Only an accelerated observer on the object would say that the front and
back were moving at the same velocity. They would not observe the
change in length contraction over time, but they would see that the
upper clock was going faster than the lower clock.
*** rD
|
| > | All right. There is no actual contraction and dilation of matter at
| > | velocity near c.
I did not pick this up on the previous reads but are you sure and for what
reasons ?
There is actual contraction for accelerated bodies,
| > | which is very clear from watching people's faces under hi G. Also
| > | there is an actual time dilation of an object in a gravitational
field.
| > | The clock on top goes faster than the clock on the bottom, and on an
| > | accelerated body,
| >
| > Not so sure of this last explain why the clock in front going at the
same
| > velocity as the one in front should show a different rate ?
| >
| > the clock in front goes faster than the clock in back.
| > |
Due to compression\contraction\velocity state at rear being higher than at
front due to finite propagation of thrust\velocity through the rod ?
| > --
| > *** rD Vacuum Energy = 4 pi duration+length ^3 perhaps
| > http://home.freeuk.com/paulps/
| > Maybe updates. The spuds, beans and onions are coming up nicely. Ooh
| > ah.{:-)
|
| If a solid object is accelerated, the internal forces (chemical
| bonding, nuclear bonding, etc.) will cause the object to maintain its
| shape, as long as the acceleration forces do not overpower the internal
| forces.
I disagree with that as it assumes an absolute rigid structure. Their is
always some effect as no structure is absolutely rigid. So a structure is
*always* modified by acceleration or deceleration.?
|
| If it maintains it's shape in its own reference frame, then as it
| accelerates it will appear to contract (or as it decelerates, it will
| appear to uncontract) in any inertial frame.
This contraction and
| decontraction occurs over a period of time so it has its own velocity
| which can be added to or subtracted from the velocity of whatever
| portion of the object is receiving the force of acceleration.
Yes, the thrust\velocity effect can propagate through the object upto c. ?
|
| Thus the front and back of the object are moving at different
| velocities, and thus the clocks are going at different speeds.
|
Good explanation, I hadn't thought of that, as the acceleration takes a
finite time to arrive at the front the back is moving faster than the front
under acceleration assuming the acceleration is being applied at the back.
Nice one {:-). This would also apply to deceleration in that the front would
be moving slower than the back assuming the deceleration was being applied
at the front. This would apply only during the period it took for the rod to
stabilise in its contracted\decontracted condition due to constant
acceleration or deceleration. To make the change in this effect occur over a
longer period the acceleration rate would also have to be accelerating or
v.v..
| Only an accelerated observer on the object would say that the front and
| back were moving at the same velocity. They would not observe the
| change in length contraction over time,
Interesting question that raises, could you observer the contraction of a
rod if you were sitting in the middle of it.
How about mirrors at front and back you shining coherent light to the front
and back. This rod is inside a spaceship just to clarify the local
environment. You would notice a change in phase between the back and front
as you accelerated from zero but this change in phase shift would stabilise
as the rods contraction became stabilised unless your rate of acceleration
was increasing. So the rod would be shorter from the middle to the back than
from the middle to the front under continuos constant acceleration ? and the
observer in the middle should be able to measure it as an actual fact. ? Its
even more interesting under continues accelerating acceleration.
*** rD
Very entertaining but I was a bit confused as I was thinking on the lines of
spaceships and only when I came back and looked at the address did I see
that its was about gravity. Are you trying to represent contraction due to
gravity being less the nearer you get to the source ?
Spoonfed
> > > Thank you but I don't subscribe to the view that simple observational
> > > limitations and mathematical perspective are of any interest but academic
> > > and I am more interested in the *actual* contraction and dilation of matter
> > > under acceleration, deceleration, velocity near c and gravitational
> > > variation.
====================
> "Spoonfed" <jonatha...@spoonfedrelativity.com> wrote in message
> news:1117602139.7...@f14g2000cwb.googlegroups.com...
> |
> | > | All right. There is no actual contraction and dilation of matter at
> | > | velocity near c.
>
> I did not pick this up on the previous reads but are you sure and for what
> reasons ?
>
>
I think I may have misunderstood what you meant by "actual."
By actual length contraction, you meant something like "What fits in a
box simultaneously in the observer's frame" and I took it to mean "What
feels uncomfortable and squished to the object itself."
When you said you were not interested in observational limitations, I
attempted to get rid of all observational limitations--this was silly.
Of course I can't remove all observational limitations from the
question, because measurement *requires* an observer. My error was in
putting the observer ON the object moving near the speed of light and
thinking that I had eliminated the "observational limitations" when in
fact, I had simply made an implicit definition of your term "actual."
The observational questions you were wanting me to avoid were
superluminal motion and other doppler related effects.
This is the sort of thing where, though the moving object CAN fit in a
small box in our reference frame, we don't observe it to have fit in
the box because doppler effects cause the object to appear greatly
elongated.
We can nearly eliminate this from our observations by thinking of our
observer as a camera, and restricting our field of view to a small
region as shown here:
http://www.spoonfedrelativity.com/files/ninephotons.swf (Start the
demo and click "show filmed region")
In this case, the observer (camera) will see the actual length
contraction, and I am changing my meaning of actual leangth here, to
mean what the observer can fit in a box.
> There is actual contraction for accelerated bodies,
> | > | which is very clear from watching people's faces under hi G. Also
> | > | there is an actual time dilation of an object in a gravitational
> field.
> | > | The clock on top goes faster than the clock on the bottom, and on an
> | > | accelerated body,
> | >
> | > Not so sure of this last explain why the clock in front going at the
> same
> | > velocity as the one in front should show a different rate ?
> | >
> | > the clock in front goes faster than the clock in back.
> | > |
>
> Due to compression\contraction\velocity state at rear being higher than at
> front due to finite propagation of thrust\velocity through the rod ?
>
Since the length is changes as the rod accelerates, the front and back
cannot be going at the same speed. I think you might be able to define
the length contraction at the front and the back of the rod and perhaps
they are different, but we are only concerned with the length
contraction between the two clocks--it would be a good approximation to
use the velocity of the center of the rod...
> | > --
> | > *** rD Vacuum Energy = 4 pi duration+length ^3 perhaps
> | > http://home.freeuk.com/paulps/
> | > Maybe updates. The spuds, beans and onions are coming up nicely. Ooh
> | > ah.{:-)
> |
> | If a solid object is accelerated, the internal forces (chemical
> | bonding, nuclear bonding, etc.) will cause the object to maintain its
> | shape, as long as the acceleration forces do not overpower the internal
> | forces.
>
> I disagree with that as it assumes an absolute rigid structure. Their is
> always some effect as no structure is absolutely rigid. So a structure is
> *always* modified by acceleration or deceleration.?
>
All I meant to assume was that the structure does not break because of
the acceleration. For instance, if someone tried to push a
space-shuttle with a toothpick the structures would probably break.
However, if you push softly enough, you WOULD be able to accelerate the
space shuttle by pushing it with a toothpick.
> |
> | If it maintains it's shape in its own reference frame, then as it
> | accelerates it will appear to contract (or as it decelerates, it will
> | appear to uncontract) in any inertial frame.
>
> This contraction and
> | decontraction occurs over a period of time so it has its own velocity
> | which can be added to or subtracted from the velocity of whatever
> | portion of the object is receiving the force of acceleration.
>
> Yes, the thrust\velocity effect can propagate through the object upto c. ?
Up to c, yes. But that does not mean that we have to stop
accelerating. It means we should switch to thinking of rapidity
v/sqrt(1-(v/c)^2) (which is proportional to momentum, p) instead of
velocity, which is proportional to p/(p^2+1). The rapidity of an
object can continue to increase towards infinity, though the velocity
is limited to the speed of light.
>
> |
> | Thus the front and back of the object are moving at different
> | velocities, and thus the clocks are going at different speeds.
> |
>
> Good explanation, I hadn't thought of that, as the acceleration takes a
> finite time to arrive at the front the back is moving faster than the front
> under acceleration assuming the acceleration is being applied at the back.
> Nice one {:-). This would also apply to deceleration in that the front would
> be moving slower than the back assuming the deceleration was being applied
> at the front.
Originally, I thought as you did that the bottom and top of the
structure should be moving at the same speed. So when I programmed the
simulation I just used the velocity of the bottom of the structure.
Then it produced this demonstration...
http://www.spoonfedrelativity.com/files/myGravity-p-change-constant.swf
Worse, I had it programmed so that the dots would disappear if they got
above the top or below the bottom of the platform--so they simply
disappeared instead of rising above the platform top--It took me a
while to figure out what was going on, that I needed to include the
change in length as part of the velocity of the top of the platform!
> This would apply only during the period it took for the rod to
> stabilise in its contracted\decontracted condition due to constant
> acceleration or deceleration. To make the change in this effect occur over a
> longer period the acceleration rate would also have to be accelerating or
> v.v..
>
As long as it keeps accelerating there is always a force coming up from
the bottom of the structure to the top... as long as there are more
phonons coming up the structure, the top and bottome will always be
moving at different rates. It doesn't require a change in
acceleration.
>
> | Only an accelerated observer on the object would say that the front and
> | back were moving at the same velocity. They would not observe the
> | change in length contraction over time,
>
> Interesting question that raises, could you observer the contraction of a
> rod if you were sitting in the middle of it.
That is my eventual goal to predict what sort of contraction would be
seen by such an observer. The demo I've made is as yet incomplete,
because the events happen at a constant rate according to the top or
bottom of the structure, but that constant rate should be affected by
time dilation when observed from outside. Also, we could set up two
versions of the problem, one where (A) the force of acceleration came
from a rocket on the platform, and one where (B) the force comes from
the frame of reference of the outside observer. I've been running into
trouble assuming (B), but perhaps assuming (A) will give me a usable
solution. (This might be exactly what I needed for my writer's block
:)
> How about mirrors at front and back you shining coherent light to the front
> and back. This rod is inside a spaceship just to clarify the local
> environment. You would notice a change in phase between the back and front
> as you accelerated from zero but this change in phase shift would stabilise
> as the rods contraction became stabilised unless your rate of acceleration
> was increasing.
I think of a constant acceleration being a series of pulses. For
instance a rocket appears to be a continuous stream of gas, but is
actually individual explosions of very tiny atoms--each explosion can
be considered to be an acceleration event, which sends energy upward in
a teeny compression wave called (if I might steal a term from solid
state phyisics and bastardize it for my own uses) a phonon, which moves
toward the top of the ship.
So it is not the change in acceleration which produces the difference
in velocity between the top and bottom of the structure, but the
production of a phonon which has accelerated the lower structure but
has not yet accelerated the upper structure. As long as these are
flowing, the bottom of the structure will be moving at a different
speed from the top.
> So the rod would be shorter from the middle to the back than
> from the middle to the front under continuos constant acceleration ? and the
> observer in the middle should be able to measure it as an actual fact. ? Its
> even more interesting under continues accelerating acceleration.
>
>
> but they would see that the
> | upper clock was going faster than the lower clock.
> |
You have question marks and I also have question marks. It is clear
from the demo as-is that the wavelength of the light gets shorter
toward the bottom, and the distance between falling objects becomes
longer toward the bottom. Of course, a person standing on the platform
would agree and outside observer (traffic cop) would also agree.
Also, any uniform distortion in the structure would also affect any
physical devices used to measure the structure (assuming that the
structure and measuring devices had the same tensile strength.)
The bottom has sped up toward the center since the light left. The top
has sped up away from the center since the light left, so the bottom is
closer than it appears, and the top is further than it appears. But if
the engines are at the bottom, then more phonons have passed through
the bottom than the top. For me, right now, there seem too many
variables to get my mind around. I'll try to focus on what I need for
the next demo.
Spoonfed
Baby steps... I need to understand gravity eventually. But I realized
that I could not approach gravity through Gedanken in the same way I
could approach acceleration. Also, I can't get acceleration to look
like gravity, because gravity comes from spheres and points, causes
tides and does not affect relative rapidities of far-away objects.
Acceleration, on the other hand, comes from forces, causes no tides,
and causes faraway objects to change their relative rapidity just as
much as nearby objects.
I could approximate gravity with acceleration, but I can create a
purely hypothetical environment where approximating acceleration with
gravity. By setting up the mass distribution properly, we can get
gravity to look exactly like acceleration. We can set up a
hypothetical infinite plane mass distribution which would result in
constant gravity toward the plane for infinite distance.
My eventual goal was to show the equivalence principle in this special
case: I think I've found (in the last half hour!) a fundamental
difference, even in this case, though: This is the Assumption (A
Constant acceleration of a space ship will exhibit a constant
acceleration according to the person on the space-ship.) vs. assumption
(B Acceleration due to constant gravity would exhibit a constant change
in rapidity according to a person on the ground) question again.
Oh yeah, this is good stuff...
*** rD
| *** rD wrote:
|
| > > > Thank you but I don't subscribe to the view that simple
observational
| > > > limitations and mathematical perspective are of any interest but
academic
| > > > and I am more interested in the *actual* contraction and dilation of
matter
| > > > under acceleration, deceleration, velocity near c and gravitational
| > > > variation.
|
| ====================
| > "Spoonfed" <jonatha...@spoonfedrelativity.com> wrote in message
| > news:1117602139.7...@f14g2000cwb.googlegroups.com...
| > |
| > | > | All right. There is no actual contraction and dilation of matter
at
| > | > | velocity near c.
| >
| > I did not pick this up on the previous reads but are you sure and for
what
| > reasons ?
| >
| >
|
| I think I may have misunderstood what you meant by "actual."
|
| By actual length contraction, you meant something like "What fits in a
| box simultaneously in the observer's frame" and I took it to mean "What
| feels uncomfortable and squished to the object itself."
My definition of actual would be something that an experimenters who were as
close as practical to an event under observation would measure as best they
could taking into consideration any effects that they should be aware of.
I'm a bit uncertain of frames as they tend to go round made up pictures in
my mind, mainly because they usually fail to adequately define the
environment they are defining and boxes can change in size due to humidity
{:-)
snip
|
| The observational questions you were wanting me to avoid were
| superluminal motion and other doppler related effects.
|
Not at all it was just heavy SR explanations that failed to explain {:-)
| This is the sort of thing where, though the moving object CAN fit in a
| small box in our reference frame, we don't observe it to have fit in
| the box because doppler effects cause the object to appear greatly
| elongated.
Doppler or observational transforms. Yes that that's the type I see as
academic
|
| We can nearly eliminate this from our observations by thinking of our
| observer as a camera, and restricting our field of view to a small
| region as shown here:
|
| http://www.spoonfedrelativity.com/files/ninephotons.swf (Start the
| demo and click "show filmed region")
|
| In this case, the observer (camera) will see the actual length
| contraction, and I am changing my meaning of actual leangth here, to
| mean what the observer can fit in a box.
Carefull with the boxes as they can be contracted as well.
|
| > There is actual contraction for accelerated bodies,
| > | > | which is very clear from watching people's faces under hi G. Also
| > | > | there is an actual time dilation of an object in a gravitational
| > field.
| > | > | The clock on top goes faster than the clock on the bottom, and on
an
| > | > | accelerated body,
| > | >
| > | > Not so sure of this last explain why the clock in front going at the
| > same
| > | > velocity as the one in front should show a different rate ?
| > | >
| > | > the clock in front goes faster than the clock in back.
| > | > |
| >
| > Due to compression\contraction\velocity state at rear being higher than
at
| > front due to finite propagation of thrust\velocity through the rod ?
| >
|
| Since the length is changes as the rod accelerates, the front and back
| cannot be going at the same speed.
Initials yes, but as the rod accommodates its length to the force of
acceleration and if this is constant the force at the front is the same as
at the rear although it may not be the same quantum of force as the force is
being propagated up the rod at up to c.
|I think you might be able to define
| the length contraction at the front and the back of the rod and perhaps
| they are different,
Only initially.
|but we are only concerned with the length
| contraction between the two clocks--
Why.
it would be a good approximation to
| use the velocity of the center of the rod...
After the initial contraction period the velocity at the front is the same
as the rear.
Hmm... I have reservations about this but need to think about it
| >
| > |
| > | Thus the front and back of the object are moving at different
| > | velocities, and thus the clocks are going at different speeds.
| > |
No I can only agree with that initially. The rod may be vibrating along its
length due to oscillations in the thrust input but this evens out the
velocity as far as the front and the rear of the rod are concerned. You can
only say that the rods velocity increases at the rear first.
| >
| > Good explanation, I hadn't thought of that, as the acceleration takes a
| > finite time to arrive at the front the back is moving faster than the
front
| > under acceleration assuming the acceleration is being applied at the
back.
| > Nice one {:-). This would also apply to deceleration in that the front
would
| > be moving slower than the back assuming the deceleration was being
applied
| > at the front.
|
| Originally, I thought as you did that the bottom and top of the
| structure should be moving at the same speed. So when I programmed the
| simulation I just used the velocity of the bottom of the structure.
| Then it produced this demonstration...
|
| http://www.spoonfedrelativity.com/files/myGravity-p-change-constant.swf
Haven't had a look yet but will come back to it latter perhaps.
|
| Worse, I had it programmed so that the dots would disappear if they got
| above the top or below the bottom of the platform--so they simply
| disappeared instead of rising above the platform top--It took me a
| while to figure out what was going on, that I needed to include the
| change in length as part of the velocity of the top of the platform!
|
| > This would apply only during the period it took for the rod to
| > stabilise in its contracted\decontracted condition due to constant
| > acceleration or deceleration. To make the change in this effect occur
over a
| > longer period the acceleration rate would also have to be accelerating
or
| > v.v..
| >
|
| As long as it keeps accelerating there is always a force coming up from
| the bottom of the structure to the top... as long as there are more
| phonons coming up the structure, the top and bottome will always be
| moving at different rates. It doesn't require a change in
| acceleration.
|
Disagree as long as the phonons are of the same strength and constant then
the velocity at the front of the rod will be, on average the same.
No, if the flow is constant the velocity at front will be the same as at the
back, on average.
|
| > So the rod would be shorter from the middle to the back than
| > from the middle to the front under continuos constant acceleration ? and
the
| > observer in the middle should be able to measure it as an actual fact. ?
Its
| > even more interesting under continues accelerating acceleration.
| >
| >
| > but they would see that the
| > | upper clock was going faster than the lower clock.
| > |
|
| You have question marks and I also have question marks. It is clear
| from the demo as-is that the wavelength of the light gets shorter
| toward the bottom, and the distance between falling objects becomes
| longer toward the bottom. Of course, a person standing on the platform
| would agree and outside observer (traffic cop) would also agree.
|
| Also, any uniform distortion in the structure would also affect any
| physical devices used to measure the structure (assuming that the
| structure and measuring devices had the same tensile strength.)
|
| The bottom has sped up toward the center since the light left. The top
| has sped up away from the center since the light left, so the bottom is
| closer than it appears, and the top is further than it appears. But if
| the engines are at the bottom, then more phonons have passed through
| the bottom than the top. For me, right now, there seem too many
| variables to get my mind around. I'll try to focus on what I need for
| the next demo.
|
--
rD *** E-field = Electric field, M-field =Magnetic field, two unbound
field effects
*** rD
|
|
| *** rD wrote:
| > "Spoonfed" <jonatha...@spoonfedrelativity.com> wrote in message
| > news:1117560437.6...@f14g2000cwb.googlegroups.com...
| > | I am more interested in the *actual* contraction and dilation of
| > | matter
| > | > under acceleration, deceleration, velocity near c and gravitational
| > | > variation.
| > |
| > |
| > | Oh and deceleration is the same as acceleration.
| > |
| > | See:
| > |
http://www.spoonfedrelativity.com/files/myGravity-contractionvelocity.swf
| > |
| >
| > Very entertaining but I was a bit confused as I was thinking on the
lines of
| > spaceships and only when I came back and looked at the address did I see
| > that its was about gravity. Are you trying to represent contraction due
to
| > gravity being less the nearer you get to the source ?
| > --
snip
|
| Baby steps... I need to understand gravity eventually. But I realized
| that I could not approach gravity through Gedanken in the same way I
| could approach acceleration. Also, I can't get acceleration to look
| like gravity, because gravity comes from spheres and points, causes
| tides and does not affect relative rapidities of far-away objects.
|
If its any help or hindrance I see it as a potential difference
gradient of 0.000030786 volts per meter but it may need rescaling and I have
yet to prove it.
| Acceleration, on the other hand, comes from forces, causes no tides,
| and causes faraway objects to change their relative rapidity just as
| much as nearby objects.
1D ver. 3D effects.
|
| I could approximate gravity with acceleration, but I can create a
| purely hypothetical environment where approximating acceleration with
| gravity. By setting up the mass distribution properly, we can get
| gravity to look exactly like acceleration. We can set up a
| hypothetical infinite plane mass distribution which would result in
| constant gravity toward the plane for infinite distance.
|
| My eventual goal was to show the equivalence principle in this special
| case: I think I've found (in the last half hour!) a fundamental
| difference, even in this case, though: This is the Assumption (A
| Constant acceleration of a space ship will exhibit a constant
| acceleration according to the person on the space-ship.) vs. assumption
| (B Acceleration due to constant gravity would exhibit a constant change
| in rapidity according to a person on the ground) question again.
| Oh yeah, this is good stuff...
|
I'm not sure I understand that.
--
rD *** E-field = Electric field, M-field =Magnetic field, two unbound
field effects
Spoonfed
in one case.
========Electricity?=================
If you are using volts, you need to know both the amount of mass of
each particle and the amount of charge on each particle. Are you using
a different non-electrical definition for volts?
============Acceleration in natural units===========
The acceleration due to gravity is (9.8 m/s^2)/(3*10^8 m/ls)=
32.67*10^-9 light seconds/s^2.
This is 32.67 nil / second^2 (notice the similarity with 32
feet/second^2), or 32.67*10^-18 nil / nanosecond^2.
=========Natural units for dv/dx========
If the particle starts from rest, and drops one meter at constant
acceleration, how fast is it going at the end?
For a Newtonian approximation, use 1/2 * a * t^2=x; v=a t
a = 9.8, x=1 ---> t=.451 --> v=4.419 m/s
This would mean that the particle accelerates 4.419 m/s per meter,
which has the painfully inadequate units of 4.419 Hz because reaching 1
meter per second has little significance.
When units disappear like this, it probably means theres a missing
connection, so lets change the problem--If the particle starts from
rest, and drops one nil at constant acceleration, how fast is it going
at the end, in units of nils/nanosecond^2?
a=32.67*10^-18 nils / nanosecond^2, x=1 nils --> t=2.47*10^8 ns
v=a t = 8.08*10^-9 nils/nanosecond = 8.08*10^-9 c.
Note c = 1 nils/nanosecond = the speed of light. Now if we consider
that the particle accelerates at 8.08*10^-9 nils/ns every nils, this
has the units of 8.08*10^-9 Hz which actually makes a good deal more
sense in this context, because reaching 1 nils per nanosecond has great
significance.
=========Calculating v[t], x[t], and t'[t] for an accelerated
object========
For the relativistic answer we have to ask that the question be
rephrased to determine between the two options.
Case 1: dp/dt is constant from the perspective of an inertial
observer.
Case 2: dp/dt is constant from the perspective of the accelerated
observer.
In either case, the quantity dv/dt will not be constant, it is the
change in momentum, p, that should be the quantity of interest because
of the impulse-momentum law.
In order to simplify things so that we can use whatever units we want,
while still maintaining a quantity calibrated to the speed of light, we
define beta, B, in terms of velocity, v and in terms of momentum, p.
Let B=v/c,
p=m B c/sqrt(1-B^2)
B=(p/mc)/sqrt([p/mc]^2+1)
For Case 1, we can simply use a constant force dp/dt= (m*c) 8.08*10^-9.
Use the chain rule to find dB/dt = (dB/dp)*(dB/dt)
Then integrate dB/dt. When I solved this(with an initial condition of
B[t=0]=0, I got B as a function of t:
---> B[t]=(at)/sqrt(1+[at]^2), where
a=1/mc*(dp/dt) = 8.08*10^-9 in this case = the time rate of change in
rapidity.
To find the position at a given time, simply integrate
x[t]=Integrate[c*B[t],t]
x[t]=(c/a) sqrt(1+[a t]^2)
This looks like a nice result. We might find later that it isn't.
To find the amount of time that passes for the object, we need to find
t'[t]
t' goes by at a rate of t*sqrt(1-B^2), which is simply the time
dilation factor times the time passed. But in our case, the time
dilation factor is changing, so we must integrate:
t'[t]=Integrate[t*sqrt(1-B[t]^2]
where B[t]=(at)/sqrt(1+[at]^2) as above.
This integral yields, though I never would have got it without
Mathematica,
t'[t]=EllipticE[ArcSin[a t], 2]/a
Which only exists as long as t is between -Pi/(4a) and Pi/(4a)
I find this to be a rather disturbing feature of the solution. If time
simply goes away periodically for accelerated particles. This might be
related in some way to the mathematics of event horizons, but it should
not actually apply to rocketships, as we will see below.
Case II: I'll approach this problem at some later time.
>
> | Acceleration, on the other hand, comes from forces, causes no tides,
> | and causes faraway objects to change their relative rapidity just as
> | much as nearby objects.
>
> 1D ver. 3D effects.
>
That is oversimplifying somewhat. A 1-D effect can be constant over a
distance or it can vary over the distance.
> |
> | I could approximate gravity with acceleration, but I can create a
> | purely hypothetical environment where approximating acceleration with
> | gravity. By setting up the mass distribution properly, we can get
> | gravity to look exactly like acceleration. We can set up a
> | hypothetical infinite plane mass distribution which would result in
> | constant gravity toward the plane for infinite distance.
> |
> | My eventual goal was to show the equivalence principle in this special
> | case: I think I've found (in the last half hour!) a fundamental
> | difference, even in this case, though: This is the Assumption (A
> | Constant acceleration of a space ship will exhibit a constant
> | acceleration according to the person on the space-ship.) vs. assumption
> | (B Acceleration due to constant gravity would exhibit a constant change
> | in rapidity according to a person on the ground) question again.
> | Oh yeah, this is good stuff...
> |
>
> I'm not sure I understand that.
> --
> rD ***
Case 1 (B): dp/dt is constant from the perspective of an inertial
observer.
Case 2 (A): dp/dt is constant from the perspective of the accelerated
observer.
In Case 2, the reason for the acceleration is a rocket on the
accelerating object. Though the observers on the rocket see and feel
the rocket expelling fuel causing them to accelerate at a constant rate
(if the rocket is designed that way) the observers in the inertial
(lab) frame see the time-dilating effects so the rocket expels a lesser
amount of fuel per unit time when it is going fast than when it is
going slow. This means that dp/dt is not constant in the lab frame.
In Case 1, the force appears constant in the lab frame. This means
that the force perceived by the accelerating object is not constant,
but growing towards infinity. This may be the reason that time
disappears in the equation for t'[t] above.
Jonathan Doolin
www.spoonfedrelativity.com
Dirk Van de moortel
"Spoonfed" <jonatha...@spoonfedrelativity.com> wrote in message news:1117899458....@z14g2000cwz.googlegroups.com...
> I decided to crosspost this since I give a fairly rigorous derivation
> in one case.
[snip]
[removed sci.physics.research - replied separately on s.p.r.
Indeed it isn't. The dimensions don't match.
Your LHS is distance, RHS is time.
See http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
I use notations:
v(t) in stead of B[t]
x(t) in stead of x[t]
T in stead of t'
I get
x(t) = 1/a ( sqrt(1+a^2 t^2)) - 1 )
or, after reintroducing c:
x(t) = c^2/a ( sqrt( 1 + (a t/c)^2 ) -1 )
>
> To find the amount of time that passes for the object, we need to find
> t'[t]
>
> t' goes by at a rate of t*sqrt(1-B^2), which is simply the time
> dilation factor times the time passed. But in our case, the time
> dilation factor is changing, so we must integrate:
>
> t'[t]=Integrate[t*sqrt(1-B[t]^2]
> where B[t]=(at)/sqrt(1+[at]^2) as above.
>
> This integral yields, though I never would have got it without
> Mathematica,
>
> t'[t]=EllipticE[ArcSin[a t], 2]/a
>
> Which only exists as long as t is between -Pi/(4a) and Pi/(4a)
I think you made a mistake there.
The integrand is t / sqrt( 1 + a^2 t^2 ).
This very easily integrates to 1/a^2 sqrt( 1 + a^2 t^2 )
I think you would have found it without Mathematica ;-)
Anyway, that result is not the standard result either:
See
http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
where you find
t'(t) = 1/a argsinh(a t)
Dirk Vdm
Spoonfed
Ooops, I left out my mathematical definition of a. I apologize because
I used a to emphasize the similarity to acceleration, but I should be
emphasizing the difference. a has units of Hz.
a = F/mc = d/dt (B/sqrt(1-B^2))
F has units kg m/s^2
m has units kg
c has units m/s
B is unitless, expressing velocity as proportion of the speed of light.
a has units 1/s
> See http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
> I use notations:
> v(t) in stead of B[t]
> x(t) in stead of x[t]
> T in stead of t'
> I get
> x(t) = 1/a ( sqrt(1+a^2 t^2)) - 1 )
> or, after reintroducing c:
> x(t) = c^2/a ( sqrt( 1 + (a t/c)^2 ) -1 )
>
> >
> > To find the amount of time that passes for the object, we need to find
> > t'[t]
> >
> > t' goes by at a rate of t*sqrt(1-B^2), which is simply the time
> > dilation factor times the time passed. But in our case, the time
> > dilation factor is changing, so we must integrate:
> >
> > t'[t]=Integrate[t*sqrt(1-B[t]^2]
> > where B[t]=(at)/sqrt(1+[at]^2) as above.
> >
> > This integral yields, though I never would have got it without
> > Mathematica,
> >
> > t'[t]=EllipticE[ArcSin[a t], 2]/a
> >
> > Which only exists as long as t is between -Pi/(4a) and Pi/(4a)
>
> I think you made a mistake there.
Oh, yes, I did make a mistake. I took the integral of sqrt(1-B[t]^2)
and instead of t*sqrt(1-B[t]^2)
> The integrand is t / sqrt( 1 + a^2 t^2 ).
> This very easily integrates to 1/a^2 sqrt( 1 + a^2 t^2 )
> I think you would have found it without Mathematica ;-)
Given an hour or two. I'm in very bad practice.
>
> Anyway, that result is not the standard result either:
Oh, good. That result goes backward and forward in time. I can stand
a distant object going backward and forward in time for an accelerated
observer, but I can't stand an accelerated object going back and forth
in time for an inertial observer.
> See
> http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
> where you find
> t'(t) = 1/a argsinh(a t)
>
> Dirk Vdm
I'll have a look at that. I'm printing it out so I can get away from
my computer.
Thanks
Jonathan
Spoonfed
Spoonfed wrote:
> > See
> > http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
> > where you find
> > t'(t) = 1/a argsinh(a t)
> >
> > Dirk Vdm
>
> I'll have a look at that. I'm printing it out so I can get away from
> my computer.
>
> Thanks
> Jonathan
Very good, you define a[T] as the acceleration of the rocket in its own
reference frame at its own time T. I was defining a with a troublesome
mathematical definition d(Gamma Beta)/dt.
By assuming the rocket's acceleration was constant in its own frame,
using your definitions, we would have
T[t]=argsin[a t] / a
x[t]=x[T[t]]
=Cosh[argsin[a T]] / a
=sqrt[1+(a t)^2]/a
By assuming the force on the rocket was constant in the lab frame,
using my definitions, I got
x[t]=c * sqrt[1+(a t)^2] / a
I'm surprised at the similarity of these solutions, given the
difference in definition of a.
I suppose I shouldn't be--there is this so-called Equivalence
Principle, after all.
But it helps me see the validity of the Equivalence Principle, assuming
I've done my math right--always a big if... ;)
Anyway, thank you for checking, Dirk. I was stuck and you've given me
a nice push.
*** rD
In one sense yes in another no sorry if that's a bit wishy washy but I'm
trying to relate velocity to volts as a potential gradient between two
dielectric motion states and amps to energy as the width of the slope. So
getting time\distance\velocity\acceleration\energy\volts\amps\gravity\mass
in one package. Please think of me as daft if you need to as I'm fairly sure
most others do {:-).
|
| ============Acceleration in natural units===========
| The acceleration due to gravity is (9.8 m/s^2)/(3*10^8 m/ls)=
| 32.67*10^-9 light seconds/s^2.
Forgive me for my slowness but the second and meter used to define 9.8
m/s^2 are themselves a derivation of c which your then dividing by ? I am
currently studying AE's EoMB (SR)? papers to try and find the justification
for relativistic velocity correction from an energy pov.
|
| This is 32.67 nil / second^2 (notice the similarity with 32
| feet/second^2), or 32.67*10^-18 nil / nanosecond^2.
Yes {:-)
|
| =========Natural units for dv/dx========
| If the particle starts from rest, and drops one meter at constant
| acceleration, how fast is it going at the end?
|
As far as gravity goes its the slope thats of main interest not the terminal
velocity. I see the slope of 9.8 m/s^2 or possibly 0.000030786 volts per
meter as the rate at which the energy from one system is transfered to
another. That may not be how its thought of conventianly as its from the
smaller mass to the larger as the final product is an increase in
mass\energy\voltage\potential to the larger. Increase the mass or proximity
to a mass and 9.8 m/s^2 increases, increase the potential slope from
0.000030786 volts per meter and 9.8 m/s^2 increases. This voltage may need
rescaling if no flaws appear in my reasoning.
| For a Newtonian approximation, use 1/2 * a * t^2=x; v=a t
| a = 9.8, x=1 ---> t=.451 --> v=4.419 m/s
|
| This would mean that the particle accelerates 4.419 m/s per meter,
| which has the painfully inadequate units of 4.419 Hz because reaching 1
| meter per second has little significance.
|
| When units disappear like this, it probably means theres a missing
| connection, so lets change the problem--If the particle starts from
| rest, and drops one nil at constant acceleration, how fast is it going
| at the end, in units of nils/nanosecond^2?
| a=32.67*10^-18 nils / nanosecond^2, x=1 nils --> t=2.47*10^8 ns
| v=a t = 8.08*10^-9 nils/nanosecond = 8.08*10^-9 c.
|
| Note c = 1 nils/nanosecond = the speed of light.
I did read in some post how nils broke down but I've not become familiar
with it yet so could you restate it please.
3*10^8 m/ls seems a bit to circular to me at this moment.
This is my definition that I work from at the moment.
Reference Environment.
Local frame defined, uniform dielectric.
Gravity = 9.807 m per s^2 = potential difference
gradient of 0.000030786 volts per meter.
The reference second (Rs), where one second (s) equals the duration of
9192631770 periods of the radiation corresponding to the transition between
the two hyperfine levels of the ground state of the
caesium 133 atom.
One meter (m) equals the path travelled by radiation in the same environment
as the second measurement in one second divided by 299792458
Not so slick as yours ? {:-) maybe yours makes more sense ?
|Now if we consider
| that the particle accelerates at 8.08*10^-9 nils/ns every nils, this
| has the units of 8.08*10^-9 Hz which actually makes a good deal more
| sense in this context, because reaching 1 nils per nanosecond has great
| significance.
Yes I've wondered about this because if velocity is gravitationaly
attractive then the gravitational dynamics between two masses is changed by
their velocities i.e. you could possible imagine a relatively small mass in
orbit being mass increased by its velocity so the larger mass started to
orbit the smaller. I think its probably safe ? to assume that
velocity\energy in a 1D form is not directly equivalent to the
velocity\energy associated with gravity in a 3D form. I'm tempted to use 2D
and 4D but I think that would generate confusion ?
I have read the rest but the maths is moving too quick for me to catch
easily so I will try and digest it over a period of days or so. I think
Dirks has come in to haggle with you on this level as he is much more
capable of debating this stuff than I at present.
Transferring data across frames is a process fraught with confusion I feel
and need some major reengineering in IMHO.
| Jonathan Doolin
| www.spoonfedrelativity.com
Sue...
I am currently studying AE's EoMB (SR)? papers to try and find the
justification
for relativistic velocity correction from an energy pov.
>>
A goalie repeatedly fires a rifle at a hocky puck to accelerate it.
When the pucks velocity equals the rifles muzzle velocity,
the process is incapabable of further accelerating the puck.
If the goalie skate after the puck, still firing, he can add
his velocity wrt the ice to the muzzle velocity and accelerate
the puck a bit more.
Now... if you accept that the puck's velocity was asymtotically
aproaching the muzzle velocity because it was getting heavier,
then you must explain how pucks loose weight when chased
by goalies. ;-)
<< So now, if we still want to maintain some meaning for relativistic
mass, then we'll have to realise that it has a directional
dependence--as if the object somehow has more mass in the direction of
its motion, than it has sideways. Evidently the idea of relativistic
mass is becoming a little more complicated than at first we might have
hoped! And this is another reason why, in the end, it's so much easier
to just take the mass to be the invariant quantity m, and to put any
directional information into a separate, matrix, factor. >>
http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html
Sue...
Dirk Van de moortel
"Sue..." <suzyse...@yahoo.com.au> wrote in message news:1118130196....@g14g2000cwa.googlegroups.com...
> <<
> I am currently studying
You are too stupid for that.
Dirk Vdm
*** rD
"Sue..." <suzyse...@yahoo.com.au> wrote in message
news:1118130196....@g14g2000cwa.googlegroups.com...
| <<
| justification
| for relativistic velocity correction from an energy pov.
| >>
|
| A goalie repeatedly fires a rifle at a hocky puck to accelerate it.
| When the pucks velocity equals the rifles muzzle velocity,
| the process is incapabable of further accelerating the puck.
|
Yes but if he stands on the puck his potential velocity is infinity or
untill he runs out of bullits.
| If the goalie skate after the puck, still firing, he can add
| his velocity wrt the ice to the muzzle velocity and accelerate
| the puck a bit more.
|
| Now... if you accept that the puck's velocity was asymtotically
| aproaching the muzzle velocity because it was getting heavier,
No I dont, it was just a reducing velocity differential that was reducing
the acceleration
| then you must explain how pucks loose weight when chased
| by goalies. ;-)
I don't think or think I need to explain why pucks loose weight ? except
under the condition of the goalie standing on the puck firing.
|
| << So now, if we still want to maintain some meaning for relativistic
| mass, then we'll have to realise that it has a directional
| dependence--as if the object somehow has more mass in the direction of
| its motion, than it has sideways. Evidently the idea of relativistic
| mass is becoming a little more complicated than at first we might have
| hoped! And this is another reason why, in the end, it's so much easier
| to just take the mass to be the invariant quantity m, and to put any
| directional information into a separate, matrix, factor. >>
| http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html
|
Ah, reletivistic mass, have not read your link yet but I will.
Mass is an energy property of a multi dimensional structure. A structure can
also contain energy in a single axial form, you cant just add them together
as if they were the same thing as to transform one energy state to the other
needs considerable amount of energy.
| Sue...
|
*** rD
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:p8Dpe.6702$OP....@news.cpqcorp.net...
Thanks Dirk but surely the amount of gain you get from study is inversely
proportional to your level of stupidity so I will only had to study a very
small amount to gain a massive amount of stupidity and be equal to you ?
--
D & R *** E-field = Electric field, M-field =Magnetic field, two unbound
field effects
sue jahn
"*** rD" <paulps...@freeuk.com> wrote in message news:11182528...@eunomia.uk.clara.net...
> Repost as previous seems to have got lost
> "Sue..." <suzyse...@yahoo.com.au> wrote in message
> news:1118130196....@g14g2000cwa.googlegroups.com...
> | <<
> | I am currently studying AE's EoMB (SR)? papers to try and find the
> | justification
> | for relativistic velocity correction from an energy pov.
> | >>
> |
> | A goalie repeatedly fires a rifle at a hocky puck to accelerate it.
> | When the pucks velocity equals the rifles muzzle velocity,
> | the process is incapabable of further accelerating the puck.
> |
>
> Yes but if he stands on the puck his potential velocity is infinity or
> untill he runs out of bullits.
Indeed. A rifle has to push the entire universe backward.
A rocket only needs to push a bit of exhaust gas backward.
>
>
> | If the goalie skate after the puck, still firing, he can add
> | his velocity wrt the ice to the muzzle velocity and accelerate
> | the puck a bit more.
> |
> | Now... if you accept that the puck's velocity was asymtotically
> | aproaching the muzzle velocity because it was getting heavier,
>
> No I dont, it was just a reducing velocity differential that was reducing
> the acceleration
No... it is because the rifle doesn't have to push the universe back
quite so far. :o)
>
> | then you must explain how pucks loose weight when chased
> | by goalies. ;-)
>
> I don't think or think I need to explain why pucks loose weight ? except
> under the condition of the goalie standing on the puck firing.
OK Fair enuff! :-)
>
> |
> | << So now, if we still want to maintain some meaning for relativistic
> | mass, then we'll have to realise that it has a directional
> | dependence--as if the object somehow has more mass in the direction of
> | its motion, than it has sideways. Evidently the idea of relativistic
> | mass is becoming a little more complicated than at first we might have
> | hoped! And this is another reason why, in the end, it's so much easier
> | to just take the mass to be the invariant quantity m, and to put any
> | directional information into a separate, matrix, factor. >>
> | http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html
> |
> Ah, reletivistic mass, have not read your link yet but I will.
>
> Mass is an energy property of a multi dimensional structure. A structure can
> also contain energy in a single axial form, you cant just add them together
> as if they were the same thing as to transform one energy state to the other
> needs considerable amount of energy.
I learned a new term today:
*Word salad*
I'm not deciphering any more of your Star Wars sripts 'till ya get clean on
these metaphysical clocks fast-moving slow-aging pet rocks. ;-)
If you can't explain it in 3 + 1 D then it is only in your mind.
Sue...
>
>
>
> | Sue...
> |
>
>
>
>
*** rD
"sue jahn" <susyse...@yahoo.com.au> wrote in message
news:42a75539$0$18640$1472...@news.sunsite.dk...
|
| "*** rD" <paulps...@freeuk.com> wrote in message
news:11182528...@eunomia.uk.clara.net...
| > Repost as previous seems to have got lost
| > "Sue..." <suzyse...@yahoo.com.au> wrote in message
| > news:1118130196....@g14g2000cwa.googlegroups.com...
| > | <<
| > | I am currently studying AE's EoMB (SR)? papers to try and find the
| > | justification
| > | for relativistic velocity correction from an energy pov.
| > | >>
| > |
| > | A goalie repeatedly fires a rifle at a hocky puck to accelerate it.
| > | When the pucks velocity equals the rifles muzzle velocity,
| > | the process is incapabable of further accelerating the puck.
| > |
| >
| > Yes but if he stands on the puck his potential velocity is infinity or
| > untill he runs out of bullits.
|
| Indeed. A rifle has to push the entire universe backward.
| A rocket only needs to push a bit of exhaust gas backward.
|
Depends on the position of the rifle and the rocket.
| >
| >
| > | If the goalie skate after the puck, still firing, he can add
| > | his velocity wrt the ice to the muzzle velocity and accelerate
| > | the puck a bit more.
| > |
| > | Now... if you accept that the puck's velocity was asymtotically
| > | aproaching the muzzle velocity because it was getting heavier,
| >
| > No I dont, it was just a reducing velocity differential that was
reducing
| > the acceleration
|
| No... it is because the rifle doesn't have to push the universe back
| quite so far. :o)
Nice idea.LOL
Sorry, for multi read 3+1D and for single axial read 1+1D I was one of those
kids that was taught to only think things others have not, bit hard going at
times and very slow as any idea has to be checked that nobody else has
thought it{:-)
|
| Sue...
|
| >
| >
| >
| > | Sue...
| > |
| >
| >
| >
| >
|
|
Spoonfed
What Dirk Van de moortel has helped me with is an equation for the
velocity, position, and proper time of a uniformly accelerated point
particle. This way, if for instance, the point particle emits fuel
particles at a uniform rate in its proper time, we can predict the
position and time where each fuel particle is released in our frame.
Also, if we assume that the point particle is the center-of-momentum of
a larger body, and this body is structurally strong enough to withstand
its the acceleration without distortion in its own frame, we can at
least approximate its apparent length in our frame as it slows down,
stops, and accelerates in the other direction.
And when I get this all accomplished, assuming I find the time, if
everything looks right, it will look very natural, and peculiarly
unimpressive.
Then I'll try to remember why I did it and I will remember that I
wanted to know if gravity is actually a result of warped spacetime, or
if it is simply action at a distance. And I'll probably realize I'm no
closer than I was before.
Paul B. Andersen
>
> A goalie repeatedly fires a rifle at a hocky puck to accelerate it.
> When the pucks velocity equals the rifles muzzle velocity,
> the process is incapabable of further accelerating the puck.
>
> If the goalie skate after the puck, still firing, he can add
> his velocity wrt the ice to the muzzle velocity and accelerate
> the puck a bit more.
>
> Now... if you accept that the puck's velocity was asymtotically
> aproaching the muzzle velocity because it was getting heavier,
> then you must explain how pucks loose weight when chased
> by goalies. ;-)
But from whence have you got the naive idea that anybody
would try to explain it in such a stupid way?
A long time ago, I wrote:
The reason why a bullet cannot accelerate a puck beyond
its own speed is that it cannot transfer kinetic energy to the puck
if it never hits it.
But a particle in an accelerator gains the same amount of
kinetic energy every time it passes through a RF-cavity,
regardless of its speed. Even when the speed of the particle
is only few mm/s below c, it gains the same amount of kinetic energy.
The synchrotron radiation comes from where the particle
trajectories are bent by a magnetic field.
The radiated energy comes from the kinetic energy of
the particles, which therefore loose some energy.
Isn't this a beautiful proof that the RF-cavities keep
putting energy into the particles?
If they didn't, where does then the radiated energy come from?
When the speed of the particle increases, the synchrotron
radiation increases. When the radiated energy is equal
to the energy the RF-cavities put into the particles,
the accelerator is in steady state and have reached
the limit of its performance.
So the RF-cavities never cease to put energy into
the particles regardless of their speed. We can measure
this energy because it is radiated in another part of the circuit.
And in case you have forgotten what an RF-cavity is,
I will remind you:
In an accelerator the charged particle is accelerated
in RF-cavities, which are a resonance cavities into which
energy is fed. The frequency is typically hundreds of MHz.
Through this cavity, there is a tube with a gap in it,
see the figure below.
--------------
| |
| cavity |
| |
| E field |
| ---> |
-------- ---> ---------
* -> v ---> drift tube
-------- ---> ---------
| ---> |
| |
| |
| |
| |
--------------
The field in this cavity is oriented such that the E-field
is parallel to the tube. The particles move in the tube.
There is no field within the tube, which is called a drift tube
because the particles move by their inertia.
But in the gap between the two tubes, there is an
alternating electric field. The phase of this alternating
field is adjusted such that it is at its maximum in the same
direction every time a particle passes the gap.
The gap us but few centimetres wide, so the time for
a particle (going close to c) to pass it is in the order
of 0.1 ns. If the frequency of the field in the cavity is 500 MHz
the particle will pass the gap in ca. 1/20 period.
So from the particle's point of view, the field will appear
to be almost static. The gained energy is simply qEd,
where q is the charge, E the field and d the width
of the gap.
Bottom line:
There is no field running after the particle.
The field is set up in the gap before the particle enters it,
and is close to static while the particle is in it.
The force on the particle is F = qE independent of its speed,
and the gained energy is thus Fd = qEd.
Every time.
You were unable to refute any of this the last time we discussed it.
Would you like another go?
And please remember the wise words you quoted:
> << So now, if we still want to maintain some meaning for relativistic
> mass, then we'll have to realise that it has a directional
> dependence--as if the object somehow has more mass in the direction of
> its motion, than it has sideways. Evidently the idea of relativistic
> mass is becoming a little more complicated than at first we might have
> hoped! And this is another reason why, in the end, it's so much easier
> to just take the mass to be the invariant quantity m, and to put any
> directional information into a separate, matrix, factor. >>
> http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html
Relativity doesn't say that the particle can't be accelerated
to c because its mass increases.
Relativity say it can't be accelerated to c and beyond because
the energy of the particle is mc^2/sqrt(1-v^2/c^2)
were m is invariant!
Paul
Spoonfed
> <<
> I am currently studying AE's EoMB (SR)? papers to try and find the
> justification
> for relativistic velocity correction from an energy pov.
> >>
>
> When the pucks velocity equals the rifles muzzle velocity,
> the process is incapabable of further accelerating the puck.
>
> If the goalie skate after the puck, still firing, he can add
> his velocity wrt the ice to the muzzle velocity and accelerate
> the puck a bit more.
>
> Now... if you accept that the puck's velocity was asymtotically
> aproaching the muzzle velocity because it was getting heavier,
> then you must explain how pucks loose weight when chased
> by goalies. ;-)
>
So, how fast does the fuel come out the back of a solid fuel rocket in
a vacuum? If you had a spaceship already going faster than that, and
it fired its rockets... Hmmm, the center of mass would still continue
at the same rate, but the fuel would slow down and the rocket would
speed up.
But if I was firing a rifle at a hockey puck moving faster than my
rifle was able to shoot, the shot would never reach the hockey puck.
> << So now, if we still want to maintain some meaning for relativistic
> mass, then we'll have to realise that it has a directional
> dependence--as if the object somehow has more mass in the direction of
> its motion, than it has sideways. Evidently the idea of relativistic
> mass is becoming a little more complicated than at first we might have
> hoped! And this is another reason why, in the end, it's so much easier
> to just take the mass to be the invariant quantity m, and to put any
> directional information into a separate, matrix, factor. >>
> http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html
>
> Sue...
Spoonfed
I think I heard a study that said that people learn the fastest when
they already know 80% of what they are being told. In high school, I
found that they spent about 80% of each year in math classes repeating
what they did the year before. In getting my Master's degree in
Physics, on the other hand, I had some classes where I had heard
perhaps 20% of information before. I found I learned faster this way,
but it was difficult to assimilate.
Up until I was working on my Master's degree, I didn't realize quite
how much difference the competence and personality of the professor
made. When you already know 80% of the information, you gain just as
much by independent study. If you only know 20% you're just as
unlikely to move toward competency as the original discoverers.
When these original discoverers discovered an idea, it generally
involved a lot of their own presumptions fitting into place, luckily or
by design. It involved a series of questions they asked themselves
over time, throughout their lives, and found answers to, whether
documented or not--answers led to more questions, each more specialized
and less connected to our mundane existence.
Each person is naturally compelled to be curious about certain
subjects, and they may develop their own jargon and mode of thinking
about those subjects. If they get their higher education in those
subjects, they will appear to be very intelligent because they will
already have an infrastructure to place the ideas they hear at school.
If they get their higher education in subjects they have not thought so
much about, they may appear to be stupid, because they are attempting
to memorize ideas which bear no relation to themselves.
Sue...
rifle was able to shoot, the shot would never reach the hockey puck. >>
Yes... a much more plausible explanation for a particle accelerators,
inability to accelerate above the speed of light than the notion that
the particle's mass is approaching infinity, requiring exponentially
more energy for a velocity increase as is sometimes offered.
If further proof is needed which explanation is correct,
one need only observe that a collider overcomes the limitation.
Sue...
Spoonfed
> | > If its any help or hindrance I see it as a potential difference
> | > gradient of 0.000030786 volts per meter but it may need rescaling and I
> have
> | > yet to prove it.
> | >
> |
Where does this number come from?
> |
> | ========Electricity?=================
> | If you are using volts, you need to know both the amount of mass of
> | each particle and the amount of charge on each particle. Are you using
> | a different non-electrical definition for volts?
>
> In one sense yes in another no sorry if that's a bit wishy washy but I'm
> trying to relate velocity to volts as a potential gradient between two
> dielectric motion states and amps to energy as the width of the slope. So
> getting time\distance\velocity\acceleration\energy\volts\amps\gravity\mass
> in one package. Please think of me as daft if you need to as I'm fairly sure
> most others do {:-).
>
Perhaps you could tell how you got the figure of 0.000030786, and give
further discussion of how two motion states create an electrical
potential gradient.
> |
> | ============Acceleration in natural units===========
> | The acceleration due to gravity is (9.8 m/s^2)/(3*10^8 m/ls)=
> | 32.67*10^-9 light seconds/s^2.
>
> Forgive me for my slowness but the second and meter used to define 9.8
> m/s^2 are themselves a derivation of c which your then dividing by ? I am
> currently studying AE's EoMB (SR)? papers to try and find the justification
> for relativistic velocity correction from an energy pov.
>
I have not read Einsteins Explanation of Medicare Benefits, nor can I
tell what you mean by relativistic velocity correction, or energy
point-of-view from context. Perhaps you could set up a
problem/solution.
> |
> | This is 32.67 nil / second^2 (notice the similarity with 32
> | feet/second^2), or 32.67*10^-18 nil / nanosecond^2.
>
> Yes {:-)
>
> |
> | =========Natural units for dv/dx========
> | If the particle starts from rest, and drops one meter at constant
> | acceleration, how fast is it going at the end?
> |
>
> As far as gravity goes its the slope thats of main interest not the terminal
> velocity. I see the slope of 9.8 m/s^2 or possibly 0.000030786 volts per
> meter as the rate at which the energy from one system is transfered to
> another.
9.8 m/s^2 is the rate at which the relative velocity increases. A volt
does not affect a neutral particle. If you have a potential of
0.000030786/meter it would accelerate a proton in one direction, and an
electron, approximately 1833 times faster in the opposite direction.
If you had a neutral hydrogen atom, it would remain stationary, but
align itself to the electric field, creating a distortion in the
electron cloud.
Perhaps you should be using Joules/meter at least, if you want to speak
of how much kinetic energy is gained by a particle of known mass as it
drops one meter or (Joules per kilogram per meter) to find energy per
unit mass of an object dropping 1 meter--this is nice because it has
units of meters/second^2.
Whereas (Volts per meter) has units of (m/s^2)*(kg/Coulomb)
> That may not be how its thought of conventianly as its from the
> smaller mass to the larger as the final product is an increase in
> mass\energy\voltage\potential to the larger. Increase the mass or proximity
> to a mass and 9.8 m/s^2 increases, increase the potential slope from
> 0.000030786 volts per meter and 9.8 m/s^2 increases. This voltage may need
> rescaling if no flaws appear in my reasoning.
>
A quick and silly nonrelativistic calculation:
dE/dx = dE/dv * dv/dt * dt/dx
= d (1/2 m v^2)/dv * d(at)/dt * dt/d(1/2 a t^2)
= (m*a*(t-t0))/t
In relativistic form
E[v]=(1/sqrt(1-(v/c)^2)-1) m c^2,
v is as calculated in an earlier post in this thread
x is also as given in an earlier post.
So I think this would not be hard to figure out this way.
> | For a Newtonian approximation, use 1/2 * a * t^2=x; v=a t
> | a = 9.8, x=1 ---> t=.451 --> v=4.419 m/s
> |
> | This would mean that the particle accelerates 4.419 m/s per meter,
> | which has the painfully inadequate units of 4.419 Hz because reaching 1
> | meter per second has little significance.
> |
> | When units disappear like this, it probably means theres a missing
> | connection, so lets change the problem--If the particle starts from
> | rest, and drops one nil at constant acceleration, how fast is it going
> | at the end, in units of nils/nanosecond^2?
> | a=32.67*10^-18 nils / nanosecond^2, x=1 nils --> t=2.47*10^8 ns
> | v=a t = 8.08*10^-9 nils/nanosecond = 8.08*10^-9 c.
> |
> | Note c = 1 nils/nanosecond = the speed of light.
>
> I did read in some post how nils broke down but I've not become familiar
> with it yet so could you restate it please.
> 3*10^8 m/ls seems a bit to circular to me at this moment.
> This is my definition that I work from at the moment.
>
nils is a nanolightsecond. It is approximately 1 foot. 3*10^8 m/ls
describes motion in a straight line, precisely 1 billion nils, or about
1 billion feet.
> Reference Environment.
> Local frame defined, uniform dielectric.
Uniform dielectric describes the electric field in a parallel plate
capacitor.
> Gravity = 9.807 m per s^2 = potential difference
> gradient of 0.000030786 volts per meter.
> The reference second (Rs), where one second (s) equals the duration of
> 9192631770 periods of the radiation corresponding to the transition between
> the two hyperfine levels of the ground state of the
> caesium 133 atom.
> One meter (m) equals the path travelled by radiation in the same environment
> as the second measurement in one second divided by 299792458
>
> Not so slick as yours ? {:-) maybe yours makes more sense ?
>
You need to clarify how you calculate the volts (Joules per Coulomb)
per meter, and probably replace it with Joules per kilogram per meter,
and assume a constant gravitational gradient instead of a constant
electric gradient.
> |Now if we consider
> | that the particle accelerates at 8.08*10^-9 nils/ns every nils, this
> | has the units of 8.08*10^-9 Hz which actually makes a good deal more
> | sense in this context, because reaching 1 nils per nanosecond has great
> | significance.
>
> Yes I've wondered about this because if velocity is gravitationaly
> attractive then the gravitational dynamics between two masses is changed by
> their velocities i.e. you could possible imagine a relatively small mass in
> orbit being mass increased by its velocity so the larger mass started to
> orbit the smaller. I think its probably safe ? to assume that
> velocity\energy in a 1D form is not directly equivalent to the
> velocity\energy associated with gravity in a 3D form. I'm tempted to use 2D
> and 4D but I think that would generate confusion ?
>
I know you have to consider conservation of momentum, and clearly you
can't attain an orbit in one dimension.
In classical dynamics, for a two-body problem, I've seen this
simplification made: you use the Center of Momentum reference frame,
and treat both bodies as though they are attracted to that center.
Perhaps the same simplification would lead to some strange results in
the relativistic case. I don't currently know.
*** rD
| > | >
| > | > If its any help or hindrance I see it as a potential difference
| > | > gradient of 0.000030786 volts per meter but it may need rescaling
and I
| > have
| > | > yet to prove it.
| > | >
| > |
|
| Where does this number come from?
It comes from the concept of the matter of the earth having a potential
difference in relation to the surrounding vacuum state and this potential
difference being expressed across the surface to vacuum as a gap and the gap
as a dielectric made of the atmosphere.
Within the gap the potential expressed as an acceleration gradient being
covered to a ratio, this ratio being transferred from an acceleration scalar
to an arbitrary voltage scalar that produced a product of a change of
0.000030786 volts per meter from the surface to a height of 22.5 meters. I
am at present trying to devise and construct an experiment that will show
that an electric field slope of ~ that value will produce the same force as
the gravity potential of the same slope and if so can be slotted into a
unified field theory.
|
| > |
| > | ========Electricity?=================
| > | If you are using volts, you need to know both the amount of mass of
| > | each particle and the amount of charge on each particle. Are you
using
| > | a different non-electrical definition for volts?
| >
| > In one sense yes in another no sorry if that's a bit wishy washy but I'm
| > trying to relate velocity to volts as a potential gradient between two
| > dielectric motion states and amps to energy as the width of the slope.
So
| > getting
time\distance\velocity\acceleration\energy\volts\amps\gravity\mass
| > in one package. Please think of me as daft if you need to as I'm fairly
sure
| > most others do {:-).
| >
|
| Perhaps you could tell how you got the figure of 0.000030786, and give
| further discussion of how two motion states create an electrical
| potential gradient.
|
See above, and as to how this is an assumption at present derived from the
assumption that all field states are formed from dilation\contraction state
differences and can be expresed as acceleration\velocity\voltage potential
difference if scaled correctly.
| > |
| > | ============Acceleration in natural units===========
| > | The acceleration due to gravity is (9.8 m/s^2)/(3*10^8 m/ls)=
| > | 32.67*10^-9 light seconds/s^2.
| >
| > Forgive me for my slowness but the second and meter used to define 9.8
| > m/s^2 are themselves a derivation of c which your then dividing by ? I
am
| > currently studying AE's EoMB (SR)? papers to try and find the
justification
| > for relativistic velocity correction from an energy pov.
| >
|
| I have not read Einsteins Explanation of Medicare Benefits,
Electrodynamics of Moving Bodies 1905 {:-) just checking my typing skills ?
| nor can I
| tell what you mean by relativistic velocity correction,
The composition of velocity formula derived out of Lorentz
| or energy
| point-of-view from context. Perhaps you could set up a
| problem/solution.
|
If you take a spaceship and accelerate it to say 0.5 c and then apply the
same amount energy and accelerate it to 1.0 c the final velocity from the
pov of the spaceship appears to be 1.0 c but relativity (or some on this
group claim relativity claims) that the final velocity is ~ 0.8 c. This
appears given other statements that an energy loss of 0.2 c is in operation
at that moment as from the spaceship pov it has put in the energy to propel
it at 1.0 c ?
| > |
| > | This is 32.67 nil / second^2 (notice the similarity with 32
| > | feet/second^2), or 32.67*10^-18 nil / nanosecond^2.
| >
| > Yes {:-)
| >
| > |
| > | =========Natural units for dv/dx========
| > | If the particle starts from rest, and drops one meter at constant
| > | acceleration, how fast is it going at the end?
| > |
| >
| > As far as gravity goes its the slope thats of main interest not the
terminal
| > velocity. I see the slope of 9.8 m/s^2 or possibly 0.000030786 volts per
| > meter as the rate at which the energy from one system is transfered to
| > another.
|
| 9.8 m/s^2 is the rate at which the relative velocity increases. A volt
| does not affect a neutral particle. If you have a potential of
| 0.000030786/meter it would accelerate a proton in one direction, and an
| electron, approximately 1833 times faster in the opposite direction.
| If you had a neutral hydrogen atom, it would remain stationary, but
| align itself to the electric field, creating a distortion in the
| electron cloud.
|
If you have three particles each composed of negative and positive
potential that sums to neutral and two are in one place and one is in
another the field state between the two partical and one particle position
is a potential difference of two state neutral and one state neutral i.e.
their is a potential difference between the two positions of one particle
which can be expressed gravitational or as a very weak electrical potential
difference if you can see voltage potential difference and gravitation
potential difference as both being an expression of rate.
Thats the nearest I can find to a description of the vacuum state at the
moment.
|
| > Gravity = 9.807 m per s^2 = potential difference
| > gradient of 0.000030786 volts per meter.
| > The reference second (Rs), where one second (s) equals the duration of
| > 9192631770 periods of the radiation corresponding to the transition
between
| > the two hyperfine levels of the ground state of the
| > caesium 133 atom.
| > One meter (m) equals the path travelled by radiation in the same
environment
| > as the second measurement in one second divided by 299792458
| >
| > Not so slick as yours ? {:-) maybe yours makes more sense ?
| >
|
| You need to clarify how you calculate the volts (Joules per Coulomb)
| per meter, and probably replace it with Joules per kilogram per meter,
| and assume a constant gravitational gradient instead of a constant
| electric gradient.
|
Yes I do and its this, Length / Duration = Velocity = Voltage
I am also trying to get energy\amperage out of duration * length.
| > |Now if we consider
| > | that the particle accelerates at 8.08*10^-9 nils/ns every nils, this
| > | has the units of 8.08*10^-9 Hz
How are you getting frequency out of acceleration ?
*** rD
If its any help I have never been formaly educated in anything {:-)
Tom Roberts
> << But if I was firing a rifle at a hockey puck moving faster than my
> rifle was able to shoot, the shot would never reach the hockey puck. >>
>
> Yes... a much more plausible explanation for a particle accelerators,
Except it simply does not apply. In particle accelerator the
accelerating E field is present within each RF cavity _before_ the
particles get there. And we _do_ measure that the particles' energy
increases by \integral qE dl each time they traverse an RF cavity,
regardless of their initial speed, up to speeds ~0.999999 c. But still,
their speed never exceeds c.
Tom Roberts tjro...@lucent.com
Ken S. Tucker
My post is a bit primitive, but it *feels* like
the energy transmitted to a particle being
accelerated is "Doppler Shifted", so if the particle
is moving away from the source of the EM-pulse,
pulse being, a laser or an accerelator, then the
energy transmitted from the Earthly ref to thy
moving particle will be Doppler down shifted.
At "c" the red shift is total.
Ken
Sue...
The E field does not wink in and out... to exist or not exist.
It results from charges moving on the cavity walls. The walls
can't communicate energy to the particle FTL. If they can,
then all this business about relativity must be a bunch of hokum
and we've wasted our money on colliders.
Sue...
*** rD
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:4Wqqe.14167$Oq7....@newssvr33.news.prodigy.com...
Tom your view is so naive its embarrassing if you accelerate something by an
effect that has a maximum transfer of rate c then you can never get more
than c out of the object you are accelerating relative to an accelerator
that's stationary. If on the other hand you stick your accelerator on the
object you are trying to accelerate the potential velocity is infinity as
ever acceleration impulse is in relation to the object that the accelerator
sees as stationary. If you cant see this you really need to be in another
group and activity. Sorry to have to spell this out to you.
Dirk Van de moortel
"*** rD" <paulps...@freeuk.com> wrote in message news:111851980...@echo.uk.clara.net...
>
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:4Wqqe.14167$Oq7....@newssvr33.news.prodigy.com...
> | Sue... wrote:
> | > << But if I was firing a rifle at a hockey puck moving faster than my
> | > rifle was able to shoot, the shot would never reach the hockey puck. >>
> | >
> | > Yes... a much more plausible explanation for a particle accelerators,
> | > inability to accelerate above the speed of light [...]
> |
> |
> | Except it simply does not apply. In particle accelerator the
> | accelerating E field is present within each RF cavity _before_ the
> | particles get there. And we _do_ measure that the particles' energy
> | increases by \integral qE dl each time they traverse an RF cavity,
> | regardless of their initial speed, up to speeds ~0.999999 c. But still,
> | their speed never exceeds c.
> |
> |
> | Tom Roberts tjro...@lucent.com
> |
>
> Tom your view is so naive its embarrassing if you accelerate something by an
> effect that has a maximum transfer of rate c then you can never get more
> than c out of the object you are accelerating relative to an accelerator
> that's stationary. If on the other hand you stick your accelerator on the
> object you are trying to accelerate the potential velocity is infinity as
> ever acceleration impulse is in relation to the object that the accelerator
> sees as stationary. If you cant see this you really need to be in another
> group and activity. Sorry to have to spell this out to you.
Maybe Tom can explain to you how you check this:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowCheck.html
Dirk Vdm
*** rD
Thanks Dirk but for one I think Tom has kill filed me and two your Fumble
page is not directly related to my above statement and three you web page is
faulty as if " how can I check this" is supposed to take the user to a place
where the facts can be checked, it does not. So I'm pleased to be include in
the list of Dirk' ,Tom's & PD's naive fumbling assertions as an immortal
poster of facts {:-) BTW I never claimed that a photon emitted from an
electron travelling at any speed would exceed c its only a very naive view
that would ever think I did. So I guess your page of others fumbles is in
fact your fumbling lack of understanding ?
Thomas Palm
This is ruled out by astronomic observation. Light and high energy
particles emitted from object moving with high velocity relative to Earth
still arrive with a speed at or near our lightspeed.
Dirk Van de moortel
"*** rD" <paulps...@freeuk.com> wrote in message news:111851980...@echo.uk.clara.net...
>
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:4Wqqe.14167$Oq7....@newssvr33.news.prodigy.com...
> | Sue... wrote:
> | > << But if I was firing a rifle at a hockey puck moving faster than my
> | > rifle was able to shoot, the shot would never reach the hockey puck. >>
> | >
> | > Yes... a much more plausible explanation for a particle accelerators,
> | > inability to accelerate above the speed of light [...]
> |
> |
> | Except it simply does not apply. In particle accelerator the
> | accelerating E field is present within each RF cavity _before_ the
> | particles get there. And we _do_ measure that the particles' energy
> | increases by \integral qE dl each time they traverse an RF cavity,
> | regardless of their initial speed, up to speeds ~0.999999 c. But still,
> | their speed never exceeds c.
> |
> |
> | Tom Roberts tjro...@lucent.com
> |
>
> Tom your view is so naive its embarrassing if you accelerate something by an
> effect that has a maximum transfer of rate c then you can never get more
> than c out of the object you are accelerating relative to an accelerator
> that's stationary. If on the other hand you stick your accelerator on the
> object you are trying to accelerate the potential velocity is infinity as
> ever acceleration impulse is in relation to the object that the accelerator
> sees as stationary. If you cant see this you really need to be in another
> group and activity. Sorry to have to spell this out to you.
Maybe Tom can explain to you how you check this:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowCheck.html
Dirk Vdm
The Ghost In The Machine
<dirkvand...@ThankS-NO-SperM.hotmail.com.telenet-ops.be>
wrote
on Sun, 12 Jun 2005 14:57:31 +0000 (UTC)
<qISqe.118626$qI2.6...@phobos.telenet-ops.be>:
Hmph. I for one have already checked that (theoretically
speaking, using the Lorentz), though I'm not sure how to
find my post now.
But it's simple enough, if one can do first- or
second-level algebra; assuming the usual three observers O,
A, and B, and two instances of the Lorentz:
x_O = (x_A - v * t_A) / sqrt(1-v^2/c^2)
t_O = (t_A - v * x_A/c^2) / sqrt(1-v^2/c^2)
x_A = (x_B - w * t_B) / sqrt(1-w^2/c^2)
t_A = (t_B - w * x_B/c^2) / sqrt(1-w^2/c^2)
substitute, and grind away, ultimately representing x_O and t_O
in terms of x_B and t_B. One will of course get
x_O = (x_B - u * t_B) / sqrt(1-u^2/c^2)
t_O = (t_B - u * x_B/c^2) / sqrt(1-u^2/c^2)
where u turns out to be (v+w)/(1+vw/c^2). The intermediate terms
get interesting but ultimately everything else cancels.
Admittedly, this isn't proof, just a good mathematical check, but
it does work. (It would look very stupid if it didn't. :-) )
As a corollary, if one substitutes w = c then u = (v+c)/(1+vc/c^2) = c.
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Tom Roberts
> Tom Roberts wrote:
>>Sue... wrote:
>>><< But if I was firing a rifle at a hockey puck moving faster than my
>>>rifle was able to shoot, the shot would never reach the hockey puck. >>
>>>
>>>Yes... a much more plausible explanation for a particle accelerators,
>>>inability to accelerate above the speed of light [...]
>>
>>Except it simply does not apply. In particle accelerator the
>>accelerating E field is present within each RF cavity _before_ the
>>particles get there. And we _do_ measure that the particles' energy
>>increases by \integral qE dl each time they traverse an RF cavity,
>>regardless of their initial speed, up to speeds ~0.999999 c. But still,
>>their speed never exceeds c.
>>Tom Roberts tjro...@lucent.com
>
> the energy transmitted to a particle being
> accelerated is "Doppler Shifted", so if the particle
> is moving away from the source of the EM-pulse,
> pulse being, a laser or an accerelator, then the
> energy transmitted from the Earthly ref to thy
> moving particle will be Doppler down shifted.
Whyever would you think that?
_EVERY_ time a particle traverses an RF cavity, the particle's
energy increases by \integral qE.dl. It does so _REGARDLESS_ of the
particle's initial velocity. So there is no experimental evidence
of "redshift" at all.
[here I fixed a typo in the previous expression; E and dl
are both 3-vectors.]
> At "c" the red shift is total.
Of course we have no charged particle that travels at c, so there
is no way to test this guess. But in the limit as v -> c, it is
simply wrong, and this has been measured for v ~ 0.999999 c.
Tom Roberts tjro...@lucnet.com
*** rD
message news:iscun2-...@sirius.athghost7038suus.net...
But this has no bearing on the physical fact that the output of a reaction
motor will theoreticaly product a reaction that will accelerate it to
infinity or untill it runs out of fuel ? We are not talking about observers
here we are in one frame that appears stationary in relation to itself
except during the acceleration pulse when the object accelerates in relation
to its previous inertial state frame. The observers are sitting on the
reaction motor with their hair flying back due to the draft {:-) Goodbye for
now cruel world and sci.physics.relativity, I will be back.
*** rD
"Dirk Van de moortel"
<dirkvand...@ThankS-NO-SperM.hotmail.com.telenet-ops.be> wrote in
message news:qISqe.118626$qI2.6...@phobos.telenet-ops.be...
Nice trick adding .research into the header so firm replies might get
rejected. I will watch out for this bit of political wangling in future.
Repeating yourself is no more convincing the second time Dirk. Why do you
not face up to the fact that you no more understand the facts of
acceleration and its result velocity than does Tom or PD. If you have any
facts post them up and don't hide behind a load of antique difficult to
access paper by people that time has shown had no understanding of the
facts like PD and Tom hiding behind his kill file. It appear you and your
companions no more understand your own mantra than the physical fact behind
it. BTW the fumble that you assigned to me was in fact PD's as the content
was just PD's naive assertion.
*** rD
"Thomas Palm" <Thoma...@chello.removethis.se> wrote in message
news:Xns967386A8942ET...@212.83.64.229...
I did not claim that propagation could be seen as more than c.
Explain to me how you can measure velocities above c with equipment thats
operation depends on procces that is limited to c.
You like the others have avoided the case of an object with a reaction type
motor fitted
being able to possible accelerate to infinity assuming enough fuel. There is
a very obvious blind spot in this group in regards to objects or space ships
being able to accelerate by reaction and so in a different manner than an
object being accelerated by a stationary accelerator. Does nobody understand
the difference ? The things that you refer to above are simple objects or
fields that have been accelerated by another system, they have no mechanism
to apply acceleration to themselves so their velocity will be obviously
limited to the rate of c so to claim that c is a limiting velocity due to
the simple fact there are no natural mechanisms to accelerate anything above
c is to be frank bloody stupid.
Dirk Van de moortel
Dirk Van de moortel
"*** rD" <paulps...@freeuk.com> wrote in message news:11186542...@damia.uk.clara.net...
>
> "Dirk Van de moortel"
> <dirkvand...@ThankS-NO-SperM.hotmail.com.telenet-ops.be> wrote in
> message news:qISqe.118626$qI2.6...@phobos.telenet-ops.be...
> |
[snip]
> |
> | Maybe Tom can explain to you how you check this:
> | http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowCheck.html
> |
> | Dirk Vdm
> |
>
> Nice trick adding .research into the header so firm replies might get
> rejected. I will watch out for this bit of political wangling in future.
>
> Repeating yourself is no more convincing the second time Dirk.
I first replied without looking at the groups, so the reply
got sent to both groups (to which *you* posted as well).
Then I realized that the reply would (and should) be
refused on the research group, so I re-sent my reply to
sci.physics.relativity only.
For some strange reason the first reply was not refused
so it appeared after all, with a moderation delay of about
4 hours.
This was no trick, but *you* just proved that you have no
idea how Usenet works either.
> Why do you
> not face up to the fact that you no more understand the facts of
> acceleration and its result velocity than does Tom or PD. If you have any
> facts post them up and don't hide behind a load of antique difficult to
> access paper by people that time has shown had no understanding of the
> facts like PD and Tom hiding behind his kill file. It appear you and your
> companions no more understand your own mantra than the physical fact behind
> it.
I have no idea what you a babbling about here.
> BTW the fumble that you assigned to me was in fact PD's as the content
> was just PD's naive assertion.
Here is my fact: you are an idiot who can't even handle the
simplest algebraic equation, and here is the proof:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowCheck.html
Dirk Vdm
*** rD
|
| "*** rD" <paulps...@freeuk.com> wrote in message
news:11186542...@damia.uk.clara.net...
| >
| > "Dirk Van de moortel"
| > <dirkvand...@ThankS-NO-SperM.hotmail.com.telenet-ops.be> wrote in
| > message news:qISqe.118626$qI2.6...@phobos.telenet-ops.be...
| > |
|
| [snip]
|
| > |
| > | Maybe Tom can explain to you how you check this:
| > |
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowCheck.html
| > |
| > | Dirk Vdm
| > |
| >
| > Nice trick adding .research into the header so firm replies might get
| > rejected. I will watch out for this bit of political wangling in future.
| >
| > Repeating yourself is no more convincing the second time Dirk.
|
| I first replied without looking at the groups, so the reply
| got sent to both groups (to which *you* posted as well).
True sloppy posting and reading of headers on my part. I will wear my best
hair shirt for two days as a punishment. Satisfied.
| Then I realized that the reply would (and should) be
| refused on the research group, so I re-sent my reply to
| sci.physics.relativity only.
| For some strange reason the first reply was not refused
| so it appeared after all, with a moderation delay of about
| 4 hours.
| This was no trick, but *you* just proved that you have no
| idea how Usenet works either.
I hope this makes you feel suitably superior but I don't think it proves
anything other than I can be a sloppy poster sometimes.
|
|
| > Why do you
| > not face up to the fact that you no more understand the facts of
| > acceleration and its result velocity than does Tom or PD. If you have
any
| > facts post them up and don't hide behind a load of antique difficult to
| > access paper by people that time has shown had no understanding of the
| > facts like PD and Tom hiding behind his kill file. It appear you and
your
| > companions no more understand your own mantra than the physical fact
behind
| > it.
|
| I have no idea what you a babbling about here.
That could be one of your difficulties. i.e. reading.
|
| > BTW the fumble that you assigned to me was in fact PD's as the content
| > was just PD's naive assertion.
|
| Here is my fact: you are an idiot who can't even handle the
| simplest algebraic equation, and here is the proof:
| http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowCheck.html
|
| Dirk Vdm
|
You are referring to this piece of unimaginative rubbish ? that was supposed
to be a reply to a cascade of spaceships that all accelerated by 0.5 c by
high efficiency reaction motors. You could not tell an idiot if you hit
yourself over the head with one until you were as clever.
| Why? Because they don't, as proven experimentally. If they did, then
| supraluminal velocities would certainly be possible (as for example, an
| electron traveling at 0.9999c in the lab frame emitting a forward
| photon at relative velocity c). This fact can be predicted, it can be
| shown, starting from the two postulates of special relativity:
| a) that the laws of physics are identical in all inertial frames;
| b) that the speed of light in a vacuum is c, regardless of inertial
| frame.
|
| It turns out that in *this* cosmos (not my cosmos), velocities add
| according to
| v(tot) = [v1 + v2]/[1 + (v1*v2/c^2)].
| This shows that when v1, v2 are much smaller than c, then v(tot) = v1 +
| v2 is a good approximation, but only an approximation. You should check
| out the answer when v1 = v2 = c.
|
| PD
You are a mental deficient of the most astounding proportions to think that
this had any relevance to the theme of the post. and until you learn to read
you will presumably continue to make a fool of yourself.
Dirk Van de moortel
"*** rD" <paulps...@freeuk.com> wrote in message news:111868468...@dyke.uk.clara.net...
>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:hVfre.119619$kI2.6...@phobos.telenet-ops.be...
> |
> | "*** rD" <paulps...@freeuk.com> wrote in message
> news:11186542...@damia.uk.clara.net...
[snip]
> |
> | > BTW the fumble that you assigned to me was in fact PD's as the content
> | > was just PD's naive assertion.
> |
> | Here is my fact: you are an idiot who can't even handle the
> | simplest algebraic equation, and here is the proof:
> | http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowCheck.html
> |
> | Dirk Vdm
> |
>
> You are referring to this piece of unimaginative rubbish ? that was supposed
> to be a reply to a cascade of spaceships that all accelerated by 0.5 c by
> high efficiency reaction motors.
no no no, little cheat... this won't work :-)
> You could not tell an idiot if you hit
> yourself over the head with one until you were as clever.
>
> | Why? Because they don't, as proven experimentally. If they did, then
> | supraluminal velocities would certainly be possible (as for example, an
> | electron traveling at 0.9999c in the lab frame emitting a forward
> | photon at relative velocity c). This fact can be predicted, it can be
> | shown, starting from the two postulates of special relativity:
> | a) that the laws of physics are identical in all inertial frames;
> | b) that the speed of light in a vacuum is c, regardless of inertial
> | frame.
> |
> | It turns out that in *this* cosmos (not my cosmos), velocities add
> | according to
> | v(tot) = [v1 + v2]/[1 + (v1*v2/c^2)].
> | This shows that when v1, v2 are much smaller than c, then v(tot) = v1 +
> | v2 is a good approximation, but only an approximation. You should check
> | out the answer when v1 = v2 = c.
> |
> | PD
>
> You are a mental deficient of the most astounding proportions to think that
> this had any relevance to the theme of the post. and until you learn to read
> you will presumably continue to make a fool of yourself.
Alas, the evidence nicely speaks for itself and there is nothing
you can do about it :-)
Dirk Vdm
*** rD
|
| "*** rD" <paulps...@freeuk.com> wrote in message
news:111868468...@dyke.uk.clara.net...
| >
| > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote
| > in message news:hVfre.119619$kI2.6...@phobos.telenet-ops.be...
| > |
| > | "*** rD" <paulps...@freeuk.com> wrote in message
| > news:11186542...@damia.uk.clara.net...
|
| [snip]
|
| > |
| > | > BTW the fumble that you assigned to me was in fact PD's as the
content
| > | > was just PD's naive assertion.
| > |
| > | Here is my fact: you are an idiot who can't even handle the
| > | simplest algebraic equation, and here is the proof:
| > |
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowCheck.html
| > |
| > | Dirk Vdm
| > |
| >
| > You are referring to this piece of unimaginative rubbish ? that was
supposed
| > to be a reply to a cascade of spaceships that all accelerated by 0.5 c
by
| > high efficiency reaction motors.
|
| no no no, little cheat... this won't work :-)
And why not I may ask and by what piece of mental gymnastics do you find my
questions a cheat.
I can guess but I will not pre-empt your possible reply the space is yours
to solve this puzzle.
Here.....
BTW I am a bit fat ugly cheat if I am one but it seems just another
unsubstantiated assertions.{:-)
Please post this evidence that you claim. I hope its not a link to the faq
as per PD as there was nothing that I could find that would experimentaly
prove a result for my question.
Ken S. Tucker
Tom Roberts wrote:
> Ken S. Tucker wrote:
> > it *feels* like
> > the energy transmitted to a particle being
> > accelerated is "Doppler Shifted", so if the particle
> > is moving away from the source of the EM-pulse,
> > pulse being, a laser or an accerelator, then the
> > energy transmitted from the Earthly ref to thy
> > moving particle will be Doppler down shifted.
> _EVERY_ time a particle traverses an RF cavity, the particle's
> energy increases by \integral qE.dl. It does so _REGARDLESS_ of the
> particle's initial velocity. So there is no experimental evidence
> of "redshift" at all.
>
> [here I fixed a typo in the previous expression; E and dl
> are both 3-vectors.]
My 1st post was prefaced "primitive", meaning it might
be difficult to justify. However Tom has provided an
integral above that I would call,
Energy(impulse) = \integral qE.dl
That energy is set in the *accelerator* frame to E=hf,
as is the energy absorbed by the accelerated particle,
per impulse set by Tom's integral.
Relative to the moving particle, by the processes of
aberration and recession the f' (frequency relative
to the receeding particle) is red-shifted, and like-
wise the relative "Energy(impulse)" is red-shifted.
In a similiar but simpler way, a particle accelerated
by a laser will have the laser red-shifted and it's
intensity reduce as the particle is accelerated.
Regards
Ken S. Tucker