- #1

kmarinas86

- 979

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If I have a photon bouncing between two mirrors, and if both mirrors travel at a velocity [itex]v[/itex] with respect to the observer, on opposite sides of the observer, it would seem that the frequency observed by an inertial observer intercepting the photon would depend on which mirror the photon is coming from. How else would Doppler ranging work? If the observer is moving toward mirror B and away from mirror A, then the frequency observed by the observer when intercepting photon would be higher if it is coming from B than from A. It would therefore appear that, in an inertial frame, the photon does not have the same frequency going one direction versus the other direction.

If mirror A is at the front of the train, mirror B at the rear of the train, then our observer who is standing beside the train tracks has a relative motion which takes him from A to B. After each time the photon bounces from mirror B, there is more energy seen leaving B that what was entering it. If there were multiple such photons bouncing back and forth at the same time, then we would see more energy traveling from B to A, than we would see going from A to B. It would seem that the matter is literally carrying the energy of the bouncing light along with it. However, the time of each trip of a photon going in the direction of the train relative to the external, standby observer would be greater than when going in the opposite direction of the train. The difference in energy (in arbitrary units) would be:

[tex]\sqrt{\frac{c+v}{c-v}}-\sqrt{\frac{c-v}{c+v}}=\frac{2v}{\sqrt{\left(c+v\right)\left(c-v\right)}}=2v\gamma[/tex]

It would seem that simply having light bouncing back and forth would act to slow down the train (when hitting mirror B) and act to speed up the train (when hitting mirror A) by a kinetic energy of [itex]2v\gamma[/itex] times the initial energy of the photon in the case where it is emitted from the same frame as the observer.

The average energy carried by each photon observed would be dependent on this:

100% = % of time going from B to A + % of time going from B to A

[tex]ct = d+vt[/tex]

[tex]c = d/t+v[/tex]

[tex]c-v = d/t[/tex]

[tex]\frac{1}{c-v} = t/d[/tex]

Given [itex]d[/itex], [itex]t\propto\frac{1}{c-v}[/itex]. Therefore, the % of time going from B to A is:

[tex]\frac{\frac{1}{c-v}}{\frac{1}{c-v}+\frac{1}{c+v}}=\frac{c+v}{2c}[/tex]

And the % of time going from A to B is:

[tex]\frac{\frac{1}{c+v}}{\frac{1}{c-v}+\frac{1}{c+v}}=\frac{c-v}{2c}[/tex]

So the average energy of each photon is scaled by:

Energy when traveling from B to A * % of time going this way + Energy when traveling from A to B * % of time going this way

[tex]\sqrt{\frac{c+v}{c-v}}\frac{c+v}{2c}+\sqrt{\frac{c-v}{c+v}}\frac{c-v}{2c}[/tex]

[tex]\frac{\left(c+v\right)^2+\left(c-v\right)^2}{2c\sqrt{\left(c+v\right)\left(c-v\right)}}[/tex]

[tex]\frac{\left(c+v\right)^2+\left(c-v\right)^2}{2c\sqrt{c^2-v^2}}[/tex]

[tex]\frac{\left(c+v\right)^2+\left(c-v\right)^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{c^2+2vc+v^2+c^2-2vc+v^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{2\left(c^2+v^2\right)}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{c^2+v^2}{c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{1+\left(v/c\right)^2}{\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\gamma\left(1+\left(v/c\right)^2\right)[/tex]

So, interestingly enough, it would seem that if one were to emit a photon from the observer's frame and have the photon bouncing between the two mirrors in a train moving close to the speed of light, the average energy of the photon would be much greater than before, largely provided by the train itself!

But are these calculations actually correct? Or did I do something wrong somewhere?

Interestingly, if I take the difference instead of the sum, that is:

Energy when traveling from B to A * % of time going this way - Energy when traveling from A to B * % of time going this way

I get:

[tex]\sqrt{\frac{c+v}{c-v}}\frac{c+v}{2c}-\sqrt{\frac{c-v}{c+v}}\frac{c-v}{2c}[/tex]

[tex]\frac{\left(c+v\right)^2-\left(c-v\right)^2}{2c\sqrt{\left(c+v\right)\left(c-v\right)}}[/tex]

[tex]\frac{\left(c+v\right)^2-\left(c-v\right)^2}{2c\sqrt{c^2-v^2}}[/tex]

[tex]\frac{\left(c+v\right)^2-\left(c-v\right)^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{c^2+2vc+v^2-c^2+2vc-v^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{4vc}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{2\frac{v}{c}}{\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]2\frac{v}{c}\gamma[/tex]

Because the speed of light forward and speed of light backwards are identical, this would simply represent a factor determining the average

If mirror A is at the front of the train, mirror B at the rear of the train, then our observer who is standing beside the train tracks has a relative motion which takes him from A to B. After each time the photon bounces from mirror B, there is more energy seen leaving B that what was entering it. If there were multiple such photons bouncing back and forth at the same time, then we would see more energy traveling from B to A, than we would see going from A to B. It would seem that the matter is literally carrying the energy of the bouncing light along with it. However, the time of each trip of a photon going in the direction of the train relative to the external, standby observer would be greater than when going in the opposite direction of the train. The difference in energy (in arbitrary units) would be:

[tex]\sqrt{\frac{c+v}{c-v}}-\sqrt{\frac{c-v}{c+v}}=\frac{2v}{\sqrt{\left(c+v\right)\left(c-v\right)}}=2v\gamma[/tex]

It would seem that simply having light bouncing back and forth would act to slow down the train (when hitting mirror B) and act to speed up the train (when hitting mirror A) by a kinetic energy of [itex]2v\gamma[/itex] times the initial energy of the photon in the case where it is emitted from the same frame as the observer.

The average energy carried by each photon observed would be dependent on this:

100% = % of time going from B to A + % of time going from B to A

[tex]ct = d+vt[/tex]

[tex]c = d/t+v[/tex]

[tex]c-v = d/t[/tex]

[tex]\frac{1}{c-v} = t/d[/tex]

Given [itex]d[/itex], [itex]t\propto\frac{1}{c-v}[/itex]. Therefore, the % of time going from B to A is:

[tex]\frac{\frac{1}{c-v}}{\frac{1}{c-v}+\frac{1}{c+v}}=\frac{c+v}{2c}[/tex]

And the % of time going from A to B is:

[tex]\frac{\frac{1}{c+v}}{\frac{1}{c-v}+\frac{1}{c+v}}=\frac{c-v}{2c}[/tex]

So the average energy of each photon is scaled by:

Energy when traveling from B to A * % of time going this way + Energy when traveling from A to B * % of time going this way

[tex]\sqrt{\frac{c+v}{c-v}}\frac{c+v}{2c}+\sqrt{\frac{c-v}{c+v}}\frac{c-v}{2c}[/tex]

[tex]\frac{\left(c+v\right)^2+\left(c-v\right)^2}{2c\sqrt{\left(c+v\right)\left(c-v\right)}}[/tex]

[tex]\frac{\left(c+v\right)^2+\left(c-v\right)^2}{2c\sqrt{c^2-v^2}}[/tex]

[tex]\frac{\left(c+v\right)^2+\left(c-v\right)^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{c^2+2vc+v^2+c^2-2vc+v^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{2\left(c^2+v^2\right)}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{c^2+v^2}{c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{1+\left(v/c\right)^2}{\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\gamma\left(1+\left(v/c\right)^2\right)[/tex]

So, interestingly enough, it would seem that if one were to emit a photon from the observer's frame and have the photon bouncing between the two mirrors in a train moving close to the speed of light, the average energy of the photon would be much greater than before, largely provided by the train itself!

But are these calculations actually correct? Or did I do something wrong somewhere?

Interestingly, if I take the difference instead of the sum, that is:

Energy when traveling from B to A * % of time going this way - Energy when traveling from A to B * % of time going this way

I get:

[tex]\sqrt{\frac{c+v}{c-v}}\frac{c+v}{2c}-\sqrt{\frac{c-v}{c+v}}\frac{c-v}{2c}[/tex]

[tex]\frac{\left(c+v\right)^2-\left(c-v\right)^2}{2c\sqrt{\left(c+v\right)\left(c-v\right)}}[/tex]

[tex]\frac{\left(c+v\right)^2-\left(c-v\right)^2}{2c\sqrt{c^2-v^2}}[/tex]

[tex]\frac{\left(c+v\right)^2-\left(c-v\right)^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{c^2+2vc+v^2-c^2+2vc-v^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{4vc}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]\frac{2\frac{v}{c}}{\sqrt{1-\left(\frac{v}{c}\right)^2}}[/tex]

[tex]2\frac{v}{c}\gamma[/tex]

Because the speed of light forward and speed of light backwards are identical, this would simply represent a factor determining the average

*net*flow of energy carried by the photon.
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