# Energy change of a photon when it bounces between two mirrors

1. Jun 17, 2011

### kmarinas86

If I have a photon bouncing between two mirrors, and if both mirrors travel at a velocity $v$ 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:

$$\sqrt{\frac{c+v}{c-v}}-\sqrt{\frac{c-v}{c+v}}=\frac{2v}{\sqrt{\left(c+v\right)\left(c-v\right)}}=2v\gamma$$

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 $2v\gamma$ 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

$$ct = d+vt$$
$$c = d/t+v$$
$$c-v = d/t$$
$$\frac{1}{c-v} = t/d$$

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

$$\frac{\frac{1}{c-v}}{\frac{1}{c-v}+\frac{1}{c+v}}=\frac{c+v}{2c}$$

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

$$\frac{\frac{1}{c+v}}{\frac{1}{c-v}+\frac{1}{c+v}}=\frac{c-v}{2c}$$

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

$$\sqrt{\frac{c+v}{c-v}}\frac{c+v}{2c}+\sqrt{\frac{c-v}{c+v}}\frac{c-v}{2c}$$
$$\frac{\left(c+v\right)^2+\left(c-v\right)^2}{2c\sqrt{\left(c+v\right)\left(c-v\right)}}$$
$$\frac{\left(c+v\right)^2+\left(c-v\right)^2}{2c\sqrt{c^2-v^2}}$$
$$\frac{\left(c+v\right)^2+\left(c-v\right)^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
$$\frac{c^2+2vc+v^2+c^2-2vc+v^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
$$\frac{2\left(c^2+v^2\right)}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
$$\frac{c^2+v^2}{c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
$$\frac{1+\left(v/c\right)^2}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
$$\gamma\left(1+\left(v/c\right)^2\right)$$

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:

$$\sqrt{\frac{c+v}{c-v}}\frac{c+v}{2c}-\sqrt{\frac{c-v}{c+v}}\frac{c-v}{2c}$$
$$\frac{\left(c+v\right)^2-\left(c-v\right)^2}{2c\sqrt{\left(c+v\right)\left(c-v\right)}}$$
$$\frac{\left(c+v\right)^2-\left(c-v\right)^2}{2c\sqrt{c^2-v^2}}$$
$$\frac{\left(c+v\right)^2-\left(c-v\right)^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
$$\frac{c^2+2vc+v^2-c^2+2vc-v^2}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
$$\frac{4vc}{2c^2\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
$$\frac{2\frac{v}{c}}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
$$2\frac{v}{c}\gamma$$

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.

Last edited: Jun 17, 2011
2. Jun 17, 2011

### Ken G

If I'm understanding your conclusion correctly, that an observer would see light bouncing back and forth inside the train gradually accumulating energy, this cannot be correct. The equations are all time reversible, but your result is not, so it must be in err. If you run time backward, your scenario claims the photon should gradually redshift, but that's just the same physical situation as reversing the direction of the train.

3. Jun 17, 2011

### kmarinas86

For the photon to accumulate energy from the train, the train must be speeding up in our frame, otherwise the photon is just bouncing back and forth between the same two energy values. This requires the train to be an energy receiver in our frame, explaining the energy gain of the photon. A time-reversed situation would have the energy going back to the fuel, stored electrical energy, or whatever was used to accelerate the train in our frame (and decelerate it in some others), and also, the time-reversed situation would have the photon contribute to this energy recovery. The time-reversed situation is hardly a likely scenario in comparison to the normal order of events.

Last edited: Jun 18, 2011
4. Jun 18, 2011

### atyy

5. Jun 18, 2011

### Ken G

Then I guess I misinterpreted what you meant when you said:
What did that mean?

6. Jun 19, 2011

### Q-reeus

kmarinas86: This thread has apparently not really reached a resolution. Without = or otherwise operators acting between each line of the algebra in #1, not sure whether a simplification or what is going on at each stage. From what I can decipher the argument boils down to whether frequency difference between forward and backward bouncing photons between a pair of mirrors in the moving train, is exactly compensated by collision rate difference at each mirror, re momentum balance as determined by a stationary observer. There is no need for any equations really. We note that the round trip time for any given photon is the same at either mirror, and that regardless of whether the incident photon has greater or lesser frequency, the reverse is true upon reversal of direction. Momentum transfer at each bounce is the vector difference - incident minus outgoing, and this is the same at each end, since subtracting a minus from a plus is just an addition of positives. Combined with equality of round trip time, we must have equal and opposite average force dp/dt acting on each mirror. Hence there is no tendency for the bouncing light to be speeding up or slowing down the train - it's just 'dead weight' so to speak. Another way of seeing this is to note that in the frame of the train there is clearly going to be equal and opposite reaction forces on each mirror, acting along the direction of motion v. It is a fact of the relativistic force transformation rules that a force acting in the direction of relative motion is invariant wrt to v - see eg eq'ns 10 on p6 at http://ozelacademy.com/OJAS_v2n2_10.pdf
Your other remark about a net energy flow is of course true, simply because all of the train's contents is moving at the relative speed v (including the center-of-energy of the bouncing photons stream). Hope this helps, or were you meaning something else?

EDIT: OK, upon a re-read of #1 it now appears your main drift is here "...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!"

Yes but only if the observer fired the photon (presumably via some kind of angled intermediate reflector setup) towards the rear mirror. Doppler shift in the frame of the train then has the photon at an enhanced frequency and thus energy. This is exactly compensated for by the initial impulse on the rear mirror acting to slow the train and thus it's KE, sure. Conversely though firing towards the forward mirror will in the train's frame mean Doppler shifted reduced frequency and thus energy, with a compensatory impulse acting to speed up the train. And because of the Doppler shift, photon energy gain in the first instance is greater than energy loss in the second instance (and vice versa for the train). In either case, system total energy is preserved.

Last edited: Jun 19, 2011