# Does photon's energy increase when i emit it from a train and get it reflected back?

Does photon's energy increase when i emit it from a train which is moving at comparable speed and get it reflected back by a mirror on a platform so that i can catch it again in the train? According to doppler effect the frequency of the photon for an observer standing near the mirror is greater to an observer in the train.So to the observer near the mirror, energy of the photon is greater than the energy observed by an observer in train.Similarly when it gets reflected back and reaches the train its frequency is larger than the initial frequency when observed by an observer in train .So obviously energy of photon is increased without doing any work....But it is against conservation of energy!!!!!!!!!!so can anyone help me with this problem ,please.....thanks in advance

• You just forgot that the doppler effect will be perceived only by a person in the same reference frame as the mirror. For an observer in the train the frequency remains the same.
– user126422
Jan 5 '17 at 1:03
• "without doing any work". Neither the train nor the mirror (even if you include the earth) has infinite mass. So they are going to exchange momentum and modify their kinetic energy. You'll want to look closely at what happens during emission/reflection/absorption. Jan 5 '17 at 1:06
• @BowlOfRed why do not you just take the limit?
– user126422
Jan 5 '17 at 1:13
• @AlbertAspect The frequency as measured by an observer on the train should go up: in a wave picture: Doppler effect of a diminishing path.
– user137289
Jan 5 '17 at 1:14
• @Pieter I am not sure what you mean. But if you are correct the frequency change has to happen at the reflection, as an observer that is not moving relative to the train at the same position than the mirror would not observe the doppler effect, because the emitter is at rest with respect to him.
– user126422
Jan 5 '17 at 1:25

From the point of view of the observer in the train, the photon is reflected off a moving mirror. That reflection will increase the energy / momentum of the photon (it can't increase the speed, obviously). In the train frame of reference, the mirror is moving towards the observer at velocity $v$: if the outgoing photon has momentum $p$, the returning one will have momentum $$p' = p\frac{c+v}{c-v}$$

• i think that the above equation cant apply to my case. in the link which you gave ,they derived in a case were the mirror is moving away from photon( i mean that photon and mirror are moving in the same direction before collision).But in my case the mirror is moving towards photon befor collision(as seen by an observer in train). so i fear that the equation will dramatically change. i think that the frequency of the photon must decrease(just the opposite case) .so i fear that the equation u ave doesnt suit my case if i am not wrong Jan 7 '17 at 0:26
• @BobBee please see the above comment Jan 7 '17 at 0:27
• It simply whether v is positive or negative. Same equations, just careful with signs. Jan 7 '17 at 4:48
• And BTW, it was a pretty good question, one has to be careful conceptually, and on the order of magnitude of the effects it's also easy to be thrown off. Jan 7 '17 at 4:49
• Think of the mirror as a baseball bat, and the photon as a ball. If the bat moves towards the ball, the ball will bounce off with more energy. And the bat experiences some recoil (lost energy). It's the same with the mirror and the photon. No free lunch here - sorry. Jan 7 '17 at 23:22

Floris is right, it does change exactly as he references from the paper. However, one does not need to look at any paper to figure it out -- it is a relatively simple physics problem if you think relativistically. I derive it below. And you do not need to involve momentum (though it's equivalent), but you can do it all from the way frequency changes from one inertial reference frame to another (i.e., special relativity, and also true in local inertial frames in gravity)

The idea is to consider the problem as two steps, and thus two changes from one reference frame to another. We'll do two transformations. It is important to do one at a time or you get confused.

We do them step by step, so it is clear there is absolutely NO MAGIC. Just straight SR transformations.

FIRST TRANSFORMATION

Frame 1: the train rest frame (TRF). It is moving wrt the mirror, but in this frame we consider the train at rest, ie, we are standing on the train

emitted frequency = f

Frame 2: the mirror rest frame (MRF). In this frame we consider it at rest.

Since the source of the emitted light is moving, there will be a Doppler shift given by

$f_2$ = frequency in MRF = (1+$\beta$)$\gamma$f == (1+z)f

where z is the blue shift, ie, $\Delta$f/f = (freq in MRF - f)/f,

and $beta$ = v/c

and the Lorentz factor $gamma$ = 1/$sqrt{(1-\beta^2)}$

as both are normally labeled.

The equation for Doppler shift is exact in SR, for motion along the line of sight. See https://en.wikipedia.org/wiki/Redshift where the equations are

1+z = (1+$\beta$)$\gamma$

and where I've take $\beta$ > 0 for the blue shift we would be seeing. (note: careful, Wiki takes the opposite, you can do it either way).

SIMPLIFICATION

Note that 1+z = (1+$\beta$)$\gamma$

1+z = $sqrt{[(1+\beta)/(1-\beta)]}$

So

$f_2$ = f (1+z) (Eq. 1)

or

$f_2$ = f $sqrt{[(1+\beta)/(1-\beta)]}$ (Eq. 2)

So that's a simple equation. We will reuse Eqs. 1 and 2 in the next steps

SECOND TRANSFORMATION

Frame 2: Mirror at rest, reflecting the light, ie, emitting light

Frequency for light reflected in Frame 2 = frequency in Frame 2 for incident light = frequency in MRF = $f_2$

Frame 3: Frame of the train when we receive the light back. In this frame the train (or us if you wish) see the source of the light, Frame 2, as approaching at velocity v

Thus at Frame 3 we see a frequency $f_3$ further blue shifted.

We can apply the same 1+z factor we applied from Eq. 1 in Eq. 2 above, to get

$f_3$ = $f_2$ $sqrt{[(1+\beta)/(1-\beta)]}$

and substituting for $f_2$ we get (with the two identical square roots multiplying) the received frequency at the train (RFT)

$f_3$ = f $[(1+\beta)/(1-\beta)]$ (Eq.3)

That is, the received frequency at the train, $f_{RFT}$ is

$f_{RFT}$ = f $[(1+\beta)/(1-\beta)]$ (Eq. 4)

Since momentum is proportional to energy and to frequency in SR, that is the same equation as in that paper referred by Floris.

Except that we derived it doing two consecutive Lorentz transformations where we used the equation for the Doppler shift in going from one frame to the next, twice.

Notice that this is the right way to do it. The mirror reflected the light, and we treated that in its rest frame. No need for other assumptions. The train received the light reflected, and it did not know that it was the light it emitted, it just saw light coming off the mirror. So it simply needed to make the transformation to get the right frequency [actually, obviously the train (or us at the train) can measure the frequency, we would not need to calculate it, and then use Eq. 4 to calculate our velocity with respect to the mirror].

CONSERVATION OF MOMENTUM AND ENERGY

Note, the paper does apply conservation of energy and momentum, at the mirror. It also states that one can obtain their slightly more complicated results (they do it for reflection at an angle, we could have also, not needed for the question at hand) simply using SR. It has to be so, and for the simple case covered in the question, it's very easy. Note also that we do not need to prove conservation of energy, we know that SR conserves energy if calculated right. Einstein and others proved that. Yes, the extra photon energy had to come from the mirror, it is also what changed its momentum by reversing direction. It's the only physical interaction.

• thanks a lot for taking your time and explaining.It was really interesting but i do have a doubt in this.you have said that extra energy of the photon came from the mirror.Then it means that the mirror should move with respect to observer on ground with a velocity after collision with photon. if i keep this mirror as one of the blades of a wind mill whose axis of rotation is perpendicular to ground then it should create energy...besides increasing the energy of photon the mirror also generates energy...i think this contradicts law of conservation of energy.. Jan 6 '17 at 2:21
• .so what i come to say is that the photon does not gain energy after colliding with the mirror Jan 6 '17 at 2:21
• @Iron man. Read the paper referred to by Floris, they calculate the same equation I get using conservation of energy and momentum. I calculate the gain in energy for the light simply from relativistic invariance which leads to the equations for transforming frequency in different coordinates, and then argue it can only come from the mirror. My derivation, and the multiple references in the paper Floris referenced get to the same conclusion. If you find anything wrong with my math I'm willing to listen. On energy conservation, the paper is more explicit, read it to see how it gets conserved Jan 6 '17 at 2:34
• And Leonard's answer also shows it. Jan 6 '17 at 2:37
• i agree with all your paper information.imagine train moves in+x direction with respect to observer on ground.before collision the mirror(imagine it to be floating) remains at rest.after collision movess in -x direction and it picks up some energy from mirror(as you said).so mirror moves in +x direction.now i am going to make a change. i am going to fix one end of the mirror to a pole perpendicular to ground. after photon hits the mirror moves in circular motion with the pole as axis. i can use this to generate energy(just like wind mill) Jan 6 '17 at 8:02

# Energy balance in elastic scattering involving photons

## Compton Scattering

First, consider the frame of reference $\mathcal{B}$ that the platform rests in.

A photon with the frequency $f^\mathcal{B}$ and momentum $p^\mathcal{B} = \frac{h}{c}f^\mathcal{B}$ will hit a mirror fixed to some resting object $\mathrm{M}$ of mass $m$. (You may imagine $\mathrm M$ to be the Earth.) The reflection will transfer a momentum of $S\approx 2p^\mathcal{B}$ to $\mathrm{M}$. After the collision the object $\mathrm{M}$ will have the velocity $$\Delta v \approx \frac{S}{m} \approx \frac{2h}{mc}f^\mathcal{B}$$ and will have some kinetic energy. To account for that energy, the reflected photon will have a reduced frequency $f'{}^\mathcal{B}$. This phenomenon is called the Compton effect and is very small for a macroscopic object $\mathrm{M}$, but very well observable if $m$ is about the mass of an electron.

This is not the energy transfer that you are interested in. It is only $$\Delta E \approx \frac{m}{2}(\Delta v)^2 \approx \frac{2h^2}{mc^2}f^\mathcal{B} \approx \frac{10^{-83}}{m}f^\mathcal{B}$$ which is very small - even compared to the energy of a customary photon ($\sim 10^{-19}\mathrm J$). Thus, in the frame of reference $\mathcal{B}$ there practically is no energy transfer and we may assume $$f'{}^\mathcal{B}\approx f^\mathcal{B}$$

## Reverse Compton scattering

In the frame of reference $\mathcal{A}$ the train is at rest. Note that the kinetic energy is not independent of the choice of reference frame. In $\mathcal{B}$ the object $\mathrm{M}$ is at rest and does not have any kinetic energy. In $\mathcal{A}$, $\mathrm{M}$ is moving with the velocity $v$ towards the observer, thus it has some kinetic energy $E_\text{kin}^\mathcal{B}\approx \frac{m}{2}v^2$.

Since the kinetic energy is not a linear function of $v$, but an approximately quadratic function, the change in energy by a boost $\Delta v$ depends on the energy that the object $\mathrm{M}$ has in the first place. (Note that $\Delta v$ is in opposite direction to $v$. Further note that $\Delta v$ is approximately the same in both frames of reference.) $$\Delta E_\text{kin}\approx \frac{m}{2}(v-\Delta v)^2-\frac{m}{2}v^2=\frac{m}{2}(-2v\Delta v + (\Delta v)^2)\approx -mv\Delta v\approx \frac{v}{c}2hf^\mathcal{B}$$

Now I still assume that $\frac{m}{2}(\Delta v)^2$ is a very small and negligible change in kinetic energy, as I did above. However, $mv\Delta v$ cannot be neglected if $v$ is near the order of magnitude of the speed of light as your problem suggests. As a result, the kinetic energy of $\mathrm{M}$ is reduced. This effect is called reverse Compton effect.

Since trains don't move with, say, a tenth of the speed of light, the reverse Compton effect is not observed in trains, but it is observed with fast moving particles in particle accelerators.

In your example, the object $\mathrm{M}$ will loose some kinetic energy and the photon will gain energy in the frame of reference $\mathcal{A}$

## Relativistic Doppler shift

The frequency of light in reference frames $\mathcal{A}$ and $\mathcal{B}$ are related by the following formulas:

For the photon moving from the train to the mirror $$f^\mathcal B = \sqrt{\frac{c + v}{c - v}}f^\mathcal A$$ and for the photon moving from the mirror to the train $$f'{}^\mathcal A = \sqrt{\frac{c + v}{c - v}}f'{}^\mathcal B$$

Combine that with $f'{}^\mathcal{B}\approx f^\mathcal{B}$ and conclude $$f'{}^\mathcal A \approx \frac{c + v}{c - v} f^\mathcal A \approx (1 + 2\frac{v}{c}) f^\mathcal A$$

Therefore the change energy of the photon is $$\Delta E_\mathrm\gamma^\mathcal A = hf'{}^\mathcal A - hf^\mathcal A \approx h\cdot2\frac{v}{c}f^\mathcal A$$

## Summary

In $\mathcal{A}$ some kinetic energy of $\mathrm M$ is transferred to the photon. In $\mathcal B$ the changes in energy are negligible, but if $m$ is very small there is a transfer of energy from the photon to the mirror.

Similar transfers of energy happen in the train when emitting and absorbing the light. In $\mathcal A$ the beam of light will transfer kinetic energy from $\mathrm M$ to the train. In $\mathcal B$ the beam of light will transfer kinetic energy from the train to $\mathrm M$.

• .that was an interesting answer.But i do have a doubt.If is consider the train moving in +x direction, to the observer in train, the mirror moves in -x direction and he is at rest. To the observer in train when the photon moving in +x direction with a given energy falls on a mirror moving in -x direction,the mirrors energy gets reduced and it moves -x direction with a much less velocity than its initiial velocity. Jan 6 '17 at 2:23
• So according to the observer the photon has done some work so its energy should decrease while getting reflected back.so i think the collision between the mirror and the photon can't explain the observed increase in energy of photon.@ Jan 6 '17 at 2:23
• When a table tennis ball hits the bat, it will perform work on the bat in the frame of reference that the bat rests in before the collision and in that frame of reference indeed the ball will loose velocity. However in the players frame of reference the bat performs work accelerating the ball. The energy for that is taken from the kinetic energy of the bat. Jan 6 '17 at 6:00
• For the photon the situation is like this: in the frame of reference that the mirror rests in before the collision the photon will have a decreased frequency after the reflection. This effect is called Compton scattering. For a macroskopic mirror you will not observe the effect, because $(\Delta v)^2$ is very small, but if your "mirror" is just an electron, $(\Delta v)^2$ will be large enough to measure the reduction in frequency and the kinetic energy of the electron (I have done this). The Compton effect is an adverse effect to the doppler shift of the photon that you are interested in. Jan 6 '17 at 6:10
• .It was really interesting but i do have a doubt in this.you have said that extra energy of the photon came from the mirror.Then it means that the mirror should move with respect to observer on ground with a velocity after collision with photon. if i keep this mirror as one of the blades of a wind mill whose axis of rotation is perpendicular to ground then it should create energy.besides increasing the energy of photon the mirror also generates energy.i think this contradicts law of conservation of energy.so final result is energy has been generated inside and outside train Jan 6 '17 at 7:40

I got it. I thought the train launches a photon at right angle. It launches it into front.

In simple words you consider a photon which is jumping between two perfect mirrors, like in a light clock. If mirrors are rigidly fixed, proper frequency (let‘s say color) of photon doesn't change. If mirrors are not rigidly fixed, photon will transfer some energy to the mirrors and mirrors will start moving (receding) from each other.

Thus, photon will be more and more redshifted at every oscillation if mirrors receding, because mirrors will gain some of it‘s energy.

If mirrors approach each other, even by inertia, they will transfer some of their kinetic energy to photon. Photon will slow them down a bit. Photon will gain some energy and will be more and more blueish at every oscillation.

If someone "brings" mirrors closer by his hands, he has to do some work.

Interesting detail is the following: if a train launches a photon at right angle in his reference frame, photon will be blueshifted at the mirror on the platform. It will be redshifted on arrival back and will appear in the same frequency. In this case we consider that observer and mirror are in motion relatively to each other at parallel patches and there is a gap between them.

Where is reciprocity of observations?

• i dont get your point .. i have considered that train and photon moves in same direction... so i fear that your answer is not answering my question...you have said that there is an interesting detail hiding....can you please say that Feb 2 '17 at 16:19
• It was really interesting but i do have a doubt in this.you have said that extra energy of the photon came from the mirror.Then it means that the mirror should move with respect to observer on ground with a velocity after collision with photon. if i keep this mirror as one of the blades of a wind mill whose axis of rotation is perpendicular to ground then it should create energy.besides increasing the energy of photon the mirror also generates energy.i think this contradicts law of conservation of energy.so final result is energy has been generated inside and outside train Feb 2 '17 at 23:50
• i have one more doubt in your answer .you have said that the photon will be blue shifted to the mirror in the platform.its fine till that. but a mirror(consider it as fixed) reflects what it gets without any change. so to the mirror reflects the photon in the same frequency it received(for observer in platform). for observer inside the train the mirror just acts as a source . so the photon obviously comes with high frequency than initial frequency. this is exactly my doubt Feb 3 '17 at 0:00