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Consider the following scenario :

From here on I would like the word "laser" to be treated as just a source of monochromatic beam that emits light . You can take any other light source(if you do not want to get caught up in unimportant details) the arguments should not differ much(at least in my opinion).

A laser is kept on a frictionless plane and switched on. The laser is emitting light in a particular direction and we know light has momentum. Now as the net force on system is zero, conservation of momentum states that the laser must also acquire a momentum in direction opposite to light.

So laser also starts moving, but where does the required kinetic energy come from? One simple answer is to say the from the source powering laser, which seems right.

Now if we take a spherically symmetrical source of light which employs same mechanism of light production as our laser above, it should be reasonable to assume it has same efficiency to light conversion as our laser.

Except now the spherical source of light moves nowhere( using conservation of momentum) and thus now it has zero kinetic energy.

But if the efficiency of light production is same where does that extra energy(that was supposed to be converted to kinetic energy) go?

Question :

Why do we have this contradiction? What am I arguing wrong here?

Edit: I have changed the title to suit the question more.

Hrishabh Nayal
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  • Do you think the energy carried by the photons in both the cases will be the same? –  Apr 03 '20 at 13:00
  • As you stated, because of energy conservation. The photons energy (and frequency) will increase because of this. –  Apr 03 '20 at 13:10
  • @FakeMod would it not? Why? As far as I can see spherical source can be considered as a limiting case of joining infinitely many lasers symmetrically opposite to each other, each with same mechanism to produce light and thus each having some unaccounted energy. Please explain if I am wrong. – Hrishabh Nayal Apr 03 '20 at 13:14
  • @FakeMod how can we say that the frequency will increase if the mechanism producing it is same. Can this be verified experimentally?(I do not think it can as the energies involved should be too low to measure) – Hrishabh Nayal Apr 03 '20 at 13:16
  • If you think about this problem classically, then the velocity of the photons should increase (because in COM frame $KE=1/2 \mu v_{\text{rel}} ^2$ and thus relative velocity should be same for all the cases). But since photons cannot change their speed, thus they would contain the extra energy in the form of frequency. To prove this, you might need to invoke special relativity, which I am not much familiar with. –  Apr 03 '20 at 13:21
  • This question isn't really answerable unless you specify exactly what is going on inside the "laser" (why is it a laser? Can you not ask this question for any powered light source?). A laser is not a magical box that emits directed light, it's a physical device. Maybe there is a spherical source inside it and the light is reflected and collimated to be emitted as a beam, with some energy loss due to reflection (see https://physics.stackexchange.com/q/539983/50583). Maybe something else is going on inside the laser. As written I don't see how one could answer this in a well-defined way. – ACuriousMind Apr 03 '20 at 13:34
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    IMO You could simplify this question. Start by removing all discussion of "efficiency." You seem to assume that a laser should be "totally efficient," but that is very far from the truth. For example, all lasers convert a substantial fraction of the input energy to heat. I think all you need to do is talk about a laser that outputs a beam of so-many milliWatts, and then compare that to a rigid assembly of two identical lasers that point in opposite directions from each other. – Solomon Slow Apr 03 '20 at 13:34
  • Maybe an analogy will help. You're wearing ice skates, standing in the middle of the ice, and you throw a ball as hard as you can, in a roughly horizontal direction. The ball will fly forward with a certain amount of momentum & kinetic energy, and because of recoil you will slide backwards with equal & opposite momentum, and much less KE than the ball since you're heavier than the ball. Now repeat the experiment, but this time you are at the edge of the ice rink, with your back braced against a rail. So now there will be no recoil and the ball will have more KE than in the 1st scenario. – PM 2Ring Apr 03 '20 at 14:14
  • @PM 2Ring well a strictly rigid wall that offers no recoil requires infinite mass and perfectly inelastic collision (practical impossibility) but however I am willing to hear arguments that show two persons with identical mass tied to each other throwing identical balls in opposite direction however I believe that case is very different from light as balls can change speed but light cannot. – Hrishabh Nayal Apr 03 '20 at 14:28
  • @ACuriousMind I have edited the question a bit. You can think of a atom(in excited state) emitting a photon instead(while coming to ground state), but we cannot "tie together" two atoms so I used laser as a more practical example. – Hrishabh Nayal Apr 03 '20 at 14:44
  • @SolomonSlow I have removed the part that was troubling you. But the problem still stands if lasers convert electricical energy to light with same efficiency then the amount of kinetic energy it transfers to lasers must also be same(for both lasers) and thus we still have a missing energy. – Hrishabh Nayal Apr 03 '20 at 14:46
  • 1. True, I should've said "almost no recoil". The rail may flex slightly. Or you could be braced against a very rigid wall, so that the entire mass of the Earth takes up the recoil. ;) 2. Light cannot change speed, but it can still change its KE, in accordance with $E=h\nu$ so if the laser moves backwards the light frequency will be Doppler shifted downwards. – PM 2Ring Apr 03 '20 at 14:53
  • @PM 2Ring If possible can you please give a answer and probably show some calculations(if they are possible which I highly doubt). The comment is too compact for me understand anything. (Sorry if I am troubling you) – Hrishabh Nayal Apr 03 '20 at 15:00
  • One problem is you assume "the efficiency of light production is same". What's the justification for this assumption? – The Photon Apr 03 '20 at 15:04
  • @The Photon I think that because they have same mechanism to produce light it should be logical to assume that they have same efficiency. Am I wrong? Please explain – Hrishabh Nayal Apr 03 '20 at 15:06
  • Re, "calculations(if they are possible which I highly doubt)" You might want to look at Wikipedia's Doppler effect page. For source and observer moving at normal (non-relativistic) speeds, it works the same for light as it works for any other wave phenomenon. As for $E=h\nu$, that is the well known Planck-Einstein relation. If you choose not to believe in those things, then this question should be closed as "off-topic/does not pertain to main-stream Physics." – Solomon Slow Apr 03 '20 at 17:28
  • @Solomon Slow Oh no! Dear me, no! When I say that I "doubt" calculations I mean that I doubt that they can be done in this problem, given the vague nature of problem(not specified mass of laser device, intensity and frequency of light ,etc). Not that I am uncertain of validity of mathematics!! That will be utterly stupid. Sorry if I mislead you to that conclusion :). And yes I know relativistic doppler effect simplifies to a classical doppler effect at speeds very slow compares to that of light. – Hrishabh Nayal Apr 04 '20 at 02:33

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Consider a laser with photon energy $E_\gamma$. This means that the photons emitted by it have an energy of $E_\gamma$ in the reference frame of the laser emitter. Since the laser emitter is on a frictionless plane, by conservation of momentum, it will move in the opposite direction of the beam (albeit very slowly, as the force applied by the beam is very small), which means that the photons emitted will have less energy in the reference frame of the plane, being redshifted according to the relativistic Doppler effect; this is where the kinetic energy of the laser emitter comes from. If we now look at the case of two identical, attached laser emitters, pointing in opposite directions, momentum is conserved when the laser assembly remains at rest, so none of the power goes toward the kinetic energy of the emitters, so the beam does not become redshifted.

Sandejo
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