Conservation of energy, hypothetical situation

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fluidistic
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Main Question or Discussion Point

I'm not sure the question belongs to classical physics, I apologize in case it's a no.
Imagine a train or any object following a straight line motion. Suppose there's no friction between the ground and the train, such that its motion would go on forever.
Now suppose that the train emits photons from its backside, parallel to the ground.
So its "losing" energy and should go slower I believe. However I remember my professor said that even though photons are massless particles at rest, they carry a momentum. Hence... I'm tempted to say that the train should in fact go faster instead of going slower.
I'm sure I'm confusing a lot of things. Can you explain me what would really happen in such an idealized situation?
 

Answers and Replies

  • #2
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The train would go faster. This is the same principle as a rocket.
 
  • #3
Nabeshin
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Conservation of momentum. When the train emits a photon, it must recoil an equal amount.

For conservation of energy, the energy that led to the creation of the photon was almost certainly not the kinetic energy of the train! Perhaps it was a battery connected to a laser or some such mechanism, but definitely not the train itself. So this line of argument is ineffective.
 
  • #4
fluidistic
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Ok thanks for the replies.
Is there a way to be sure the photon wasn't emitted due to a loss of kinetic energy?
What would happen in such a case? Is it even possible?
I'm not even sure it makes sense to talk about a loss of KE since the train would gain some KE from the momentum of the photon... and hence it looks like possible for the train to keep a stable velocity and emitting photons, which is impossible. So it'd lose KE after all I guess, even though it emits photons, which also looks like an impossibility.
 
  • #5
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Actually, this depends on the frame of reference you are using.

Let's say that we are generating photons by annihilating positrons and electrons, and that each time we are sending one photon straight forward and one straight behind. In the rest frame the momentum of the forward and backward photons are equal, so the train stays at rest and therefore the KE of the train is unchanged.

Now, imagine from a reference frame in which the train is moving. The forward photon is blueshifted so it carries more momentum forward than the backward photon carries. So by conservation of momentum the trains momentum must decrease, but the train's velocity cannot change, so the momentum decreases due to the reduction in the mass of the train, therefore the KE is also reduced in this frame for the same reason.
 
  • #6
fluidistic
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Actually, this depends on the frame of reference you are using.

Let's say that we are generating photons by annihilating positrons and electrons, and that each time we are sending one photon straight forward and one straight behind. In the rest frame the momentum of the forward and backward photons are equal, so the train stays at rest and therefore the KE of the train is unchanged.

Now, imagine from a reference frame in which the train is moving. The forward photon is blueshifted so it carries more momentum forward than the backward photon carries. So by conservation of momentum the trains momentum must decrease, but the train's velocity cannot change, so the momentum decreases due to the reduction in the mass of the train, therefore the KE is also reduced in this frame for the same reason.
Thanks a lot. Yeah during the day I had in mind that the mass of the train has to decrease... unfortunately my physics knowledge is too little to make any equation saying what I'm thinking regarding this problem.
I get the idea, that's what matter.
Problem solved!
 

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