Light Bulb in a Box

  • #1

Homework Statement


[/B]
A box is traveling with a certain velocity in space. Inside the box is a light bulb, which is capable of releasing one photon of light. The light bulb harnesses the energy from the box; it does not require an outside power source but instead uses the energy on the system.

If the box is traveling with a velocity ν1 and then releases one photon of light (but does not change the total energy of the system since the light stays within the box) does the velocity of the box change?

* This is not a homework question but rather just individual thought. Thank you for any help!


Homework Equations



E = ½ Mv2[/B]

Ephoton ≠ 0


The Attempt at a Solution


[/B]
Before the release of a photon: ETotal = ½M(v1)2

After the release of a photon: ETotal = ½M(v2)2 + Ephoton

Since Ephoton ≠ 0 and since ETotal is conserved, v2 < v1.

Thank you again!
 
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Answers and Replies

  • #2
lekh2003
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I'm not sure, but I think this is impossible. You are using energy from a photon to increase the energy of a box which provided the energy in the first place. It reminds me of an attempt at a perpetual motion machine and I'm sure this violates the laws of thermodynamics.
 
  • #3
I'm not sure, but I think this is impossible. You are using energy from a photon to increase the energy of a box which provided the energy in the first place. It reminds me of an attempt at a perpetual motion machine and I'm sure this violates the laws of thermodynamics.

A small amount of energy of the box is being converted to the energy of the photon. The total energy is not increasing but rather being moved around. The system's energy is conserved.
 
  • #4
lekh2003
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A small amount of energy of the box is being converted to the energy of the photon. The total energy is not increasing but rather being moved around. The system's energy is conserved.
Ok, I think I get it. I'm curious how you are going to transfer the velocity into a photon. Sure, your calculations make sense, but how is this ever going to work or be used in the real world?

From my thinking, the velocity will decrease. I can't think of how this is at all useful.
 
  • #5
Ok, I think I get it. I'm curious how you are going to transfer the velocity into a photon. Sure, your calculations make sense, but how is this ever going to work or be used in the real world?

From my thinking, the velocity will decrease. I can't think of how this is at all useful.

It's not... at all. This is definitely not applicable, just like there is not or will be a box in space with a light bulb that can produce a single photon by only converting kinetic energy to light. But thank you for your response. This was used to clarify a foggy concept.
 
  • #6
lekh2003
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It's not... at all. This is definitely not applicable, just like there is not or will be a box in space with a light bulb that can produce a single photon by only converting kinetic energy to light. But thank you for your response. This was used to clarify a foggy concept.
I'm happy I could help o_O.
 
  • #7
I'm happy I could help o_O.


Sorry to restart the discussion, but considering all of the established theoreticals, what if the photon shot from the right side of the box to the left and hit the side of the box? Light has momentum, right? Wouldn't that affect the motion of the box in addition to the change in kinetic energy?

Thanks.
 
  • #8
haruspex
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it does not require an outside power source but instead uses the energy on the system.
There is no way to do this. In the frame of reference of the box, there is no KE.
 
  • #10
lekh2003
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There is no way to do this. In the frame of reference of the box, there is no KE.
Yes that is another limitation to your hypothetical. There is no way to accurately define the velocity of the box and whether it actually takes energy to move the box, so can the hypothetical energy really be used?
Light has momentum, right? Wouldn't that affect the motion of the box in addition to the change in kinetic energy?
Yes, it is possible to use light as a source of moving an object (although limited). Look here: https://en.m.wikipedia.org/wiki/Solar_sail
 
  • #12
That is not using the KE of the box to emit the photon. Maybe that is not what you meant, but it is how I read your question.

Right.. Originally that was the aim of the question. However, as you pointed out, that is not possible. Not using KE makes more sense and is how I formed the question but while I was typing I added the KE conversion in there to try to make more sense (but it just added confusion).

Thanks.
 

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