Light Bulb in a Box: Velocity Change After Release of Photon?

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Homework Help Overview

The discussion revolves around a hypothetical scenario involving a box traveling in space that contains a light bulb capable of emitting a photon. The original poster questions whether the box's velocity changes upon the release of a photon, considering the conservation of energy within the system.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of energy conservation when a photon is emitted from the box. Some express skepticism about the feasibility of the scenario, likening it to a perpetual motion machine. Others question how the velocity of the box could be affected by the emission of light, particularly considering the momentum of the photon.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have provided insights into the conservation of energy and momentum, while others have raised concerns about the practicality and theoretical underpinnings of the scenario. There is no explicit consensus, but the conversation appears to be clarifying certain concepts for the participants.

Contextual Notes

Participants note the limitations of the hypothetical scenario, including the challenges of defining kinetic energy in the box's frame of reference and the implications of using light as a means of propulsion. There is a recognition that the scenario may not be applicable in real-world physics.

Matt Callicott
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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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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.
 
lekh2003 said:
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.
 
Matt Callicott said:
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.
 
lekh2003 said:
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.
 
Matt Callicott said:
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.
 
lekh2003 said:
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.
 
Matt Callicott said:
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.
 
haruspex said:
There is no way to do this. In the frame of reference of the box, there is no KE.

I just did some searching on google... I found this similar question used as a derivation of E=mc^2

http://www.adamauton.com/warp/emc2.html
 
  • #10
haruspex said:
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?
Matt Callicott said:
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
 
  • #11
Matt Callicott said:
I just did some searching on google... I found this similar question used as a derivation of E=mc^2

http://www.adamauton.com/warp/emc2.html
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.
 
  • #12
haruspex said:
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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