Capturing Photons in a Reflective Ball

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Discussion Overview

The discussion revolves around the theoretical possibility of capturing photons within a perfectly spherical and reflective ball, particularly through the use of a Faraday isolator. Participants explore the implications of continuously pumping light into the sphere, the mass changes associated with photons, and the potential outcomes of such a scenario.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that if the inside of a sphere is perfectly reflective and a Faraday isolator is used, it might be possible to capture photons indefinitely.
  • Others argue that the mass of the sphere would increase as photons are continuously pumped into it, contingent on the sphere remaining still.
  • A later reply questions the relevance of light having mass in this context, suggesting that photons have no rest mass but can contribute to the energy of the system.
  • Some participants discuss the implications of photons affecting the mass of the sphere, with references to Einstein's equation E=mc² and the concept of relativistic mass.
  • There are conflicting views on whether the sphere could be filled with photons without allowing any to escape, with some suggesting that a perfect reflector could theoretically allow this.
  • One participant speculates about the consequences of the sphere becoming so full of photons that it might collapse into a black hole, while others express skepticism about the feasibility of such a scenario.
  • Technical details regarding the momentum and energy of photons are discussed, with some participants providing mathematical formulations to support their claims.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the feasibility of capturing photons in the sphere or the implications of doing so. Multiple competing views remain regarding the mass changes and the physical principles involved.

Contextual Notes

Limitations include assumptions about the perfect reflectivity of materials, the conditions under which mass changes occur, and the unresolved nature of the mathematical implications discussed.

  • #31
xArcherx said:
Ball 1 is traveling from Point A to Point B with momentum p. Ball 2 is traveling from Point B to Point A and so it has the momentum -p in relation to Ball 1. The total momentum is...
|p| + |-p| = 2p
It's p+(-p)=0.

xArcherx said:
...but talking of photons I just can't see it. E = pc but E = hv also, where v = c/λ.
This should be E=|p|c. It only tells you the relationship between the energy and the magnitude of the momentum.
 
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  • #32
It's p+(-p)=0

Even when you are looking for the total momentum?
Where if ball 1 is 2 kg and ball 2 is 4 kg, both are traveling 10m/s then...

Momentum of ball 1 is 2 * 10 = 20
Momentum of ball 2 is 4 * 10 = 40
Total momentum should be 60 kg.m/s

If p means the ball is traveling from point A to point B then -p should simply mean the ball is traveling from point B to point A. Since we wouldn't have a negative mass, then we must have a negative velocity. A negative velocity would simply mean reverse direction just as a negative acceleration means a deceleration.

This should be E=|p|c. It only tells you the relationship between the energy and the magnitude of the momentum.

So does this mean that if we had two photons traveling toward each other from opposite directions, at the point of collision their momentums would stack and therefore so would their energy?
 
  • #33
xArcherx said:
It's p +(-p)=0
Even when you are looking for the total momentum?
Just because you're looking for total momentum. Momentum is a vector (I have written it in bold for this reason) and it's additive.
Where if ball 1 is 2 kg and ball 2 is 4 kg, both are traveling 10m/s then...

Momentum of ball 1 is 2 * 10 = 20
Momentum of ball 2 is 4 * 10 = 40
Total momentum should be 60 kg.m/s
If they travel in the same direction; if they travel in opposite directions, the total momentum is the difference.
If p means the ball is traveling from point A to point B then -p should simply mean the ball is traveling from point B to point A. Since we wouldn't have a negative mass, then we must have a negative velocity. A negative velocity would simply mean reverse direction just as a negative acceleration means a deceleration.
It's for this reason that the two momentums have opposite signs.
So does this mean that if we had two photons traveling toward each other from opposite directions, at the point of collision their momentums would stack and therefore so would their energy?
What do you mean with "stack"?
Their momentums are always opposite, not only where they "collide", and their energies are always different than zero (as their sum as well).
 
  • #34
What I mean by stack is...

If E = |p|c
Then if we have two photons, each will have it's own momentum. So this would give us...

Et = (|p| + |p|)c
 
  • #35
i like the emitter on the inside. i thought of the same thing a while back called it a light sink , wondered why it couldn't fill with light but i think losses in the reflection might be a large problem . I just didnt know if photons interact or just go aroung each other. I was wondering if you could confine them in a magnetic loop with a swept entry point then once inside they just accumulate going around following the field . even if you skewed the light entering you sphere by only a couple of angstroms along the x and y-axis it would travel so fast along the inside of the shere it would come right back out , I see that's why your imitter was internal and all surfaces reflective i just this you would have power loss absorbtions in the atoms of the glass i don't know. i would like to see this work . However even if it worked perfectly i wouldn't want to break the glass after you spent x amount of time filling it up with photons , could get a little toasty. Wear sun screen...
 

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