Capturing Photons in a Reflective Ball

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The discussion centers on the theoretical possibility of capturing photons within a perfectly spherical, reflective ball using a Faraday isolator. Participants explore whether continuously pumping light into the sphere would increase its mass, noting that while photons have no rest mass, their energy contributes to the system's relativistic mass. The conversation also touches on the implications of reaching a point where the sphere can no longer accept more photons, speculating about potential outcomes like black hole formation. Ultimately, the consensus is that while the sphere could theoretically gain mass from the energy of the photons, practical limitations and the laws of physics impose constraints on this scenario. The complexities of photon behavior and energy conservation are key points of contention in the discussion.
  • #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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