Capturing Photons in a Reflective Ball

[nuncoop]

Is it possible to capture photons within a perfectly spherical/reflective ball by aiming a faraday isolator within the ball?

 

[nuncoop]

Maybe I wasn't clear, my idea is to coat the inside of a sphere with a perfectly mirror-like substance, and aim a Faraday isolator inside the sphere. If light has mass, and I were able eliminate any chance of photon-leakage, could I not continue to pump more and more light into the sphere? Someone tell me why this wouldn't work..

 

[jtbell]

Why do you think light having or not having mass would have anything to do with the answer?

 

[nuncoop]

I forgot to include the second part of my question :P. Would the sphere become more massive if photons were continuously pumped into it?

 

[jwalker1196]

The light and therefore the mass is coming from somewhere, and that somewhere is getting lighter by definition. If that somewhere is within the sphere already, then the sphere would maintain its mass.

 

[nuncoop]

Ok, what if I were to collect the photons from outside the sphere and move them within the sphere? Can this be done?

 

[jwalker1196]

Let us pretend you could. Then the sphere would indeed get more massive. Now, if you were asking could this be done, I would have to say no. You'd have to open the sphere to put them in, allowing for the escape of photons already inside.But I'll do you one better. Let's say the inside of the sphere is perfectly reflective. Let's also put a hypothetical "photon emitter" in the sphere. This emitter is itself a perfect 1-way reflector, on its outside. Now, requiring power, we run a wire from the emitter through the sphere wall. This wire and its exit point shall also be coated in 100%-reflective material. Now we have an emitter which will continue to do so as long as it has power, filling a sphere ad infinitum.

The sphere becomes more and more massive. The loss is from outside the sphere, so we aren't breaking any Laws of Conservation. The real question is, what happens when the sphere becomes so full of photons that it cannot accept any more?

 

[cesiumfrog]

Yes, if you captured a pulse of light inside a perfect mirror-box, the box would be heavier than before. Not sure quite how your Faraday isolator is going to help (don't they just absorb the escaping photons? In that case you might as well ask: if I heat a potato, whether by using a laser or microwave or any other means, does it get heavier? It does, but the change is negligible).

 

[PatPwnt]

Let us pretend you could. Then the sphere would indeed get more massive. Now, if you were asking could this be done, I would have to say no. You'd have to open the sphere to put them in, allowing for the escape of photons already inside.


But I'll do you one better. Let's say the inside of the sphere is perfectly reflective. Let's also put a hypothetical "photon emitter" in the sphere. This emitter is itself a perfect 1-way reflector, on its outside. Now, requiring power, we run a wire from the emitter through the sphere wall. This wire and its exit point shall also be coated in 100%-reflective material. Now we have an emitter which will continue to do so as long as it has power, filling a sphere ad infinitum.

The sphere becomes more and more massive. The loss is from outside the sphere, so we aren't breaking any Laws of Conservation. The real question is, what happens when the sphere becomes so full of photons that it cannot accept any more?

Then all the light in the universe is in the sphere?

 

[505Kelvin]

...
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.
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The sphere becomes more and more massive. The loss is from outside the sphere, so we aren't breaking any Laws of Conservation. The real question is, what happens when the sphere becomes so full of photons that it cannot accept any more?

Singularity?

 

[Gear300]

The real question is, what happens when the sphere becomes so full of photons that it cannot accept any more?

It would be awesome if it formed a black hole (I'm not using much reasoning here but it would be awesome).

 

[lightarrow]

I forgot to include the second part of my question :P. Would the sphere become more massive if photons were continuously pumped into it?

As others have written, the answer is yes, but at the condition that the sphere remains still: if a still body acquire an energy E, then its mass increases of E/c^2 (this is the correct meaning of Einstein's famous equation E = mc^2), whatever the way it acquires energy (for example, even giving it a spin.)
So, in the caso of photons, to be nitpicking, we should add the condition that they enters (for example) from two diametrically opposite holes on the sphere, so that there is no variation in the sphere's momentum.

 

[xArcherx]

There is one problem. Photons have no mass.

E^2 = (mc^2)^2 + (pc)^2

m = 0

E^2 = (0*c^2)^2 + (pc)^2
E^2 = (0)^2 + (pc)^2
E^2 = (pc)^2
E = pc.

Energy equals the momentum multiplied by the speed of light.

So the energy of a photon is in its momentum, not mass. This can be broken down further...

E = hv
h = Planck's Constant
v = frequency

E = pc then E/c = p but with E = hv then...

p = (hv)/c

Now to get v we use λ (wave length).

v = c/λ

then

p = (h(c/λ))/c

So by knowing a particular photon's wavelength we can use Planck's constant and the speed of light to find the momentum. The momentum multiplied by c will get you the energy of the photon.

Now although there is no mass involved, when an electron absorbs a photon, the electron also takes on the photons momentum. This increase in momentum results in can increase in relativistic mass based upon the equation...

p = γmc rewritten as m = p/(γc)

So there is only an increase in relativistic mass that would be very slight (even with a massive amount of photons). This increase only exists as long as the electron are at the higher energy state which they don't tend to stay long as they will release the energy received as another photon.

 

[lightarrow]

There is one problem. Photons have no mass.

E^2 = (mc^2)^2 + (pc)^2

m = 0

But two photons traveling in opposite direction have:

E^2 = (mc^2)^2 + (pc)^2

p = 0 (because you have to add two equal and opposite momentum) so:

E^2 = (mc^2)^2 --> m = E/c^2

That is: the system of the two photons have mass.

 

[dst]

But two photons traveling in opposite direction have:

E^2 = (mc^2)^2 + (pc)^2

p = 0 (because you have to add two equal and opposite momentum) so:

E^2 = (mc^2)^2 --> m = E/c^2

That is: the system of the two photons have mass.

Invalid application. Each photon has no rest mass and energy equal to (pc)^2. With 2 photons in opposite directions the energy is (pc)^2 + (-pc)^2 = 2(pc)^2.

 

[Fredrik]

It's definitely the "E" that matters when you measure the mass of the sphere filled with light, not the "m". One way to see that the photons affect the result when you weigh the sphere is this: Imagine a photon bouncing up and down between the top and the bottom. The effect of gravity is to redshift the photons on the way up and blueshift them on the way down. This means that they will hit the bottom harder (with more momentum) than they hit the top, and that pushes the sphere down.

If the sphere doesn't explode when the pressure becomes high, then it would eventually have to collapse to a black hole, but any realistic material would of course break long before that.

 

[lightarrow]

Invalid application. Each photon has no rest mass and energy equal to (pc)^2. With 2 photons in opposite directions the energy is (pc)^2 + (-pc)^2 = 2(pc)^2.

Didn't understood what you said; furthermore, you mean to add energy squared?

 

[Keys]

I've thought of such a device in high school, the problem is finding a 100% reflective material. The construction is not a problem if you abandon these closed sphere or box ideas. All you need is 2 parabolic mirrors of different sizes. Position the mirrors facing each other so that the foci are in the same location. Then begin pumping light into the larger mirror from behind the smaller mirror. This device could also produce a laser like beam if a small hole is made in the center of either mirror.

 

[xArcherx]

(pc)^2 + (-pc)^2 = 2(pc)^2.

Thanks for this bit of info. It helps with my project. I had to take a second look though, lol. About the adding of a negative but I then clued in with the squares. Just so that I do understand...

you get 2(pc)^2 because (-pc)^2 = (pc)^2 and so...

(pc)^2 + (-pc)^2 = (pc)^2 + (pc)^2 = 2(pc)^2

 

[lightarrow]

Thanks for this bit of info. It helps with my project. I had to take a second look though, lol. About the adding of a negative but I then clued in with the squares. Just so that I do understand...

you get 2(pc)^2 because (-pc)^2 = (pc)^2 and so...

(pc)^2 + (-pc)^2 = (pc)^2 + (pc)^2 = 2(pc)^2

And which is the physical meaning of the quantity (pc)^2 + (-pc)^2?

 

[Fredrik]

Didn't understood what you said; furthermore, you mean to add energy squared?

And which is the physical meaning of the quantity (pc)^2 + (-pc)^2?

He's adding energy squared, not because of its physical meaning, but because you calculated the energy squared of a system of two photons incorrectly, (and then took the square root and solved for the mass).

But two photons traveling in opposite direction have:

E^2 = (mc^2)^2 + (pc)^2

p = 0 (because you have to add two equal and opposite momentum) so:

E^2 = (mc^2)^2 --> m = E/c^2

That is: the system of the two photons have mass.

You can't get the total energy squared by first adding the momenta and then inserting the result into the formula for energy squared. The sum of the squares of the two energies is what he said. The sum of the energies is $\sqrt{m^2c^4+p^2c^2}+\sqrt{m^2c^4+(-p)^2c^2}$. The contributions from the momenta clearly do not cancel each other.

 

[xArcherx]

m_r = m_0/(1-v^2/c^2)^2

m_r = relativistic mass
m_0 = rest mass

If the rest mass is 0 then regardless of velocity, the relativistic mass is still 0.

Because m = 0 then E^2 = (pc)^2. You then can't use E^2/c^4 = m to find the mass. You can't find the mass of something when the energy is entirely momentum based.

For something like an electron, that is a different matter. Electrons have a rest mass and so they have a relativistic mass due to their velocity. This means that they have both mass and momentum and so to find the energy of an electron you use the full equation...E^2 = (mc^2)^2 + (pc)^2

 

[lightarrow]

He's adding energy squared, not because of its physical meaning, but because you calculated the energy squared of a system of two photons incorrectly, (and then took the square root and solved for the mass).

But two photons traveling in opposite direction have:

E^2 = (mc^2)^2 + (pc)^2

p = 0 (because you have to add two equal and opposite momentum) so:

E^2 = (mc^2)^2 --> m = E/c^2

That is: the system of the two photons have mass.

You can't get the total energy squared by first adding the momenta and then inserting the result into the formula for energy squared. The sum of the squares of the two energies is what he said. The sum of the energies is $\sqrt{m^2c^4+p^2c^2}+\sqrt{m^2c^4+(-p)^2c^2}$. The contributions from the momenta clearly do not cancel each other.

Sorry Fredrik, (maybe it's the hot) I don't understand where is my mistake.
Which is the momentum of the system of the two photons traveling in opposite direction? It should be 0.
Which is the energy? Energy is additive.

$$E = E_1\ +\ E2\ =\ \sqrt{m^2c^4+p^2c^2}\ +\ \sqrt{m^2c^4+(-p)^2c^2}\ =\ |cp|\ +\ |cp|\ =\ 2|cp|\ \neq\ 0.$$

If you agree up to here, then I ask how you apply (if you do) the equation

$$E^2\ =\ M^2c^4\ +\ P^2c^2$$

to the entire system, where M and P are system's mass and momentum.
Thank you.

Or you mean that my mistake was to have used the same letter "m" to mean a single photon's mass and the system's mass? In this case I'm sorry to have generated confusion.

 

[Fredrik]

This looks like a mistake to me:

But two photons traveling in opposite direction have:

E^2 = (mc^2)^2 + (pc)^2

p = 0 (because you have to add two equal and opposite momentum) so:

E^2 = (mc^2)^2 --> m = E/c^2

That is: the system of the two photons have mass.

It looks like you're adding the momenta first, and then inserting the result into a formula that tells us the energy of a particle with a given mass and momentum. You're doing it right in #23.

Regarding the "relativistic mass" of the combined system: You can assign a relativistic mass to any system with energy simply by writing E=mc² just because the m defined this way must have dimensions of mass. When the system is a massive particle, this m turns out to be the same as the relativistic mass defined in a more meaningful way, and if the system is something else, like a massless particle or, as in this case, two massive particles, you just take this to be the definition of the relativistic mass.

This is kind of pointless, so I'd rather just talk about the energy of the system instead of its mass. But I guess it makes some sense to do this because a box with a bunch of photons in it is definitely heavier than the same box when it's empty, and I assume that the extra mass we measure when we weigh the box is equal to E/c². (I haven't seen a proof, but I'd be very surprised if there isn't one).

 

[lightarrow]

Regarding the "relativistic mass" of the combined system: You can assign a relativistic mass to any system with energy simply by writing E=mc² just because the m defined this way must have dimensions of mass. When the system is a massive particle, this m turns out to be the same as the relativistic mass defined in a more meaningful way, and if the system is something else, like a massless particle or, as in this case, two massive particles, you just take this to be the definition of the relativistic mass.

This is kind of pointless, so I'd rather just talk about the energy of the system instead of its mass. But I guess it makes some sense to do this because a box with a bunch of photons in it is definitely heavier than the same box when it's empty, and I assume that the extra mass we measure when we weigh the box is equal to E/c². (I haven't seen a proof, but I'd be very surprised if there isn't one).

I thought that what I wrote was such a proof: if you add photons to the box so that it remains still = total momentum acquired from the box is zero = total momentum of the photons injected is zero --> the system of injected photons have mass M = E/c^2 where E is the total energy added with the photons.

 

[xArcherx]

But two photons traveling in opposite direction have:

E^2 = (mc^2)^2 + (pc)^2

p = 0 (because you have to add two equal and opposite momentum) so:

E^2 = (mc^2)^2 --> m = E/c^2

That is: the system of the two photons have mass.

For photons, m = 0 so E = pc. Now if you were two have a p + (-p) scenario then you get E = 0. Essentially this is two photon waves canceling each other out (destructive interference). Since p = h/λ then -p = h/-λ. p + (-p) = h/λ + h/-λ = h/λ + (-h/λ) = 0

 

[jtbell]

For photons, m = 0 so E = pc. Now if you were two have a p + (-p) scenario then you get E = 0

No. Energy is a scalar, not a vector. Strictly speaking, it's related to the magnitude of the momentum: $E = | \vec p | c$ for a photon, or $E = \sqrt {(|\vec p|c)^2 + (m_0 c^2)^2}$ for a particle that has (rest) mass.

 

[xArcherx]

So it's |p| + |-p| where |-p| = |p| (of course) and so you get E = (2p)c ?
Also, isn't momentum a vector?

 

[jtbell]

So it's |p| + |-p| where |-p| = |p| (of course) and so you get E = (2p)c ?

Yes.

Also, isn't momentum a vector?

Yes, so the total momentum is p + (-p) = 0.

 

[xArcherx]

I can see the use of absolutes when referring to macroscopic objects of mass.

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

...but talking of photons I just can't see it. E = pc but E = hv also, where v = c/λ.

p = E/c
p = hv/c
p = h(c/λ)/c
p = h/λ

Then -p = h/-λ and so h/λ + (h/-λ) = h/λ - h/λ = 0

λ is the wavelength of the photon and would have to be negative to get a negative momentum. So if the wave of a photon were to be represented as sin(x) then would having a negative wavelength mean -sin(x)? Or would it simply be a wave propagating in the opposite direction?

 

[Fredrik]

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.

...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.

 

[xArcherx]

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?

 

[lightarrow]

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).

 

[xArcherx]

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

 

[bscfrnd]

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...