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Photon capture

  1. Oct 17, 2007 #1
    Is it possible to capture photons within a perfectly spherical/reflective ball by aiming a faraday isolator within the ball?
     
  2. jcsd
  3. Oct 17, 2007 #2
    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..
     
  4. Oct 17, 2007 #3

    jtbell

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    Why do you think light having or not having mass would have anything to do with the answer?
     
  5. Oct 17, 2007 #4
    I forgot to include the second part of my question :P. Would the sphere become more massive if photons were continuously pumped into it?
     
  6. Oct 17, 2007 #5
    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.
     
    Last edited: Oct 17, 2007
  7. Oct 17, 2007 #6
    Ok, what if I were to collect the photons from outside the sphere and move them within the sphere? Can this be done?
     
  8. Oct 17, 2007 #7
    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?
     
    Last edited: Oct 17, 2007
  9. Oct 17, 2007 #8
    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).
     
    Last edited: Oct 17, 2007
  10. Oct 19, 2007 #9
    Then all the light in the universe is in the sphere?
     
  11. Jun 27, 2008 #10
    Singularity?
     
  12. Jun 27, 2008 #11
    It would be awesome if it formed a black hole (I'm not using much reasoning here but it would be awesome).
     
  13. Jun 28, 2008 #12
    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.
     
  14. Jun 28, 2008 #13
    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.
     
  15. Jun 30, 2008 #14
    But two photons travelling 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.
     
  16. Jun 30, 2008 #15

    dst

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    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.
     
  17. Jun 30, 2008 #16

    Fredrik

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    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.
     
  18. Jul 1, 2008 #17
    Didn't understood what you said; furthermore, you mean to add energy squared?
     
  19. Jul 1, 2008 #18
    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.
     
  20. Jul 1, 2008 #19
    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
     
  21. Jul 2, 2008 #20
    And which is the physical meaning of the quantity (pc)^2 + (-pc)^2?
     
  22. Jul 2, 2008 #21

    Fredrik

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    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).

    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 [itex]\sqrt{m^2c^4+p^2c^2}+\sqrt{m^2c^4+(-p)^2c^2}[/itex]. The contributions from the momenta clearly do not cancel each other.
     
  23. Jul 2, 2008 #22
    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
     
  24. Jul 3, 2008 #23
    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 travelling in opposite direction? It should be 0.
    Which is the energy? Energy is additive.

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

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

    [tex]E^2\ =\ M^2c^4\ +\ P^2c^2[/tex]

    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.
     
    Last edited: Jul 3, 2008
  25. Jul 3, 2008 #24

    Fredrik

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    This looks like a mistake to me:
    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).
     
  26. Jul 4, 2008 #25
    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.
     
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