Photons are bosons so, in principle, they are perfectly happy to exist together in the same state.
But is there a limit on how many can exists in a finte volume? Does QED or vacuum breakdown eventually kick in?
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Are you asking whether there's a limit, or whether there's a density beyond which the state becomes more complicated? – Jan 18 '18 at 03:03
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See kugelblitz – PM 2Ring Jan 18 '18 at 04:59
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What’s the complicated state that you are thinking about @BenCrowell ? Something beyond $a^{\dagger} |0\rangle$? – SuperCiocia Jan 18 '18 at 22:51
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The only fundamental limit is that eventually the energy density gets high enough that you form a black hole. In principle, you could achieve an unbounded number density by decreasing the energy of each photon while increasing the number (and so keeping the energy density fixed, preventing the formation of a black hole).
Chris
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Photons propagate at the speed of light. How would they form a black hole? Wouldn't it violate the conservation principles? – flippiefanus Jan 18 '18 at 04:52
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1@flippiefanus When the stress-energy density in a region is high enough a black hole forms. What conservation law do you think would be violated? – PM 2Ring Jan 18 '18 at 05:05
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2@flippiefanus Black holes made from only radiation even have a name: kugelblitz. – Chris Jan 18 '18 at 11:44
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@MaDrung No, a black hole is massive, so it can't move at the speed of light. For any collection of photons that merge to form a black hole, there is some inertial frame in which the total momentum of the photons is zero. When the photons form a black hole, its speed will be zero in that reference frame. – Chris Jan 19 '18 at 07:48
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But what about the case where some photons move in one way, but a lot more of them move in the other and are alone sufficient to form a kugelblitz? Then it would actually move? – MaDrung Jan 19 '18 at 07:51
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1@MaDrung It would move, but not at the speed of light. In the instance you are talking about, there is some reference frame where the fewer photons individually have much higher energy, and in that frame the black hole is created at rest. – Chris Jan 19 '18 at 07:53
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Sorry when you say that ‘Black holes are massive’, how does that reconcile with the photons having no mass? – SuperCiocia Jan 21 '18 at 18:46
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@SuperCiocia A collection of things can have a different mass than its constituents. Not to mention by the time the black hole forms, it's not made of photons- a black hole can have charge associated with a long-range field, but besides that a black hole is a black hole. There's no difference between a black hole formed of photons and a black hole formed of baryonic matter. – Chris Jan 21 '18 at 21:35
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Yes but mass of bound system is the mass of individual constituents - binding energy . The individual constituents being the photons, which are massless, the only contribution to the mass of the black hole is the binding energy - what will it be? – SuperCiocia Jan 21 '18 at 21:55
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@SuperCiocia In special relativity, that is true. It is not the case in general relativity. Even if it were, the black hole has no constituents. – Chris Jan 21 '18 at 22:01
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1@SuperCiocia You cannot say in general that the mass of a bound system is the mass of its individual constituents minus the binding energy. In GR you can have (unstable) gravitationally bound systems of just photons. Such a system attracts objects gravitationally in its rest frame (ergo it has mass), but it is bound and its constituents have no mass. So clearly that formula doesn't work. – Chris Jan 21 '18 at 23:10
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Is it bound because of the tightly curved spacetime? But if the Kugelblitz were to move around, what would its effective mass be? You're saying it's not going to be moving at the speed of light, so it has to have a finite "mass". Where would I get its effective mass from? – SuperCiocia Jan 21 '18 at 23:50
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@SuperCiocia This is getting to be a bit long for a comment thread, so if you have any more questions after this, please ask a new question. You can link it to me here if you want me to take a stab at it. Anyway, it gets its mass from the energy of the photons that went into making it. – Chris Jan 22 '18 at 01:34
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I actually found two answers:
1) Kugelblitz, the black hole thing explained by Chris in his answer.
2) When the energy density becomes comparable to $\frac{m_e c^2}{\lambda^3_{\textrm{Compton}}}$, QED vacuum breakdown occurs. Photons will create particle-antiparticle pairs. In order for these to be real, some other particle needs to present to conserve momentum.
SuperCiocia
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I think you may be confusing two-photon pair production with one photon pair production. The latter indeed requires the presence of a nucleus to conserve momentum, the former does not. – Selene Routley Jan 18 '18 at 22:53
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Wait but two incoming photons have a "centre of mass frame" moving at $c$, whereas the outgoing massive particles have one at $v<c$? – SuperCiocia Jan 18 '18 at 23:35
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Yes, if the light is a unidirectional beam. But not in general: if the photons are in a cavity, the COM frame is at rest relative to the cavity. – Selene Routley Jan 18 '18 at 23:40