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GRB Question

  1. Nov 8, 2003 #1

    selfAdjoint

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    It's embarassing to have to ask this, but I have never seen this issue discussed. Is there an upper limit on the energy of the gamma rays from the gamma ray bursts? Namely 1.022 MeV? Because any gamma ray of that energy or greater can and will produce pairs of electrons and positrons by supplying its energy to the quantum vacuum. In accelerator experiments a high energy photon cant get more than a few centimeters without doing this.
     
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  3. Nov 8, 2003 #2
    Gamma ray bursts have been observed up past the TeV range.

    I didn't realize the range in accelerators was so short, but if so, it's because it's interacting with other particles. A photon in pure vacuum can't pair-produce real particles (this violates conservation of energy-momentum). Space is a very good vacuum, better than anything we can produce.

    Photons in GRBs do lose energy through higher-order interaction with microwave background photons, infrared photons in galaxies, etc. But they still have plenty of energy left when they get here.
     
    Last edited: Nov 8, 2003
  4. Nov 9, 2003 #3
    If you think that's embarrasing wait till you start getting hair growing in "funny" places and your voice changes.
     
  5. Nov 9, 2003 #4

    selfAdjoint

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    Uh, no. The photon energy and momentum go away and are replaced by the particles' energy (mass and kinetic) and momentum.All conservation laws obeyed. And those experimental chambers are evacuated; who wants uncontrolled interactions? So what is the explanation of the TeV gammas' success in crossing astronomical distances?

    And zooby, at my age I worry more about losing hair than gaining it.
     
  6. Nov 9, 2003 #5
    Incorrect. This is well-known; it's the same reason why pair annihilation in vacuum can't produce a single photon, but must always produce two.

    Consider the physics in the center-of-momentum frame of the produced pair: the net momentum will be zero, and the net energy will be 2γmc2, where m is the mass of one of the produced particles.

    For energy and momentum to be conserved, the photon that produced them would have to have had energy 2γmc2, and zero momentum. A photon cannot have zero momentum, and in fact the momentum of a photon with energy 2γmc2 must exactly equal 2γmc (from E=pc).

    (If you had two photons, then they could each have energy γmc2 and momentum γmc, but in opposite directions, so the net momentum would be zero, and the conservation laws would be obeyed.)

    Anyway, just think about it: you're claiming that there is some energy cutoff for photons in vacuum. But the state of a single particle in vacuum must be Lorentz invariant: if the photon is above the cutoff in one frame, it's below the cutoff in another. From symmetry, it can't pair-produce in one frame but not another.

    It's only when you have other particles around, that can define a preferred frame, that you can have real pair production: e.g., if the photon has energy above some cutoff as measured in the rest frame of another particle (or system of particles, like a photon gas) that it's exchanging energy and momentum with. (Or, instead of looking at it from a symmetry perspective, you can use the previous argument that there has to be another particle around for pair production to obey the conservation laws.)

    The experimental chambers are not perfect vacuum, as I said.
     
    Last edited: Nov 9, 2003
  7. Nov 9, 2003 #6

    selfAdjoint

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    I'm not going to give up yet. Take the rest frame to be that of the point where production occurs. In THIS frame the sum of momenta has to be zero. The incoming photon has a momentum. The outgoing particles aren't standing still in this frame they are moving away. Their momentum relative to their own center of mass is indeed zero, But what balances the incoming momentum of the photon? The two particles center of mass has a momentum relative to the point of production and that balances the photon momentum. The track is not a tee but a vee.

    I am just not convinced by your statement that pair produiction is always mediated by some other matter. Do you know the "Two photon" decays in QCD and their explanation?
     
  8. Nov 9, 2003 #7
    A rest frame isn't defined by just a point, it's defined by a state of motion. We can take my frame to have an origin centered on the point where production occurs, too. Which frame are you talking about? A rest frame at rest with respect to what?

    But that's the frame I used!

    I still don't know what frame you're working in. Above, you said that the sum of the momenta of the system was zero in this frame; now you say that the produced particles' center of mass has a nonzero momentum. But a frame in which the net momentum is zero is a frame in which the center of mass of the two particles has zero momentum.

    Anyway, go ahead and just write down both the relativistic energy and momentum conservation equations in whatever frame you're working in, and you will find that you simply cannot satisfy them both.

    Fine. Then point out the error in either the conservation proof or the symmetry proof. You ignored them both and presented your own, non-mathematical handwaving argument. If you want to claim that everything balances, prove it. I can dig up some supporting references if you want, but they just say the same thing I already did.

    And for the record, pair production from a photon doesn't have to be mediated by matter; it can be mediated by, say, another photon (in the inverse of pair annihilation). But it has to be mediated by some other real particle.

    Note also that this argument applies specifically to pair production from a single massless particle. A massive particle can pair-produce in vacuum.

    No. What is a "two photon decay" in QCD, and does it have anything to do with pair production from a single photon?
     
  9. Nov 10, 2003 #8
    We have been busy DickT?..I see the connection, and we have discussed this elsewhere, in superstringtheory.com

    In my 'original' question:Particle "Virtual" question at this location:http://www.superstringtheory.com/forum/partboard/index6.html

    The answer was in the Question! I do see that Patricia is re-evaluating the superstringtheory site, it remains to be seen if there is going to be a worthwhile!

    There has been an embarrassing amount of really interesting discussions, and Selfadjoint/DickT? you may wish to ponder the 'original' Question and how you responded to it..the process action-reaction in the responses you gave gives a good account of your current knowledge "in the context of this emmbarrassing question you raise here in PF":wink:
     
  10. Nov 10, 2003 #9

    selfAdjoint

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    Ambitwister, I concede, and I did learn something. I used a comoving frame that passes through the spacetime point of the supposed interaction and has a velocity equal to that of the center of motion of the output particles. Then in that frame we have the photon momentum pc, say along the x-axis, and the output momentum is zero since the frame coincides always with the output center of momentum. So momentum cannot be conserved and the interaction is impossible.

    This of course is what you said, but I had to explain it to myself.

    The "two photon decay" should have been the "two photon interaction", which I think is what you mentioned.
     
  11. Nov 10, 2003 #10
    Same here. Haven't you noticed your voice getting a little more gravelly, and hair growing in your ears?
     
  12. Nov 19, 2003 #11

    Nereid

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    Mean-free path of TeV (+) gammas?

    Inter-galactic space sure is a good vacuum, many orders of magnitude better than anything we can create here on Earth. However, it's not perfect.

    Besides, there's the CMB everywhere.

    Put all this together and some folk felt (feel?) that really high energy gammas can't possibly reach us from beyond ~100Mpc (there's even a name for this, which I can't remember just now). Trouble is, to the extent that space sources of such can be identified, blazers and their ilk seem to be prime suspects (along with supernovae shock fronts).

    For sure, GLAST (and Auger?) will make things a lot clearer, but find lots new too.

    Incidently, some models of GRBs suggest copious production of gammas with energies of 1,000 TeV and above. Enough energy in a single c-speed particle to produce observable GR effects on the curvature of space-time?
     
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