B Rest Energy and a 1 million solar mass black hole

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Suppose there is a 1 million solar mass black hole. A 1.022MeV photon near a nucleus produces an electron-positron pair a distance=A from the event horizon. The positron’s vector is directly “up” away from the singularity. It doesn’t have escape velocity. It reaches an apogee at distance=B from the event horizon, then falls directly towards the singularity. Later another 1.022MeV photon near a nucleus at distance=1/2A from the event horizon produces another pair. The electron’s vector is directly “up” away from the singularity and is in fact the same initial vector as the positron. The electron doesn’t have escape velocity and reaches apogee at distance=C from the event horizon at the same time as a collision occurs with the positron. Will the annihilation photons have more, equal, or less energy than 1.022MeV combined when measured from the center of mass frame?
 
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You do realize that A=B and C=A/2 the way you have set this up, right?
 
A=B and C=A/2
do you mean A=B and C=A/2 or A=~B and C=~A/2?
 
Relative to what observer?
Suppose each production has 2* rest energy of electron in center of mass. (edit: 2* just the electron)
 
I didn't know if the distances between A and B might be very short when compared to the distance to the singularity.
 
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I didn't know if the distances between A and B might be very short when compared to the distance to the singularity.
The distance to the singularity has nothing to do with it. Also that distance is not physically meaningful anyway. A=B because of the energy you chose, and same with C=A/2.

Can you see that now?
 
Can you see that now?
I don't understand, I thought I specified at first the positron is moving away from the singularity at production point A, then next, without escape velocity from the black hole, it reaches an apogee with respect to the singularity at point B. How can these 2 distances from the center of the black hole be the same?
 
To calculate what happens at point C, can I just specify point B, leaving point A out of the equation?
 
What is the kinetic energy of the positron when it is first created at A?
I see. I suppose it's easier to say C = 1/2B as well.
 
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I see. I suppose it's easier to say C = 1/2B as well.
Yes. So now going back to this:

The electron doesn’t have escape velocity and reaches apogee at distance=C from the event horizon at the same time as a collision occurs with the positron. Will the annihilation photons have more, equal, or less energy than 1.022MeV combined when measured from the center of mass frame?
Do you see the answer now?
 
Intuition tells me the answer is more, but I have no idea.
 
In fact intuition tells me there'd be so much energy you wouldn't get just photons, but I also have no idea about this point either...
 
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distances from the center of the black hole
There is no such thing; the "center" of the black hole (the locus ##r = 0##) is not a place in space. It's a moment of time, which is to the future of all events inside the hole.
 
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Intuition tells me the answer is more, but I have no idea.
Yes, it is more. Since the infalling positron has KE the system of the positron and electron have more than 1.022 MeV energy
 
There is no such thing; the "center" of the black hole (the locus r=0r=0r = 0) is not a place in space. It's a moment of time, which is to the future of all events inside the hole.
Ah, thank you I misspoke in that post since all the distances in the original post refer to distances to the event horizon.
 
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A=B because of the energy you chose
I actually think the problem as stated in the OP is underspecified, because, as I pointed out in previous posts, energy is relative. The energy needs to be specified relative to some observer who is co-located with the electron-positron pair when it is created (and this needs to be done separately for each pair).

If the energy of the electron-positron pair created at A is specified relative to an observer who is momentarily at rest at altitude A, then I agree that A = B (since the created electron and positron will both be momentarily at rest at altitude A as well). But the OP did not specify that (though he might have intended to).
 
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If the energy of the electron-positron pair created at A is specified relative to an observer who is momentarily at rest at altitude A, then I agree that A = B
That is what I assumed was intended.
 
I am saying a positron at apogee with no orbital velocity falls from distance B from the event horizon to distance C from the event horizon which equals 1/2B, where the positron collides with an electron just reaching its apogee, also with no orbital velocity.
 
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Yes, that is what I understood
 
To prove the answer is “more” is there a simple formula to find the kinetic energy of the positron at point C with respect to the electron?
 

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