Rest Energy and a 1 million solar mass black hole

In summary, the conversation discusses a scenario where a 1 million solar mass black hole is producing electron-positron pairs at various distances from the event horizon. The positron's vector is directly "up" away from the singularity and reaches an apogee before falling towards the singularity. Another pair is produced at a distance 1/2A from the event horizon and the electron's vector is also directly "up" with the same initial vector as the positron. The electron reaches its apogee at distance C at the same time as a collision occurs with the positron. It is asked whether the annihilation photons will have more, equal, or less energy than 1.022 MeV combined when measured from the center
  • #1
metastable
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53
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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  • #2
You do realize that A=B and C=A/2 the way you have set this up, right?
 
  • #3
Dale said:
A=B and C=A/2

do you mean A=B and C=A/2 or A=~B and C=~A/2?
 
  • #4
metastable said:
A 1.022MeV photon

1.022 MeV relative to what observer? Energy is relative.

metastable said:
another 1.022MeV photon

Relative to what observer?
 
  • #5
PeterDonis said:
Relative to what observer?

Suppose each production has 2* rest energy of electron in center of mass. (edit: 2* just the electron)
 
  • #6
metastable said:
do you mean A=B and C=A/2 or A=~B and C=~A/2?
Why would I mean A=~B and C=~A/2 when I said A=B and C=A/2? I meant exactly what I said.
 
  • #7
I didn't know if the distances between A and B might be very short when compared to the distance to the singularity.
 
  • #8
metastable said:
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?
 
  • #9
Dale said:
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?
 
  • #10
metastable said:
Suppose each production has 2* rest energy of electron in center of mass.

This doesn't help because you haven't specified the state of motion of the center of mass.
 
  • #11
To calculate what happens at point C, can I just specify point B, leaving point A out of the equation?
 
  • #12
metastable said:
How can these 2 distances from the center of the black hole be the same?
What is the kinetic energy of the positron when it is first created at A?
 
  • #13
Dale said:
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.
 
  • #14
metastable said:
I see. I suppose it's easier to say C = 1/2B as well.
Yes. So now going back to this:

metastable said:
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?
 
  • #15
Intuition tells me the answer is more, but I have no idea.
 
  • #16
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...
 
  • #17
metastable said:
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.
 
  • #18
metastable said:
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
 
  • #19
PeterDonis said:
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.
 
  • #20
Dale said:
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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  • #21
metastable said:
all the distance in the original post refer to distances to the event horizon

Ok, that clarifies things.
 
  • #22
PeterDonis said:
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.
 
  • #23
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.
 
  • #25
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?
 
  • #26
Why don’t you explicitly derive the formula. Then you can decide if it is simple or not.
 
  • #27
I’m sure it’s fiendishly complicated but if I’m not mistaken there are only 2 variables... mass of the singularity and the distance B from the event horizon...
 
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  • #28
Roughly how many equations do you think I’ll have to combine? I assume step 1 is converting the 1 million solar masses of the black hole into electron volts... or convert the mass of the positron into solar masses...
 
  • #29
You will also want to calculate the amount of KE gained by falling a distance A/2. Once you have that then you can transform to the center of momentum frame to see what the energy is.
 
  • #30
Can I simply derive the value of g at distances B & 1/2B then treat the fall from B to 1/2 B as falling towards a normal planet?
 
  • #31
Dale said:
A=B because of the energy you chose, and same with C=A/2.

That works for the energy but not for the momentum. Something is missing here.
 
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  • #32
metastable said:
Can I simply derive the value of g at distances B & 1/2B then treat the fall from B to 1/2 B as falling towards a normal planet?
As long as B/2 is much larger than the Schwarzschild radius that is a fine approximation.
 
  • #33
Thanks for your answer DrStupid, but the quote is misattributed- its says "metastable" made that quotation but it was Dale.
 
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  • #34
DrStupid said:
That works for the energy but not for the momentum. Something is missing here.
Assuming that the atom is very massive then the atom can receive any of the momentum without significantly impacting the energy. That is what I assumed.
 
  • #35
Dale said:
As long as B/2 is much larger than the Schwarzschild radius that is a fine approximation.
If I'm particularly interested in distances "close" to the event horizon, what corrections will I have to make to the "falling halfway to a planet surface" approximation?
 

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