Dale
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Why don’t you explicitly derive the formula. Then you can decide if it is simple or not.
That works for the energy but not for the momentum. Something is missing here.A=B because of the energy you chose, and same with C=A/2.
As long as B/2 is much larger than the Schwarzschild radius that is a fine approximation.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?
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.That works for the energy but not for the momentum. Something is missing here.
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?As long as B/2 is much larger than the Schwarzschild radius that is a fine approximation.
Will these corrections generally increase or decrease the expected KE relative to the over-simplified "falling halfway to a planet surface" approximation as distance B decreases?need to solve the geodesic equations for a massive particle falling in a Schwarzschild spacetime
I don't know, I would have to do the calculation to find out. In all likelihood those calculations have already been done somewhere, but I don't have a reference ready.Will these corrections generally increase or decrease the expected KE relative to the over-simplified "falling halfway to a planet surface" approximation as distance B decreases?
Sorry. I fixed it.Thanks for your answer DrStupid, but the quote is misattributed- its says "metastable" made that quotation but it was Dale.
That makes sense. Within the given accuracy of four significant digits for the total energy, everything heavier than hydrogen should do the job.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.
That can be easily fixed by having the electron-positron pair produced by a pair of photons instead of a single photon. The photons would just have to be such that they had zero total linear momentum relative to an observer momentarily at rest at the given altitude.That works for the energy but not for the momentum.
If you are restricting yourself to events outside the horizon, the distribution of mass inside the horizon does not matter at all. The only thing that matters is that there is vacuum outside the horizon and that the spacetime is spherically symmetric. The total mass is the only relevant parameter in this regime.Can the black hole be modeled as having its mass in a thin shell at the event horizon?
Huh? Where did this come from?during the core collapse of a supernova, we won't see matter at the core "falling upwards" towards an over-density at the forming event horizon.
What you need are the equations describing geodesic (free-fall) motion in the Schwarzschild spacetime geometry. Most GR textbooks cover this; I believ Sean Carroll's online lecture notes do as well:I’m still just looking at how to correctly model the expected KE when distance B is relatively close to the event horizon.
Around equationI believ Sean Carroll's online lecture notes do as well:
I told you how to correctly model it. You have to solve the geodesic equation.I’m still just looking at how to correctly model the expected KE when distance B is relatively close to the event horizon.