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Originally Posted by Yakk
m*a/(1-(v/c)^2)^.5 = -K/r^2
m/K * a * r^2 = -(1-(v/c)^2)^(1/2)
Let c = 1 by choice of units (ie, time seconds and distance in light-seconds -- always a good idea when working in relativity!)
r`` * r^2
---------------- = (-K)/m
(1-(r`^2))^(1/2)
This looks like a non-linear second order differential equation. I think you are fucked? My knowledge of this area of math is very sparse.
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Ok, I guess I'll let the c = 1 thing slide even though I forsee problems doing so... :shrugs:
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I'm wondering -- wouldn't there be time and space dialations/contractions as well, which are missing from your equation? This could either simplify or complicate the situation. I'd have to hit up my relativity-knowing friend.
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The time dialations aren't significant to the particles. The only contraction necessary from an outside observer is the mass dialation which is represented by m over gamma.
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Second, initial conditions will matter. It wouldn't be that hard to solve the position/accelleration/velocity if the two objects where moving in a perfect orbit around each other, or so I'd guess...
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Like I said, they start from rest at a distance
d apart.
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I also wonder if conservation of mass/energy might be of help here... From your perspective, as you see someone falling towards you, they are losing potential 'gravitic' energy and gaining kinetic energy. Isn't this a net zero change in their energy, and hence a net zero change in their mass?
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I don't think so. You contradicted yourself. The net energy is not zero because relative to one object, the other is moving rather fast. But still, 2 x KE of one will be the energy of the system at that instant. Don't we just love infinities???