4.3 The Orbit Equations of a Massive Particle |
169 |
Fig. 4.8 Schwarzschild geometry: Enlargement in the region around the stable minimum of the effective
potential√ for
= 2 L2 = 24m2
as r → 0. In Figs. 4.7 and 4.8 we have plotted the effective potential for L2 = 24m2, namely for a value of the angular momentum that is just above the stability bound. For this value√ of the parameters the potential starts developing√ both a minimum r+ = 6(2 + 2)m = 20.4853m and a maximum r− = 6(2 − 2)m = 3.51472m. As predicted from our general discussion, the maximum r− corresponding to the unstable circular orbit falls in the interval ]3m, 6m[, while the minimum r+ corresponding to the stable orbit is larger than 6m. What is physically important to note is that r+ grows rapidly with angular momentum and it is already far away from the Schwarzschild emiradius at these small values of L2.
Energy of a Particle in a Circular Orbit Let us now calculate the energy of a test particle in motion on a circular orbit and compare our result with the Newtonian theory. We start from (4.3.10) that on a circular orbit (r˙ = 0) reduces to:
1 |
E 2 = |
1 |
− |
m |
+ |
L2 |
mL2 |
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2r2 − |
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(4.3.33) |
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2 |
2 |
r |
r3 |
If the orbit is extremal, namely if r = r± we have the relation (4.3.17) between the radius and the angular momentum that can be solved by:
L2 = |
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mr2 |
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(4.3.34) |
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r |
− |
3m |
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Inserting (4.3.34) into (4.3.33) we obtain: |
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E (r) = |
√ |
r − 2m |
(4.3.35) |
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r(r − 3m) |
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which yields the general relativistic formula for the energy of a test particle in circular motion around a spherically symmetric massive body like the Sun or the Earth. It is very instructive to make a post-Newtonian development of this formula and compare it with the purely Newtonian result. What is the small expansion parameter for the post-Newtonian approximation? This is to be established in the natural units we have adopted. At G = c = 1 both r and the Schwarzschild emiradius m are measure