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Apr3-12, 10:13 AM
Sci Advisor
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P: 5,080
Quote Quote by clamtrox View Post
Don't do that! The term you added already contains the second term. There are many ways of writing the same thing, and I think everyone benefits if we don't try to confuse things further by using angular momentum density in one term and velocity in the next.

So I will stick to my earlier assumptions that particle velocity is relativistic, but the mass of the gravitating object is not: this means we don't get a circular orbit, and you can't use the "usual" formalism that one uses with for example Mercury precession.

The equation for acceleration to order vē/cē is
[itex] \mathbf{a} = -\frac{GM\hat{\mathbf{r}}}{r^2} + \frac{4GM(\hat{\mathbf{r}} \cdot \dot{\mathbf{r}}) \dot{\mathbf{r}}}{c^2 r^2} - \frac{GMv^2 \hat{\mathbf{r}}}{c^2 r^2} [/itex]

If you have difficulties with vectors, it's very easy to write it without them: Let me define
[itex] a_r = \mathbf{a} \cdot \hat{\mathbf{r}} [/itex] and
[itex] a_{\perp} = |\mathbf{a} - a_r \hat{\mathbf{r}}| [/itex] and likewise for velocity v. Then
[itex] a_r = -\frac{GM}{r^2} (1 + \frac{v^2-4 v_r^2}{c^2}) = -\frac{GM}{r^2} (1 + \frac{v_{\perp}^2-3 v_r^2}{c^2}) [/itex] and
[itex]a_{\perp} = \frac{4 GM v_r v_{\perp}}{c^2 r^2} [/itex]
Did that make it clearer?
And all this simplifies, for a high speed flyby, at closest approach (where radial v=0) and all acceleration is radial:

a = -Gm/r^2 (1+v^2/c^) =(appx) -gamma^2 Gm/r^2