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From N. Straumann, General Relativity

Exercise 4.9: Calculate the radial acceleration for a non-geodesic circular orbit in the Schwarzschild spacetime. Show that this becomes positive for $r>3GM$. This counter-intuitive result has led to lots of discussions.

This is one of those problems where I have absolutely no clue what to do. Since it says non-geodesic, I can't use any of the usual equations. I don't know what equation to solve. Maybe I solve $\nabla_{\dot\gamma}\dot\gamma=f$ with $f$ some force that makes $\gamma$ non-geodesic. But I don't know where to go from there if that's the way to do it.

Also any specific links to discussions?

Any help would be greatly appreciated.

EDIT: So I tried solving $\nabla_u u=f$ with the constraints $\theta=\pi/2$, $u^\theta=0$ and $u^r=0$. lionelbrits has explained I must also add $\dot u^r=0$ to my list. This all leads to $$(r_S-2Ar)(u^\varphi)^2+\frac{r_S}{r^2}=f^r$$ ($A=1-r_S/r$, notation is standard Schwarzschild) The problem with this is that the $u^\varphi$ term is negative for $r>3m$. So somewhere a sign got screwed up and for the life of me I don't know where it is. A decent documentation of my work: http://www.texpaste.com/n/a6upfhqo, http://www.texpaste.com/n/dugoxg4a.

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  • $\begingroup$ You can find a general path with $\dot r = 0, \theta = \pi/2$. Then, yes, you just calculate that acceleration term. $\endgroup$ – Jerry Schirmer Dec 5 '14 at 21:07
  • $\begingroup$ correct. Any curve should have unit length. $\endgroup$ – Jerry Schirmer Dec 5 '14 at 21:33
  • $\begingroup$ @Jerry Schirmer What do I do about the $2\Gamma^r_{tt}(u^t)^2$ term? I'm not sure what $u^t$ is in a non-geodesic prescription. ⟨u,u⟩=−1 still holds, right? $\endgroup$ – Ryan Unger Dec 5 '14 at 21:33
  • $\begingroup$ @JerrySchirmer What I have so far: texpaste.com/n/a6upfhqo. I feel like I should be able to express $(u^t)^2$ in terms of $(u^\varphi)^2$ using $\langle u,u\rangle=-1$, right? $\endgroup$ – Ryan Unger Dec 5 '14 at 21:47
  • $\begingroup$ Yes, you solve for $u^t$ in terms of $u^\varphi$ using the norm of the velocity vector being -1. Basically this just means that you have suitably chosen an affine parameter. $\endgroup$ – lionelbrits Dec 5 '14 at 22:32
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The equivalence principle tells us that we can evaluate $\nabla_u u$ in a co-moving reference frame and that for geodesics we should find no acceleration (to the occoupants of an elevator in free-fall, the contents seem to be experiencing no acceleration). Therefore, if we evaluate this when we are not along a geodesic (elevator sitting on earth), we find that it is not zero. Because it is a vector, if it is non-zero in one frame, it must be non-zero in another. In other words, yes, $f^r$ is what you have to calculate. The ingredient that you are missing is that $r=\mathrm{const}$ for a circular orbit implies that $\dot{u}^r = 0$. This is not a local thing, it is simply because you are forcing the orbit to be circular.

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  • $\begingroup$ So by doing this in texpaste.com/n/dugoxg4a I get $0=-r_S(u^\varphi)^2-\frac{r_S}{r^2}+2Ar(u^\varphi)^2+f^r$. I think I screwed up a sign because when I rearrange to get $\frac{r_S}{r^2}+(u^\varphi)^2(r_S-2Ar)=f^r$, it is actually negative for $r>3m$. Any idea what's up? $\endgroup$ – Ryan Unger Dec 6 '14 at 13:23
  • $\begingroup$ Is that a covariant/contravariant issue? $\endgroup$ – lionelbrits Dec 6 '14 at 13:43
  • $\begingroup$ You haven't defined $A$ yet. $\endgroup$ – lionelbrits Dec 6 '14 at 14:03
  • $\begingroup$ I don't think so. $r_S-2Ar=6m-2r$. Obviously this is negative when it should be positive. On the other hand this is not a simple overall sign flip because the term $r_S/r^2$ is positive when it should be. So $f^r$ can either be positive or negative depending on $(u^\varphi)^2$ and $m$ it seems. $\endgroup$ – Ryan Unger Dec 6 '14 at 14:08
  • $\begingroup$ $A$ is defined in the OP and in the links in the above discussion. $\endgroup$ – Ryan Unger Dec 6 '14 at 14:12

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