Non-geodesic circular orbit?

Question

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.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user 0celo7

Comments

I would assume by radial acceleration the radial component of acceleration is meant. I'm assuming this is centripetal, hence why I set $u^r=0$. By proper acceleration do you mean $dr/dt$?

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user 0celo7

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If $u^r = 0$ as a fact, then $\dot u^r = 0$, also.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user lionelbrits

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@lionelbrits Calculus doesn't work that way. A ball bas zero velocity at the top of its path but always has 9.8 m/s^s acceleration. Or in GR, $\Gamma=0$ in a normal coordinate system but $\text{Riem}\sim\partial \Gamma\ne0$. How would you answer the problem then? I posted the full question.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user 0celo7

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A circular orbit has $r = \mathrm{const}$. No matter how many derivatives of $r$ you take with respect to $\tau$, zero you will get. I suggest that you factor your workings into your question because it is one vote away from being closed (I would personally keep it open).

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user lionelbrits

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Let us continue this discussion in chat.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user 0celo7

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I want to say $\dot u^r = 0$...

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user lionelbrits

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@lionelbrits But the question statement says it is positive.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user 0celo7

Answer 1

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.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user lionelbrits

Comments

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?

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user 0celo7

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Is that a covariant/contravariant issue?

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user lionelbrits

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You haven't defined $A$ yet.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user lionelbrits

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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.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user 0celo7

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$A$ is defined in the OP and in the links in the above discussion.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user 0celo7

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$r_s - 2 A r = r_s - 2r + 2 r_s$. Your units look screwy but I'm not sure what $u^\varphi$ is. In any case, $(u^\varphi)$ must be positive because you are squaring it. Nevermind, you meant the factor multiplying that term.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user lionelbrits

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$r_S=2m$ ($m=GM$), so $r_S-2Ar=6m-2r$. $u^\varphi$ is the $\varphi$ component of velocity, which ought to be a constant of motion. Yes, I am also concerned about the units.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user 0celo7