# Non-geodesic circular orbit?

+ 1 like - 0 dislike
40 views

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

asked Dec 5, 2014
Most voted comments show all comments
You can find a general path with $\dot r = 0, \theta = \pi/2$. Then, yes, you just calculate that acceleration term.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user Jerry Schirmer
correct. Any curve should have unit length.

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

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

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user lionelbrits
@lionelbrits Ok, I think I'm close now: texpaste.com/n/dugoxg4a

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

## 1 Answer

+ 2 like - 0 dislike

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
answered Dec 6, 2014 by (110 points)
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
Is that a covariant/contravariant issue?

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user lionelbrits
You haven't defined $A$ yet.

This post imported from StackExchange Physics at 2015-01-06 22:42 (UTC), posted by SE-user lionelbrits
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
$A$ 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
$r_s - 2 A r = r_s - 2r + 2 r_s$. 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
$r_S=2m$ ($m=GM$), 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

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.