# Schwarzschild Circular Orbits Problem

• AuraCrystal

## Homework Statement

Find the linear velocity of a particle in a circular orbit of radius $R$ in the Schwarzschild geometry as would be measured as by a stationary observer stationed at one point on the orbit. (It's problem 10 in chapter 9 of Hartle, if that helps)

## Homework Equations

The Scwarzschild metric, obviously,
$ds^2=- \left (1-\frac{2M}{r} \right )dt^2 + \left (1-\frac{2M}{r} \right )^{-1} dr^2 +r^2(d \theta^2 + \sin ^2 \theta d \phi ^2)$
And in the book they derive the equation (for a circular orbit of radius R)
$\Omega \equiv \frac{d \phi}{d t}= \frac{M}{R^3}$

## The Attempt at a Solution

OK, so in this problem, obviously $\frac{dr}{d \bar{t}}=\frac{d\theta}{d \bar{t}}=0$, where $\bar{t}$ is the time measured in the stationary observer's frame. Ok so we know the angular velocity measured from the far away observer's frame:
$\Omega \equiv \frac{d \phi}{d t}= \frac{M}{R^3}$
since the stationary observer is, by definition not moving (w.r.t. to our far-away observer), we can easily find $dt/d\bar{t}$ from the metric. We then have $\frac{d \phi}{d \bar{t}}.$ My question is, after we get the velocity 3-vector (i.e. the velocity w.r.t. to the stationary orbit), what do we do and why?

Can you write down the 4-velocity of the particle in orbit and the 4-velocity of the stationary observer?

Can you write down the 4-velocity of the particle in orbit and the 4-velocity of the stationary observer?
Of course, it's just:
$$u_{observer} = ( dx^{\alpha}_{observer}/d \bar{t} )$$
$$u_{particle} = ( dx^{\alpha}_{particle}/d \tau )$$
Where α ranges from 0 to 3. (0 is the time component w.r.t. to the observer at ∞).

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Of course, it's just:
$$u_{observer} = ( dx^{\alpha}_{observer}/d \bar{t} )$$
$$u_{particle} = ( dx^{\alpha}_{particle}/d \tau )$$
Where α ranges from 0 to 3. (0 is the time component w.r.t. to the observer at ∞).

Okay, now I have Hartle in front of me. I meant also the specific components of the 4-velocities with respect to Schwarzschild coordinates, i.e., (9.16) for the hovering observer and (9.47) together with (9.48) for the orbiting particle.

What do you get when you divide the t components of the 4-velocities you wrote above? What do you get when you divide the t components of the 4-velocities from the equations I gave?

^So you want me to write out the components?

$u_{particle}=(\frac{dt}{d \tau} , 0, 0, \frac{d \phi_{particle}}{d \tau}) = ( \sqrt{1-\frac{3m}{R}}, 0, 0, \frac{M}{R^3} )$ (from 9.46, 9.47, and 9.48)
And:
$u_{observer}=(\frac{dt}{d \bar{t}} , 0, 0, 0)$
From the metric:
$-(d \bar{t})^2=- \left (1-\frac{2M}{r} \right )dt^2 + \left (1-\frac{2M}{r} \right )^{-1} dr^2 +r^2(d \theta^2 + \sin ^2 \theta d \phi ^2)$
since $(d \bar{t})^2= - ds^2$. Of course, $d \phi = d \theta = dr = 0$ since the observer is stationary, so :
$(d \bar{t})^2=\left (1-\frac{2M}{r} \right )dt^2$
Hence, rearranging,
$\left (\frac{dt}{d \bar{t}} \right ) ^2 = \left (1-\frac{2M}{r} \right ) ^ {-1}$
i.e.,
$\frac{dt}{d \bar{t}} = \left (1-\frac{2M}{r} \right ) ^ {-1/2}$
so that
$u_{observer}=(\frac{dt}{d \bar{t}} , 0, 0, 0) = \left ( \left (1-\frac{2M}{r} \right ) ^ {-1/2}, 0, 0, 0 \right )$

$u_{particle}=(\frac{dt}{d \tau} , 0, 0, \frac{d \phi_{particle}}{d \tau}) = ( \sqrt{1-\frac{3m}{R}}, 0, 0, \frac{M}{R^3} )$ (from 9.46, 9.47, and 9.48)

I am again without Hartle, but shouldn't u^t above be $\left( 1-3m/R \right)^{-1/2}$?

What is $d \bar{t} / d \tau$ in Schwarzschild coordinates? What is $d \bar{t} / d \tau$ in special relativity?

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^Probably. Typo.

Also:
$d \bar{t} / d \tau = \frac{dt}{d \tau} \frac{d \bar{t}}{dt} = \left ( 1 - \frac{3M}{R} \right ) ^ {-1/2} \left (1-\frac{2M}{R} \right ) ^ {1/2}$

And in SR, isn't it just the gamma-factor?

^Probably. Typo.

Also:
$d \bar{t} / d \tau = \frac{dt}{d \tau} \frac{d \bar{t}}{dt} = \left ( 1 - \frac{3M}{R} \right ) ^ {-1/2} \left (1-\frac{2M}{R} \right ) ^ {1/2}$

And in SR, isn't it just the gamma-factor?

Yes. What happens when set these expressions equal?

^So you just solve for v?

Why can you do that? Is it b/c the observer at R is in a local inertial frame?

And if so, can't you just use $v^2 = \eta_{ij} v^{i} v^{j}$ where $\eta_{ij}$ is the flat space metric (in spherical coordinates) and you sum over 1 to 3 and $v^{i}=\left(\frac{dr}{d \bar{t}}, \frac{d \theta}{d \bar{t}}, \frac{d \phi}{d \bar{t}}\right)$?

...Am I being an idiot here? xD

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^So you just solve for v?

Why can you do that? Is it b/c the observer at R is in a local inertial frame?

Which observer at R? Both observers have constant $r = R$. The hovering observer H has non-zero 4-acceleration and the orbiting observer O has zero 4-acceleration, so O is inertial and H is not. But, as you suspect, a local inertial coordinate (LIC) can be used to demonstrate the result.

Let $p$ be an event of coincidence for O and H. Let $\left\{ x^{\mu '} \right\}$ be an LIC that has the following properties:

1) $p$ is at the origin;

2) H is at rest at $p$;

3) at $p$, O moves in the positive $x^{1 '}$ direction.

More about 2). Even though H has non-zero 4-acceleration, 2) is possible, but, because of the non-zero 4-acceleration, H will not be at rest away from $p$. I used H instead of O for 2), because the final is not dependent on one of the observers having zero 4-acceleration.

Let $\mathbf{u}$ and $\mathbf{w}$ be the 4-velocities of H and O respectively. In the LIC, $\mathbf{u}_p = \left( 1, 0, 0, 0 \right)$ and $\mathbf{w}_p = \left( \gamma, \gamma v, 0, 0 \right)$. Consequently, in the LIC,
$$\gamma = - g_{\mu \nu} u_p^\mu w_p^\nu .$$
Because $g_{\mu \nu} u_p^\mu w_p^\nu$ is a coordinate-invariant scalar, $\gamma = - g_{\mu \nu} u_p^\mu w_p^\nu$ in any coordinate system in both special and general relativity! This useful result gives the physical (not coordinate) between any two coincident observers. Use Schwarzschild spherical coordinates and this expression to do the problem. In coordinate-free notation,
$$\gamma = - g \left( \mathbf{u}_p , \mathbf{w}_p \right)$$
And if so, can't you just use $v^2 = \eta_{ij} v^{i} v^{j}$ where $\eta_{ij}$ is the flat space metric (in spherical coordinates) and you sum over 1 to 3 and $v^{i}=\left(\frac{dr}{d \bar{t}}, \frac{d \theta}{d \bar{t}}, \frac{d \phi}{d \bar{t}}\right)$?

This works in Minkowski spacetime, but I don't see how to use this here, and it doesn't work for general curvilinear coordinates in Minkowski spacetime. In Hartle's problem, we can constuct a LIC around $p$, and then use this LIC to construct spherical coordinates locally around $p$, but there won't be an obvious relationship between the local spherical coordinates and Schwarzschild spherical coordinates.
Am I being an idiot here?
No.