# Some general formulae for circular orbits in symmetric spacetimes

1. Apr 17, 2013

### pervect

Staff Emeritus
Consider the equatorial plane of a spherically symmetric space-time. Then we can write the metric in the equatorial plane (theta=pi/2) in terms of three coordinates - [t, r, phi]

ds^2 = -f(r) dt^2 + g(r) dr^2 + h(r) dphi^2

For the Schwarzschild metric we can write:
$$f = c^2(1 -\frac{2 G M}{c^2 r}) \quad g = 1 / (1 -\frac{2 G M}{c^2 r}) \quad h = r^2$$

For the isotropic Schwarzschild metric we can write:
$$f = c^2 \left( \frac{1-\frac{GM}{2 c^2 r}}{1+\frac{GM}{2 c^2 r}} \right)^2 \quad g = (1 + \frac{GM}{2 c^2 r})^4 \quad h = r^2 (1 + \frac{GM}{2 c^2 r})^4$$

For the non-cartesian PPN apprxomiation to isotropic Schwarzschild we can write:
$$f = c^2 \left( 1 - \frac{2GM}{c^2 r} + \frac{2 G^2 M^2}{c^4 r^2} \right) \quad g = \left( 1 + \frac{2GM}{c^2 r} \right) \quad h = r^2 \left( 1 + \frac{2GM}{c^2 r} \right)$$

The geodesic equation for r is
$$\frac{d^2r}{d \tau^2} + \frac{(\frac{df}{dr})}{ 2g} (\frac{d^2t}{d \tau^2}) + \frac{(\frac{dg}{dr})}{ 2g} (\frac{d^2 r}{d \tau^2}) + \frac{(\frac{dh}{dr})}{ 2g} (\frac{d^2\phi }{d \tau^2}) = 0$$

Various useful quantities are (the answers all appear to be sensible to me in Schwarzschild coordinates):

The magnitude of the proper acceleration of a static observer (constant r) as measured with an onboard accelerometer

$$c^2 \frac{(\frac{d f}{d r})}{2 f \sqrt{g}}$$

The coordinate acceleration (d^2 r / dt^2) of a geodesic observer who starts out at rest, (dr/dt=0, dphi/dt=0):

$$-\frac{(\frac{d f}{d r})}{2g}$$

The coordinate angular velocity $d \phi / dt$
$$\sqrt{\frac{(\frac{d f}{d r})}{(\frac{d h}{d r})}}$$

The proper angular velocity $d \phi / d \tau$

$$c \frac{(\frac{d f}{d r}) } { f (\frac{d h}{d r}) - h (\frac{d f}{d r})}$$

coordinate and proper orbital periods - 2 pi over the respective angular velociteis

The orbital period as measured by a co-located static observer
*correction*
$$\left( \frac{2 \pi}{c} \right){ \sqrt{f (\frac{d h}{d r}) / (\frac{d f}{d r}) } }$$

circumference of the circle with coordinate values r
$$2 \pi \sqrt{h}$$

orbital velocity measured by a colocated static observer
$$c \sqrt{ \frac{h (\frac{d f}{d r}) }{ f (\frac{d h}{d r})}}$$

The concept of velocity from "remote points of view" is not well-defined in GR, due to the problems of parallel tranpsorting velocities.

http://math.ucr.edu/home/baez/einstein/node2.html

though various ad-hoc approaches are possible.

Note: The factors of c are annoyig to keep tract of - I took the simple approach of making sure the units come out correctly.

Last edited: Apr 17, 2013
2. Apr 17, 2013

### PAllen

Nice. But .. what do you mean by coordinate acceleration of a static observer? Even for circular orbit shouldn't dr/dt and the second derivative both be zero?

3. Apr 17, 2013

### Mentz114

Useful. There is a small typo - the equatorial plane is $\theta=\pi/2$, I think.

4. Apr 17, 2013

### pervect

Staff Emeritus
That was wrongly worded - I meant the coordinate acceleration of a geodesic observer who was initially at rest. I'll fix it up, if I can.

And add the missing minus sign...

Last edited: Apr 17, 2013
5. Apr 17, 2013

### pervect

Staff Emeritus
Fixed - thanks

6. Apr 17, 2013

### pervect

Staff Emeritus
Some sanity check values, for the Schwarzschild metric:

proper acceleration of a static observer:
$$\frac{GM}{r^2 \sqrt{1-2GM/c^2r}}$$

notes: matches wiki

coordinate acceleration of a geodesic observer who was initially at rest
$$\frac{GM}{r^2} \left(1 - \frac{2GM}{c^2 r} \right)$$

coordinate angular velocity $d \phi / dt$
$$\sqrt{\frac{GM}{r^3}}$$

(notes: matches Newtonian)

proper angular velocity $d \phi / d \tau$
$$\sqrt{\frac{GM}{r^2 \left(r - 3GM/c^2\right) }}$$

notes: goes to infinity at photon sphere

orbital period measured by co-located static observer
(correction!)

$$2 \pi \sqrt{1-2GM/c^2r)}\sqrt{\frac{r^3}{GM}}$$

notes: time dilated from coordinate period, goes to infinity at the event horizon

orbital velocity measured by co-located static observer
$$\sqrt{\frac{GM}{r-2GM/c^2}}$$

matches Wiki

7. Apr 17, 2013

### pervect

Staff Emeritus
One more correction, for the orbital period of a static observer, the time dilation was done wrongly. (It's fixed)

The results seem to match wiki's http://en.wikipedia.org/w/index.php?title=Circular_orbit&oldid=540533714

for the Schwarzschild metric. It's non-intuitive that $d \phi / dt$ remains finite at the event horizon (this scared me for a bit!) but this seems to match wiki's results.

I should add that circular orbits are impossible below the photon sphere at r=3GM/c^2, so applying it at this altitude is wrong in any event, even if the results don't appear to "blow up".

Last edited: Apr 17, 2013
8. Apr 17, 2013

### Mentz114

I'm not sure that $d \phi / dt$ represents anything physically meaningful. It is an approximation that would break down anywhere near the horizon, like r < 3m. There's no doubt though that $(d\phi/d\tau)/(dt/d\tau)$ gives the Newtonian value.

9. Apr 17, 2013

### pervect

Staff Emeritus
Another correction, but this one is too late for me to fix. Perhaps someone else can do it.

$$\frac{d^2r}{d \tau^2} + \frac{(\frac{df}{dr})}{ 2g} (\frac{d^2t}{d \tau^2}) + \frac{(\frac{dg}{dr})}{ 2g} (\frac{d^2 r}{d \tau^2}) + \frac{(\frac{dh}{dr})}{ 2g} (\frac{d^2\phi }{d \tau^2}) = 0$$

it should really read (fixes sign error and some typos)

$$\frac{d^2r}{d \tau^2} + \frac{(\frac{df}{dr})}{ 2g} \left(\frac{dt}{d \tau} \right) ^2 + \frac{(\frac{dg}{dr})}{ 2g} \left(\frac{d r}{d \tau} \right)^2 - \frac{(\frac{dh}{dr})}{ 2g} \left(\frac{d \phi }{d \tau} \right)^2 = 0$$

Symbolically, this is just the geodesic equation, with the non-zero Christoffel symbols for the given metric written out explicitly. The symbolic version is:

$$\frac{d^2r}{d \tau^2} + \Gamma^r{}_{tt} \left( \frac{dt}{d \tau} \right) ^2 + \Gamma^r{}_{rr} \left( \frac{d r}{d \tau} \right)^2 + \Gamma^r{}_{\phi \phi} \left( \frac{d \phi }{d \tau} \right)^2 = 0$$

thus
$$\Gamma^r{}_{tt} = \frac{(\frac{df}{dr})}{ 2g} \quad \Gamma^r{}_{rr} = \frac{(\frac{dg}{dr})}{ 2g} \quad \Gamma^r{}_{\phi \phi} = - \frac{(\frac{dh}{dr})}{ 2g}$$

Last edited: Apr 18, 2013
10. Apr 19, 2013

### Agerhell

Very nice. I could probably rearrange the terms of the second expression of the last post above and get the expression myself, but it would be useful to know the expressions for $dt/d\tau$ for isotropic Schwarzschild and for the low order expansion of isotropic Schwarzschild (the PPN- approximation). The expression for the velocity of light in coordinate time in the PPN-approximation would be interesting to have to. For some reason people seem to "pick and choose", take the velocity of light and the proper time from anisotropic coordinates, but get the orbits from isotropic coordinates, which seems a bit odd to me.

11. Apr 19, 2013

### pervect

Staff Emeritus
For a circular orbit

*note missing square root in previously posted expression*

$$\frac{d \phi}{d \tau} = c \sqrt{\frac{(\frac{d f}{d r}) } { f (\frac{d h}{d r}) - h (\frac{d f}{d r})}}$$
$$\frac{d t}{d \tau} = c \sqrt{\frac{(\frac{d h}{d r}) } { f (\frac{d h}{d r}) - h (\frac{d f}{d r})}}$$

Utilizing the facts that $\frac{dr}{d \tau} = \frac{d^2r}{d \tau^2} = 0$ for a circular orbit, this comes from solving the following two equations.

$$(1) \quad -f \left( \frac{d t}{d \tau} \right)^2 + h \left( \frac{d \phi}{d \tau} \right) ^2 = -c^2$$

$$(2) \quad \frac{(\frac{df}{dr})}{ 2g} \left(\frac{dt}{d \tau} \right) ^2 - \frac{(\frac{dh}{dr})}{ 2g} \left(\frac{d \phi }{d \tau} \right)^2 = 0$$

If you multiply both sides of eq (1) by $d \tau$^2, you can see this eq (1) is just the metric equation.

Eq (2) is the geodesic equaton for r, symbolically:

$$\Gamma^r{}_{tt} \left( \frac{dt}{d \tau} \right) ^2 +\Gamma^r{}_{\phi \phi} \left( \frac{d \phi }{d \tau} \right)^2 = 0$$

The equation of light can be solved for by simply setting ds^2 = 0 in the metric - rearranging this will give the coordinate velocity .

In the radial direction (dr/dt)^2 = f/g. In the $\phi$ direction, (d$\phi$/dt)^2 = f/h.

12. Apr 19, 2013

### pervect

Staff Emeritus
*corrected version*
Consider the equatorial plane of a spherically symmetric space-time. Then we can write the metric in the equatorial plane (theta=pi/2) in terms of three coordinates - [t, r, $\phi$]

$$ds^2 = -f(r)\, dt^2 + g(r)\, dr^2 + h(r)\, d\phi^2$$

For the Schwarzschild metric we can write:
$$f = c^2(1 -\frac{2 G M}{c^2 r}) \quad g = 1 / (1 -\frac{2 G M}{c^2 r}) \quad h = r^2$$

For the isotropic Schwarzschild metric we can write:
$$f = c^2 \left( \frac{1-\frac{GM}{2 c^2 r}}{1+\frac{GM}{2 c^2 r}} \right)^2 \quad g = (1 + \frac{GM}{2 c^2 r})^4 \quad h = r^2 (1 + \frac{GM}{2 c^2 r})^4$$

For the non-cartesian PPN apprxomiation to isotropic Schwarzschild we can write:
$$f = c^2 \left( 1 - \frac{2GM}{c^2 r} + \frac{2 G^2 M^2}{c^4 r^2} \right) \quad g = \left( 1 + \frac{2GM}{c^2 r} \right) \quad h = r^2 \left( 1 + \frac{2GM}{c^2 r} \right)$$

We will need only one additional equation (other than the metric) to solve for circular orbits - the r component of the geodesic equation:

$$\frac{d^2r}{d \tau^2} + \Gamma^r{}_{tt} \left( \frac{dt}{d \tau} \right) ^2 + \Gamma^r{}_{rr} \left( \frac{d r}{d \tau} \right)^2 + \Gamma^r{}_{\phi \phi} \left( \frac{d \phi }{d \tau} \right)^2 = 0$$

Computing the values of the Christoffel symbols:
$$\Gamma^r{}_{tt} = \frac{(\frac{df}{dr})}{ 2g} \quad \Gamma^r{}_{rr} = \frac{(\frac{dg}{dr})}{ 2g} \quad \Gamma^r{}_{\phi \phi} = - \frac{(\frac{dh}{dr})}{ 2g}$$

we write:

$$\frac{d^2r}{d \tau^2} + \frac{(\frac{df}{dr})}{ 2g} \left(\frac{dt}{d \tau} \right) ^2 + \frac{(\frac{dg}{dr})}{ 2g} \left(\frac{d r}{d \tau} \right)^2 - \frac{(\frac{dh}{dr})}{ 2g} \left(\frac{d \phi }{d \tau} \right)^2 = 0$$

Note that the first and second derivatives of r with respect to time are zero for a circular orbit. Additionally, $d^2 t / d \tau^2 = d^2 \phi / d \tau^2 = 0$, so $dt/d\tau$ and $d\phi / d\tau$ are constants of motion for a circular orbit.

The metric equation implies that

$$-f(r) \, \left( \frac{dt}{d\tau} \right) ^2 + g(r) \, \left( \frac{dr}{d\tau} \right)^2 + h(r) \, \left( \frac{d \phi}{d\tau} \right)^2 = -c^2$$

This, combined with the geodesic equation, enables us to solve for all quantities of interest.

Various useful quantities:

The magnitude of the proper acceleration of a static observer (constant coordinates) as measured with an onboard accelerometer

$$c^2 \frac{(\frac{d f}{d r})}{2 f \sqrt{g}}$$

The coordinate acceleration (d^2 r / dt^2) of a geodesic observer who starts out at rest, (dr/dt=0,$d\phi/dt=0$) is

$$-\frac{(\frac{d f}{d r})}{2g}$$

The coordinate angular velocity $d \phi / dt$
$$\sqrt{\frac{(\frac{d f}{d r})}{(\frac{d h}{d r})}}$$

The proper angular velocity $d \phi / d \tau$
$$\frac{d \phi}{d \tau} = c \sqrt{\frac{(\frac{d f}{d r}) } { f (\frac{d h}{d r}) - h (\frac{d f}{d r})}}$$

Time dilation for the orbiting observer, $dt / d\tau$
$$\frac{d t}{d \tau} = c \sqrt{\frac{(\frac{d h}{d r}) } { f (\frac{d h}{d r}) - h (\frac{d f}{d r})}}$$

Coordinate and proper orbital periods are just2 pi over the respective angular velocities

The orbital period as measured by a co-located static observer is:
$$\left( \frac{2 \pi}{c} \right){ \sqrt{f (\frac{d h}{d r}) / (\frac{d f}{d r}) } }$$

The set of points with coordinate values r=constant is a circle. It's circumference is:

$$2 \pi \sqrt{h}$$

The orbital velocity measured by a co-located static observer is:
$$c \sqrt{ \frac{h (\frac{d f}{d r}) }{ f (\frac{d h}{d r})}}$$

The concept of velocity from "remote points of view" is not well-defined in GR, due to the problems of parallel tranpsorting velocities.

http://math.ucr.edu/home/baez/einstein/node2.html

though various ad-hoc approaches (mostly based on specific coordinate systems) are possible.

Note: The factors of c are annoyig to keep tract of - I took the simple approach of making sure the units come out correctly.