# Circular Orbits of a Black Hole

1. Apr 8, 2008

### mysearch

In part, this thread is a tangential extension of an earlier thread entitled Questions on Effective Potential. However, this thread now moves the discussion in the direction of interpreting the results of the effective potential in terms of the circular orbital velocity, as derived from the Schwarzschild metric.

The difference between the 2 effective potential (Veff) plots attached to this post is not really that important to this new thread, the main point to highlight is that the shape of the max/min curve is essentially identical, although the vertical scales differ. Both Veff curves correspond to a given value of angular momentum [L=7.42E12]. The value was selected as suitable based on the assumption the L>3.4642GMm/c. The max and min points of the curve correspond to a quasi-stable inner orbit and a stable outer orbit, while the graphs allude to the approximate value, the actual values are said to be calculated from the equations:

$$r_{outer} = \frac {a^2}{Rs}\left(1+\sqrt{1-\frac{3Rs^2}{a^2}\right)$$

$$r_{inner} = \frac {a^2}{Rs}\left(1-\sqrt{1-\frac{3Rs^2}{a^2}\right)$$

where $$a = \frac{L}{mc}$$

See 3rd attachment shows a graph against different values of normalised [L]
Details of the derivation can be found on the Wikipedia site:
http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity

For the example value of angular momentum [L=4.2GMm/c] the outer and inner radius, as a ratio to the Schwarzschild radius [Rs], are 6.9 and 1.92 respectively. However, one of the questions being raised to the PF forum concerns the velocity that corresponds to these radii. Clearly, at these radii, the orbit is subject to relativistic effects due to gravity and orbital velocity. How these effects apply is assumed to depend on the observer:

1. Distant [dt]
2. Stationary/shell at [r]
3. Orbiting at [r] $$d\tau$$

As a generalisation, the assumption is that distant observer is essentially unaffected by gravity or velocity, while the stationary (shell) observer is affected by gravity. However, the orbiting observer is affected by both gravity and velocity. So the first question is:

Who measures the radii [r] calculated from the equations?

The answer to this question is required in order to determine the frame of reference of the angular momentum. For a circular orbit, the normal assumption is that [L=mvr]. Therefore, by rearranging, we can determine a velocity [v=L/mr]. However, if we apply this equation to both radii produced above, the outer orbital velocity [v=9.10E7], while the implication is that the inner velocity would exceed the speed of light [c] at [v=3.28E8]. Taking another approach, we might try solve the Schwarzschild metric for $$[d\phi/dt]$$ and $$[d\phi/d\tau]$$, which is assumed to correspond to the distant and orbiting observers. The following solutions are assumed, although only the $$[d\phi/dt]$$ case has been checked:

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

$$v_o = r\left(\frac {d\phi}{dt}\right)=c\sqrt{\frac{Rs}{2r}}$$

Again, using the value in the specific example cited leads to a value of v=8.35E7 for the outer orbit, as perceived by the orbiting observer and the shell observer. This assumption is based on the fact that these two observers sit at the same radius, i.e. same gravity effects, and therefore are only separated by velocity. Velocity under special relativity is assumed to be invariant to both observers. The velocity with respect to the distant observer appears to come out at [v=8.05E7]. If these equations are applied with [r=inner radius] then the respective velocities are [1.78E8] and [1.53E8] and so, at least, fall below the speed of light [c]. However, focusing on just the values calculated for the outer orbit by $$[d\tau]$$ and [dt] using the original assumption [L=mvr], we then appear to get an inconsistency to the value of [L=7.42E12] derived from the relationship [L=N*GMm/c] where [N=4.2] in this example, i.e.

V=8.35E7; L=mvr=6.82E12
V=8.05E7; L=mvr=6.56E12

So how do you correlate angular moment [L] to velocity and who measures it?

Initially, I assumed that although the distant observer is in flat spacetime, other results from the Schwarzschild metric suggest that measurements taken by this observer are distorted by the gravity and velocity effects acting on spacetime at [r]. This leaves the other 2 observers sitting at the same radius [r], although they are separated by the orbital velocity. Now relativity seem to suggest that time with respect to the distant observer will run slower, while distant in the radial direction will be expanded due to gravity. The addition of a relativistic velocity would cause time for the orbiting observer to appear to run even slower with respect to both the distant and shell observer, while the velocity causes the circumference distant to be contracted maintaining the invariance of velocity. However:

Would this change the perception of [r] based on $$r=C/2\pi$$?

Sorry if this might appear as a very convoluted set of questions, but would appreciate any help in trying to understand exactly how these issues are resolved by the standard text. Thanks

2. Apr 8, 2008

### mysearch

Missing attachments

Not sure what happen to the attachments. They were there on preview. Here goes again:

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3. Apr 11, 2008

### Jorrie

Your equations and interpretations are correct as far as I can establish.

You asked:
It does not matter in Schwarzschild coordinates, as long as one understand it as a circumferential radius $r=C/2\pi$ and the observer is either static in the field, or in orbit around the body. The length contractions and time dilations cancel out in the measurement of the circumference C.

I think both are correct. Any Schwarzschild transverse velocity $$(r d\phi/dt)$$ transforms by a factor $$1/\sqrt{1-Rs/r}$$ to the get the local transverse velocity.

-J

4. Apr 13, 2008

### mysearch

Response to Jorrie

Apologises for not responding sooner, but I didn’t appear to get any email notification. I agree that the velocity equations do appear to be supported by derivation from the Schwarzschild metric. However, in my mind, who measures a given radius [r], which is then input into the equation [L=mvr] would seem to be important to determining the frame of reference of the angular momentum. One of the problems I am having is that I could not get the value of [L] calculated from [mvr] to correlate to radius [r] or velocity [v] of any of the observers cited.

To summarise, at one level, the effective potential (Veff) defines an outer radius [r], which you can then calculate a velocity [v]. However, if you input these values into [L=mvr] you don’t get the same value of [L] implied by the effective potential.

5. Apr 14, 2008

### Jorrie

You must remember that effective potential is an energy and orbital energy is observer dependent. The fact that $L/m = r^2 d\phi/d\tau$ tells us that as well, because $d\tau/dt$ varies with observer in Schwarzschild coordinates.

The Schwarzschild radial parameter r is however observer independent, by its definition as a circumferential radius.

-J

6. Apr 14, 2008

### yuiop

Hi mysearch,

I tend to agree with Jorrie's comments. R should always be calculated as circumference/(2Pi). If a circular orbit is available the circumference can be measured by a local observer who measures the velocity of an orbiting sattelite and then multiplies the orbital velocity by the orbital period. This figure will agree with the circumference (and the implied radius) by an observer at any other altitude calculated from his measurement of the orbital velocity and period of the same sattelite. Where a circular orbit is not available (below the photon orbit radius) circumference can be measured (in principle) by a local ruler. Trying to measure radius by timing vertiacal signals or direct vertical measurement with a local ruler will give very inconsistent results.

Using L=mvr for the angular momentum seemed to cause problems with the energy associated with angular motion at infinity and may well be causing the problem you are having here. Probably best to stick with $d\tau/dt$ as Jorrie suggested.

7. Apr 14, 2008

### mysearch

Who measures the radii [r] calculated from these equations?

I agree that effective potential is observer dependent, our previous discussions have shown that Veff is a function of angular momentum [L], mass [m] and radius [r] and if we assumed that [L=mvr] for a circular orbit, then Veff is also a function of [v,m, r]. However, I am not sure about the statement:

I would have thought the circumference would be observer dependent as it depends on the velocity of observer, i.e. relativistic velocity will cause length contraction in the direction of motion. Picking up on Kev’s statement:

I agree, but want to highlight a caveat for verification. A stationary shell’ observer sits at the same radius [r] as the orbiting observer. While both will agree on the orbital velocity [v] due to the invariance of velocity under special relativity, they disagree on the orbit time and circumference. As such, both have a different perception of the radius [r]. I believe we can predict the effects on each of my observers by applying one or both of the following assumptions as appropriate to their position:

o As velocity [v] approaches the speed of light [c], time slows and length contracts in the direction of motion, at least, with respect to a stationary observer.

o On approaching a gravitational mass [M], time slows and length expands in the direction of gravitational pull, as a function of radius [r], at least, with respect to a distant` observer.

If so, then all observers measure a different radius [r]. As indicated, velocity will be invariant between the shell and orbiting observer, but this is not true for the distant observer who will determine a different velocity. Presumably, mass [m] is also affected by the relativistic effects of velocity [v]? I am not sure if there is any relativistic effect on mass [m] due to gravity? Now while the following equation for Veff is not one normally used, I believe it is equivalent in form:

$$Veff = \left[1/2mv_o^2-\frac{GMm}{r} \left(1+\frac{vo^2}{c^2}\right)\right]$$

So can any of our observers plug in their values of [m,v & r] and determine a Veff curve that has max/min that correlates to the values predicted by the equations:

$$r_{outer} = \frac {a^2}{Rs}\left(1+\sqrt{1-\frac{3Rs^2}{a^2}\right)$$

$$r_{inner} = \frac {a^2}{Rs}\left(1-\sqrt{1-\frac{3Rs^2}{a^2}\right)$$

where $$a = \frac{L}{mc}$$

As such, we may have we arrived back to the question in my first posting relating to the outer and inner radii forwarded by the two equations produced by the quadratic solution in the Wikipedia reference, i.e.

Who measures the radii [r] calculated from these equations?

8. Apr 14, 2008

### Jorrie

They disagree on orbit time but agree on the circumference, because surprisingly they don't agree on velocity.

The distant observer measures circular orbital velocity at r as $v_d = \sqrt{m/r}$

The shell observer measures circular orbital velocity as $v_s = \sqrt{m/(r-2m)}$

The orbiting observer measures circular orbital velocity as $v_o = \sqrt{m/(r-3m)}$

simply due to their respective proper times $d\tau/dt$ that differ. Remember that the observers are in curved spacetime where they cannot simply measure a Doppler shift and so determine their relative velocity. They are forced to lay a set of static rods, say of proper length L each, along the orbit and measure velocity relative to that using their own clocks. That fixes the circumference for everybody, I would think.

-J

9. Apr 15, 2008

### mysearch

Thanks Jorrie, let me check out your equations

10. Apr 15, 2008

### mysearch

Puzzling?

I have found a derivation of sorts in Taylor & Wheeler, which supports your statement regarding the shell observed velocity. I have not had time to work through the details; but will assume it is correct. However, I haven't yet resolved the apparent anomaly of why the separation between the orbiting observer and the shell observer is not explained by special relativity.

Just to restate this issue, I assumed that because these observers were both positioned at the same radius [r] the gravitational effects on time and space must be equal, leaving only the effects of special relativity to be taken into account. While the velocity of the orbiting observer causes an additional time dilation, it would also cause a comparable length contract of the orbit circumference. So while time is running slower, the circumference distant is shorter, leading to the velocity being invariant between these two observers. While, I have yet to resolve this anomaly in my mind, it would be very useful to clarify a few point on your statement of the equations:

I am assuming these equations are in geometric units, but are the distant observer velocity $$r d\phi/dt$$ and orbiting velocity $$r d\phi/d\tau$$ still meant to correspond to:

$$v_o = r\left(\frac {d\phi}{dt}\right)=c\sqrt{\frac{Rs}{2r}}$$

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

If so, are the following equations equivalent in geometric units (?):

The first equation seems compatible in form to the distant observer:

$$c\sqrt{\frac{Rs}{2r}} = c\sqrt{\frac{2GM/c^2}{2r}} = c\sqrt{\frac{GM}{rc^2}} -> \sqrt{\frac{M}{r}}$$

I haven’t actually derived this equation from the Schwarzschild metric yet, but noted its equivalence to the classical equation:

$$\frac{mv_o^2}{r} = \frac{GMm}{r^2}$$

$$v_o = c\sqrt{\frac{GM}{r}} = c\sqrt{\frac{Rs}{2r}}$$

However, had trouble in matching the orbit velocity equations, i.e. not sure where the 3M comes from:

$$c\sqrt{\frac{Rs}{2r-Rs}} = c\sqrt{\frac{2GM/c^2}{2r-2GM/c^2}} = c\sqrt{\frac{GM}{rc^2-GM}} -> \sqrt{\frac{M}{r-M}} (?)$$

Finally, want to confirm the equivalence of the form of the shell velocity equation:

$$v_s = \sqrt{m/(r-2m)} = c\sqrt{\frac{GM/c^2}{r-2GM/c^2}} = c\sqrt{\frac{Rs}{2(r-Rs)}} (?)$$

I recognise that I now need to sit down and work through the issues, but would appreciate any help/references, which explain the apparent anomalies.

11. Apr 15, 2008

### Jorrie

Yes. If you stick to geometric units, so that Rs=2m, it's easy to spot the correspondence.

Start with the distant observer's orbital velocity (the Schwarzschild coordinate circular orbital speed) $v_d = \sqrt{m/r}$ and multiply it by $dt/d\tau$, where $d\tau$ represents the propertime of the orbiting observer, giving (in geometric units):

$$v_o^2 = \frac{m/r}{1-2M/r-v_d^2} = \frac{m}{r-3m}$$

Remember this v_o is based on a a fixed number of meter sticks, laid out along the orbit and is not length contracted, because we are working with the time dilated clock of the orbiting observer.

-J

Last edited: Apr 15, 2008
12. Apr 15, 2008

### Jorrie

On second thoughts, this may not be the proper way of expressing the orbiting observer's own observed velocity, because it means $v_o \rightarrow \infty$ as $r \rightarrow 3m$.

I suppose as the orbiting and static shell observers flash past each other, they can determine their relative speed and both should obtain

$$v_o^2 = \frac{m}{r-2m}$$

If this is correct, then (as 'mysearch' has hinted) the orbiting observer's time is dilated relative to the shell observer's time by a factor

$$\frac{d\tau_o}{d\tau_s}=\sqrt{1-v_o^2} =\sqrt{1-\frac{m}{r-2m}} =\sqrt{\frac{r-3m}{r-2m}}$$

One must then conclude that the orbiting observer's own measurement of the circumference of the orbit is Lorentz contracted and no longer 'r', but rather

$$r_o = r \sqrt{\frac{r-3m}{r-2m}}$$

approaching zero as r approaches 3m (the 'light radius'), which sounds right.(?)

However, this does not detract from the fact that the Schwarzschild radial parameter remains 'r' for all the equations of the metric.

13. Apr 16, 2008

### mysearch

Radius [r] in the metric?

Jorrie, thanks for both inputs. I am hoping to get some time today to work through all the issues you’ve raised in my mind, but will try to respond as soon as possible. However, your last sentence raised one immediate train of thought:

Just showing a simple form of the metric for reference, I wanted to clarify some basic assumptions:

$$c^2 d\tau = c^2\left(1-\frac{Rs}{r}\right)dt^2-\left(1-\frac{Rs}{r}\right)dt^{-1}dr^2-r^2d\phi^2$$

It would seem that we have 4 key variables that define the context of the observer in question, albeit described in very simplified terms in the following summary:

[dt]: far-away time in essentially flat spacetime
$$[d\tau]$$: wristwatch time of the observer
[dr]: associated with radial velocity via [dt] or $$[d\tau]$$
$$[d\phi]$$: associated with angular velocity via [dt] or $$[d\tau]$$

Therefore, depending on how you manipulate the basic metric into any given solution, e.g.

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

I would have though the “Schwarzschild radial parameter [r]” would be the radial distance as measured by the observer in question? If so, in the example above, the radius [r] would correspond to that measured by the orbiting observer?

14. Apr 16, 2008

### Jorrie

Haven't you got a spurious $dt^{-1}$ in the second coefficient? I think it should read:

$$c^2 d\tau = c^2\left(1-\frac{Rs}{r}\right)dt^2-\left(1-\frac{Rs}{r}\right)^{-1}dr^2-r^2d\phi^2$$
Correct.
No, AFAIK, the “Schwarzschild radial parameter [r]” is defined by the metric as the same for all static observers, at least.

Last edited: Apr 16, 2008
15. Apr 16, 2008

### mysearch

Sorry about the typo, don't seem to have edit access to correct it.

So if I say that r=3*Rs, the orbit velocity $$[v_o]$$ would be 0.44c?
Who perceives this velocity?
Who perceives the orbiting observer to be at a radius r=3Rs?

Last edited: Apr 16, 2008
16. Apr 16, 2008

### yuiop

As Jorrie mentioned, R is determined by a static shell observer who measures the circumference and assumes R= C/(2Pi). Any other static shell observer at any other altitude will agree with that measurement. The same is not true for orbiting observers.

As for your equation:

for r = (3/2)*Rs (the photon orbit) I get

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{3Rs-Rs}} = 0.707c$$

using the other equation for coordinate time rather than proper time

$$v_o = r\left(\frac {d\phi}{dt}\right)=c\sqrt{\frac{Rs}{2r}}$$

for r = (3/2)*Rs (the photon orbit) I get:

$$v_o = r\left(\frac {d\phi}{dt}\right)=c\sqrt{\frac{Rs}{3Rs}} = 0.577c$$

neither of which seem quite right for a photon.

[EDIT} The observer at infinity perceives the coordinate speed of light at the photon orbit radius to be:

$$c_0 \sqrt{1-(2Rs)/(3Rs)} = 0.577 c_0$$ so the second equation seems OK.

The first equation where dtau would presumably be zero for a photon does not seem to make sense.

Last edited: Apr 16, 2008
17. Apr 16, 2008

### Jorrie

The equation that you gave actually also contains an error that I didn't spot initially:

Should read:

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2(r-Rs)}}$$

The corrected value is then 0.5c and it is measured by the shell static observer, who also knows that the the orbit is at r=3Rs, because he measures the circumference by his meter rods.

18. Apr 16, 2008

### yuiop

Using your corrected equation and r= (3/2)Rs for the photon orbit radius

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2(r-Rs)}} = c$$

indicating dtau is the time measured by a local static shell observer rather than the time measured by the clock of an orbiting observer.

19. Apr 16, 2008

### Jorrie

Hi kev, yep - if dtau was for the orbiting observer, r should also have been as determined by the orbiting observer, I think.

I have to be off - laters...

20. Apr 16, 2008

### mysearch

Kev, just wanted to clarify a couple of points on one of your recent posts. I was using the following equation for a mass object:

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

Therefore, if I insert r=3Rs, the equation becomes

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{6Rs-Rs}}$$

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{1}{5}} = 0.447c$$

One of Jorrie’s recent posts appears to question this equation by quoting:

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2(r-Rs)}}$$

As far as I can see, the equation I have used is derived from the Schwarzschild metric when dividing through by $$d\tau$$, i.e. it is the orbiting velocity. While I thought the equation quoted by Jorrie related to the shell’s observer perception of the orbiting velocity? While I conceded this equation is supported by standard text, I am in the process of trying to check why.

$$v_s = \sqrt{m/(r-2m)} = c\sqrt{\frac{GM/c^2}{r-2GM/c^2}} = c\sqrt{\frac{Rs}{2(r-Rs)}} (?)$$

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