# Orbital Velocity at r=3M?

1. Oct 31, 2007

### Jorrie

I understand that light can be in an unstable orbit at Schwarzschild radial coordinate r=3M (geometric units) around a Schwarzschild black hole. I also understand that the local* circular orbital speed of a massive particle around the hole is

$$v_o = r\frac{d\phi}{dt} = \sqrt{\frac{M}{r(1-2M/r)}}$$

which is unstable for r <= 6M. The equation suggests that the local orbital speed equals the speed of light at r = 3M, which (sort of) confirms the orbit of light there.

If this is correct, my question is: if a particle in roughly circular orbit slowly spirals in from 3M < r < 6M, will it only approach the speed of light as it passes r = 3M or can it actually reach it?

*With local I mean as measured by an observer static at radial coordinate r.

2. Oct 31, 2007

### George Jones

Staff Emeritus
Such a worldline isn't a circle at r = 3M. An observer (static or not) at any event outside (and on the event horizon, and even inside) of a black hole always measures the local speed of a material particle to be less than the local speed of a light.

Your post brings up another interesting point. In order for a particle to spiral down from conditions dr/dt = 0 and dr/dphi =/= 0 to an unstable exactly circular orbit with r near 3M, it would have to start with r *very* large. The closer the r of the unstable circle is to to 6M (below), the closer the starting r is to 6M (above). As r for the unstable circle approaches 3M, the starting r approaches infinity.

3. Oct 31, 2007

### pervect

Staff Emeritus
What makes you think that $$r \frac{d\phi}{dt}$$ is the local circular orbital speed?

You've neglected to account for the time dilation of the static observer relative to the observer at infinity, I think.

You can regard $r d \phi$ as the local distance, because the circumference of a black hole is 2 pi r, but you can't regard dt as the local time. dt is coordinate time, not local time.

Last edited: Oct 31, 2007
4. Oct 31, 2007

### Jorrie

Oops, yes, I should have written in terms of dphi/dtau:

$$v_o = r\frac{d\phi}{d\tau} = \sqrt{\frac{M}{r(1-2M/r)}}$$

But the rest of formula is correct for the local observer, not so?

For the observer at infinity the orbital speed at r is just $$v_{inf} = \sqrt{M/r}[/itex], I think. Last edited: Oct 31, 2007 5. Oct 31, 2007 ### pervect Staff Emeritus Unless I'm calculating something incorrectly (unfortunately possible), as one approaches r0=3m, the total energy required approaches infinity. I get E=1 (bare escape to infinity) occurring at r0=4m, for r0<4m, the object has to have more than enough energy to escape to infinity. It takes a little work to get the hang of it, but I find the orbital simulator at http://www.fourmilab.ch/gravitation/orbits/ useful to get some hands-on idea of the orbital behavior. One first has to pick a value of L and M. L=3.462 and m=1 is close to but above the critical value of L of sqrt(12) for m=1. Doing the math or looking at http://www.fourmilab.ch/gravitation/orbits/figures/roots.gif [Broken] indicates that this value of L results in an orbit of r near 6m, and yields slow-decaying spirals of the sort asked for. Also, objects here can only inspiral, they can't outspiral, as E<1. One uses the simulator by picking L and m, and then clicking on the effective potential diagram (not on the orbital pane!) to "launch" the particle at some particular value of r. Last edited by a moderator: May 3, 2017 6. Oct 31, 2007 ### pervect Staff Emeritus I'm getting something different. Unfortunately, at the moment I'm getting two different answers by two different routes, and it's getting late. I think I mispoke when I called r dphi/dtau a velocity, though. It should actually be a celerity (proper distance/ proper time) because tau is the proper time of the orbiting particle. 7. Oct 31, 2007 ### George Jones Staff Emeritus Yes, this is right. I even have this as an exercise for some stuff I wrote on black holes (see below)! Duh! I rushed my answer, so as not to miss my bus to work. I wrote my own Schwarzschild orbital simulator as a Java applet. It doesn't show effective potential or allow the user to play with E and L. It takes as input an r value and spatial velocity (speed and angle with respect to "horizontal") with respect to the orthonormal frame of a static observer, and then ppolts the orbit. I also wrote a set of exercises to illustrate different types of motion. But this spiral is nowhere near circular at r = 3M. In fact, for this spiral, I get v = sqrt(5/8) = 0.79 [edit: dropped a couple of sqrt's] with respect to a static observer at r = 3M. Last edited: Oct 31, 2007 8. Oct 31, 2007 ### George Jones Staff Emeritus Yes. Using $r = 6M$ in this gives $v_0 = 1/2$. To compute the speed at $r = 3M$, use the conserved quantity E twiddle (lose the tiddle from now on) given in pervect's link. From this link, and letting $\tau_p$ and $\tau_s$ be the proper times of the spiraling particle and the static observer at $r =$ something. [tex] \begin{equation*} \begin{split} E &= \left(1 - \frac{2M}{r} \right) \frac{dt}{d\tau_p}\\ &= \left(1 - \frac{2M}{r} \right) \frac{dt}{d\tau_s} \frac{d\tau_s}{d\tau_p}\\ &= \left(1 - \frac{2M}{r} \right) \left(1 - \frac{2M}{r} \right)^{-\frac{1}{2}} \gamma, \end{split} \end{equation*}$$

where $\gamma = \left( 1 - v^2 \right)$ and $v$ is the speed of the particle with respect to a static observer.

Now consider the limiting case where $E$ for a particle in orbit at $r = 6M$ equals $E$ for a spiraling particle at $r = 3M$.

$$\begin{equation*} \begin{split} \sqrt{1 - \frac{2M}{6M}} \gamma_0 &= \sqrt{1 - \frac{2M}{3M}} \gamma\\ \frac{2}{3} \gamma_0^2 &= \frac{1}{3} \gamma^2\\ 1 - v_0^2 &= 2 \left(1 - v^2 \right)\\ v &= \sqrt{\frac{1 + v_0^2}{2}} \end{split} \end{equation*}$$

Using $v_0 = 1/2$ gives $v = \sqrt{5/8} = 0.79$.

Last edited: Oct 31, 2007
9. Oct 31, 2007

### Jorrie

Yea, I was a bit perplexed for a while, because I calculated the v = 0.79 as well.

I suppose one can change the scenario so that the static observer at r=3M gives the particle a transverse boost to close to v=1, but can obviously never achieve v=1, due to the infinite energy required.

I also made these sums before and asked a rather stupid question here, but the discussion was quite interesting, thanks George.

10. Oct 31, 2007

### Jorrie

I followed the straightforward route of calculating the specific total energy E/m in the Schwarzschild frame for a circular orbit at r=6M, v=0.408 (v_local=0.5) from:

$$E/m = \frac{1-2M/r}{\sqrt{1-2M/r-v^2}} = 0.943$$

Using this energy, I extracted v for r=3 and got v=0.456 in the Schwarzschild frame. This converts to v_local=0.791 in the local (r=3M) frame.

Last edited: Oct 31, 2007
11. Oct 31, 2007

### Chris Hillman

Jorrie: ditto pervect, you made several errors. I have posted on this topic in the past so you might try to find my discussions.

The single worst error you committed was this: I have repeatedly warned (with hopefully clear and detailed examples) that failure to recognize the existence of multiple distinct operationally significant notions of "distance in the large", and thus (angular) "velocity in the large", even in flat spacetime, inevitably leads to confusion. Your post is only one examples of hundreds we have seen at PF alone.

You might look at [post=1484927]this post[/post], where I gathered links to some of my PF posts dealing with various topics related to black holes.

Last edited: Oct 31, 2007
12. Oct 31, 2007

### Jorrie

Chris, I have read most of what you wrote and will refresh again.

I think the issue in this particular instance is not the Schwarzschild radial parameter r, which is well defined, but the angular or transverse velocity 'at large'. Maybe the mention of r dphi/dtau was superfluous in this post. What is crucial here is the instantaneous transverse (circular orbital) velocity of a particle measured locally by a stationary observer at Schwarzschild radial parameter r in a Schwarzschild vacuum. We seem to have converged here on:

$$v_{o(r)} = \sqrt{\frac{M}{r(1-2M/r)}}$$

Given the definitions, is this the correct equation for it? If it is correct, the energy situation is more or less trivial.

Last edited: Oct 31, 2007
13. Oct 31, 2007

### Chris Hillman

Do you mean the velocity measured by a static observer using his rocket engine to hover at $r=r_0$, as an object in a circular orbit at the same radius whizzes by very close to his position? If so, yes, that would circumvent the "distance in the large" issue, since all such notions of distance agree for sufficiently small distances.

14. Oct 31, 2007

### Jorrie

Yep, that, or the observer being inertial and instantaneously static at r_o.

As a side issue, is $v_{o(r)} = r_o d\phi/d\tau$ in such a simple scenario in the Schwarzschild vacuum?

Last edited: Oct 31, 2007
15. Oct 31, 2007

### pervect

Staff Emeritus
Yep, that's exactly what we want to compute.

16. Oct 31, 2007

### pervect

Staff Emeritus
For the conserved quantity $E_p$ of a circular orbit of radius $r_0$ (I will also drop the twiddles), I get

$$E_p = \sqrt{\left( 1 - \frac{2\,m}{r_0} \right) \left(1 + \frac{L^2}{r_0^2} \right) }$$

with
$$L^2 = \frac{r_0^2}{\frac{r_0}{m}-3}$$

being the precondition for a circular orbit. This gives

$$E_p = \sqrt { \left( 1-2\,{\frac {m}{{ r_0}}} \right) \left( 1+ \frac{1} { \frac { r_0}{m}-3 } \right) } = \frac{1-\frac{2\,m}{r_0}}{\sqrt{1-\frac{3\,m}{r_0}}}$$

Evaluating this at r0=6m gives Ep=.9428, in agreement with Jorrie's specific calculation for that radius.
But I don't quite see how to get general agreement with Jorrie's results, yet.

I find it convenient to rewrite George's result as

$$\gamma = \frac{1}{\sqrt{1-v^2}} = \frac{E_p}{E_s}$$

which can be solved for v

$$v = \sqrt{1-\frac{1}{\gamma^2}} = \sqrt{1-\left( \frac{E_s}{E_p} \right)^2}$$

Here $E_s$ is the energy of a static observer at$r=r_s$, equal to
$$\sqrt{1-\frac{2\,m}{r_s}}$$. Even though the static observer isn't following a geodesic, E_s is constant.

Taking an alternate route, $$r \frac{d\phi}{d\tau}$$ should be equal to the celerity, and this should be given by L/r_s.

Given that
$$cel = \frac{v}{\sqrt{1-v^2}}$$

we can solve for v as a function of celerity
$$v = \frac{cel}{\sqrt{1+cel^2}}$$

We shouldn't expect this route to give the same result as the previous one, though, because it computes only the "orbital" component of the relative velocity, and ignores the components due to the infalling radial velocity.

Last edited: Oct 31, 2007
17. Oct 31, 2007

### pervect

Staff Emeritus
A geometrical motivation for why

$$\gamma = E_p / E_s$$

Consider two 4-velocities u1 and u2, in flat spacetime. Then $u1 \cdot u2 = \gamma$ (for simplicity we adopt a +--- metric signature to avoid minus signs) where $\gamma^2 = 1/(1-v^2)$, v being the magnitude of the relative velocity.

We can basically extend this result to the locally flat space-time. Call the 4-velocity of our orbiting observer up, the 4-velocity of the static observer us. Then we expect that $\gamma = us \cdot up$ in the local tangent space just as it was in flat space-time.

What are the conserved quantites that we call E_s and E_p, the energy per unit mass of a static observer, and the energy per unit mass of the orbiting observer?

They are just

E_s = us_0
E_p = up_0

We have utilized the fact that for a unit mass, the energy-momentum 4-vector P^i is just the 4-velocity u^i, and similarly P_i is just u_i. The conserved energy is therefore P_0 = u_0.

Because the metric is diagonal, and the only non-zero component of us is us_0, we can write:

$$\gamma = us \cdot up = g^{00 }E_s E_p$$

But we also have $g^{00}$ us_0 us_0 = $g^{00}$ E_s E_s = 1, because us is a 4velocity and the norm of a 4-velocity with a +--- metric signature is +1.

Thus $g^{00}$ E_s = 1/E_s, and it immediatly follows that
$$\gamma = g^{00}[/itex] E_s E_p = E_p/E_s Last edited: Nov 1, 2007 18. Nov 1, 2007 ### Jorrie Interesting manipulations, but I think both methods give only the transverse component of the local velocity, because they are both based on L. Numerically, the results are the same, it appears. I suppose the radial velocity component at r=3m might be small in comparison to the transverse component, but I guess it will not stay like that for an in-spiral to the horizon. Would it not approach "all radial" just outside the horizon? 19. Nov 1, 2007 ### pervect Staff Emeritus The first method actually involves E, not L, so it gives the total velocity, and should be identical to George's method with minor notational differences. The only reason L enters the picture in the first method is to add the contribution of the orbital velocity to the total energy to compute E. I thought your method was also based on E, but I don't quite follow how you computed E, perhaps that is where the discrepancy arises. I compute L via the fact that d(veff)/dr = 0 for a circular orbit, and given L and r, I compute E from the effective potential. The effective potential is given by the webpage I cited, you can also find it in MTW. The second method does give only the orbital component of the velocity. Last edited: Nov 1, 2007 20. Nov 1, 2007 ### Jorrie Yep, I did use the energy method and it gave a result numerically the same as George's and your method. I now understand that the answer is the resultant velocity, including the radial component. I goofed when I said that the road through L (and by implication celerity) will give the same answer; I agree it does not. I computed E for a circular orbit ($dr/dt=d \theta /dt=0$) in the Schwarzschild frame from MTW's eq. (25.18), [tex] \frac{dt}{d\tau} = \frac{E/m}{1-2M/r} = \frac{1}{\sqrt{1-2M/r - r^2(d\phi/dt)^2}}$$

which gave me my

$$E/m = \frac{1-2M/r}{\sqrt{1-2M/r-v_o^2}}$$

However, there is a lot of insight 'embedded' in the method you followed. Tx for the input and the patience.

Last edited: Nov 1, 2007
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