# Orbital velocities in the Schwartzschild geometry

by espen180
Tags: geometry, orbital, schwartzschild, velocities
 P: 835 I'm trying to use the tensor formulation of GR to calculate the velocity of a particle in a circular orbit around a black hole. Here is the work I have done so far. What concerns me is that I end up getting zero velocity when applying the metric to the differential equations I get from the geodesic equation. I wonder if I have made a miscalculation, but I am unable to find any, so maybe there is a misunderstanding on my part. Any help is appreciated.
 Mentor P: 5,903 Compare what you have with page 177 from http://books.google.com/books?id=uG9...page&q&f=false.
 PF Patron Sci Advisor Emeritus P: 5,311 There seems to be a problem with signs in (13), because both terms are negative-definite. Using (12), and choosing t=0 to coincide with $\tau=0$, you can set $t=\beta \tau$, where $\beta$ is a constant. Let's also set $\omega=d\phi/der t$. Then (13) gives $\omega=\pm i\sqrt{2m/r^3}$, where $m=r_s/2$. This seems sort of right, since it coincides with Kepler's law of periods. However, it's imaginary due to the sign issue. It would also surprise me if Kepler's law of periods was relativistically exact when expressed in terms of the Schwarzschild coordinates, but maybe that's the case. If you can fix the sign problem, then you seem to have the right result in the nonrelativistic limit of large r. You might then want to check the result in the case of r=3m, where I believe you should obtain lightlike circular orbits.
P: 1,568

## Orbital velocities in the Schwartzschild geometry

 Quote by bcrowell There seems to be a problem with signs in (13), because both terms are negative-definite.
You are right, his equation (13) is in the "not-even-wrong" category. It is easu to see that since the correct Lagrangian, for the simplified case he's considering is:

$$L=(1-r_s/r)\frac{dt^2}{ds^2}-r^2\frac{d\phi^2}{ds^2}$$
 P: 835 @George Jones , bcrowell Thanks for pointing that out, and thank you for the reference. I traced the sign error back to a differentiation error when calculating the Christoffel symbol. As for writing $t=\beta\tau$ I don't see how that will help since t does not appear in the other equations. I must be missing something. I will try to arrive at a result and do the "reality checks" you mentioned. @Starthaus Thanks for your input. I am afraid I don't know how to arrive at or what to do with the Lagrangian. From it's appearance it looks just like the metric, so L=1 here, I imagine? EDIT: I arrived at $$\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{r}\sqrt{\frac{r_s}{r-r_s}}$$. The units match, but I doubt this is correct, since letting r=3m gives $$\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{\sqrt{2m}}$$ while my intuition tells me it should be c.
P: 1,568
 Quote by espen180 EDIT: I arrived at $$\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{r}\sqrt{\frac{r_s}{r-r_s}}$$. The units match, but I doubt this is correct, since letting r=3m gives $$\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{\sqrt{2m}}$$ while my intuition tells me it should be c.
No, the correct equations are:

$$\frac{d^2t}{ds^2}=0$$

and

$$\frac{d^2\phi}{ds^2}=0$$

Hint: in your writeup you made $$dr=d\theta=0$$, remember? You need to think what that means.
P: 835
 Quote by starthaus No, the correct equations are: $$\frac{d^2t}{ds^2}=0$$ and $$\frac{d^2\phi}{ds^2}=0$$ Hint: in your writeup you made $$dr=d\theta=0$$, remember? You need to think what that means.
Yes, I arrived at these as well when I made the assumptions you mentioned. As for their meaning, I interpret it as circular (constant radial coordinate) motion around the equator ($\theta=\pi/2$) with constant velocity. In addition, there was a third equation I arrived at,

$$\frac{c^2r_s}{r^2}\left(\frac{\text{d}t}{\text{d}\tau}\right)^2-r\left(\frac{\text{d}\phi}{\text{d}\tau}\right)^2=0$$

which, when I used the substitution

$$\left(\frac{\text{d}t}{d\tau}\right)^2=1+\frac{r^2}{c^2}\left(\frac{\te xt{d}\phi}{\text{d}\tau}\right)^2$$

which I got from the metric, gave me

$$\frac{c^2r_s}{r^2}+(r_s-r)\left(\frac{\text{d}\phi}{\text{d}\tau}\right)^2=0$$

which solves to

$$\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{r}\sqrt{\frac{r_s}{r-r_s}}$$

So now I am unsure about where I made my mistake.
P: 1,568
 Quote by espen180 Yes, I arrived at these as well when I made the assumptions you mentioned. As for their meaning, I interpret it as circular (constant radial coordinate) motion around the equator ($\theta=\pi/2$) with constant velocity. In addition, there was a third equation I arrived at, $$\frac{c^2r_s}{r^2}\left(\frac{\text{d}t}{\text{d}\tau}\right)^2-r\left(\frac{\text{d}\phi}{\text{d}\tau}\right)^2=0$$
There are only two independent equations, the ones I mentioned to you.
The third Lagrange equation, exists only if r is variable and its correct form would have been:

$$\frac{r_s}{r^2}\frac{dt^2}{ds^2}-2r\frac{d\phi^2}{ds^2}=0$$

But you made $$dr=0$$ (this is why I gave you the hint), so the third equation does not exist. This is the root of your errors.
 P: 835 I don't understand why it shouldn't exist. I derived the general case, then assumed dr=0 and substituted that into the equations. dr doesn't even appear in that equation anymore, do I don't see how it has any influence. EDIT: Where does the factor of 2 come from there? I somehow didn't appear in my calculations. For the velocity, I obtained $$v=\frac{\text{d}\phi}{\text{d}\tau}r=c\sqrt{\frac{r_s}{r-r_s}}=\sqrt{\frac{2GM}{r-\frac{2GM}{c^2}}}$$ My intuition says that this is wrong by a factor of $\sqrt{2}[/tex], since then it would give v=c at [itex]r=\frac{3GM}{c^2}$, but I don't see how that factor dissapeared.
P: 1,568
 Quote by espen180 I don't understand why it shouldn't exist. I derived the general case, then assumed dr=0 and substituted that into the equations. dr doesn't even appear in that equation anymore, do I don't see how it has any influence.
You started with the metric that has $$dr=0$$. therefore all your attempts to differentiate wrt $$r$$ should result in null terms. Yet, you clearly differentaite wrt $$r$$ in your derivation and this renders your derivation wrong.

 EDIT: Where does the factor of 2 come from there? I somehow didn't appear in my calculations.
From the Euler-Lagrange equations.

 For the velocity, I obtained $$v=\frac{\text{d}\phi}{\text{d}\tau}r=c\sqrt{\frac{r_s}{r-r_s}}=\sqrt{\frac{2GM}{r-\frac{2GM}{c^2}}}$$ My intuition says that this is wrong by a factor of $\sqrt{2}[/tex], since then it would give v=c at [itex]r=\frac{3GM}{c^2}$, but I don't see how that factor dissapeared.
Yes, it is very wrong.
From the correct equation $$\frac{d^2\phi}{ds^2}=0$$ you should obtain (no surprise):

$$\frac{d\phi}{ds}=constant=\omega$$

The trajectory is completed by the other obvious equation

$$r=R=constant$$

You get one more interesting equation, that gives u the time dilation. Start with:

$$ds^2=(1-r_s/R)dt^2-(Rd\phi)^2$$ and you get:

$$\frac{dt}{ds}=\sqrt{\frac{1+(R\omega)^2}{1-r_s/R}}$$

or:

$$\frac{ds}{dt}=\sqrt{1-r_s/R}\sqrt{1-\frac{(R\omega)^2}{1-r_s/R}$$

The last equation gives you the hint that:

$$v=\frac{R\omega}{\sqrt{1-r_s/R}}$$

The last expression is what you were looking for.
 P: 835 The only thing I said is that I have constant r. If you have a function, say f(x)=x2, you can say that x is fixed at three, but you can still get the slope at x=3 by differentiating wrt x. As for the last equation you posted. I don't doubt its validity, but I don't see how it will get me anywhere, since neither v nor $$\omega$$ are known. In fact, since $$v=\omega R$$, v cancels on both sides.
P: 1,568
 Quote by espen180 The only thing I said is that I have constant r. If you have a function, say f(x)=x2, you can say that x is fixed at three, but you can still get the slope at x=3 by differentiating wrt x.
Umm, no. If you did things correctly, then you'd have realised that $$dr=d\theta=0$$ reduces the metric to :

$$ds^2=(1-r_s/R)dt^2-R^2d\phi^2$$

So, your Christoffel symbols need to reflect that. They don't.

 As for the last equation you posted. I don't doubt its validity, but I don't see how it will get me anywhere, since neither v nor $$\omega$$ are known.
Yet, the result is extremely important since it tells you that the orbiting object has constant angukar speed and the trajectory is a circle.

 In fact, since $$v=\omega R$$, v cancels on both sides.
No, $$v$$ is not $$\omega R$$.
P: 835
 Quote by starthaus Umm, no. If you did things correctly, then you'd have realised that $$dr=d\theta=0$$ reduces the metric to : $$ds^2=(1-r_s/R)dt^2-R^2d\phi^2$$ So, your Christoffel symbols need to reflect that. They don't. Yet, the result is extremely important since it tells you that the orbiting object has constant angukar speed and the trajectory is a circle. No, $$v$$ is not $$\omega R$$.
How do you define v?

Of course the trajectory is a circle. I imposed that restriction by setting r=constant after deriving the general case geodesic equations. The equations saying d2t/ds2=0 and d2φ/ds2=0 are neccesary consequences.

What I am seeking is an expression which gives the orbital velocity as a function of r. I define $$v=\frac{\text{d}\phi}{\text{d}\tau}r$$, and except for the factor $$\sqrt{2}$$ my result reduces to the Newtonian formula at large r, which makes me beleive my derivation is valid, the erronous factor $$\sqrt{2}$$ notwithstanding.
P: 1,568
 Quote by espen180 .I define $$v=\frac{\text{d}\phi}{\text{d}\tau}r$$, and except for the factor $$\sqrt{2}$$ my result reduces to the Newtonian formula at large r, which makes me beleive my derivation is valid, the erronous factor $$\sqrt{2}$$ notwithstanding.
You don't get to "define", you need to "derive" :

$$\frac{ds}{dt}=\sqrt{1-r_s/R}\sqrt{1-\frac{(R\omega)^2}{1-r_s/R}$$

The last equation gives you the hint that:

$$v=\frac{R\omega}{\sqrt{1-r_s/R}}$$

since, in GR:

$$\frac{ds}{dt}=\sqrt{1-r_s/r}\sqrt{1-v^2}$$
 P: 835 And is v, in your case, measured by an observer from infinity? You have to give a definition. $$v=\frac{\text{d}\phi}{\text{d}\tau}r$$ and $$\frac{\text{d}\phi}{\text{d}t}r$$, for example, aren't the same, so you have to specify. Aside from that, how can your equation be used to calculate the orbital period as a function of r only?
P: 1,568
 Quote by espen180 And is v, in your case, measured by an observer from infinity? You have to give a definition. $$v=\frac{\text{d}\phi}{\text{d}\tau}r$$ and $$\frac{\text{d}\phi}{\text{d}t}r$$, for example, aren't the same, so you have to specify.
You can do it all by yourself by remembering that $$\frac{d\phi}{d\tau}=\omega$$ (see the derivation from the Euler-Lagrange equation)

 Aside from that, how can your equation be used to calculate the orbital period as a function of r only?
$$\phi=\omega \tau$$. Make $$\phi=2\pi$$
The orbital period is not a function of r.
P: 835
 Quote by starthaus You can do it all by yourself by remembering that $$\frac{d\phi}{d\tau}=\omega$$ (see the derivation from the Euler-Lagrange equation)
For circular motion, it is easy to obtain the relationship $$v=\omega r$$. You get it directly from the definition of the radian. Do you claim $$v=\frac{\text{d}\phi}{\text{d}\tau} r$$ is not a valid definition of v? If so, please explain.

Please link to the Euler-Lagrange derivation, and I'll do my best to understand it.

 Quote by starthaus $$\phi=\omega \tau$$. Make $$\phi=2\pi$$ The orbital period is not a function of r.
How do you arrive at that conclusion? It is obvious that the orbital period is a function of r. That's why Mercury's orbital period is shorter that Earth's.
P: 1,568
 Quote by espen180 For circular motion, it is easy to obtain the relationship $$v=\omega r$$. You get it directly from the definition of the radian.
Not in GR. You are fixated on galilean physics. I am sorry, untill you get off your fixations, I can't help you.

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