Orbital velocities in the Schwartzschild geometry

starthaus

Not in GR. You are fixated on galilean physics. I am sorry, untill you get off your fixations, I can't help you.

espen180

Then please explain the situation in GR.

starthaus

What do u think I've been doing for you starting with post 4?

starthaus

Certainly true since neither the Earth nor Mercury move in circles. You are trying to analyze circular motion. The rules of elliptical motion don't apply. You can't force your preconceptions on solving the problem.

espen180

So if Earth and Mercury were moving in circular paths they would have the same orbital period? How does that work? A circle is just a spacial case of an ellipse anyway.





As for your earlier equation:
[tex]v=\frac{r\omega}{\sqrt{1-r_s/r}}[/tex]



If I use this definition with my calculations I get

[tex]v=c\sqrt{\frac{\frac{r_s}{2}}{\left(r-\frac{r_s}{2}\right)\left(1-\frac{r_s}{r}\right)}}=\sqrt{\frac{GM}{\left(r-\frac{GM}{c^2}\right)\left(1-\frac{2GM}{rc^2}\right)}}[/tex]

If I let [tex]r=\frac{3GM}{c^2}[/tex] I get [tex]v=\sqrt{\frac{3}{2}}c[/tex] where it is expected to be v=c.

So my calculations must be wrong since it produces that factor of [tex]\sqrt{\frac{3}{2}}[/tex].


The equations I based my calculations off of are identical to the ones given in George Jones' reference in post #2, so I don't get what's wrong here. I also don't understand your argument that the equations don't exist.

yuiop

The substitution you got from the metric (equation 15 in your document) contains a typo that causes an error to propogate.

You state "From the Shwarschild metric we can , by imposing dr/ds=0 and theta=pi/2 obtain the relation":

[tex] ds^2 = dt^2 - r^2 d\phi^2 \;\;\; (15)[/tex]

where I am using units of G=c=1 and ds to mean proper time of the test particle.

That should read:

[tex] ds^2 = (1-2M/r) dt^2 - r^2 d\phi^2 \;\;\; (15)[/tex]

which gives:

[tex] \left(\frac{dt}{ds}\right)^2 = \frac{1}{(1-2M/r)}\left(1 + r^2 \left(\frac{d\phi}{ds}\right)^2\right) \;\;\; (16)[/tex]

I am not quite sure how you got equation (13) from (10) but there seems to be a problem there somewhere when cos(pi/2)=0 and dr/ds=0.

yuiop

I introduced the idea that L=1 in several other threads so perhaps I should clarify a little. This is valid for massive particles and is obtained directly from the metric as:

[tex]1 = \alpha \dot{t}^2 - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2[/tex]

We can of course multiply both sides by some multiple and obtain a different constant on the left. Eg if I use a multiple of 1/2 then:

[tex]1/2 = (\alpha \dot{t}^2 - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2)/2[/tex]

I can now declare L=1/2 and proceed from there if I wish. The important thing is that I arrive at an equation in a form with a constant on one side.

For a massless particle such as a photon, ds=0 and I can write the metric as:

[tex]0 = \alpha - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2[/tex]

and in this case I can use L=0. I can also obtain non-zero values of L here by adding a constant to both sides. Note that for a masslesss particle, the over-dot means the derivative with respect to coordinate time (t) rather than proper time (s). Obtain a new constant of motion for the angular velocity of a photon in the Schwarzschild metric by taking the partial differential of the right hand side with respect to [tex]\dot{\phi}[/tex]. Using the constant, solve for dr^2/dt^2 and differentiate both sides with respect to r and divide both sides by 2 to obtain the radial acceleration of a light particle. For a circular orbit d^2(r)/d(t)^2 = 0 and by setting the acceleration to zero and solving for r, the result that the photon radius is r=3M comes out very simply and clearly.

espen180

Thank you very much for pointing that error out for me.

As for (13), it was derived from (9), not (10). (10) reduces to 0=0 when the restrictions are imposed.

Updated article.

I corrected the error you pointed out, and another regarding a constant factor in one of the Christoffel symbol's entries.

The new (19), though, still does not do what it's supposed to, giving infinite velocity at the photosphere radius.

Regarding Starthaus' last equation in post #10, doesn't the two angular terms indicate that v=ωr , being the same as for flat spherical coordinates?

yuiop

Yes..sort of.. so that w=v/r, but you must bear in mind that starthaus uses a definition of w=d(theta)/ds and not the normal Newtonian w=d(theta)/dt. In spherical coordinates in SR, ds includes a time dilation factor of 1/sqrt(1-v^2/c^2) and in Schwarzschild coordinates ds includes an additional factor of 1/(sqrt(1-2M/r) so it is no longer the same as the SR case.

You have to be careful when using ds rather than dt. In the example below for something moving at the speed of light, you should expect to obtain an infinite velocity r*d(phi)/0 rather than c.

Bearing that in mind an infinite result for (19) in the case of a photon, might well be a correct result, as it is in terms of proper time.

I am a bit short of time and that is why post #25 is a bit cryptic, but it might be interesting to carry out the calculations and obtain coordinate velocity = c for a photon at r=3m using the method described.

P.S. There still seems to be a problem with (13) derived from (9) but I do not have time to check that at the moment.

yuiop

I have had another look at your revised document and checked all the calculations from (9) onwards and they all seem to be correct. Your final result can be simplified to:

[tex]\frac{rd\phi}{ds} = c \sqrt{\frac{GM}{rc^2 - 3GM}[/tex]

which gives the correct result that the velocity with respect to proper time (ds) of a particle orbiting at r=3GM/c^2 is infinite.

If you convert the above equation to local velocity as measured by a stationary observer at r by using [tex]ds = dt'\sqrt{1-(rd\phi)^2/(cdt')^2}[/tex] you get:

[tex]\frac{rd\phi}{dt'} = c \sqrt{\frac{GM}{rc^2 - 2GM}[/tex]

which gives the result that the local velocity of a particle orbiting at r=3GM/c^2 is c.

If you convert the equation to coordinate velocity using [tex]dt = dt'\sqrt{1-2M/(rc^2)}[/itex] you get:

[tex]\frac{rd\phi}{dt} = c \sqrt{\frac{GM}{r}[/tex]

which is the same as the Newtonian result (if you use units of c=1).

All the above are well known solutions, so it seems all is in order with your revised document (from (9) onwards anyway - I have not checked the preceding calculations).

starthaus

It is just as wrong as I explained to you in post 12. If you did things correctly, then you'd have realised that [tex]dr=d\theta=0[/tex] reduces the metric to :

[tex]ds^2=(1-r_s/R)dt^2-R^2d\phi^2[/tex]

where [tex]R[/tex] is a constant. This is important in the correct derivation of the Christoffel symbols, since you should get a lot more "zeroes" than you have in your writeup.

So, your Christoffel symbols need to reflect that. They don't. You need to recalculate the Christoffel symbols based on the correct metric.

yuiop

I think Espen makes it clear that he is deriving the general case in section 1 using Christoffel symbols and then in section 2 analyses the specific case of pure circular motion. Espen's updated document contains the corrected metric (15) for the special case.

starthaus

Section 1 has no bearing on the subject.

This is not how things are done. You start with the correct metric, you construct the associated metric tensor and you calculate the appropriate Christoffel symbols. If you do all this correctly you will get two (not three) very simple equations that are identical to the ones already shown in post 6.
Covariant derivatives and lagrangian methods, if done correctly, should produce the same results.


Yet, it contains the wrong Christoffel coefficients associated with the metric (15). This is the source of his errors as explained already in post 12.

espen180

@#27:
Thank you very much. I will study these in greater detail.

How can the general case be irrelevant to the special case?


The main problem I have with accepting this is that those two equations alone cannot give an expression of the angular velocity of a circular orbit at a paticular radius. If I am mistaken on this point, please show a derivation of such an expression from those two equations.

starthaus

As I explained to you several times, you need to start with the appropriate metric. Do that and you'll get the correct results. Until you do that, you will continue to get just errors.

yuiop

I will expand on the above to flesh out this alternative proof that the photon sphere in the Schwarzschild metric is r=3M.

Starting with Schwarzschild metric and assuming orbital motion in a plane about the equator such that [itex]\theta = \pi/2[/itex] and [itex]d\theta = 0[/itex]

[tex]ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2[/tex]

[tex]\alpha=1-\frac{2m}{r}[/tex]

For a massless particle travelling at the speed of light, ds=0 so in this case:

[tex]L = 0 =\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2[/tex]

Divide both sides by dt^2:

[tex]L = 0 =\alpha -\frac{1}{\alpha}\frac{dr^2}{dt^2}-r^2\frac{d\phi^2}{dt^2}[/tex]

The metric is independent of [itex]\phi[/itex] and t, so the constant associated with the angular velocity of a photon is obtained by finding the partial derivative of L with respect to [itex]d\phi/dt[/itex]

[tex]\frac{\delta{L}}{\delta(d\phi/dt)} = r^2 \frac{d\phi}{dt} = H_c [/tex]

where [itex]H_c[/itex] is the specific form of the constant for angular velocity that applies to a massless particle.

Substitute this constant into the metric for a massless particle above to obtain:

[tex]0 =\alpha -\frac{1}{\alpha}\frac{dr^2}{dt^2}-\frac{H_c^2}{r^2}[/tex]

and solve for (dr/dt)^2:

[tex]({dr}/{dt})^2 = (1-2M/r)(1-2M/r -H_c^2/r^2) [/tex]

Differentiate the above with respect to r and divide by 2 (this is the same as differentiating dr/dt with respect to t, but is much quicker and simpler) to obtain the radial acceleration of a photon in the metric:

[tex]\frac{d^2r}{dt^2}= \frac{H_c^2}{r^4}(r-3M) [/tex]

Setting d^2r/dt^2 = 0 (which is true for a circular orbit) and solving for r gives r=3M as the circular orbit radius of a particle travelling at the speed of light.

starthaus

How can this be since in the line above you declared [tex]L=0[/tex]?

Is this again some sort of numerology that defies the rules of calculus?