Orbital velocities in the Schwartzschild geometry

[espen180]

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.

 

[George Jones]

Compare what you have with page 177 from

http://books.google.com/books?id=uG...rasinski+relativity&cd=10#v=onepage&q&f=false.

 

[bcrowell]

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.

 

[starthaus]

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}

 

[espen180]

@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.

 

[starthaus]

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.

 

[espen180]

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{\text{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.

 

[starthaus]

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.

 

[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.

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 r=\frac{3GM}{c^2}, but I don't see how that factor dissapeared.

 

[starthaus]

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 r=\frac{3GM}{c^2}, but I don't see how that factor dissapeared.

<br /> <br /> Yes, it is very wrong.<br /> From the correct equation \frac{d^2\phi}{ds^2}=0 you should obtain (no surprise):<br /> <br /> \frac{d\phi}{ds}=constant=\omega<br /> <br /> The trajectory is completed by the other obvious equation<br /> <br /> r=R=constant<br /> <br /> You get one more interesting equation, that gives u the time dilation. Start with:<br /> <br /> ds^2=(1-r_s/R)dt^2-(Rd\phi)^2 and you get:<br /> <br /> \frac{dt}{ds}=\sqrt{\frac{1+(R\omega)^2}{1-r_s/R}}<br /> <br /> or:<br /> <br /> \frac{ds}{dt}=\sqrt{1-r_s/R}\sqrt{1-\frac{(R\omega)^2}{1-r_s/R}<br /> <br /> The last equation gives you the hint that:<br /> <br /> v=\frac{R\omega}{\sqrt{1-r_s/R}}<br /> <br /> The last expression is what you were looking for.

 

[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.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.

 

[starthaus]

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 realized 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.

 

[espen180]

Umm, no. If you did things correctly, then you'd have realized 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 believe my derivation is valid, the erronous factor \sqrt{2} notwithstanding.

 

[starthaus]

.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 believe 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}

 

[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.


Aside from that, how can your equation be used to calculate the orbital period as a function of r only?

 

[starthaus]

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.

 

[espen180]

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.

\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.

 

[starthaus]

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, until you get off your fixations, I can't help you.

 

[espen180]

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

Then please explain the situation in GR.

 

[starthaus]

Then please explain the situation in GR.

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

 

[starthaus]

For circular motion, ...



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.

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]

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.

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 spatial case of an ellipse anyway.As for your earlier equation:
v=\frac{r\omega}{\sqrt{1-r_s/r}}
If I use this definition with my calculations I get

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)}}

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

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

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{\text{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.

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":

ds^2 = dt^2 - r^2 d\phi^2 \;\;\; (15)

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

That should read:

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

which gives:

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

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]

From it's appearance it looks just like the metric, so L=1 here, I imagine?

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:

1 = \alpha \dot{t}^2 - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2

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:

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

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:

0 = \alpha - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2

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 \dot{\phi}. 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]

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?

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.

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.

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]

...
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.

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:

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

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 ds = dt&#039;\sqrt{1-(rd\phi)^2/(cdt&#039;)^2} you get:

\frac{rd\phi}{dt&#039;} = c \sqrt{\frac{GM}{rc^2 - 2GM}

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 dt = dt&#039;\sqrt{1-2M/(rc^2)}[/itex] you get:<br /> <br /> \frac{rd\phi}{dt} = c \sqrt{\frac{GM}{r}<br /> <br /> which is the same as the Newtonian result (if you use units of c=1).<br /> <br /> 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]

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.

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

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

where R 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]

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

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

where R 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.

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]

I think Espen makes it clear that he is deriving the general case in section 1 using Christoffel symbols

Section 1 has no bearing on the subject.

and then in section 2 analyses the specific case of pure circular motion.

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.

Espen's updated document contains the corrected metric (15) for the special case.

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.

Section 1 has no bearing on the subject.

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

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.

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]

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

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]

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

0 = \alpha - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2

and in this case I can use L=0.
...
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 \dot{\phi}. 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.

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 \theta = \pi/2 and d\theta = 0

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

\alpha=1-\frac{2m}{r}

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

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

Divide both sides by dt^2:

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

The metric is independent of \phi 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 d\phi/dt

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

where H_c 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:

0 =\alpha -\frac{1}{\alpha}\frac{dr^2}{dt^2}-\frac{H_c^2}{r^2}

and solve for (dr/dt)^2:

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

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:

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

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 traveling at the speed of light.

 

[starthaus]

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 \theta = \pi/2 and d\theta = 0

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

\alpha=1-\frac{2m}{r}

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

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

Divide both sides by dt^2:

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

The metric is independent of \phi 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 d\phi/dt

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

.
.

How can this be since in the line above you declared L=0?

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

 

[espen180]

You mean:

\frac{\delta{L}}{\delta(d\phi/dt)} =-2 r^2 \frac{d\phi}{dt}?

How can this be since in the line above you declared L=0?

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

It's true that there is a differentiation error, but that doesn't invalidate the proof since the result was unchanged.

\frac{H_c^2}{4r^4}(r-3M)=0

still gives

r=3M

 

[starthaus]

It's true that there is a differentiation error, but that doesn't invalidate the proof since the result was unchanged.

\frac{H_c^2}{4r^4}(r-3M)=0

still gives

r=3M

No, it is not a "differentiating error", it is numerology.Apparently you share the same misconception with kev, that constants can be differentiated resulting into non-zero algebraic expressions.

 

[yuiop]

It's true that there is a differentiation error, but that doesn't invalidate the proof since the result was unchanged.

\frac{H_c^2}{4r^4}(r-3M)=0

still gives

r=3M

I was aware that

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

should be

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

but since the above equation is effectively

0 = 2 r^2 \frac{d\phi}{dt}

it is perfectly valid to divide both sides by 2. As long as the substitution is done correctly (as Espen has shown) it does not matter what multiple you use. The Lagrangian does not tell you the numerical value of the constants, it just tells you that they are in fact constants. I am not a mathematician so I can not explain it rigorously. I am just observing how others do it.

 

[starthaus]

I was aware that

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

should be

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

but since the above equation is effectively

0 = 2 r^2 \frac{d\phi}{dt}

it is perfectly valid to divide both sides by 2. As long as the substitution is done correctly (as Espen has shown) it does not matter what multiple you use.

that is a minor error.
Why don't you address the major error?

 

[yuiop]

that is a minor error.
Why don't you address the major error?

It's not.

 

[starthaus]

It's not.

So, according to you differentiating 0 produces a polynomial ? You really need to learn calculus.

 

[yuiop]

So, according to you differentiating 0 produces a polynomial ? You really need to learn calculus.

Are you saying that you can not differentiate the left hand side of ax^2 + bx + c = 0 because the right hand side is zero?

 

[starthaus]

Are you saying that you can not differentiate the left hand side of ax^2 + bx + c = 0 because the right hand side is zero?

If you do that , you get something as meaningless as you have been getting in your "derivation', i.e. 2ax+b=0.

You declared L=0. Otherwise, you can't execute all the manipulations you have been executing. So, you can't change your mind and differentiate it and obtain a non-zero result. This is calculus 101.

 

[espen180]

If you do that , you get something as meaningless as you have been getting in your "derivation', i.e. 2ax+b=0.

You declared L=0. Otherwise, you can't execute all the manipulations you have been executing. So, you can't change your mind and differentiate it and obtain a non-zero result. This is calculus 101.

This makes sense. But still, even though you say the derivations are erronous, they are producing correct results, so there must be something compensating for the errors. For example, you say that post #8 that I am wrong when obtaining three differential equations, but how do you then explain that they give the correct orbits?

 

[Al68]

This makes sense. But still, even though you say the derivations are erronous, they are producing correct results...

Imagine that. Someone can technically violate someone else's "rules" of calculus and get the correct result? Seriously, that's done routinely. And as has been pointed out in other threads, there are often ways to prevent the "broken rule" from rendering the result incorrect.

 

[starthaus]

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

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:

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

.

Nope, your "differentiation" skills are still bad, in addition to the errors I flagged down earlier you are missing a whole bunch of terms in the above differentiation.Show your steps and you'll find out what terms are missing.
You simply put in the desired answer by hand. According to Al68 and espen180, the glaring errors don't matter, it is important that you got the "expected" result.

 

[Al68]

According to Al68 and espen180, the glaring errors don't matter, it is important that you got the "expected" result.

I didn't say errors don't matter. I said textbook calculus rules were broken routinely. I meant on purpose, not by "error".

Sorry if that wasn't clear.

 

[yuiop]

Nope, your "differentiation" skills are still bad, in addition to the errors I flagged down earlier you are missing a whole bunch of terms in the above differentiation.Show your steps and you'll find out what terms are missing.
You simply put in the desired answer by hand. According to Al68 and espen180, the glaring errors don't matter, it is important that you got the "expected" result.

I have to admit you are right and I made a mistake in the calculations. I do not claim to have any calculus skills and simply use mathematical software (e.g. http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=calculus&s2=differentiate&s3=advanced. ) Unfortunately I plugged in the wrong expression:

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

instead of:

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

and got a seemingly correct solution by a bizarre self cancelling double error (GIGO). I will try and fix that. You claim to be good at calculus and have an understanding of the Euler-Lagrange formalism for the equations of motion, so maybe you could demonstrate the correct solution to us?

 

[yuiop]

You claim to be good at calculus and have an understanding of the Euler-Lagrange formalism for the equations of motion, so maybe you could demonstrate the correct solution to us?

Seeing as how Starthaus has failed to rise to the challenge (presumably because the solution is not in one of his textbooks) I will make this second attempt to find an alternative direct derivation of the coordinate orbital velocity and orbital radius of a photon.

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

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

\alpha=1-\frac{2m}{r}

Divide both sides by \alpha dt^2 and rearrange so that the constant (1) is on the LHS:

L = 1 =\frac{1}{\alpha}\frac{ds^2}{dt^2} -\frac{1}{\alpha^2}\frac{dr^2}{dt^2}-\frac{r^2}{\alpha}\frac{d\phi^2}{dt^2}


The metric is independent of \phi and t, so there is a constant associated with coordinate angular velocity (H_c) which is obtained by finding the partial derivative of L with respect to d\phi/dt

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

The metric is independent of s and t, so there is a constant associated with time dilation (K_c) which is obtained by finding the partial derivative of L with respect to ds/dt

\frac{\delta{L}}{\delta(ds/dt)} = \frac{1}{\alpha} \frac{ds}{dt} = K_c

Substitute these constants into the equation for L and solve for (dr/dt)^2:

({dr}/{dt})^2 = (1-2M/r)^2 -(1-2M/r)^3 (K_c^2 + H_c^2/r^2)

Differentiate the above with respect to r and divide by 2 to obtain the general coordinate radial acceleration of a freefalling particle in the metric:

a = \frac{d^2r}{dt^2}= \frac{\alpha^2 H_c^2}{r^4}(r-5M) +\frac{M}{r^2}(2\alpha-3K_c^2\alpha^2)

Special case of an orbiting photon:

For a photon, ds=0 and therefore K_c=0 and for a circular orbit dr/dt =0. Substituting these values into the equation for (dr/dt)^2 and solving for H_c^2 gives:

H_c^2 = r^2/\alpha

For a circular orbit we also require the acceleration to be zero. Substituting H_c^2 = r^2/\alpha and K_c=0 into the equation for the general coordinate radial acceleration (a) gives:

0 = \frac{\alpha}{r^2}(r-5M) +\frac{2M\alpha}{r^2}

and solving for r gives r=3M.

Still using a = 0 and K_c=0 and substituting the expanded version of H_c[/tex] back into the equation for (a) gives:<br /> <br /> 0 = \frac{d\phi^2}{dt^2}(r-5M) +\frac{2M}{r^2}(1-2M/r)<br /> <br /> Solve the above for the coordinate tangential velocity of an orbiting photon:<br /> <br /> \frac{r d\phi}{dt} = \sqrt{\frac{-2M}{r}\frac{(r-2M)}{(r-5M)}}<br /> <br /> At r=3M this gives:<br /> <br /> \frac{r d\phi}{dt} = \frac{1}{\sqrt{3}}<br /> <br /> which is in agreement with the equation I obtained in #27 based on Espen&#039;s derivation:<br /> <blockquote data-attributes="" data-quote="kev" data-source="post: 2761498" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> kev said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> If you convert the equation to coordinate velocity using dt = dt&amp;#039;\sqrt{1-2M/(rc^2)} </div> </div> </blockquote>

you get:<br /> <br /> \frac{rd\phi}{dt} = c \sqrt{\frac{GM}{r}

<br /> The local velocity according to a stationary observer at r=3M is:<br /> <br /> \frac{r d\phi}{dt} \frac{dt}{dt&amp;#039;} = \frac{r d\phi}{dt&amp;#039;} = \frac{1}{\sqrt{3}}\frac{1}{\sqrt{1-2M/r}} = 1

 

[starthaus]

Seeing as how Starthaus has failed to rise to the challenge (presumably because the solution is not in one of his textbooks)

No, it is because I am tired of correcting your errors and hacks.

 

[espen180]

After reading post #27, the similarity between the Newtonian expression and the one I arrived at makes it very difficult for me to believe it is incorrect. It also predicts that the photon sphere should be at r=3GM. Post #48 seems to confirm my belief here.I tried to use the same approach to derive the coordinate acceleration of a particle dropped from rest at r relative to a stationary observer also at r. The result was
\frac{\text{d}^2r}{\text{d}t^2}=-\left(\frac{GM}{r^2}-\frac{2G^2M^2}{r^3c^2}\right)
which goes to 0 as r approaches the Schwartzschild radius, I'm unsure what to make of that. It also changes sign when r<2GM, so I doubt its validity in that region. Still, it approximates the Newtonian expression at large r, differing only by about 1.4 ppb at Earth's surface.

Does it look correct? If neccesary, I can post my derivation.