Orbital velocities in the Schwartzschild geometry

[yuiop]

Proof that H is not a function of r.

Starting with the textbook geodesic equation derived without any assumptions of constants:

\frac{d^2\phi}{ds^2} = \frac{-2}{r} \frac{dr}{ds}\frac{d\phi}{ds} - 2 \cot \theta \frac{d\phi}{ds} \frac{d\theta}{ds}

Seehttp://sites.google.com/site/espen180files/Schwartzschild.pdf?attredirects=2"

For orbital motion in a plane with d\theta/ds=0 and \theta = \pi/2[/tex] this reduces to:<br /> <br /> \frac{d^2\phi}{ds^2} = -\frac{2}{r} \frac{dr}{ds}\frac{d\phi}{ds}<br /> <br /> \Rightarrow \frac{d}{ds} \left(\frac{d\phi}{ds} \right) = -\frac{2}{r} \frac{dr}{ds}\frac{d\phi}{ds}<br /> <br /> \Rightarrow \frac{d}{dr} \left(\frac{d\phi}{ds} \right) = -\frac{2}{r} \frac{d\phi}{ds}<br /> <br /> \Rightarrow r^2 \frac{d}{dr} \left(\frac{d\phi}{ds} \right) = -2r \frac{d\phi}{ds}<br /> <br /> \Rightarrow r^2 \frac{d}{dr} \left(\frac{d\phi}{ds} \right) = -\frac{d}{dr}(r^2) \frac{d\phi}{ds} + C_{(r)}<br /> <br /> where C_{(r)} is a constant with respect to r.<br /> <br /> \Rightarrow r^2 \frac{d}{dr} \left(\frac{d\phi}{ds} \right) + \frac{d}{dr}(r^2) \frac{d\phi}{ds} = C_{(r)}<br /> <br /> \Rightarrow \frac{d}{dr} \left( r^2\frac{d\phi}{ds} \right) = C_{(r)}<br /> <br /> \Rightarrow \frac{d}{dr} \left(H) = C_{(r)}<br /> <br /> Therefore H is a constant with respect to r for a particle in freefall under all circumstances.<br /> <br /> QED.

 

[starthaus]

Nonsense. It is ridiculous that you want to correct D'inverno. Go read the page 196 of his "well-known" GR book. Where on Earth do they claim r=r(s) isn't a part of spherical coordinates? You're really clueless about basics of physics.

AB

I am not correcting, D'Iverno, I am correcting your error. We are not talking about any r(s), we are talking about your mixinng up coordinate r with trajectory r=r(\phi).

 

[starthaus]

Proof that H is not a function of r.

Starting with the textbook geodesic equation derived without any assumptions of constants:

\frac{d^2\phi}{ds^2} = \frac{-2}{r} \frac{dr}{ds}\frac{d\phi}{ds} - 2 \cot \theta \frac{d\phi}{ds} \frac{d\theta}{ds}

Seehttp://sites.google.com/site/espen180files/Schwartzschild.pdf?attredirects=2"

For orbital motion in a plane with d\theta/ds=0 and \theta = \pi/2[/tex] this reduces to:<br /> <br /> \frac{d^2\phi}{ds^2} = -\frac{2}{r} \frac{dr}{ds}\frac{d\phi}{ds}

<br /> <br /> You should recognize the above equation as nothing but the second Euler-Lagrange equation:<br /> <br /> \frac{d}{ds}(r^2*\frac{d\phi}{ds})=0<br /> <br /> i.e.<br /> <br /> r^2\frac{d\phi}{ds}=H<br /> <br /> Indeed:<br /> <br /> 0=\frac{d}{ds}(r^2*\frac{d\phi}{ds})=2r\frac{dr}{ds}\frac{d\phi}{ds}+r^2\frac{d^2\phi}{ds^2}<br /> <br /> resulting into:<br /> <br /> \frac{d^2\phi}{ds^2}=-\frac{2}{r}\frac{dr}{ds}\frac{d\phi}{ds}<br /> <br /> So, your &quot;proof&quot; uses the conclusion, meaning that it is invalid.<br /> <br /> (PS: If you want to learn how this ODE is solved in mainstream mathematics, it is really done by using variable separation, not the method that you used)<br /> <br /> Look, kev<br /> <br /> We&#039;ve been round and round about your approach in solving this kind of problems, you just pick the <b>results</b> from the proper methods (the geodesic or the Euler-Lagrange equations) because you can&#039;t derive the equations from scratch, you combine them with your own approach to differentiation and you get the same results as the ones <b>already</b> obtained through rigorous math. <br /> Your type of approach has been copied by <b>espen180</b> and , if you look at his paper, has led him way into the left field. One look at eqs (51) thru (58) and you can see that he&#039;s hopelessly bogged down without any hope of discovering the correct solution. The correct solution takes only about 5 lines (see posts 53 or 430). Why don&#039;t you spend all this effort in defending your approach by learning one of two the proper methods? It would be infinitely more productive. Then, you can try helping espen180 out of the mess he&#039;s gotten into.

 

[Altabeh]

I am not correcting, D'Iverno, I am correcting your error. We are not talking about any r(s), we are talking about your mixinng up coordinate r with trajectory r=r(\phi).

Yes you're correcting him because he doesn't do anything in the page 196 of his book but supporting my view over the issue. You want me write out the exact sentence from his book here, huh!? LOL.

The coordinate r here is the same thing as the trajectory r=r(\phi) and all you're trying to do is nothing but walking through a dead-end escape ruote towards not standing corrected. Another nonsense of yours so try something else!

AB

 

[starthaus]

Yes you're correcting him because he doesn't do anything in the page 196 of his book but supporting my view over the issue. You want me write out the exact sentence from his book here, huh!? LOL.

Stop trying to deflect the discussion from your errors.
No, I am correcting you, d'Iverno is just a red herring. Since you don't know the difference between coordinate
r and trajectory r=r(\phi) you are hardly in any position to "teach".

The coordinate r here is the same thing as the trajectory r=r(\phi)

Repeating the same errors is not the way to demonstrate your knowledge. r and \phi are coordinates, r=r(\phi) represents a connection between the two coordinates, so it is a trajectory. Please find a different thread to troll.

 

[espen180]

Stop trying to deflect the discussion from your errors.
No, I am correcting you, d'Iverno is just a red herring. Since you don't know the difference between coordinate
r and trajectory r=r(\phi) you are hardly in any position to "teach".




Repeating the same errors is not the way to demonstrate your knowledge. r and \phi are coordinates, r=r(\phi) represents a connection between the two coordinates, so it is a trajectory. Please find a different thread to troll.

As qbert said earlier, there is only one independent coordinate, the proper time. All the others are dependent coordinates, or functions if you will, that obey the geodesic equation.

 

[starthaus]

As qbert said earlier, there is only one independent coordinate, the proper time. All the others are dependent coordinates, or functions if you will, that obey the geodesic equation.

Doesn't change the fact that r=r(\phi) represents a trajectory, not a coordinate.
Aren't you going to clean up the mess in (51)-(58) in your paper? Your time would much better spent by fixing the errors in your paper, especially since you have been already shown the correct the solutions. BTW: your eq (30) is still wrong.

 

[Altabeh]

Stop trying to deflect the discussion from your errors.
No, I am correcting you, d'Iverno is just a red herring. Since you don't know the difference between coordinate
r and trajectory r=r(\phi) you are hardly in any position to "teach".

No! You stop hacking and hacking more! Once you turned out to be finished in your post "251" about your old fallacy, you have been wasting our time by your other nonsense here since then and it has started to be like a job of distracting minds from believing that kev's solution is fine through making nonsense claims/wishful thinking! So if you think you're not correcting him, I'm going to qoute directly from his book:

"... Then (15.20) can be integrated directly to give

r^2\dot{\phi}=h,

where h is a constant. This is conservation of angular momentum (compare with (15.6) and note that, in the equatorial plane, the spherical polar coordinate r is the same as the plane polar coordinate R).
"

Eq. (15.6) from the book is:

"R^2\dot{\phi}=h,"

and

the equation (15.8) exactly shows that

"R=R(\phi)."

You could have simply said that "I made a mistake and sorry for confusing you" but rather kept fudging until now that you're double finished. You should first read books to gain the basic background of the required issue coming up here and then jump into the discussions. I read D'inverno completely twice years ago and you should take a swipe at doing the same thing some day.

Repeating the same errors is not the way to demonstrate your knowledge. r and \phi are coordinates, r=r(\phi) represents a connection between the two coordinates, so it is a trajectory. Please find a different thread to troll.

That is crystal clear the one who attempts to jump into discussions without even having an asked-for background knowledge is nobody but you and by now you're seen to be a "troller" if wanting to keep making nonsense anymore.

AB

 

[Altabeh]

Doesn't change the fact that r=r(\phi) represents a trajectory, not a coordinate.

This is finished by now. Read the above post and find another hack to go for. We are not here to waste our time on your repeated nonsense claims.

AB

 

[Altabeh]

You should recognize the above equation as nothing but the second Euler-Lagrange equation:

\frac{d}{ds}(r^2*\frac{d\phi}{ds})=0

i.e.

r^2\frac{d\phi}{ds}=H

Indeed:

0=\frac{d}{ds}(r^2*\frac{d\phi}{ds})=2r\frac{dr}{ds}\frac{d\phi}{ds}+r^2\frac{d^2\phi}{ds^2}

resulting into:

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

So, your "proof" uses the conclusion, meaning that it is invalid.

Your undersatnding of mathematics is really leaky. It is so simple to provide a "proof" that such constant is not your hack H(r,\phi) (as cannot be ever found in any textbook.) Using the fact that r=r(s) along the geodesic, and that the tangent vector is necessarily non-null by looking at the inner product u^a\xi_a=const. and that the condition of co-observer being "momentarily at rest" is point-wise in the sense that along the geodesic it is not correct to put dr/ds=0 but at any event separately, we conclude that in the radial motion,

\frac{d}{ds}(r^2\dot{\phi})=0=\frac{dr}{ds}\frac{d}{dr}(r^2\dot{\phi})=0

\Rightarrow \frac{d}{dr}(r^2\dot{\phi})=0;

thus you "hack" reduces to

H=H(\phi).

And it is now all up to you to use this method to prove for us that the constant is actually H and what you've been trying to say is nothing but nonsense.

AB

 

[starthaus]

but at any event separately, we conclude that in the radial motion,

\frac{d}{ds}(r^2\dot{\phi})=0=\frac{dr}{ds}\frac{d}{dr}(r^2\dot{\phi})=0

Err, wrong. The above doesn't even make sense.

\Rightarrow \frac{d}{dr}(r^2\dot{\phi})=0;

Basic calculus says that it doesn't follow.

Pure nonsesne. For example, in the equatorial plane, the spherical polar coordinate r is the same as plane polar coordinate R=R(\phi)

AB

r is a coordinate while contrary to your fallacious claims R=R(\phi) is not a "plane polar coordinate" but rather a trajectory. You are in no position to "teach" since you don't even know the difference between coordinates and trajectories. Please stop trolling this thread.

 

[yuiop]

So, your "proof" uses the conclusion, meaning that it is invalid.

(PS: If you want to learn how this ODE is solved in mainstream mathematics, it is really done by using variable separation, not the method that you used)

My proof in #511 contains no initial assumption that H is a constant with respect to r. By careful step by step analysis with all steps shown, I have demonstrated that your assertion that H is a function of r is false. Altabeh, qbert and Eespen180 have also gone to pains to prove this to you and yet you still stand by your false assertion.

It has also been demonstrated by a number of people in this thread ( https://www.physicsforums.com/showpost.php?p=2784292&postcount=248" ) , that your assertion that dr/dt=0 \Rightarrow d^2r/dt^2=0 is false.

I have also proven step by stephttps://www.physicsforums.com/showpost.php?p=2790954&postcount=477"that your assertion that "equation (57) in Espen's document is wrong", is yet another false assertion on your part.

You also keep making the false assertion that equation (30) in Espen's document is wrong, simply because you do not understand the context, while everyone else in this thread does.

Your contributions to this thread is just a list of false assertions, red herrings and unhelpful vague statements that the work of other contributers are wrong without any indication of why it wrong or how to fix it. Your claim to be the "teacher" of the likes of Altabeh and espen180, when their knowledge of GR, physical intuition and mathematical ability is way beyond yours, is just laughable.

 

[starthaus]

My proof in #511 contains no initial assumption that H is a constant with respect to r.

Sure it does, your "proof" is circular.

It has also been demonstrated by a number of people in this thread, that your assertion that dr/dt=0 \Rightarrow d^2r/dt^2=0 is false.

Basic calculus says that you are wrong.

I have also proven step by step that your assertion that "equation (57) in Espen's document is wrong", is yet another false assertion on your part.

No, you haven't. (51)-(57) is a mess. If it weren't, then espen180 would have completed (58). He can't , because of the mess he's gotten into following your tips.

You also keep making the false assertion that equation (30) in Espen's document is wrong, simply because you do not understand the context, while everyone else in this thread does.

Err, no, equation (30) is definitely wrong. I even pointed you to the correct equation in my blog but you kept claiming that (30) is correct as is. espen180 (and you) are missing a factor of (3\alpha(r)/\alpha(r_0)-2) where r_0 is the initial drop distance.

Your contributions to this thread is just a list of false assertions, red herrings and unhelpful vague statements that the work of other contributers are wrong without any indication of why it wrong or how to fix it. Your claim to be the "teacher" of the likes of Altabeh and espen180, when their knowledge of GR, physical intuition and mathematical ability is way beyond yours, is just laughable.

Ad-hominems will not prove your point, quite the opposite.

 

[yuiop]

No, you haven't. (51)-(57) is a mess. If it weren't, then espen180 would have completed (58). He can't , because of the mess he's gotten into following your tips.

Espen's work is an independent work that he is trying to do the derivation without any reference to constants. All he done is ask me to proof read it, which I have done and privately we have now sorted out the problems behind the scenes and succeeded in completing his objective. We would have done it a lot sooner, if it was not for your false leads such as equation (30) and (57) being wrong, when they are not.

If you read Espen's document with any care, you would notice that (57) is derived completely from the Schwarzschild metric and equation (49), without any reference to equations (51)-(56).

Sloppy and wrong again.

 

[yuiop]

@Starthaus. I see in #523 you have quietly stopped defending your assertion that H is a function of r. Does that mean you are finally convinced that you were wrong about that?

Basic calculus says that you are wrong.

Everyone elses's basic calculus says you are wrong. See the post by George.

Sure it does, your "proof" is circular.

It is only circular if it is obvious that:

\frac{d}{ds}(r^2*\frac{d\phi}{ds})=0 \Rightarrow \frac{d}{dr}(H) =0

but it seems it was not obvious to you (although it is to everyone else), so posted the step by step explanation for you in post #511.

 

[starthaus]

Espen's work is an independent work that he is trying to do the derivation without any reference to constants. All he done is ask me to proof read it, which I have done and privately we have now sorted out the problems behind the scenes and succeeded in completing his objective. We would have done it a lot sooner, if it was not for your false leads such as equation (30) and (57) being wrong, when they are not.

Sure they are, you need to pay attention.

If you read Espen's document with any care, you would notice that (57) is derived completely from the Schwarzschild metric and equation (49), without any reference to equations (51)-(56).

Sloppy and wrong again.

What I've been telling you is that all equations, starting with (51) and ending with (57) are wrong. This explains why neither you, nor espen180 have been able to complete the trivial equation (58), even after I showed you how to do it exactly 250 posts ago.

 

[espen180]

Alright! I have now finished the calculation, thanks to kev's help and support!

Here is the newest document:
http://sites.google.com/site/espen180files/Schwartzschild2.pdf?attredirects=0&d=1"

I have removed the sections 2-4 which dealt with special cases, focusing on the general, unrestricted case. I abandoned the calculation I was currently working on (with ridicculously large expressions etc) for a simpler approach. Please take a look.

Edit: Updated with angular accelerations.

 

[atyy]

Alright! I have now finished the calculation, thanks to kev's help and support!

Here is the newest document:
http://sites.google.com/site/espen180files/Schwartzschild2.pdf?attredirects=0&d=1"

I have removed the sections 2-4 which dealt with special cases, focusing on the general, unrestricted case. I abandoned the calculation I was currently working on (with ridicculously large expressions etc) for a simpler approach. Please take a look.

Edit: Updated with angular accelerations.

At the start of section 2, should d(theta)/dt=0 be d(theta)/d(tau) instead?

Also, I don't understand the comment "turning to the metric", just before Eqn 15, since the metric is Eqn 1?

 

[atyy]

Also, I don't understand the comment "turning to the metric", just before Eqn 15, since the metric is Eqn 1?

I see. It's not the metric alone that you use to get Eqn 15. It's the metric plus the geodesic equation which implies that gab.dxa/dtau.dxb/dtau is a constant on a geodesic (following say Eq 2.4 and 2.5 of http://www.blau.itp.unibe.ch/lecturesGR.pdf ; also see Altabeh's post #342)

 

[Altabeh]

Err, wrong. The above doesn't even make sense.



Basic calculus says that it doesn't follow.

Tell us which basic calculus supprts your nonsense here! LOL! I see you're completely bogged down with basics of calculus. Work hard!

r is a coordinate while contrary to your fallacious claims R=R(\phi) is not a "plane polar coordinate" but rather a trajectory. You are in no position to "teach" since you don't even know the difference between coordinates and trajectories. Please stop trolling this thread.

Nonsense. Read the page 196 of D'inverno to see why you were blindly correcting the author.

"... Then (15.20) can be integrated directly to give

r^2\dot{\phi}=h,

where h is a constant. This is conservation of angular momentum (compare with (15.6) and note that, in the equatorial plane, the spherical polar coordinate is the same as the plane polar coordinate R).
"

Eq. (15.6) from the book is:

"R^2\dot{\phi}=h,"

and

the equation (15.8) exactly shows that

"R=R(\phi)."

Is you labelling us a "troller" another dead-end escape route towards not standing corrected? If so, then I don't want to say "you stop trolling" because I see that you're double finished by now.

AB

 

[espen180]

At the start of section 2, should d(theta)/dt=0 be d(theta)/d(tau) instead?

Also, I don't understand the comment "turning to the metric", just before Eqn 15, since the metric is Eqn 1?

Thanks for pointing out, I've fixed it now.

I said "turning to the metric" instead of "from (1) we obtain" since I used the matric in a different form. (spacetime element vs. matrix).

 

[Altabeh]

Alright! I have now finished the calculation, thanks to kev's help and support!

Here is the newest document:
http://sites.google.com/site/espen180files/Schwartzschild2.pdf?attredirects=0&d=1"

I have removed the sections 2-4 which dealt with special cases, focusing on the general, unrestricted case. I abandoned the calculation I was currently working on (with ridicculously large expressions etc) for a simpler approach. Please take a look.

Edit: Updated with angular accelerations.

Good job espen. Way to go!

AB

 

[Anthony]

I think the OP might be making life difficult for himself. With any form of variational problem, you should try to use Noether's theorem to integrate your equations up, rather than try to go from the Euler-Lagrange equations themselves. This will invariably make your life easier. Your Lagrangian density reads:

\mathcal{L} = \left( 1- \frac{1}{r}\right) \left( \frac{\mathrm{d} t}{\mathrm{d} \lambda}\right)^2 - \left( 1- \frac{1}{r}\right)^{-1} \left( \frac{\mathrm{d} r}{\mathrm{d} \lambda}\right)^2 - r^2 \left( \frac{\mathrm{d} \theta}{\mathrm{d} \lambda}\right)^2 - r^2 \sin^2\theta \left( \frac{\mathrm{d} \phi}{\mathrm{d} \lambda}\right)^2

using natural units. It is clear that the vector fields:

\frac{\partial}{\partial t} \quad \textrm{and}\quad \frac{\partial}{\partial \phi}

are Killing, so Noether's theorem integrates up two of the Euler-Lagrange equations for you and gives you two constants of motion:

\left(1-\frac{1}{r}\right) \frac{\mathrm{d} t}{\mathrm{d}\lambda} = \mathrm{const} \,(=E) \quad \textrm{and} \quad r^2 \sin^2\theta \frac{\mathrm{d} \phi}{\mathrm{d} \lambda} = \mathrm{const} \,(=h) \qquad (*)

i.e. on a given geodesic, these quantities remain unchanged. Similarly, since \partial_\lambda \mathcal{L}=0, we know \mathcal{L} remains constant, and we set it to {+1,-1,0} depending on whether you're interested in timelike, spacelike or nulll geodesics. Call this constant k. Note that all our ODEs are now 1st order. Setting \theta = \pi/2 (validity can be deduced from the \theta E-L equation) and using (*) in \mathcal{L}=k gives the ODE:

\left( \frac{\mathrm{d} r}{\mathrm{d}\lambda}\right)^2 = E^2 - \left( 1-\frac{1}{r}\right) \left(k + \frac{h^2}{r^2}\right)

If you'd prefer to parameterise your geodesics using \phi, use the second of the constraints in (*) again and you get:

\frac{h^2}{r^4} \left( \frac{\mathrm{d} r}{\mathrm{d} \phi}\right)^2 = E^2 - \left( 1-\frac{1}{r}\right) \left( k + \frac{h^2}{r^2}\right)

If you'd prefer to do all this using tensors, just apply Noether's theorem in the form: if L_V g=0 (i.e. V is a Killing vector) then V^\mu \dot{x}^\nu g_{\mu\nu} = \mathrm{const}.

 

[espen180]

Good job espen. Way to go!

AB

Thanks! :)

And thanks for your hard work!

 

[Altabeh]

I think the OP might be making life difficult for himself. With any form of variational problem, you should try to use Noether's theorem to integrate your equations up, rather than try to go from the Euler-Lagrange equations themselves. This will invariably make your life easier. Your Lagrangian density reads:

\mathcal{L} = \left( 1- \frac{1}{r}\right) \left( \frac{\mathrm{d} t}{\mathrm{d} \lambda}\right)^2 - \left( 1- \frac{1}{r}\right)^{-1} \left( \frac{\mathrm{d} r}{\mathrm{d} \lambda}\right)^2 - r^2 \left( \frac{\mathrm{d} \theta}{\mathrm{d} \lambda}\right)^2 - r^2 \sin^2\theta \left( \frac{\mathrm{d} \phi}{\mathrm{d} \lambda}\right)^2

using natural units. It is clear that the vector fields:

\frac{\partial}{\partial t} \quad \textrm{and}\quad \frac{\partial}{\partial \phi}

are Killing, so Noether's theorem integrates up two of the Euler-Lagrange equations for you and gives you two constants of motion:

\left(1-\frac{1}{r}\right) \frac{\mathrm{d} t}{\mathrm{d}\lambda} = \mathrm{const} \,(=E) \quad \textrm{and} \quad r^2 \sin^2\theta \frac{\mathrm{d} \phi}{\mathrm{d} \lambda} = \mathrm{const} \,(=h) \qquad (*)

i.e. on a given geodesic, these quantities remain unchanged. Similarly, since \partial_\lambda \mathcal{L}=0, we know \mathcal{L} remains constant, and we set it to {+1,-1,0} depending on whether you're interested in timelike, spacelike or nulll geodesics. Call this constant k. Note that all our ODEs are now 1st order. Setting \theta = \pi/2 (validity can be deduced from the \theta E-L equation) and using (*) in \mathcal{L}=k gives the ODE:

\left( \frac{\mathrm{d} r}{\mathrm{d}\lambda}\right)^2 = E^2 - \left( 1-\frac{1}{r}\right) \left(k + \frac{h^2}{r^2}\right)

If you'd prefer to parameterise your geodesics using \phi, use the second of the constraints in (*) again and you get:

\frac{h^2}{r^4} \left( \frac{\mathrm{d} r}{\mathrm{d} \phi}\right)^2 = E^2 - \left( 1-\frac{1}{r}\right) \left( k + \frac{h^2}{r^2}\right)

If you'd prefer to do all this using tensors, just apply Noether's theorem in the form: if L_V g=0 (i.e. V is a Killing vector) then V^\mu \dot{x}^\nu g_{\mu\nu} = \mathrm{const}.

Exactly! This simply is based on the proposition given in post #389 and by this perfect explanation we are done here.

AB

 

[Anthony]

I'm glad it was of some use (I don't have the will power to find post #389).

 

[Redbelly98]

It's on page 25, if you are using the forum default of 16 posts per page.