Orbital velocities in the Schwartzschild geometry

[espen180]

Well, the only way to proove you wrong is to get you to perform the computations.

You're unwilling to substansiate your claim, then. Either that, or you know it won't lead anywhere (a false assertion). Either way, you're breaking the rules.

 

[starthaus]

You're unwilling to substansiate your claim, then. Either that, or you know it won't lead anywhere (a false assertion). Either way, you're breaking the rules.

H(r,\phi)=r^2\frac{d\phi}{ds} (1)

and

\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r}) (2)

Eliminate \frac{d\phi}{ds} between (1) and (2) and show that the resultant expression for H(r) does not depend on r.
Your challenge is to show that H(r) is not a function of r

 

[yuiop]

It is an exercise for you and kev, all set up in post 377. See which one of you can finish it first.

The result is inconclusive as Espen has already explained. If you have a proof, show your hand or shut up.

 

[starthaus]

The result is inconclusive as Espen has already explained. If you have a proof, show your hand or shut up.

H(r,\phi)=r^2\frac{d\phi}{ds} (1)

and

\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r}) (2)

Eliminate \frac{d\phi}{ds} between (1) and (2) and show that the resultant expression for H(r) does not depend on r.
Your challenge is to show that H(r) is not a function of r. Do you think you can do this simple calculation all by yourself? Or is it that you have already done it and you know that it contradicts your claims?

 

[yuiop]

Then , I recommend you get it or that you read my blog and stop imagining errors where there aren't any.

I have already shown that your statement in your blog that for radial motion, K is function of r is a false assertion. In the interests of intellectual honesty, please make it clear that you accept that you made a mistake there.

 

[starthaus]

I have already shown that your statement in your blog that for radial motion, K is function of r is a false assertion. In the interests of intellectual honesty, please make it clear that you accept that you made a mistake there.

We are not talking radial motion, we are talking general orbits.


H(r,\phi)=r^2\frac{d\phi}{ds} (1)

and

\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r}) (2)

Eliminate \frac{d\phi}{ds} between (1) and (2) and show that the resultant expression for H(r) does not depend on r.
Your challenge is to show that H(r) is not a function of r

 

[yuiop]

H(r,\phi)=r^2\frac{d\phi}{ds} (1)

and

\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r}) (2)

Eliminate \frac{d\phi}{ds} between (1) and (2) and show that the resultant expression for H(r) does not depend on r.
Your challenge is to show that H(r) is not a function of r. Do you think you can do this simple calculation all by yourself? Or is it that you have already done it and you know that it contradicts your claims?

Nope, we have done it and seen that the result is inconclusive and does not clearly support either side of the dispute.

If you can provide a clear proof from the above that supports your argument, then please do so, or you are just making unsupported assertions.

 

[starthaus]

Nope, we have done it and seen that the result is inconclusive and does not clearly support either side of the dispute.

No, the result isn't "inconclusive", it is quite conclusive. You have one step to calculate, why aren't you doing it?


H(r,\phi)=r^2\frac{d\phi}{ds} (1)

and

\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r}) (2)

Eliminate \frac{d\phi}{ds} between (1) and (2) and show that the resultant expression for H(r) does not depend on r.
Your challenge is to show that H(r) is not a function of r

 

[yuiop]

We are not talking radial motion, we are talking general orbits.

In your blog you make the clearly make the assertion that K is a function of r in the context of purely radial motion, and I have proved you wrong in that context.

Do you now agree that in the context of purely radial motion, that K=(1-2M/r)(dt/ds) is NOT a function of r, even though the equation for K superficially appears to contain the variable (r)?

 

[starthaus]

In your blog you make the clearly make the assertion that K is a function of r in the context of purely radial motion, and I have proved you wrong in that context.

Do you now agree that in the context of purely radial motion, that K is a function is NOT a function of r?

Err, so you don't understand the challenge or you understand it and you don't want to do a simple calculation? We are talking about H, not K , kev. We are talking arbitrary orbits (that's where your hack is at its ugliest) not about radial motion.

 

[yuiop]

Err, so you don't understand the challenge or you understand it and you don't want to do a simple calculation? We are talking about H, not K , kev. We are talking arbitrary orbits (that's where your hack is at its ugliest) not about radial motion.

I am specifically talking about a false assertion in your blog.

 

[starthaus]

You seem to be implying that you have your independent solution and you are gently guiding Espen and Altabeh towards it, but I doubt you have own solution or derivation, because it is not in the textbooks

True, it isn't in any textbook.

for you to copy it from.

False. Try downloading the second attachment from https://www.physicsforums.com/blog.php?b=1957

 

[yuiop]

@Starthaus: As I said before "you have to prove that dH/dr \ne 0"

 

[starthaus]

I am specifically talking about a false assertion in your blog.

1. First of, I assert that "k may be a function of r". In the radial motion it isn't, in the case of arbitrary orbits it is.

2. Second off, we are talking about your hacky method of deriving the equations of motion.

3. While talking about your hacky method, we are unravelling your claim that H can be treated under differentiation wrt r as if it were a constant. The exercise from post 377 is set up to disprove your assertion.

4. You must have calculated H by now and you already know that your method is wrong, this is why you want to divert the discussion.

5. My derivation (two files at https://www.physicsforums.com/blog.php?b=1957 ) does NOT use any assumption about the nature of K and/or H. As such, as opposed to your method, my method is fully rigorous.

 

[yuiop]

False. Try downloading the second attachment from https://www.physicsforums.com/blog.php?b=1957

I have seen it before.

You have a general equation in terms of proper time. - Not what we are looking for,

You have the special case of circular motion in coordinate time. - Not what we are looking for.

You have the special case of purely radial motion in coordinate time. - Not what we are looking for.

We are looking for the general (radial & angular) equation in terms of coordinate time in a single equation as I have produced in my derivation. I think you don't want to produce it because your either not sure how to, or you know that it will be the same as my general result and you won't be able to continue implying my result is wrong.

 

[starthaus]

@Starthaus: As I said before "you have to prove that dH/dr \ne 0"

So, I see you don't want to do a simple calculation:

H(r)=r^2\sqrt{\frac{\frac{d^2r}{ds^2}+\frac{m}{r^2}}{r-3m}

Show that the above is not a function of r.

 

[yuiop]

1. First of, I assert that "k may be a function of r". In the radial motion it isn't, in the case of arbitrary orbits it is.

OK. We have finally established that the presence of the variable (r) in the equation for K does not have to imply that K is a function of (r).

3. While talking about your hacky method, we are unravelling your claim that H can be treated under differentiation wrt r as if it were a constant. The exercise from post 377 is set up to disprove your assertion.

Your exercise or proof relies on the fact that when H is differentiated wrt (r) the right hand side contains the variable (r). We have now established that that is not sufficient proof that H is a function of (r). Try another tack.

 

[starthaus]

OK. We have finally established that the presence of the variable (r) in the equation for K does not have to imply that K is a function of (r).

For radial motion. NOT for arbitrary orbits. Try reading for comprehension.

Your exercise or proof relies on the fact that when H is differentiated wrt (r) the right hand side contains the variable (r). We have now established that that is not sufficient proof that H is a function of (r). Try another tack.

Post 416 says that you are wrong.

 

[yuiop]

3. While talking about your hacky method, we are unravelling your claim that H can be treated under differentiation wrt r as if it were a constant. The exercise from post 377 is set up to disprove your assertion.

Your exercise is inconclusive. That leaves the only the rigourous argument that H is NOT a function of r as the one provided by Altabeh in relation to Killing vectors and conserved quantities.

 

[starthaus]

Your exercise is inconclusive.

Post 416 shows plainly that you are wrong.

 

[espen180]

So, I see you don't want to do a simple calculation:

H(r)=r^2\sqrt{\frac{\frac{d^2r}{ds^2}+\frac{m}{r^2}}{r-3m}

Show that the above is not a function of r.

Results cannot be reached from that alone, kev has been trying to tell you this repeatedly, simply because when \frac{d^2r}{ds^2} is written out, you recover the expression

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

 

[yuiop]

Well let's see what your general equation for acceleration in terms of coordinate time is and see if it differs in way from mine. See post #415.

 

[starthaus]

Results cannot be reached from that alone, kev has been trying to tell you this repeatedly, simply because when \frac{d^2r}{ds^2} is written out, you recover the expression

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

LOL, of course it can, don't you understand how to eliminate a cariable between two equations even after all the steps were done for you?

 

[starthaus]

Well let's see what your general equation for acceleration in terms of coordinate time is and see if it differs in way from mine. See post #415.

The results are identical, it is your method that is wrong. Do you need 200+ posts to understand iwhy?

 

[espen180]

LOL, of course it can, don't you understand how to eliminate a cariable between two equations even after all the steps were done for you?

Dear me.

As explained, the result reduses to the original expression for H. In other words, there is no easily accessible information about H in that equation which establishes whether or not it remains constant along a geodesic.

 

[starthaus]

If my solution is incorrect, then so is your solution in your blog because it is easy to prove they are equaivalent using only simple algebra. Since you seem to have difficulties with algebra I will do it for you.

The solution given in section (11) of your blog is:

\frac{d^2r}{dt^2} = -\frac{M}{r^2} \alpha \left(3\frac{(1-2M/r)}{(1-2M/R)} -2 \right)

Correct, see , you found it. So, why are you claiming that I do not provide for such a solution? Besides, mine is not dependent on speed, as yors is. You even made the effort to prove that your solution is identical to mine.

 

[starthaus]

Dear me.

As explained, the result reduses to the original expression for H. In other words, there is no easily accessible information about H in that equation which establishes whether or not it remains constant along a geodesic.

No, it does not "reduce" since it was derived from You know the difference, don't you? One look at the expression and you can say that it is impossible not to be a function of r. If you don't believe it, calculate \frac{dH}{dr} and compare with 0. Do you think you can do this calculation all by yourself?

 

[yuiop]

The solution given in section (11) of your blog is:

\frac{d^2r}{dt^2} = -\frac{M}{r^2} \alpha \left(3\frac{(1-2M/r)}{(1-2M/R)} -2 \right)

Correct, see , you found it. So, why are you claiming that I do not provide for such a solution?

Your solution is only valid for d\phi =0 . It is a limited case. You have spent several threads proclaiming the limited or special cases are "wrong" and that only fully generalised equations are "right". Your solution is a subset of my more fully general solution.

My general solution that is valid for any values of d\phi was given in post #277 as :

which after a bit of algebra simplifies to:

\Rightarrow \frac{d^2r}{dt^2}= \alpha\left(<br /> r \frac{d\phi^2}{dt^2}-\frac{M}{r^2}+<br /> \frac{3M}{\alpha^2 r^2} \frac{dr^2}{dt^2}\right)

This is the fully general form of acceleration in Schwarzschild coordinates in terms of coordinate time and gives Espen something to compare his results against in his latest document.

You have so far failed to provide an alternative solution to my fully general solution. Does that mean you agree my final result is correct?

Besides, mine is not dependent on speed, as yors is. You even made the effort to prove that your solution is identical to mine

It is trivial to convert one form to the other. You are conveniently forgetting that it was I that taught you how you could determine the velocity of a falling particle from the height of the apogee (R) where dr/dt=0.

 

[yuiop]

No, it does not "reduce" since it was derived from You know the difference, don't you? One look at the expression and you can say that it is impossible not to be a function of r. If you don't believe it, calculate \frac{dH}{dr} and compare with 0. Do you think you can do this calculation all by yourself?

I am pretty sure you can't.

 

[starthaus]

Your solution is only valid for d\phi =0 . It is a limited case. You have spent several threads proclaiming the limited or special cases are "wrong" and that only fully generalised equations are "right". Your solution is a subset of my more fully general solution.

My general solution that is valid for any values of d\phi was given in post #277 as :

The solution is correct, the derivation is NOT.

You have so far failed to provide an alternative solution to my fully general solution.

This is a trivial exercise.

I have already proven that :

\frac{d^2r}{ds^2}=-\frac{m}{r^2}+(r-3m)(\frac{d\phi}{ds})^2

Combine the above with the first Euler-Lagrange equation

\alpha\frac{dt}{ds}=K

and, without any hacky assumptions about K you will obtain that :

\frac{d^2r}{dt^2}=(r-3m)(\frac{d\phi}{dt})^2-(\frac{\alpha}{K})^2\frac{m}{r^2}+\frac{2m}{\alpha*r^2}(\frac{dr}{dt})^2

Please do not waste another 50 posts trying to prove that the above is incorrect, I can assure you that it is.

You are conveniently forgetting that it was I that taught you how you could determine the velocity of a falling particle from the height of the apogee (R) where dr/dt=0.

LOL

 

[yuiop]

@Starhaus: If you want to prove that H is a function of (r) all you have to do is show that you can obtain a different (but correct) solution, to the one I obtained by assuming that H is NOT a function of (r). Think you can do it?

[EDIT] Just seen your last post. You can't. Therefore H is a NOT a function of (r). I rest my case.

 

[starthaus]

I am pretty sure you can't.

I agree, you can't prove \frac{dH}{dr}=0

 

[starthaus]

@Starhaus: I f you want to prove that H is a function of (r) all you have to do is show that you can obtain a different (but correct) solution, to the one I obtained by assuming that H is NOT a function of (r). Think you can do it?

Post 430. Doesn't use any of your hacks. All you need to know is how to apply the Euler-Lagrange equations and chain differentiation correctly.

 

[espen180]

Post 430.

doesn't address the issue.

 

[starthaus]

doesn't address the issue.

LOL, if you don't understand the derivation, just say so and I'll explain it to you as I have done in the past. I told you that I'm not using any of the hacky assumptions about H. All you need to know is the Euler-Lagrange formalism and chain differentiation.

 

[espen180]

LOL, if you don't understand the derivation, just say so and I'll explain it to you as I have done in the past. I told you that I'm not using any of your hacky assumptions about H.

The post doesn't mention H. Also, I don't use it. I'm using the geodesic equations, as I have throughout the thread.

 

[starthaus]

@Starhaus: If you want to prove that H is a function of (r) all you have to do is show that you can obtain a different (but correct) solution, to the one I obtained by assuming that H is NOT a function of (r). Think you can do it?

[EDIT] Just seen your last post. You can't. Therefore H is a NOT a function of (r). I rest my case.

LOL, the solutions are equivalent. You need to do a little algebra to prove it. Please don't waste another 50 posts trying to find imaginary errors again. There aren't any.

 

[starthaus]

The post doesn't mention H. Also, I don't use it. I'm using the geodesic equations, as I have throughout the thread.

...but kev does. This is the point.

Look, espen

The geodesic formalism and the Euler-Lagrange formalsim are two sides of the same coin. I suggest that you learn the latter since it produces the same exact results with a lot less work and fewer chances of making errors. Since your derivations have shown to be late and error prone, switching formalisms would be an improvement.

 

[espen180]

Since your derivations have shown to be late and error prone, switching formalisms would be an improvement.

Being "late" has nothing to do with it (why "late"? Is there a deadline?)

 

[yuiop]

I agree, you can't prove \frac{dH}{dr}=0

You can't prove \frac{dH}{dr} \ne 0

 

[yuiop]

LOL, the solutions are equivalent. You need to do a little algebra to prove it. Please don't waste another 50 posts trying to find imaginary errors again. There aren't any.

If the solutions are equivalent and if your solution is correct, then by simple logic I have obtained the correct result by using the assumption that H is NOT a function of (r). If my assumption was incorrect I would have obtained a wrong result. Since I did not, it stands to reason that H is NOT a function of (r).

 

[starthaus]

Being "late" has nothing to do with it (why "late"? Is there a deadline?)

But being late and wrong has a lot to do with the usefulness of your derivations. Case and point: your equation (30) is still wrong despite repeated feedback.

 

[starthaus]

If the solutions are equivalent and if your solution is correct, then by simple logic I have obtained the correct result by using the assumption that H is NOT a function of (r). If my assumption was incorrect I would have obtained a wrong result.

You obtained the correct solution by accident. Your derivation is sttill a hack.

 

[qbert]

even though it's late in the game --

i think a point has come up which hasn't been answered very well. So I'll
take a stab at it. You need to be careful when using the calculus of
variations to keep in mind which variables are dependent and which are
independent. In the current case, the only independent variable is the
worldline coordinate (t or s). Everything else is a function of (say t) t.
That means H can be at most a function of t. period. since we know
dH/dt = 0 it is the constant function. meaning a real, honest to goodness,
constant.

If I've expressed myself clearly -- this is enough to prove the result.
The rest is for the formula lovers out there.
S'pose L = L(q_i(t), \dot{q}_i(t)) then by the E-L eqns we know
\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right) <br /> = \frac{\partial L}{\partial q_j}[/itex]<br /> <br /> Now suppose that L is cyclic in one variable say q_k. That is<br /> \frac{\partial L}{\partial q_k} = 0 &lt;br /&gt; = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_k} \right)<br /> <br /> So that \frac{\partial L}{\partial \dot{q}_k} = C<br /> Now suppose to the contrary that C were really a function of the &quot;variables&quot;<br /> ie C = f(q_i).<br /> <br /> Then we get<br /> \frac{\partial f}{\partial q_j} = \frac{\partial^2 L}{\partial q_j \partial \dot{q}_k}&lt;br /&gt; = \frac{\partial^2 L}{\partial \dot{q}_k \partial q_j}<br /> <br /> = \frac{\partial }{\partial \dot{q}_k} \left( \frac{\partial L}{\partial q_j} \right)&lt;br /&gt; = \frac{\partial}{\partial \dot{q}_k}\left( \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} \right) = \frac{\partial}{\partial \dot{q}_j}\left( \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_k} \right) = \frac{\partial}{\partial \dot{q}_k}\left( 0 \right) = 0.

 

[Dale]

The geodesic formalism and the Euler-Lagrange formalsim are two sides of the same coin. I suggest that you learn the latter since it produces the same exact results with a lot less work and fewer chances of making errors.

I would recommend everyone learn both. Since they produce the same results it gives a way to check the results and be a little more confident that there isn't an error.

You obtained the correct solution by accident.

That seems to happen so often with kev that I am not convinced they are accidents. I think he has an intuitive style that just doesn't mesh well with your more analytical style, but I think that you are wrong to dismiss repeatedly correct intuition as a hack. Particularly intuition about something as esoteric as GR. If you really think that kev is getting right answers by luck then you should be asking him for stock tips etc.

 

[espen180]

But being late and wrong has a lot to do with the usefulness of your derivations. Case and point: your equation (30) is still wrong despite repeated feedback.

Then show what you thnk it should be.

 

[starthaus]

Then show what you thnk it should be.

You can find it https://www.physicsforums.com/blog.php?b=1957 , first attachment.

 

[espen180]

You can find it https://www.physicsforums.com/blog.php?b=1957 , first attachment.

What about the word "special" in "special case" don't you get?

 

[espen180]

I'm getting closer to a solution, but the expressions are ridicculously big, so doing the calculus/algebra/proofreading is a pain.

Here's how far I've gotten:
http://5554229043997450163-a-1802744773732722657-s-sites.googlegroups.com/site/espen180files/Schwartzschild.pdf?attachauth=ANoY7crvgkJaf2YSb9wipuKnvEoacyFDok-MWepVPK4tZrlLjK8If_HxjS_lzaZVPS-0StjbvJLQU3Bvx-t_oZcneyRG3Orj_V5UXTUFarMpmWYhgtEJNgxcRpypcyted9W0Qa7Re2gn-2ib6A0f-WJV5jukKhp_IfJhNoCnhscy-m2uBhjTO2cHlLX45JEfEI8kNFatSDfwlnBKGeus9Ufx5OdK13A-sw%3D%3D&attredirects=1" (Section 5, page 7)

I'm somewhat worried that I have an r-derivative raised to the 4th power. I can't see how it would disappear later on, either, so maybe I have an error somewhere.

 

[yuiop]

\frac{d^2r}{ds^2}=-\frac{m}{r^2}+(r-3m)(\frac{d\phi}{ds})^2

Combine the above with the first Euler-Lagrange equation

\alpha\frac{dt}{ds}=K

and, without any hacky assumptions about K you will obtain that :

\frac{d^2r}{dt^2}=(r-3m)(\frac{d\phi}{dt})^2-(\frac{\alpha}{K})^2\frac{m}{r^2}+\frac{2m}{\alpha*r^2}(\frac{dr}{dt})^2

Let's see if I am following your derivation correctly because you have glossed over an important detail. (See step 3).

First we use a fairly lengthy series of chain and product rules and substitutions to relate d^2r/ds^2 to d^2r/dt^2 and obtain:

\frac{d^2r}{dt^2} = \frac {d^2r}{ds^2} \frac{ds^2}{dt^2} - \frac{d}{dr}\left( \frac{dt}{ds}\right) \frac {dr^2}{ds^2} \frac{ds^3}{dt^3} \qquad \qquad (1)

Now the values of dr/ds and d^2r/ds^2 are already given and we substitute these into the above equation to obtain:

\frac{d^2r}{dt^2} = \left ((r-3M)\frac{d\phi^2}{ds^2} - \frac {M}{r^2} \right ) \frac{\alpha^2}{K^2} - \frac{d}{dr}\left( \frac{K}{\alpha}\right) \frac {dr^2}{ds^2} \frac{\alpha^3}{K^3} \qquad \qquad (2)

Next we need to evalute the (d/dr)(K/\alpha) expression on the right:

\frac{d}{dr}\left( \frac{K}{\alpha}\right) \Rightarrow \frac{d}{dr}\left( \frac{K}{1-2M/r}\right) \Rightarrow \frac{-2KM}{r^2(1-2M/r)^2 } \Rightarrow -\frac{2KM}{r^2 \alpha^2} \qquad \qquad (3)

Note that we have to treat K as NOT being a function of (r) when differentiating wrt (r).

Substitute (3) back into (2) to obtain:

\frac{d^2r}{dt^2} = \left ((r-3M)\frac{d\phi^2}{ds^2} - \frac {M}{r^2} \right ) \frac{\alpha^2}{K^2} - \left( -\frac{2KM}{r^2 \alpha^2}\right)\frac {dr^2}{ds^2} \frac{\alpha^3}{K^3} \qquad \qquad (4)

and simplify using (\alpha/K) = (ds/dt):

\frac{d^2r}{dt^2} = (r-3M)\frac{d\phi^2}{dt^2} - \frac{\alpha^2}{K^2} \frac {M}{r^2} + \frac{2M}{ \alpha {r^2}} \frac {dr^2}{dt^2} \qquad \qquad (5)

which is the same as your result and the same as the result I derived earlier in the thread.