Orbital velocities in the Schwartzschild geometry

[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.