Orbital velocities in the Schwartzschild geometry

[yuiop]

You may have made a mistake somewhere, since you have the mass squared instead of just the mass.

You are dead right Espen. I made a typo in post #203 by entering mass squared instead of just mass for the deriviative. The typo crept in when transfering my result on paper to latex and the error has propogated into subsequent posts. Sorry for any confusion. This is the corrected version with the correction following the line in red about half down:

==============================

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

$ds^2=\alpha dt^2-dr^2/\alpha -r^2d\phi^2$ where $\alpha=(1-2m/r)$

Solve for (dr/ds):

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

The well known constants of motion are:

$K = \alpha(dt/ds)$ and $H = r^2 (d\phi/ds)$

Insert these constants into the equation for (dr/ds):

$$\frac{dr}{ds} = \sqrt{K^2 - \alpha - \alpha \frac{H^2}{r^2}}$$

At this point I accidently entered m^2 instead of just m for the differentiation in the original version.

Differentiate both sides with respect to s:

$$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$

(The easiest way to carry out the above differentiation is to square both sides and carry out the differentiation with respect to r rather than s and then divide the final result by 2.)

Re-insert the full form of H back into the equation:

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

This is the same as the result that I derived form Espen's solution. The two derivations arrive at the same result using different methods. This final general form is much more compact and useful than any other solution previously provided in this thread.

For radial motion only, $$(d\phi/ds)^2=0$$ and the radial acceleration is:

$$\frac{d^2r}{ds^2} = -\frac{m}{r^2}$$

This is the same as a result as quoted by mathpages http://www.mathpages.com/rr/s6-07/6-07.htm (first equation.)

For circular motion the radius is constant and so the radial acceleration $(d^2r)/(ds^2)$ must be zero and in this limited case:

$$0 = -\frac{m}{r^2} + \frac{d\phi^2}{ds^2}(r-3m)$$

Solve for the angular velocity in terms of proper time:

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

When r=3m (the photon orbit):

$$d\phi/ds = \infty$$

This is the expected result because for a particle moving at the speed of light, ds=0.

This is not the only mistake, the whole "method" is invalid, resulting into bogus results. Can you figure out at what step, in his "derivation", does kev introduce his error? It is pretty gross, so it shouldn't be difficult to spot.

I made a simple typo. There is no need to go over the top. If you had carried out the differentiation yourself you would have seen that it was simply a typo. I do not believe my method is invalid.

 

[starthaus]

You are dead right Espen. I made a typo in post #203 by entering mass squared instead of just mass for the deriviative. The typo crept in when transfering my result on paper to latex and the error has propogated into subsequent posts. Sorry for any confusion. This is the corrected version with the correction following the line in red about half down:

==============================

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

$ds^2=\alpha dt^2-dr^2/\alpha -r^2d\phi^2$ where $\alpha=(1-2m/r)$

Solve for (dr/ds):

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

The well known constants of motion are:

$K = \alpha(dt/ds)$ and $H = r^2 (d\phi/ds)$

Insert these constants into the equation for (dr/ds):

$$\frac{dr}{ds} = \sqrt{K^2 - \alpha - \alpha \frac{H^2}{r^2}}$$

At this point I accidently entered m^2 instead of just m for the differentiation in the original version.

Yes, but this is not the most serious mistake you made.

Differentiate both sides with respect to s:

$$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$

This is where you made your blunder. Not only that you are differentiating incorrectly but your final result is also incorrect.
Your error is that you are differentaiting wrt $$s$$ by differentiating $$\sqrt{K^2 - \alpha - \alpha \frac{H^2}{r^2}}$$ wrt $$r$$ followed by multiplying by $$\frac{dr}{ds}$$. While $$H$$ and $$K$$ are constants wrt $$s$$, they both depend on $$r$$, so your approach to differentiation is totally bogus. This is not the first time you are making this mistake, I pointed it out to you in another thread. The net result of this bungle is that $$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$, is wrong. So, everything that you derive from it, is also totally wrong. For example, your $$\frac{d\phi}{ds}$$ is totally wrong. I pointed out to you the correct formula.

 

[starthaus]

The well known constants of motion are:

$K = \alpha(dt/ds)$ and $H = r^2 (d\phi/ds)$

Insert these constants into the equation for (dr/ds):

$$\frac{dr}{ds} = \sqrt{K^2 - \alpha - \alpha \frac{H^2}{r^2}}$$Differentiate both sides with respect to s:

$$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$

Incorrect. Your differentiation failed.

For circular motion the radius is constant and so the radial acceleration $(d^2r)/(ds^2)$ must be zero and in this limited case:

$$0 = -\frac{m}{r^2} + \frac{d\phi^2}{ds^2}(r-3m)$$

Solve for the angular velocity in terms of proper time:

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

Also wrong. (because it is derived from your other incorrect formula)

I made a simple typo. There is no need to go over the top. If you had carried out the differentiation yourself you would have seen that it was simply a typo. I do not believe my method is invalid.

No, it isn't a "simple typo". Your whole method is invalid due to the differentiation blunder you made earlier.

 

[Altabeh]

that $$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$, is wrong.

Seconded. Think about where you have made an error and I sure know you'll figure it out.

AB

 

[Altabeh]

I made a simple typo.

Of course it is not as much fallacious as the other bungles we have had not only in this thread but in all thread where you've been attacked. You'll become more practiced in this zone by the passing of time.

AB

 

[yuiop]

... that $$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$, is wrong. So, everything that you derive from it, is also totally wrong. For example, your $$\frac{d\phi}{ds}$$ is totally wrong. I pointed out to you the correct formula.

Seconded. Think about where you have made an error and I sure know you'll figure it out.

You both seem to think that my final (corrected) solution $$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$ is wrong, but you have not clearly stated what you think the correct solution should be. This is not the homework forum and giving hints rather than facts is just irritating.

Please clearly state what you think the general solution

$$\frac{d^2r}{ds^2} =$$,

and the special case for a circular orbit when dr/dt=0

$$\frac{d\phi}{ds} =$$

should be.

 

[starthaus]

You both seem to think that my final (corrected) solution $$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$ is wrong, but you have not clearly stated what you think the correct solution should be.

Your approach is unusable, so it can't be fixed. The fact that on top of this you add basic errors in calculating the differential doesn't help either.
The correct solution uses the Euler-Lagrange equations, or, their equivalent, the geodesic equations. You need to learn one of the two formalisms.

This is not the homework forum and giving hints rather than facts is just irritating.

You have to learn how to derive the solutions yourself. The correct answers and their correct derivations have been already given to you , in this thread, yet you insisted my solution was wrong. You need to go back and read them.

 

[Altabeh]

You both seem to think that my final (corrected) solution $$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$ is wrong, but you have not clearly stated what you think the correct solution should be. This is not the homework forum and giving hints rather than facts is just irritating.

Please clearly state what you think the general solution

$$\frac{d^2r}{ds^2} =$$,

Of course I'm supposed to point at the error and sorry for the pretermission:

In your post 211, you exactly report the correct result:

Re-insert the full form of H back into the equation:

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

This is the same as the result that I derived form Espen's solution. The two derivations arrive at the same result using different methods. This final general form is much more compact and useful than any other solution previously provided in this thread.

Well something is wrong with this as you can check http://www.astro.umd.edu/~miller/teaching/astr498/lecture10.pdf" to make sure why. I mean let's see how we can retrieve $$H$$ in this equation. You clearly know that

$$H = r^2 (d\phi/ds).$$

So it's simple to see

$$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + H^2(1-3m/r)/r^7.$$

But what is it that makes the error? There got to be a differentiation problem. I don't have time to check. But I'm going to review your method and see what's the cause of this problem. One thing to recall is that you should be able to check the derivate of $$\alpha$$ in the radical must produce a term including $$dr/ds$$.

AB

 

[starthaus]

(from post 53)

The general equation is:

$$-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0$$

If you make $$r=R$$ in the above, this means the cancellation of the terms in $$dr/ds$$ , so you get the correct equation for circular orbital motion.

If you make $$d\phi/ds=0$$ you get the correct equation for radial motion.

 

[Altabeh]

If you make $$r=R$$ in the above, this means the cancellation of the terms in $$dr/ds$$ , so you get the correct equation for circular orbital motion.

If you make $$d\phi/ds=0$$ you get the correct equation for radial motion.

I assume this doesn't make use of the fallacy $$dr/ds=0\Rightarrow d^2r/ds^2=0.$$

AB

 

[starthaus]

I assume this doesn't make use of the fallacy $$dr/ds=0\Rightarrow d^2r/ds^2=0.$$

AB

No, there is no implication on $$d^2r/ds^2=0.$$. Contrary to your repeated insinuations, none of my solutions uses the above implication. .

 

[Al68]

This is not the homework forum and giving hints rather than facts is just irritating.

Seconded. And forum rules place the burden for substantiating claims on the person making the claim. Making a claim, then leaving it "as an exercise" for another to show why it's true may be appropriate for the homework forum, but not here. It's not only obnoxious, it renders the thread useless to others by unnecessarily stretching it out to hundreds of posts.

 

[Altabeh]

No, there is no implication on $$d^2r/ds^2=0.$$. Contrary to your repeated insinuations, none of my solutions uses the above implication. .

Because of your repeated void emphasis on such nonsense in this thread, it came to our attention that maybe the motivation behind an unnecessary direct show-off in your post 219 was to hint at the use of it while its validity being questioned. Well that would be awesome if I could know how you manage to get rid of $$d^2r/ds^2$$ appearing in the expansion of the first term on the LHS of your "equation" if we put $$dr/ds=0$$!

AB

 

[starthaus]

Because of your repeated void emphasis on such nonsense in this thread, it came to our attention that maybe the motivation behind an unnecessary direct show-off in your post 219 was to hint at the use of it while its validity being questioned. Well that would be awesome if I could know how you manage to get rid of $$d^2r/ds^2$$ appearing in the expansion of the first term on the LHS of your "equation" if we put $$dr/ds=0$$!

AB

-For radial motion, it's pretty trivial, if you know how to differentiate. If you do the differentiation and the reduction of like terms correctly you should be getting $\frac{d^2r}{ds^2}=-\frac{m}{r^2}$

-For circular orbits, why would you expect such a silly thing? It is trivial, if $\frac{dr}{ds}=0$ in the general equation of motion, then $\frac{d\phi}{dt}=\sqrt{\frac{m}{r^3}}$.

 

[yuiop]

... For circular orbits, why would you expect such a silly thing? It is trivial, if $$\frac{dr}{ds}=0$$ in the general equation of motion, then $$\frac{d\phi}{dt}=\sqrt{\frac{m}{r^3}}$$.

I asked you clearly state what you think $$d\phi/ds$$ should be and for some reason you have decided to state what $$d\phi/dt$$ is instead.
Not to worry. We can soon sort that out.

$$\frac{d\phi}{dt}=\sqrt{\frac{m}{r^3}}\Rightarrow \frac{d\phi}{ds}=\frac{dt}{ds}\sqrt{\frac{m}{r^3}}$$

This implies:

$$(d\phi/ds)^2=\frac{m}{R^3}\frac{dt^2}{ds^2}$$

This contradicts the result you gave in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53"(The post that you keep telling everyone to refer to for the correct solution for circular motion) where you said:

$$-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0$$

If you make $$r=R$$ in the above, this means the cancellation of the terms in $$dr/ds$$ and if you giving you

$$(d\phi/ds)^2=\frac{m}{R^3}$$

I can only guess at how you arrived at this erronous result in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53"because you have not defined what R is, and the phrase "and if you giving you" is not a meaningful expression in English.

However, I will try and decipher and unravel your mess for you. In previous discussions we have usually used "R" to mean the radius where dr/dt=0 which can mean the apogee for radial motion or the radius of a circular orbit.

Now I guess by the expression "the cancellation of the terms in $$dr/ds$$" you mean that the terms containing $$dr/ds$$ go to zero when dr/ds=0. There is a strong implication that you are using the erronous $dr/ds=0\Rightarrow d^2r/ds^2=0$ hack as Altabeh puts it, when you make this assumption. Never the less I will continue with your line of reasoning.

Using

$$-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0$$

and setting both terms containing dr/ds to zero, the expression becomes:

$$0-(2m/r^2(dt/ds)^2-0-2r(d\phi/ds)^2)=0$$

$$\Rightarrow -2m/r^2(dt/ds)^2 + 2r(d\phi/ds)^2=0$$

$$\Rightarrow 2r(d\phi/ds)^2 = -2m/r^2(dt/ds)^2$$

$$\Rightarrow (d\phi/ds)^2 = -m/r^3(dt/ds)^2$$

It would seem that you made an error in calculating the equation for circular motion in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53" and you should probably admit that you made a mistake there and stop telling people to refer to that post for the correct equation for circular motion.

If there is any doubt about whether you made a blunder or not, it is easy to see that in your latest post you have declared:

$$\frac{d\phi}{dt}=\sqrt{\frac{m}{r^3}}$$

while in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53" you declared:

$$\frac{d\phi}{ds}=\sqrt{\frac{m}{r^3}}$$

This implies that ds = dt, which is a clear contradiction to your statement in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53" where you concluded that:

$$ds^2=(1-3m/R)dt^2$$

Since the two posts contradict each other, would you care to explain to your "students" which post you blundered in?

 

[yuiop]

In your post 211, you exactly report the correct result:

Thank you

... something is wrong with this as you can check
...
$$H = r^2 (d\phi/ds).$$

So it's simple to see

$$\frac{d^2r}{ds^2} = -\frac{m}{r^2} + H^2(1-3m/r)/r^7.$$

This is my check:

$$H = r^2 (d\phi/ds) \Rightarrow H^2 = r^4 (d\phi/ds)^2 \Rightarrow (d\phi/ds)^2 = H^2/r^4$$

Given:

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

$$\Rightarrow \frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$

$$\Rightarrow \frac{d^2r}{ds^2} = -\frac{m}{r^2} + H^2(1-3m/r)/r^3$$

You seem to have a small error with the power of r.
If we can agree on the power of r, does the rest of my derivation seem reasonable?

 

[starthaus]

I asked you clearly state what you think $$d\phi/ds$$ should be and for some reason you have decided to state what $$d\phi/dt$$ is instead.

What is the problem? I gave you the correct answer.
$$\frac{d\phi}{ds}$$ has been given to you in another thread already, more than a month ago. Rememeber the thread where I was discussin determination of acceleration via geodesic methos with Dalespam, you can find it there . You can also find the answer in posts 139. Just go back and read the thread.

I can only guess at how you arrived at this erronous result in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53"because you have not defined what R is, and the phrase "and if you giving you" is not a meaningful expression in English.

Err, no. Instead of trying to find errors where they don't exist, try understanding the posts.

Now I guess by the expression "the cancellation of the terms in $$dr/ds$$" you mean that the terms containing $$dr/ds$$ go to zero when dr/ds=0. There is a strong implication that you are using the erronous $dr/ds=0\Rightarrow d^2r/ds^2=0$ hack as Altabeh puts it, when you make this assumption. Never the less I will continue with your line of reasoning.

There is already established knowledge that you don't know basic calculus. No need to repeat the same fumbles over and ove.

Using

$$-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0$$

and setting both terms containing dr/ds to zero, the expression becomes:

$$0-(2m/r^2(dt/ds)^2-0-2r(d\phi/ds)^2)=0$$

$$\Rightarrow -2m/r^2(dt/ds)^2 + 2r(d\phi/ds)^2=0$$

$$\Rightarrow 2r(d\phi/ds)^2 = -2m/r^2(dt/ds)^2$$

$$\Rightarrow (d\phi/ds)^2 = -m/r^3(dt/ds)^2$$

It would seem that you made an error in calculating the equation for circular motion in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53" and you should probably admit that you made a mistake there and stop telling people to refer to that post for the correct equation for circular motion.

Congratulations! You managed to get the equation of motion for circular orbits. Now, from the above , you can easily obtain :

$$\frac{d\phi}{dt}=\sqrt{\frac{m}{r^3}}$$

Post 53 contained a typo that I corrected in several posts since. So, you are just harping about a typo. I hope that you have understood by now that the general equation of motion in post 53 is correct. There is a lot to learn from it since both radial and circular orbit motion can be (and have been) derived from it. Instead of harping about imagined errors (or a typo), try learning the derivation.

 

[starthaus]

Thank you


This is my check:

$$H = r^2 (d\phi/ds) \Rightarrow H^2 = r^4 (d\phi/ds)^2 \Rightarrow (d\phi/ds)^2 = H^2/r^4$$

Given:

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

$$\Rightarrow \frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{H^2}{r^4}(r-3m)$$

$$\Rightarrow \frac{d^2r}{ds^2} = -\frac{m}{r^2} + H^2(1-3m/r)/r^3$$

You seem to have a small error with the power of r.
If we can agree on the power of r, does the rest of my derivation seem reasonable?

No, it doesn't. This has been explained to you repeatedly. Hint : H is a function of r. Try remembering it when you bungle the differentiation process.

 

[Altabeh]

-For circular orbits, why would you expect such a silly thing? It is trivial, if $\frac{dr}{ds}=0$ in the general equation of motion, then $\frac{d\phi}{dt}=\sqrt{\frac{m}{r^3}}$.

Speaking of your nonsense claims, this is another edition of your hacks about "how we can get a correct result via a leaky interpretation." I asked you a question simply and you got me answered by your silly logic that contains $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ as an "always true" proposition in it. The first term in your equation includes $$d^2r/ds^2$$ and this can't be made zero by any means unless using the old fallacy mentioned above. Find another hack to go for.

AB

 

[Altabeh]

Thank you

My pleasure.

This is my check:

$$H = r^2 (d\phi/ds) \Rightarrow H^2 = r^4 (d\phi/ds)^2 \Rightarrow (d\phi/ds)^2 = H^2/r^4$$

Nothing wrong with this!

Given:

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

No this is not the second equation you see in http://www.astro.umd.edu/~miller/teaching/astr498/lecture10.pdf" . (In fact yours lacks a factor $$1/r^4$$ in the second term on the right hand side.) I don't know why you insist using this equation but definitely it leads you to a wrong equation. Can we have another start with the equation established in that paper?

AB

 

[yuiop]

In the interests of intellectual integrity, I have to admit to my own errors and my earlier claim that the third Euler-Lagrange equation quoted by Starthaus:

$$-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0$$

... is not the same as the solution obtained by Espen, is mistaken. My earlier calculation contained a sign error that caused the dr/ds term in the Euler-Lagrange equation to differ from Espen's solution by a factor of three. Here is the (hopefully) correct derivation of one from the other:

Given:

$$-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0$$

Carrying out the differentiation of the $-d/dr(1/\alpha)(dr/ds)^2$ term gives $2m/(\alpha^2 r^2)(dr/ds)^2$ and the expression becomes:

$$-d/ds(1/\alpha*2dr/ds)- 2m/ r^2 (dt/ds)^2 - 2m/(\alpha^2 r^2)(dr/ds)^2 + 2r (d\phi/ds)^2) = 0$$

Mutliply both sides by $-\alpha/2$

$$(\alpha/2)* d/ds(1/\alpha*2dr/ds) + m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r \alpha(d\phi/ds)^2) = 0$$

Carrying out the differentiation of the $d/ds(1/{\alpha}*2dr/ds)$ term gives $((2/{\alpha}) (d^2r/ds^2) + d/ds(1/{\alpha})*2(dr/ds))$ and the expression becomes:

$$(\alpha/2)* ( (2/{\alpha}) (d^2r/ds^2) + (d/ds(1/{\alpha})*2(dr/ds) )+ m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

Now $d/ds(1/{\alpha})*2(dr/ds) \Rightarrow d/dr(1/{\alpha})*2(dr/ds)^2$ and carrying out the differentiation of this term gives $-4m/(\alpha^2 r^2)(dr/ds)^2$ and the expression becomes:

$$\Rightarrow (\alpha/2)* ( (2/\alpha) d^2r/ds^2 -4m/(\alpha^2 r^2)(dr/ds)^2 ) + m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

$$\Rightarrow d^2r/ds^2 -2m/(\alpha r^2)(dr/ds)^2 + m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

$$\Rightarrow d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

At this point the messy equation given by Starthaus is now in the cleaner form given by Espen. The multiple steps required to put the equation given by Starthaus (presumably quoted from a textbook) into a cleaner form leaves plenty of room for error and unfortunately I made a sign error the first time around.

This form can further simplified by noting that:

$$(dt/ds)^2 = (1+(1/\alpha)(dr/ds)^2+(r d\phi/ds)^2)/\alpha$$

can be directly obtained from the Schwarzschild metric and after substituting this expression for (dt/ds)^2 into the Espen/Euler-Lagrange solution above, further simple algebraic manipulations result in this condensed (but still valid for both radial and orbital motion) version:

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

This is the same as the result I obtained more directly by my method in https://www.physicsforums.com/showpost.php?p=2781228&postcount=211":

Re-insert the full form of H back into the equation:

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

Having shown that the same final result is obtained by three different methods, should give reasonable confidence in this final result.

 

[espen180]

(30) is wrong.
(37) is as wrong as before.

Going back to this post, do you disagree that $$\frac{dt}{d\tau}=\left(\frac{d\tau}{dt}\right)^{-1}$$ ?

 

[yuiop]

...
Err, no. Instead of trying to find errors where they don't exist, try understanding the posts.

The error in post #53 does exist and I pointed out your blunder in post #113 . See below:

$$0-(2m/r^2(dt/ds)^2-0-2r(d\phi/ds)^2)=0$$

which solves to:

$$(d\phi/ds)^2 = \frac{m}{R^3} (dt/ds)^2$$

and not as you claim:

$$(d\phi/ds)^2 = \frac{m}{R^3}$$

All you had to was use basic algebra to derive the equation of circular motion from the Euler-lagrange equation you got from a textbook and you failed.

Even after I pointed out in #113 how you bungled the equation for circular motion in #53 you never admitted the error (were you hoping no one would notice?) and you were still telling people in post #136 to refer to post #53 for the correct equation for circular orbits. See below:

$$d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$
...
A nice consequence of the above general equation is that you can derive the equations of motion for

-circular orbits (see post 53)

-radial motion by making $$\frac{d\phi}{ds}=0$$

... You can also find the answer in posts 139. Just go back and read the thread.

Congratulations. You finally got the correct solution in #139 after I explained to you how to do it in #113.

...
There is already established knowledge that you don't know basic calculus. No need to repeat the same fumbles over and ove.

This is yet another personal attack. We have already established that your basic physics and algebra leaves a lot to be desired and for someone who claims to be a calculus guru you have made some major blunders in that department too. Altabeh and myself had to go to great lengths to explain to you why the "Starthaus calculus fallacy" that $dr/ds=0\Rightarrow d^2r/ds^2=0$ is indeed a fallacy. Hopefully you understand now.

No, it doesn't. This has been explained to you repeatedly. Hint : H is a function of r. Try remembering it when you bungle the differentiation process.

This is another example of where your grasp of the physics and calculus involved leaves a lot to be desired. Rolfle2 (and others) went to great lengths in another thread to explain to you why H is a constant and not a function of r.

See for example these posts in the other thread:

https://www.physicsforums.com/showpost.php?p=2737150&postcount=217
https://www.physicsforums.com/showpost.php?p=2737098&postcount=214
https://www.physicsforums.com/showpost.php?p=2737890&postcount=262

or just about any of these in this list:

https://www.physicsforums.com/search.php?searchid=2193373

You still don't seem to get it. Do we really need to go over all that again? Hint: There is a clue in the phrase "H is a constant".

P.S. Could we stick to the substance of the posts (i.e the physics and maths) rather than personal attacks as per the forum rules in future? I am quite willing to have a truce if you agree to stick to the spirit of the forum guidelines and stop making statements like "There is already established knowledge that you don't know basic calculus".

 

[yuiop]

$$H = r^2 (d\phi/ds) \Rightarrow H^2 = r^4 (d\phi/ds)^2 \Rightarrow (d\phi/ds)^2 = H^2/r^4$$

Given:

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

No this is not the second equation you see in http://www.astro.umd.edu/~miller/teaching/astr498/lecture10.pdf" . (In fact yours lacks a factor $$1/r^4$$ in the second term on the right hand side.) I don't know why you insist using this equation

Hi Altabeh,
I have been sticking with this solution because I have obtained it 3 different ways starting with:

1) The geodesic method derived by Espen.
2) The Euler-Lagrange equation quoted by Starthaus.
3) My own derivation given in https://www.physicsforums.com/showpost.php?p=2781228&postcount=211".

... but definitely it leads you to a wrong equation. Can we have another start with the equation established in that paper?

I have great respect for your work and opinion in these matters, but to be fair, you have not established the equation in that paper. You have simply stated (without derivation) that:

$$\frac{d^2r}{ds^2} + \frac{M}{r^2} - (1-3M/r) u_{\phi}^2/r^3 = 0$$

without defining what you mean by the $u_{\phi}$ variable. (Note that I am using units of c=1 and $ds = d\tau$)

In the context of the four vector $u = (u_{t}, u_{r}, u_{\theta}, u_{\phi})$ we would normally take $u_{\phi}$ to mean $d{\phi}/ds$ but you seem to be using a different notation.

In your third equation (in the Newtonian limit) you define $u_{\phi}$ as the "specific angular momentum".

Now in the Newtonian context, angular momentum ($L_{\phi}$) is defined as:

$$L_{\phi} = I * \omega = I*(d\phi/dt)$$

where I is the moment of inertia.

For a point particle the moment of inertia (I) is defined as $m_or^2$

This gives:

$$L_{\phi} = m_o r^2*(d\phi/dt)$$

The specific angular momentum ($u_{\phi}$) is the angular momentum per unit rest mass so it follows that:

$$u_{\phi} = L_\phi/m_o = r^2*(d\phi/dt)$$

and since in the Newtonian limit $dt \approx ds$ this can be stated as:

$$u_{\phi} = r^2*(d\phi/ds)$$

in the context that you are using it.

(It can also be noted that $u_{\phi}$ now has the same definition as the conserved Lagrange constant of angular motion that we were calling H earlier.)

Now if we revisit your original equation using this definition of $u_{\phi}$ we obtain:


$$\frac{d^2r}{ds^2} + \frac{M}{r^2} - (1-3M/r)* r^4*(d\phi^2/ds^2)/r^3 = 0$$

$$\Rightarrow \frac{d^2r}{ds^2} = -\frac{M}{r^2} + (1-3M/r)*r* (d\phi^2/ds^2)$$

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

which is the same as the equation I obtained if we agree on the definition of $u_{\phi}$ I have given here.

 

[Altabeh]

Speaking of your nonsense claims, this is another edition of your hacks about "how we can get a correct result via a leaky interpretation." I asked you a question simply and you got me answered by your silly logic that contains $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ as an "always true" proposition in it. The first term in your equation includes $$d^2r/ds^2$$ and this can't be made zero by any means unless using the old fallacy mentioned above. Find another hack to go for.

AB

Actually I see that starthaus does not like to stand corrected; so I go on and correct his "logic" to make his method look better. There is no need to resort to the nonsense $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ for a cicular motion and all we have to do is to say "there is no motion along radius and motion is, like, frozen radially so that both radial velocity and radial acceleration are equivalently zero". Problem resolved.

AB

 

[Altabeh]

Hi Altabeh,
I have great respect for your work and opinion in these matters, but to be fair, you have not established the equation in that paper. You have simply stated (without derivation) that:

$$\frac{d^2r}{ds^2} + \frac{M}{r^2} - (1-3M/r) u_{\phi}^2/r^3 = 0$$

without defining what you mean by the $u_{\phi}$ variable. (Note that I am using units of c=1 and $ds = d\tau$)

In the context of the four vector $u = (u_{t}, u_{r}, u_{\theta}, u_{\phi})$ we would normally take $u_{\phi}$ to mean $d{\phi}/ds$ but you seem to be using a different notation.

In your third equation (in the Newtonian limit) you define $u_{\phi}$ as the "specific angular momentum".

Now in the Newtonian context, angular momentum ($L_{\phi}$) is defined as:

$$L_{\phi} = I * \omega = I*(d\phi/dt)$$

where I is the moment of inertia.

For a point particle the moment of inertia (I) is defined as $m_or^2$

This gives:

$$L_{\phi} = m_o r^2*(d\phi/dt)$$

The specific angular momentum ($u_{\phi}$) is the angular momentum per unit rest mass so it follows that:

$$u_{\phi} = L_\phi/m_o = r^2*(d\phi/dt)$$

and since in the Newtonian limit $dt \approx ds$ this can be stated as:

$$u_{\phi} = r^2*(d\phi/ds)$$

in the context that you are using it.

(It can also be noted that $u_{\phi}$ now has the same definition as the conserved Lagrange constant of angular motion that we were calling H earlier.)

Now if we revisit your original equation using this definition of $u_{\phi}$ we obtain:


$$\frac{d^2r}{ds^2} + \frac{M}{r^2} - (1-3M/r)* r^4*(d\phi^2/ds^2)/r^3 = 0$$

$$\Rightarrow \frac{d^2r}{ds^2} = -\frac{M}{r^2} + (1-3M/r)*r* (d\phi^2/ds^2)$$

[tex] \Rightarrow $$\frac{d^2r}{ds^2} = -\frac{M}{r^2} + \frac{d\phi^2}{ds^2}(r-3M)$$

which is the same as the equation I obtained if we agree on the definition of $u_{\phi}$ I have given here.

I apologize for confusing my own definition of $$u_{\phi}$$ with the one given in that paper and your equation. The reason that I told you that your equation lacks a factor $$1/r^4$$ in the second term on the RHS was that I made use of the standard definition for $$u_{\phi}$$ i.e.,

$$u_{\phi}=d\phi/ds$$

so that

$$\frac{d^2r}{ds^2} = -\frac{M}{r^2} + {u^2_{\phi}}(1-3M/r)/r^3 ,$$

was considered by me to be

$$\frac{d^2r}{ds^2} = -\frac{M}{r^2} + \frac{d\phi^2}{ds^2}(r-3m)/r^4 ,$$

which seems different from yours because of the given reason. Sorry for the inconvenience.

But did you ignore my work in post #132!?

AB

 

[yuiop]

I apologize for confusing my own definition of $$u_{\phi}$$ with the one given in that paper and your equation.

Thanks Altabeh. You are a true gentleman and a scholar for resolving the discrepancy without a protracted argument. I wish there were more like you here

But did you ignore my work in post #132!?

AB

Of course not. I have looked at it several times and seen no obvious logical flaws with it. I have reviewed it again in more detail now that we have resolved some issues / definitions in this thread and it seems flawless and it also introduces some new viewpoints that are of interest.

The first Euler-Lagrange equation,

$$\frac{d}{ds}[(1-2m/r)\dot{t}]=0$$

if integrated would have the following simple solution:

$$E=\frac{1-2m/r}{\sqrt{1-3m/r}}=\frac{1}{\gamma_g^2\sqrt{1-3m/r}}=const.$$

To wit, the energy of the particle is a conserved quantity. Now what about the energy per unit mass of particle with respect to the hovering observer? Let such energy be denoted by$$\gamma$$ (with $$c=1$$), then projecting the 4-momentum of the particle onto the 4-velocity of the observer gives

$$\gamma=u^au_a={E}\gamma_g.$$

Recall that the theory of special relativity portrays $$\gamma=1/\sqrt{1-v^2}$$ to hold, when $$c=1$$, between any two inertial frames. So using the equation for $$E$$ above and solving this for the orbital velocity $$v$$ yields

$$v=(m/r)^{1/2}\gamma_g.$$

And we are done. Clearly, putting $$r=3m$$ leads to $$v=1=c$$ which stands for the orbital speed of photons.

This is a nice result and rigorously derived. I obtain from your solution here that the local velocity $V_R$ of a particle in a circular orbit (radius =R) according to a stationary/ hovering observer at R is:

$$V_R/c = \sqrt{1-\alpha (1-3M/R)}$$

which correctly gives the expected result that the local velocity is c when R=3M.

I also note that you obtained

$$L=\sqrt{\frac{mr}{1-3m/r}}.$$

I extract the following from your equation.

$$L=\sqrt{\frac{mr}{1-3m/r}} = r^2 \frac{d\phi}{ds}$$

$$\Rightarrow \frac{d\phi}{ds} = \frac{1}{r^2}\sqrt{\frac{mr}{1-3m/r}}$$

$$\Rightarrow \frac{d\phi}{ds} = \sqrt{\frac{(m/r^2)}{(r-3m)}}$$

If this is correct then we are agreement because this is the same as the result I obtained in post #211:

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

Also wrong...

All we have to do now, is figure out why Starthaus thinks we both got the wrong result. We can however note that he has failed to explicitly state what he thinks $d\phi/ds$ should be, despite being asked to do so.

 

[yuiop]

Solve for the angular velocity in terms of proper time:

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

Also wrong...

Is it really?

You have stated that for a circular orbit (R=r) the following is true:

$$ds^2=(1-3m/R)dt^2$$

You agreed in #227 that for a circular orbit the following is also true:

$$\frac{d\phi}{dt}=\sqrt{\frac{m}{r^3}}$$

Now by using the chain rule:

$$\frac{d\phi}{ds} = \frac{dt}{ds} \sqrt{\frac{m}{r^3}}$$

Inserting your equation for dt/ds then gives:

$$\frac{d\phi}{ds} = \frac{1}{\sqrt{1-3m/r}} \sqrt{\frac{m}{r^3}}$$

$$\Rightarrow \frac{d\phi}{ds} = \sqrt{\frac{(m/r^2)}{(r-3m)}}$$

So using your own equations, I have proved that my equation is not wrong.

Perhaps you meant to say my equation is RIGHT but you do not like the way I derived it.
Is English your second language?

 

[yuiop]

Actually I see that starthaus does not like to stand corrected; so I go on and correct his "logic" to make his method look better. There is no need to resort to the nonsense $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ for a cicular motion and all we have to do is to say "there is no motion along radius and motion is, like, frozen radially so that both radial velocity and radial acceleration are equivalently zero". Problem resolved.

AB

I agree. For the special case of a circular orbit, both radial velocity and radial acceleration are zero. The trouble is that Starhaus thinks that by the rules of "elementary calculus" that $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ is a universal truth and must therefore also apply to a particle with purely radial motion only, at the apogee of the trajectory. This is a surprising mistake for someone that claims to be a calculus expert and he has not yet admitted it was a calculus blunder on his part.

I have also shown him how he does not need to use the "Starthaus calculus fallacy" to obtain the equation for circular motion in post #211 when I said:

For circular motion the radius is constant and so the radial acceleration $(d^2r)/(ds^2)$ must be zero and in this limited case:

and I also made it clear it is a limited case and can not be extended to purely radial motion. On a matter of common courtesy, he has still not thanked me for the free calculus lesson.

 

[starthaus]

Speaking of your nonsense claims, this is another edition of your hacks about "how we can get a correct result via a leaky interpretation." I asked you a question simply and you got me answered by your silly logic that contains $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ as an "always true" proposition in it. The first term in your equation includes $$d^2r/ds^2$$ and this can't be made zero by any means unless using the old fallacy mentioned above. Find another hack to go for.

AB

It is your problem that you don't know that $$\frac{d}{ds}(\frac{dr}{ds})=0$$ for $$\frac{dr}{ds}=0$$. I do not know why you have so much difficulty with this elementary subject.

 

[starthaus]

Going back to this post, do you disagree that $$\frac{dt}{d\tau}=\left(\frac{d\tau}{dt}\right)^{-1}$$ ?

Of course not, I am simply saying that both (30) and (37) are wrong, you need to figure out why.

 

[starthaus]

In the interests of intellectual integrity, I have to admit to my own errors and my earlier claim that the third Euler-Lagrange equation quoted by Starthaus:... is not the same as the solution obtained by Espen, is mistaken.

Good , it only took you 150 posts and several hints to understand that.

My earlier calculation contained a sign error that caused the dr/ds term in the Euler-Lagrange equation to differ from Espen's solution by a factor of three. Here is the (hopefully) correct derivation of one from the other:

Given:

$$-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0$$

Carrying out the differentiation of the $-d/dr(1/\alpha)(dr/ds)^2$ term gives $2m/(\alpha^2 r^2)(dr/ds)^2$ and the expression becomes:

$$-d/ds(1/\alpha*2dr/ds)- 2m/ r^2 (dt/ds)^2 - 2m/(\alpha^2 r^2)(dr/ds)^2 + 2r (d\phi/ds)^2) = 0$$

Mutliply both sides by $-\alpha/2$

$$(\alpha/2)* d/ds(1/\alpha*2dr/ds) + m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r \alpha(d\phi/ds)^2) = 0$$

Carrying out the differentiation of the $d/ds(1/{\alpha}*2dr/ds)$ term gives $((2/{\alpha}) (d^2r/ds^2) + d/ds(1/{\alpha})*2(dr/ds))$ and the expression becomes:

$$(\alpha/2)* ( (2/{\alpha}) (d^2r/ds^2) + (d/ds(1/{\alpha})*2(dr/ds) )+ m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

Now $d/ds(1/{\alpha})*2(dr/ds) \Rightarrow d/dr(1/{\alpha})*2(dr/ds)^2$ and carrying out the differentiation of this term gives $-4m/(\alpha^2 r^2)(dr/ds)^2$ and the expression becomes:

$$\Rightarrow (\alpha/2)* ( (2/\alpha) d^2r/ds^2 -4m/(\alpha^2 r^2)(dr/ds)^2 ) + m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

$$\Rightarrow d^2r/ds^2 -2m/(\alpha r^2)(dr/ds)^2 + m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

$$\Rightarrow d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

At this point the messy equation given by Starthaus is now in the cleaner

It isn't messy, I simply left the simple computations to you to finish. It isn't my problem that it took you 160 posts and multiple hints to figure it out. I am glad that you finally figured it out.

can be directly obtained from the Schwarzschild metric and after substituting this expression for (dt/ds)^2 into the Espen/Euler-Lagrange solution above, further simple algebraic manipulations result in this condensed (but still valid for both radial and orbital motion) version:

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

This is the same as the result I obtained more directly by my method

...except that :

-your"method" is an invalid hack that shows crass ignorance of basic calculus

-your "method" is incapable of deriving something as simple s $$\frac{d\phi}{ds}$$. Let's see you do it.

 

[starthaus]

You still don't seem to get it. Do we really need to go over all that again? Hint: There is a clue in the phrase "H is a constant".

Since you obviously don't know how $$H$$ was derived (you don't understand the Euler-Lagrange) formalism), you don't understand that the corret statement is "H does not depend on s but it does depend on r, so kev's attempt of differentiating wrt r by considering H a constant is a mistake"

P.S. Could we stick to the substance of the posts (i.e the physics and maths) rather than personal attacks

Sure we can. If you stopped trying to cover your mistakes by trying silly attacks.

I am quite willing to have a truce if you agree to stick to the spirit of the forum guidelines and stop making statements like "There is already established knowledge that you don't know basic calculus".

Start with understanding what $$H=r^2\frac{d\phi}{ds}$$. It doesn't mean "H is constant". In any calculus book. OK?

 

[starthaus]

Rolfle2 (and others) went to great lengths in another thread to explain to you why H is a constant and not a function of r.
.

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

 

[starthaus]

Your post #53 is very suspect. Let's go through it.



If you work out the equations of motion for a photon (which by definition always has ds=0) then when you divide both sides by ds^2 you end up with:

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

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

which is undefined on the left and a bunch of infinite terms on the right. This is a very shaky foundation for a rigorous derivation.

LOL. You really need to take a class in calculus. If you want the lightlike metric, you only need to set $$ds=0$$. If you manage to do this correctly, you will be rewarded by getting:

$$\alpha dt^2-dr^2/\alpha=0$$

$$0-(2m/r^2(dt/ds)^2-0-2r(d\phi/ds)^2)=0$$

which solves to:

$$(d\phi/ds)^2 = \frac{m}{R^3} (dt/ds)^2$$

and not as you claim:

$$(d\phi/ds)^2 = \frac{m}{R^3}$$

You can add this to the list of basic algebra blunders you have already made in this thread.

Congratulations for finally managing to substitute correctly $$\frac{dr}{ds}=0$$ in the general Euler-Lagrange formula I've shown you. Yes, you managed to find a typo , it is $$\frac{d\phi}{dt}=\sqrt{m/r^3}$$, not $$\frac{d\phi}{ds}=\sqrt{m/r^3}$$. I made a cut and paste error, it is quite obvious, no need to keep harping on it over 200 posts.