Orbital velocities in the Schwartzschild geometry

[espen180]

What is the puropse of this nonsensical request? The derivation is based on the fact that (see post 430)

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

Nonsensical? Write it out (also write out ds/dt using the metric) and then compare the expressions.

 

[yuiop]

kev,

I understand that you did not understand calculus very well and you are desperately trying to cover your mathematical blunder. So, let's try an "intuitive" explanation, what you "proved" above is that the angular speed depends on radius. Do you grasp the enormity of your error?

You are wrong. For a particle in freefall with non zero angular momentum and dr/dt not equal to zero, the angular velocity is indeed a function of r.

 

[starthaus]

You are wrong. For a particle in freefall with non zero angular momentum and dr/dt not equal to zero, the angular velocity is indeed a function of r.

Prove it.

 

[espen180]

Prove it.

Mercury's orbital period is less than Pluto's. QED.

 

[starthaus]

Mercury's orbital period is less than Pluto's. QED.

This is certainly true along the geodesics, where $$r=r(\phi)$$. This is certainly not true for $$H=r^2\frac{d\phi}{ds}$$ where $$r$$ and $$\phi$$ are independent of each other . See my post 481 to qbert.

 

[espen180]

This is certainly true along the geodesics, where $$r=r(\phi)$$. This is certainly not true for $$H=r^2\frac{d\phi}{ds}$$ where $$r$$ and $$\phi$$ are independent of each other . See my post 481 to qbert.

When did we stop discussing geodesics?

 

[starthaus]

Nonsensical? Write it out (also write out ds/dt using the metric) and then compare the expressions.

You can do it all by yourself, you seem very good at these "turn the crank" exercises.

 

[espen180]

This is certainly true along the geodesics, where $$r=r(\phi)$$. This is certainly not true for $$H=r^2\frac{d\phi}{ds}$$ where $$r$$ and $$\phi$$ are independent of each other . See my post 481 to qbert.

This got me thinking.

Starthaus, do you view H and K as initial conditions do you look at their change along a geodesic?

 

[starthaus]

When did we stop discussing geodesics?

We didn't. What you keep missing is the following: there are cleaner ways of deriving the equations of motion that do employ the hacks that kev is using. See for example the derivations of the equations of motion for arbitrary planar orbits (post 53 for derivatives wrt proper time and post 430 for derivatives wrt coordinate time).
We have the geodesic method and its perfect equivalent, the lagrangian method. We don't need any hacks that borrow at will the solutions of the lagrange equations and mix them with methods of differentiation that defy the basic rules of calculus.

 

[starthaus]

This got me thinking.

Starthaus, do you view H and K as initial conditions

You know ODE's , right? You know what the constants signify, right?

When I wrote the Euler-Lagrange equation:

$$\frac{d}{ds}(\alpha*\frac{dt}{ds})=0$$

and I put down the obvious solution:

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

what did I claim $$K$$ was?When I wrote the Euler-Lagrange equation:

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

and I put down the obvious solution:

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

what did I claim $$H$$ was?

All I have been telling kev, for hundreds of posts is "Don't borrow the above solutions only to stick them into expressions derived from the metric followed by your hacky ways of differentiating. There are much cleaner ways of deriving the solutions."

 

[espen180]

You know ODE's , right? You know what the constants signify, right?

When I wrote the Euler-Lagrange equation:

$$\frac{d}{ds}(\alpha*\frac{dt}{ds})=0$$

and I put down the obvious solution:

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

what did I claim $$K$$ was?


When I wrote the Euler-Lagrange equation:

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

and I put down the obvious solution:

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

what did I claim $$H$$ was?

In the equations of motion!
How about giving a straight answer?

 

[qbert]

kev is trying to hack his derivation by differentiating as if $$H$$ and $$K$$ are constant everywhere. This is obviously not true. To wit, in your example $$f(r)=r^2\frac{d\phi}{ds}$$ so, $$\frac{df}{dr}$$ isn't zero.

Sorry, apparently, I haven't made myself clear.

$$\frac{d}{ds}\left( r^2\frac{d\phi}{ds} \right) =0$$
in the present context means

$$r(s)^2 \frac{d\phi}{ds}(s) = r_0^2 \right. \frac{d\phi}{ds}\left|_{s=0}$$

The number on the right hand side is a number. it isn't a function of \phi or r
or anything. it's a number. If you take the derivative of a number wrt anything
it is 0. everywhere.


second. What you're doing in the piece about f(r) confuses me.
let me write down explicitly how my brain parses this -- i'll leave
in (s) to remind me something is a function of s:
$$f(r(s)) = r(s)^2 \dot{\phi}(s) \Rightarrow \frac{df}{dr(s)}(s) = 2r(s)\dot{\phi}(s)$$
which literally makes no sense. How do you
even define a $\frac{d}{dr}$ derivative in this case?
(Remember r isn't an INDEPENDENT VARIABLE here. By inverting the proposed (or once
we have it, the real) solution r = r(s) to give us s = s(r), we could use
$$\frac{d}{dr} = \frac{ds}{dr} \frac{d}{ds}$$
but of course that gives us a 0 anyway.

third. On Hacks. /read this at your own peril -- i won't
debate it but it doesn't get said enough here/

There is a longstanding physics tradition. If it works, it's right. This drives
mathies and mathy-inclinded physicists nuts. But the real truth is, physics
is a way of understanding the real world. There doesn't need to be a
logical derivation if the result is right. it's not math. there are many ways
to skin a cat. especially with something as complicated as a g.r. cat.

just because the derivation is flawed doesn't mean the result is wrong
either : http://xkcd.com/759/.

 

[starthaus]

What you're doing in the piece about f(r) confuses me.
let me write down explicitly how my brain parses this -- i'll leave
in (s) to remind me something is a function of s:
$$f(r(s)) = r(s)^2 \dot{\phi}(s) \Rightarrow \frac{df}{dr(s)}(s) = 2r(s)\dot{\phi}(s)$$
which literally makes no sense. How do you
even define a $\frac{d}{dr}$ derivative in this case?

Simple, you need to re-read my post 481 to you and the limitations of your method. The short of it is that it applies only along the geodesics.

There is a longstanding physics tradition. If it works, it's right.

That's bad. I do not ascribe to this philosophy.

just because the derivation is flawed doesn't mean the result is wrong
either : http://xkcd.com/759/.

I didn't say that the results were wrong, I said that kev's methods are wrong.
Now, I enjoy very much interacting with you but I have to convince espen180 that he still has errors in the document that started it all. He has the roight approach but he keeps getting the wrong results. Maybe because he mixes correct mathematics with an occasional kev hack.

 

[starthaus]

Nonsensical? Write it out (also write out ds/dt using the metric) and then compare the expressions.

Ok, since you don't seem inclined to do it, I did it. Here is he summary: your (57) is still wrong. To convince yourself that this is the case, make $$\frac{d\phi}{dt}=0$$. You should recover your (correct) equation (44). This is not the case.

 

[espen180]

Ok, since you don't seem inclined to do it, I did it. Here is he summary: your (57) is still wrong. To convince yourself that this is the case, make $$\frac{d\phi}{dt}=0$$. You should recover your (correct) equation (44). This is not the case.

But (57) is not the complete expression.

 

[starthaus]

But (57) is not the complete expression.

Then either:

-write out (58) and make $$\frac{d\phi}{dt}=0$$

or

-make $$\frac{d\phi}{dt}=0$$ in (57) and you'll still see that you got things wrong.

I think your errors start with the mess at eq (51)-(52).

 

[Altabeh]

Err, $$r$$ and $$\phi$$ are independent variables

Pure nonsesne. For example, in the equatorial plane, the spherical polar coordinate $$r$$ is the same as plane polar coordinate $$R=R(\phi)$$ which you seem to have forgotten from my post #335 where I stress that

Look, $$H=r^2\dot{\phi}$$ is a "constant of motion" and is to be considered in the equatorial plane so that by definition the spherical polar coordinate $$r$$ is the same as the plane polar coordinate $$R=R(\phi).$$

I even gave you the name of a source, but you didn't manage to learn. Even worse, your basec knowledge could have helped you remember from the equations of the orbits of test particles that we put $$u(\phi)=1/r(\phi)=u$$. Why do you insist on your onw nonsense claims and hacks all the time?

AB

 

[Altabeh]

This is certainly true along the geodesics, where $$r=r(\phi)$$. This is certainly not true for $$H=r^2\frac{d\phi}{ds}$$ where $$r$$ and $$\phi$$ are independent of each other.

Nonsense. H here comes from geodesic equations (if not consider the general approach in getting it using "Killing vectors" and the proposition in post #389), which is only valid in its form along geodesics thus the second part of your post contradicts the first; making a nonsense out of the whole of it. LOL!

See my post 481 to qbert.

There is nothing in that post to be inferred about your "hacky" claim above. Get a clue on where your fallacy arises from.

AB

 

[starthaus]

Pure nonsesne. For example, in the equatorial plane, the spherical polar coordinate $$r$$ is the same as plane polar coordinate $$R=R(\phi)$$ which you seem to have forgotten from my post #335 where I stress that

Err, wrong . $$R=R(\phi)$$ is a trajectory , not a coordinate.
Once you express $$r$$ as a function of $$\phi$$ it becomes a trajectory.

Even worse, your basec knowledge could have helped you remember from the equations of the orbits of test particles that we put $$u(\phi)=1/r(\phi)=u$$. Why do you insist on your onw nonsense claims and hacks all the time?

Why do you persist in the same basic error of mixing $$r$$ as a coordinate with $$r(\phi)$$ which is a trajectory?

 

[Altabeh]

We didn't. What you keep missing is the following: there are cleaner ways of deriving the equations of motion that do employ the hacks that kev is using. See for example the derivations of the equations of motion for arbitrary planar orbits (post 53 for derivatives wrt proper time and post 430 for derivatives wrt coordinate time).
We have the geodesic method and its perfect equivalent, the lagrangian method. We don't need any hacks that borrow at will the solutions of the lagrange equations and mix them with methods of differentiation that defy the basic rules of calculus.

Yeah, I agree! Speaking of your old fallacy, $$dr/ds=0\Rightarrow d^2r/ds^2=0$$, which you corrected it serenely in your post #251, and the fact that along timelike geodesics you can introduce a "momentarily at rest" co-observer whose velocity is zero at "each" event but yet his acceleration isn't, which you seem to deny it without even daring to take a look at the sources I provided you with days ago, and now recently another nonsense in your post 495, and above all else persisting on them while knowing that they are nothing but hacks and fallacies, all I can say is that you're just trying to not stand corrected after making mistakes and this will not get us to believe in your nonsense ever. Take a swipe at correcting yourself from time to time.

Err, wrong . $$R=R(\phi)$$ is a trajectory , not a coordinate.
Once you express $$r$$ as a function of $$\phi$$ it becomes a trajectory.

Why do you persist in the same basic error of mixing $$r$$ as a coordinate with $$r(\phi)$$ which is a trajectory?

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

 

[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 [itex]\theta = \pi/2[/tex] this reduces to:

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

$$\Rightarrow \frac{d}{ds} \left(\frac{d\phi}{ds} \right) = -\frac{2}{r} \frac{dr}{ds}\frac{d\phi}{ds}$$

$$\Rightarrow \frac{d}{dr} \left(\frac{d\phi}{ds} \right) = -\frac{2}{r} \frac{d\phi}{ds}$$

$$\Rightarrow r^2 \frac{d}{dr} \left(\frac{d\phi}{ds} \right) = -2r \frac{d\phi}{ds}$$

$$\Rightarrow r^2 \frac{d}{dr} \left(\frac{d\phi}{ds} \right) = -\frac{d}{dr}(r^2) \frac{d\phi}{ds} + C_{(r)}$$

where $C_{(r)}$ is a constant with respect to r.

$$\Rightarrow r^2 \frac{d}{dr} \left(\frac{d\phi}{ds} \right) + \frac{d}{dr}(r^2) \frac{d\phi}{ds} = C_{(r)}$$

$$\Rightarrow \frac{d}{dr} \left( r^2\frac{d\phi}{ds} \right) = C_{(r)}$$

$$\Rightarrow \frac{d}{dr} \left(H) = C_{(r)}$$

Therefore H is a constant with respect to r for a particle in freefall under all circumstances.

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 [itex]\theta = \pi/2[/tex] this reduces to:

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

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.

(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)

Look, kev

We've been round and round about your approach in solving this kind of problems, you just pick the results from the proper methods (the geodesic or the Euler-Lagrange equations) because you can't derive the equations from scratch, you combine them with your own approach to differentiation and you get the same results as the ones already obtained through rigorous math.
Your type of approach has been copied by espen180 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'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'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'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.