Orbital velocities in the Schwartzschild geometry

[yuiop]

...meaning that \frac{dH_c}{dt}=0 . Not that \frac{dH_c}{dr}=0, as you and kev incorrectly keep claiming.
The fact that H_c is a conserved quantity (and so is K_c) , does not in any way preclude them being functions of r as both quantities obviously are. You only need to look at their algebraic expressions.

As a particle falls its proper time advances, but the conserved quantities remain constant. By definition as the particle falls r is also changing but the conserved quanties remain constant and therefore they are constant with respect to s and r. This is very simple logic that you can not bury under any amount of mathematical symbols. It is not generally explicitly stated in the textbooks, presumably because they assume this would be self evident to the average reader.

 

[yuiop]

Of course they are , I have been telling you this for 5 weeks since we started discussing thie subject in the thread dealing with orbital acceleration. I even cited the exact paragraph and equation numbers. With one notable exception, the Lagrangian (11.31) in Rindler is incorrect, so I corrected it in post 53.

Rindler gives the the third Euler-Lagrange equation as:

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

and all you have done is change \delta /\delta{r}(1/\alpha) to d /dr(1/\alpha) but this makes no material difference because (1/\alpha) only contains the single variable (r) so the partial differential of (1/\alpha) wrt r is the same the differential with respect to r.

Am I, (or Rindler) missing something?

 

[yuiop]

Anyway, thanks again for the posts and the spirit of your discussion with me. I see I'm in a minority here, so it could well be my own mental block, and I'll get some sleep now and study your posts again.

You don't seem to be making any distinction between normal full differentials of a function and partial differentials of a multi variable function. I think this may be where we differ.

 

[starthaus]

Rindler gives the the third Euler-Lagrange equation as:

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

and all you have done is change \delta /\delta{r}(1/\alpha) to d /dr(1/\alpha) but this makes no material difference because (1/\alpha) only contains the single variable (r) so the partial differential of (1/\alpha) wrt r is the same the differential with respect to r.

Am I, (or Rindler) missing something?

Yes, you are missing something. Rindler incorrectly writes:

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

You should be extremely familiar with the above equation, you wasted about 50 posts in this thread trying to prove that it is incorrect, remember?

Did you finally break down and bought the book?

 

[starthaus]

As a particle falls its proper time advances, but the conserved quantities remain constant. By definition as the particle falls r is also changing but the conserved quanties remain constant and therefore they are constant with respect to s and r.

Constant in time does not mean independent of r. You and Altabeh are co-mingling two totally unrelated concepts.

This is very simple logic that you can not bury under any amount of mathematical symbols. It is not generally explicitly stated in the textbooks, presumably because they assume this would be self evident to the average reader.

Words, words and more words that are clearly disproved by the underlying math. Why don't you dp the math and actually prove your claim that {tex]H[/tex] does not depend on r. To make things interesting, start with the case of arbitrary orbits. To help you out, start from:

H=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 does not depend on r.

 

[starthaus]

Clearly (contrary to the claims of Starthaus) they have treated K as a constant here. They can not directly differentiate the expression for (dr/ds) by s, because the variable s does not appear in the expression. Presumably they used implicit differentiation with respect to s but this is quite a lengthy procedure.

They used the same exact procedure I am using in my blog and they are definitely avoiding your "simplifying" hack.

A much simpler method is to differentiate (dr/ds)^2 with respect to r (instead of s) and divide the result by 2.

... a hack that produces the correct answer by accident but as math goes it is clearly incorrect.

The end result is the same and this demonstrates that K is not only independent of s but also of r, or an incorrect result would have been obtained.

Just in case anyone is wondering, I can provide a proof that:

\frac{d}{ds}\left(\frac{dr}{dt}\right) = \frac{1}{2}*\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right)

It is obvious that the above math is false. Show the steps and you'll find out your error.

 

[yuiop]

Clearly (contrary to the claims of Starthaus) they have treated K as a constant here. They can not directly differentiate the expression for (dr/ds) by s, because the variable s does not appear in the expression. Presumably they used implicit differentiation with respect to s but this is quite a lengthy procedure. A much simpler method is to differentiate (dr/ds)^2 with respect to r (instead of s) and divide the result by 2. The end result is the same and this demonstrates that K is not only independent of s but also of r, or an incorrect result would have been obtained.

Just in case anyone is wondering, I can provide a proof that:

\frac{d}{ds}\left(\frac{dr}{dt}\right) = \frac{1}{2}*\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right)

It is obvious that the above math is false. Show the steps and you'll find out your error.

There is typo in the last equation which is clear from the text above it.
I meant to say I can provide a proof that:

\frac{d}{ds}\left(\frac{dr}{ds}\right) = \frac{1}{2}*\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right)

but well spotted

 

[starthaus]

There is typo in the last equation which is clear from the text above it.
I meant to say I can provide a proof that:

\frac{d}{ds}\left(\frac{dr}{ds}\right) = \frac{1}{2}*\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right)

but well spotted

The text above it is an attempt to give credibility to your hacky solution. There is no wonder that the book authors did not ascribe to your "simplifying" method and did the derivation in a rigorous way..

 

[yuiop]

The text above it is an attempt to give credibility to your hacky solution. There is no wonder that the book authors did not ascribe to your "simplifying" method and did the derivation in a rigorous way..

Please remember that conjecture about another poster's motives and snidey comments or attempts to "put down" another member are against the forum rules. Stick to the maths and physics.

 

[yuiop]

Yes, you are missing something. Rindler incorrectly writes:

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

Naughty Rindler.

 

[starthaus]

Please remember that conjecture about another poster's motives and snidy comments or attempts to "put down" another member are against the forum rules. Stick to the maths and physics.

You should practice what you preach.

 

[yuiop]

You should be extremely familiar with the above equation, you wasted about 50 posts in this thread trying to prove that it is incorrect, remember?

Nearly as many as you spent trying to show how my final result was flawed, until I demonstrated to you that my result and your result were algebraically the same.

 

[starthaus]

Nearly as many as you spent trying to show how my final result was flawed,

...not the final result, the derivation is what is flawed.

until I demonstrated to you that my result and your result were algebraically the same.

...the difference being that you get yours through a hack while I derived mine in a rigorous way.

 

[Altabeh]

...meaning that \frac{dH_c}{dt}=0 . Not that \frac{dH_c}{dr}=0, as you and kev incorrectly keep claiming.
The fact that H_c is a conserved quantity (and so is K_c) , does not in any way preclude them being functions of r as both quantities obviously are. You only need to look at their algebraic expressions.

Another nonsense. According to the fact that you also have supported it all along in this thread after post #251, which was incorrectly pictured to be in support of your old fallacy,

y(r,t)=constant \Rightarrow \partial y/\partial r=0,\ \partial y/\partial t=0

because the right hand side is simply a "constant" and if differentiated wrt any of variables, then should generate zero if you've not completely forgotten basic calculus. From this and the fact that

r^2d{\phi}/ds=H=const.

by your hack we must have

dH/dr=0=2rd{\phi}/ds

which means either r or d{\phi}/ds must be zero which is in either status a nonsense. Likewise we can discuss this for the energy of particle. You're already finished and this is another shot at escaping from standing corrrected. Find another hack!

AB

 

[Altabeh]

The fact that H_c is a conserved quantity (and so is K_c) , does not in any way preclude them being functions of as both quantities obviously are. You only need to look at their algebraic expressions.

When you even don't know what a conserved quantity is, then please do not distract students' minds. Only energy (if considered to be a non-constant) remains invariant under time translations and no such thing can be ever defined for the other conserved quantities if they're not constants. Heck that you don't even know of basics of Physics.

AB

 

[Altabeh]

...the difference being that you get yours through a hack while I derived mine in a rigorous way.

Kev's method leads to your result and there is any chasm but the fact that you're trying to belittle him through nonsense claims and your wishful thinking. I quote once again Zz's post here:

Thread has been reopened, but it is still under moderation.

I would strongly recommend those participating to stick with the physics discussion and cease with the personal attack and trying to belittle other participants in here. There will be serious infractions forthcoming.

If you notice another member providing false information, REPORT IT. If you take it upon yourself to tackle it and it turns ugly, you bear the same responsibility in its escalation.

Zz.

Try to learn something that you seem to be unfamiliar with. Use the books I provided you with. Nowhere in any books you can ever find they say "constant wrt t or s". What it can be inferred about "constant" is just that the quantity is a constant.

AB

 

[starthaus]

From this and the fact that

r^2d{\phi}/ds=H=const.

by your hack we must have

dH/dr=0=2rd{\phi}/ds

which means either r or d{\phi}/ds must be zero which is in either status a nonsense.

Incorrect. the correct statement is

\frac{d H}{ds}=0. This comes from integrating the Euler-Lagrange equation,

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

wrt s resulting into H(r,\phi)=r^2\frac{d\phi}{ds}.

Likewise we can discuss this for the energy of particle. You're already finished and this is another shot at escaping from standing corrrected. Find another hack!

AB

Err, no.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

 

[espen180]

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

Since your expressions also include the double derivative of r, that "excercise" is flawed. All you accomplish is finding a relationship between the angular velocity and the radial distance, velocity and acceleration, but nothing that shows H varies throughout the fall.

 

[starthaus]

Since your expressions also include the double derivative of r, that "excercise" is flawed. All you accomplish is finding a relationship between the angular velocity and the radial distance,

Wrong, the angular velocity gets eliminated between (1) and (2).

velocity and acceleration, but nothing that shows H varies throughout the fall.

Err, you got this backwards, the exercise it to prove that H(r) does not depend on r. Since you stepped in, try proving it, you've been provided (as always) all the tools.
This going to be a good challenge for you, especially considering the fact that H(r,\phi)=r^2\frac{d\phi}{ds}.

 

[yuiop]

Hi Altabeh, I managed to find the book on google here: http://books.google.co.uk/books?id=...resnum=8&ved=0CDEQ6AEwBw#v=onepage&q&f=false"

On page 209 of the linked book they give equation (9.35) as:

\frac{dr^2}{ds^2} = c^2(K^2-1) +\frac{2GM}{r}

They then differentiate equation (9.35) with respect to s and then divide through by (dr/ds) (which is effectively the same as differentiating (dr/ds) by s) to obtain:

\frac{d^2r}{ds^2} = -\frac{GM}{r^2}

Clearly (contrary to the claims of Starthaus) they have treated K as a constant here. They can not directly differentiate the expression for (dr/ds) by s, because the variable s does not appear in the expression. Presumably they used implicit differentiation with respect to s but this is quite a lengthy procedure. A much simpler method is to differentiate (dr/ds)^2 with respect to r (instead of s) and divide the result by 2. The end result is the same and this demonstrates that K is not only independent of s but also of r, or an incorrect result would have been obtained.

Just in case anyone is wondering, I can provide a proof that:

\frac{d}{ds}\left(\frac{dr}{ds}\right) = \frac{1}{2}*\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right)

The text above it is an attempt to give credibility to your hacky solution. There is no wonder that the book authors did not ascribe to your "simplifying" method and did the derivation in a rigorous way..

Here is the proof:

\frac{1}{2}\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right) =<br /> \frac{1}{2}\frac{ds}{dr} \frac{d}{ds} \left(\frac{dr^2}{ds^2}\right)= <br /> \frac{1}{2}\frac{ds}{dr} \frac{d}{ds} \left(\frac{dr}{ds} \frac{dr}{ds}\right)= <br /> \frac{1}{2}\frac{ds}{dr} \left(\frac{d^2r}{ds^2}\frac{dr}{ds}+ \frac{dr}{ds} \frac{d^2r}{ds^2}\right)= <br /> \frac{1}{2}\frac{ds}{dr} \left(2\frac{d^2r}{ds^2}\frac{dr}{ds}\right)=<br /> \frac{d^2r}{ds^2}= <br /> \frac{d}{ds}\left(\frac{dr}{ds}\right)

of my claim that:

\frac{d}{ds}\left(\frac{dr}{ds}\right) = \frac{1}{2}\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right)

Now to carry out the differentiation of:

\frac{dr^2}{ds^2} = c^2(K^2-1) +\frac{2GM}{r}

with respect to r.

First substitute f(r) for (2GM/r) to make it clear that the last term is a function of r (and that K is not a function of r):

\frac{dr^2}{ds^2} = c^2(K^2-1) +f(r)

Now carry out the implicit differerentiation of the expression wrt (r):

\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right) = f&#039;(r)

Reinsert the full form of the function f and carry out the explicit differentiation:

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

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

\Rightarrow \frac{1}{2}\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right) = \frac{-GM}{r^2}\right)

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

This is the same result as obtained by the authors of the book. It is a bit lengthier than it need to be, because I wanted to make it crystal clear that I was not assuming K to be a function of r. The authors do not make it clear what method they use, but it seems to me that it is not possible to differentiate equation (9.35) directly wrt (s). If Starthaus can demonstrate (with the same clarity that I have used), how to obtain the result the authors obtained without diiferentiating with respect to (r) at some intermediate step, then I am willing to stand corrected.

What I have demonstrated here is that I can obtain the same result as the Hobson and Lasenby by differentiating with respect to (r) while assuming K is not a function of r. Since the result is the same, this implies that K is not a function of r or s.

 

[starthaus]

Here is the proof:

\frac{1}{2}\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right) =<br /> \frac{1}{2}\frac{ds}{dr} \frac{d}{ds} \left(\frac{dr^2}{ds^2}\right)= <br /> \frac{1}{2}\frac{ds}{dr} \frac{d}{ds} \left(\frac{dr}{ds} \frac{dr}{ds}\right)= <br /> \frac{1}{2}\frac{ds}{dr} \left(\frac{d^2r}{ds^2}\frac{dr}{ds}+ \frac{dr}{ds} \frac{d^2r}{ds^2}\right)= <br /> \frac{1}{2}\frac{ds}{dr} \left(2\frac{d^2r}{ds^2}\frac{dr}{ds}\right)=<br /> \frac{d^2r}{ds^2}= <br /> \frac{d}{ds}\left(\frac{dr}{ds}\right)

of my claim that:

\frac{d}{ds}\left(\frac{dr}{ds}\right) = \frac{1}{2}\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right)

Err, this is overly complicated. The proof is much simpler:

\frac{1}{2}\frac{d}{dr}(\frac{dr}{ds})^2=\frac{1}{2}\frac{d}{ds}\frac{d}{dr}(\frac{dr}{ds})^2\frac{ds}{dr}=<br /> \frac{1}{2}*2\frac{dr}{ds}\frac{d^2r}{ds^2}\frac{ds}{dr}=\frac{d^2r}{ds^2}

Now to carry out the differentiation of:

\frac{dr^2}{ds^2} = c^2(K^2-1) +\frac{2GM}{r}

with respect to r.

Where di you pull this expression from?

First substitute f(r) for (2GM/r) to make it clear that the last term is a function of r (and that K is not a function of r):

\frac{dr^2}{ds^2} = c^2(K^2-1) +f(r)

Now carry out the implicit differerentiation of the expression wrt (r):

\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right) = f&#039;(r)

Reinsert the full form of the function f and carry out the explicit differentiation:

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

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

\Rightarrow \frac{1}{2}\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right) = \frac{-GM}{r^2}\right)

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

This is the same result as obtained by the authors of the book. It is a bit lengthier than it need to be, because I wanted to make it crystal clear that I was not assuming K to be a function of r. The authors do not make it clear what method they use, but it seems to me that it is not possible to differentiate equation (9.35) directly wrt (s). If Starthaus can demonstrate (with the same clarity that I have used), how to obtain the result the authors obtained without diiferentiating with respect to (r) at some intermediate step, then I am willing to stand corrected.

I'll be more than happy to do that after you answer the question above.

 

[yuiop]

Now to carry out the differentiation of:

\frac{dr^2}{ds^2} = c^2(K^2-1) +\frac{2GM}{r}

Where di you pull this expression from?

If you looked at the quote in the post you responded to, I have already stated where I got it from:

... I managed to find the book on google here: http://books.google.co.uk/books?id=...resnum=8&ved=0CDEQ6AEwBw#v=onepage&q&f=false"

On page 209 of the linked book they give equation (9.35) as:

\frac{dr^2}{ds^2} = c^2(K^2-1) +\frac{2GM}{r}

 

[starthaus]

If you looked at the quote in the post you responded to, I have already stated where I got it from:

can you explain, in your own words, how it is derived?

 

[yuiop]

can you explain, in your own words, how it is derived?

You said if I told you where I got the equation from, you would show how to differentiate it with respect to s instead of r. Now you seem to be back peddaling on your promise. I can show how it is obtained from derivations I have already done, but unless you are claiming the equation given by the authors of the book is incorrect, that is an unecessary diversion. So are you going to stick to your promise?

 

[starthaus]

You said if I told you where I got the equation from, you would show how to differentiate it with respect to s instead of r. Now you seem to be back peddling[/color] on you promise.

I am not "peddling" anything. :LOL:
And I am not back peddaling. I have already uploaded the solution I promised you in my blog, Since you subscribe to my blog, you should have received notification of the update.

I can show how it is obtained from derivations I have already done, but unless you are claiming the equation given by the authors of the book is incorrect, that is an unecessary diversion. So are you going to stick to your promise?

Now, explain to me where does your new starting point equation come from?

\frac{dr^2}{ds^2} = c^2(K^2-1) +f(r)

 

[yuiop]

... I have already uploaded the solution I promised you in my blog, Since you subscribe to my blog, you should have received notification of the update.

I don't remember subscribing to your blog. I occasionally look at it, because for some reason you insist on forcing readers of this forum to look at your blog, rather than answering questions directly in the threads.

Now, explain to me where does your new starting point equation come from?

\frac{dr^2}{ds^2} = c^2(K^2-1) +f(r)

I have already told you it came from a textbook. Since you want to stall on this point I can also obtain it by going back to my derivation in post #211 of this tread:

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}}

Now for radial motion, H=0 so the last equation can be written as:

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

\Rightarrow \frac{dr^2}{ds^2} = K^2 - \alpha

\Rightarrow \frac{dr^2}{ds^2} = K^2 - (1 -2GM/r)

\Rightarrow \frac{dr^2}{ds^2} = (K^2 - 1) + 2GM/r

which is the same as the equation given by the authors of the textbook.

and if we define a function f such that f(r) = (2GM/r) we recover the form you have quoted (using units of c=1):

\Rightarrow \frac{dr^2}{ds^2} = (K^2 - 1) + f(r)

Will you now show how you differentiate the above expression wrt (s) without differentiating wrt (r) in an intermediate step, as this is what you have claimed the authors of the book have done?

 

[espen180]

Wrong, the angular velocity gets eliminated between (1) and (2).

Err, you got this backwards, the exercise it to prove that H(r) does not depend on r. Since you stepped in, try proving it, you've been provided (as always) all the tools.
This going to be a good challenge for you, especially considering the fact that H(r,\phi)=r^2\frac{d\phi}{ds}.

That doesn't change anything. All you end up with is the radial acceleration in terms of H rather than dø/ds.

 

[yuiop]

Let's not waste time and take it for granted that you can obtain:

\frac{d^2r}{ds^2} = \frac{-GM}{r^2}\right)

by assuming K is contant wrt (s) and differentiating (dr/ds) wrt (s).

Unless you prove a flaw in my calculations in #380, I have demonstrated I can obtain the same result by assuming K is a constant wrt (r) and differentiating (1/2)(dr/dt)^2 wrt (r).

This proves that for radial motion, K is constant wrt (s) AND (r).

This means that your claim in your blog document "General Euler-Lagrange Solution for Calculating Coordinate Acceleration", (which is actually the solution for radial motion only - you messed up the titles) that:

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

where, obviously k may be a function of r and possibly t but not of s

is a fallacy, because I have shown k is not a function of r for radial motion. Are you going to correct your blog?


From further qualitative analysis I have come to the conclusion that

for radial motion only when (H = 0) that:

K is contant wrt (s,t and r)

and (\phi) is constant by definition under these conditions,

and for circular motion only. when (K = constant) that:

H is constant wrt (s,t and \phi)

and (r) is constant by definition under these conditions.

This leaves open what happens when a falling particle has both non-zero radial and non-zero angular motion at the same time. My initial hunch is that in relativity, energy and momentum are not individually conserved as in Newtonian physics, but are conserved as a pair in the form of momentum-energy. I think the same thing is happening here. The angular momentum and energy are conserved as a pair.

One thing to recall is that when we take the partial derivatives of L with respect to a given variable, obtain K and H we are by definition treating the "other variables" as constants and so we have not proved anything about what happens when we allow the other variables to vary.

I am not "peddling" anything. :LOL:
And I am not back peddaling.

If you are going to pick on spelling, try and get it right. it should be "pedaling" or sometimes "pedalling" (American?) but not with two d's.

 

[Altabeh]

Incorrect. the correct statement is

\frac{d H}{ds}=0.

Nonsense. I think now we have to also take care of "semantics". This is just because you're only dealing with basics of physics and an ODE which says nothing about constants of motion but the fact that their derivative with respect to s is zero. The scenario of "Killing vectors" calls the shouts here and they exactly describe what such constants are and why they are "constant". During motion, such quantities remain constant and that's the reason for their name.

This comes from integrating the Euler-Lagrange equation,

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

wrt s resulting into


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

It doesn't. It comes from "Killing vectors" and the following proposition:

Let \xi^a be a Killing vector field and let \gamma be a geodesic with tangent u^a. Then \xi_au^a is constant along \gamma.

Err, no.


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

Nonsense. It's really the first time that I see someone makes use of hacks like K(r,\phi) or H(r,\phi). Read the above proposition and show that it holds. Then we can discuss where your big fallacies arise.

AB

 

[starthaus]

That doesn't change anything. All you end up with is the radial acceleration in terms of H rather than dø/ds.

Err, no, you get that H is a function of r. Which means that kev's method of calculating differentials is all wrong, something that we've spent 100+ posts in trying to explain to him.