Orbital velocities in the Schwartzschild geometry

[Altabeh]

Hc and Kc do indeed contain the variable r (as do H and K in your blog) but it does not mean that these constants are a function of r, because these constants contain other variables that are dedendent on r and change in such a way that the constants remain constant for any value of r. If a function remains unchanged for any value of r then the function is not a function of r. Let me give you a very simple example. Let us say we have a function f defined as f = r/r. This is not a function of r because the value of f is 1 for any value of r. This is an obvious example, but it is not always so obvious. Let us say we have another variable s defined as s=2r and a function g defined as g = 3+s/r. The function g is not a function of r because the function g always evaluates to 5 for any value of r and g is in fact a constant.

I don't like the examples, but definitely your reasoning is completely correct!

AB

 

[yuiop]

Pretty much what kev and Altabeh have been saying (I haven't checked algebra details, but certainly their big picture is correct, and from experience Altabeh makes very, very few algebraic errors in PF):

"It follows from the geodesic equations that L is constant. In fact, on the worldline of a particle in free fall, ds2 = gab.dxa.dxb, by definition, so L = gab.(dxa/ds)(dxb/ds) = 1. ... worldline of a photon is also given by the geodesic equations with gab.(dxa/ds).(dxb/ds)= 0. The parameter s does not here have the interpretation of time: it is called an affine parameter and can be replaced by any linear function of s." From p 27 of http://people.maths.ox.ac.uk/nwoodh/gr/gr03.pdf

Thanks for the link atyy.

Looks like a LOT of useful material in that document. Are you the author?

 

[atyy]

Thanks for the link atyy.

Looks like a LOT of useful material in that document. Are you the author?

Nope, I am a clueless biologist. Woodhouse's notes are just a free source I have found useful in my own self-study. He gives lots of the orbits in a Schwarzschild spacetime later on in the same set of notes.

 

[espen180]

Thanks for the link atyy.

Looks like a LOT of useful material in that document.

I agree. The person who wrote this seems very pedagogically inclined.

 

[yuiop]

kev, since you calculated $$\frac{d^2r}{d\tau^2}$$ for arbitrary orbits a few pages back. I'm aiming for the same thing, and found

$$\frac{d^2r}{d\tau^2}=-\frac{GM}{r^2}+\left(r-\frac{3GM}{c^2}\right)\left(\frac{d\phi}{d\tau}\right)^2$$

OK, this seems in agreement with what we obtained here:

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

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

so, so far so good!

I'm currently working on $$\frac{d^2r}{dt^2}$$. I'll post my derivation once I'm done.

It would be nice to see an alternative derivation for the general equation of the acceleration of a particle in freefall, in coordinate time.

 

[Altabeh]

]

This is false.

1. The lagrangian $$L$$ depends on both $r$ and $\dot{r}$.

Correct.

2. Your expression $$L$$ depends on $$H_c$$ and $$K_c$$. Since both $$H_c$$ and $$K_c$$ are clear function of $$r$$ your attempt at differentiatin $$L$$ as if it weren't a function of $$r$$ is incorrect.

Completely nonsense and without a physical/mathematical basis. Read the sources given to you to get to see how $$H_c$$ and $$K_c$$ are both derived to be CONSTANTS and indeed they each correspond to a conserved quantity (respectively angular momentum and energy of particle):

$$m_0r^2\dot{\phi}=m_0H_c=const.$$

is, for instance, the angular momentum ($$m_0$$ being the mass of particle). But let's dig through the details of how to get $$K_c$$ and why it is a constant.

From the invariance of Killing vector fields along a symmetry axis, or

$$\xi_au_a=const.$$

where $$\xi^a$$ are the contravariant components of the Killing vector field and $$u_a$$ is a geodesic tangent, for the Schwartzschild metric with the only two non-null normalized Killing vectors $$\xi=(1,0,0,0)$$ and $$\eta=(0,0,0,1)$$ which correspond altogether to time-independence and axial symmetry of the spacetime, we get (because of time-independence)

$$\xi^au_0=(1,0,0,0).(u_0,...,u_3)=u_0=const.$$

which means $$u_0=\dot{x_0}=g_{00}\dot{x}^0=const.$$ Now let's take $$\dot{x}^0=ct$$ and with $$g_00=1-2m/r$$ one would immediately obtain

$$u_0=(1-2m/r)c\dot{t}=cK_c.$$

On the other hand, $$p_0=m_0\dot{x}_0=m_0u_0$$ where $$p_0$$ is the time component of the four-momentum and again $$m_0$$ is the mass of particle and in a flat spacetime it is obvious that $$p_0=E/c,$$ with $$E$$ being the energy. Thus

$$E=p_0c=m_0u_0c=km_0c^2.$$

And this is the total energy for motion in a Schwarzschild metric.

Please do not attempt to collect nonsense claims and rather read books and use information provided here to understand things sounding to be at a higher level than your knowledge. If you persist on nonsense, I'll have to report your inutile posts.

3. This is easily provable to be false .I have already shown that, according to your very own definition:


$$H_c=\frac{r^2}{\alpha} \frac{d\phi}{dt}=\frac{r^2}{1-2m/r}*\sqrt{\frac{m}{r^3}}$$

So, your statement "these constants contain other variables that are dedendent on r and change in such a way that the constants remain constant for any value of r. " is easily proven false.

The same nonsense as above which has no mathematical/physical knowledge behind. Read the following books and stick with physics not your own wishful thinking:

Hobson M., Efstathiou G., Lasenby A. General relativity.. an introduction for physicists (CUP, 2006, pp 205-209.

A. Papapetrou, Lectures on GR, 1974, pp 70-73.

AB

 

[starthaus]

The same nonsense as above which has no mathematical/physical knowledge behind. Read the following books and stick with physics not your own wishful thinking:

This is precisely the $$H_c$$ definition used by kev in his derivation.

 

[starthaus]

Not quoted from a textbook eh? here is post #53 again:Compare this to http://books.google.co.uk/books?id=...w#v=onepage&q=lagrange schwarzschild&f=false"

Your equations are the exactly the same as equations 11.31, 11.32 and 11.33 in Rindler's book, with the same odd use of parentheses, the same introduction of the variable [itex]\omega[/tex] even though it is never used later.

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.

 

[starthaus]

Correct.
Completely nonsense and without a physical/mathematical basis. Read the sources given to you to get to see how $$H_c$$ and $$K_c$$ are both derived to be CONSTANTS and indeed they each correspond to a conserved quantity (respectively angular momentum and energy of particle):

$$m_0r^2\dot{\phi}=m_0H_c=const.$$

...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.

 

[yuiop]

The same nonsense as above which has no mathematical/physical knowledge behind. Read the following books and stick with physics not your own wishful thinking:

Hobson M., Efstathiou G., Lasenby A. General relativity.. an introduction for physicists (CUP, 2006, pp 205-209.

AB

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:

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

 

[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:

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

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):

$$\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right) = f'(r)$$

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:

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

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:

$$\frac{1}{2}\frac{d}{dr}(\frac{dr}{ds})^2=\frac{1}{2}\frac{d}{ds}\frac{d}{dr}(\frac{dr}{ds})^2\frac{ds}{dr}= \frac{1}{2}*2\frac{dr}{ds}\frac{d^2r}{ds^2}\frac{ds}{dr}=\frac{d^2r}{ds^2}$$

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):

$$\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right) = f'(r)$$

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 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?

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