Gravitational force and acceleration in General Relativity.

[starthaus]

Starthaus, I must say that this is the part that troubles me most about the potential approach. It is indeed more elegant, but this field approximation step makes it very suspicious to me. Particularly in light of the results for the rotating reference frame where we found that obtaining the correct answer with the potential approach depended critically on whether you were using a strong-field approximation or a weak-field approximation. We should have been able to tell, from the beginning, that we needed to use the strong-field approximation.

First things first.

1. Do you agree that post 1 of the OP is a mess?
2. That the results are put in by hand, none of them is derived? kev claims that he sent the derivation to you and DrGreg, is this true? (the derivation cannot be valid given the mistakes in the post).
3. That the only correct formula is the one for proper acceleration a_0=\frac{GM}{r^2}(1-\frac{GM}{rc^2})^{-1/2}?.
5. Since kev did not derive the above for proper acceleration until very late under my guidance, it isn't clear if he din't simply lucked out in the OP.

Now, to the justification of the derivation based on the strong field:

6. Rindler gives the justification for this approach in chapter 9.6.
7. Rindler uses it for deriving proper acceleration for rotation in 9.7.
8. Rindler uses it for deriving proper acceleration in a gravitational field in 11.2.
9. What troubled me was something different, the fact that he uses the approach by equating the strong field metric

ds^2=e^{2 \Phi/c^2} dt^2-...

with the Newtonian approximation for the weak field ds^2=(1-\frac{2m}{r})dt^2-.... I wrote to him about this (in conjunction to criticizing his circular derivation of the equations of accelerated motion in 3.7). He (Rindler) got very defensive and said (textually) that he prefers simpler proofs to the more rigorous ones, so I let the issue drop.

Can you explain more about that step? When is the weak field approximation safe,

See point 9 above. Both fields are required.

under what conditions is the strong field approximation required, and when would even the strong field approximation fail?

Both fields are required (see point 9 , above). To my knowledge, Rindler's approach works for all the examples.

Without that information it seems that we have to do the brute-force approach anyway, just to check if the field approximations were good.

Not necessarily. Rindler's approach gives the correct answers. I had to resort to this simpler approach because there was no hope in teaching kev how to approach the problem from the Lagrangian angle. The other approach that always works is the covariant derivative. So, we have three different approaches that work.
For the rotation case we have the method described in Nikolic's (Demystifier) paper. I much prefer that approach as I demonstrated in the attempt to teach kev. So, we really have 4 different ways to solving the rotating motion. There are more, there is an excellent chapter on the equations of rotating frames in Moller's "Theory of RElativity".

 

[starthaus]

I can do a derivation based on the exact Schwarzschild solution that gives equations that agree exactly with equations in #1

Really, does this include correcting the blunder about a_0=a\gamma^3? I just taught you how to do the derivations starting from the Schwarzschild solution and you are all ready to do it? Then, let's see it. For once, do a derivation rather than put expressions in by hand.

and disagree significantly with your conclusion based on a field aproximation.

Let's see it. Drop the bluster and do the derivations. Once you publish your results, I suggest that you contact Rindler.

 

[Dale]

Now, to the justification of the derivation based on the strong field:

6. Rindler gives the justification for this approach in chapter 9.6.
7. Rindler uses it for deriving proper acceleration for rotation in 9.7.

Thanks, this is helpful. I am looking at the whole chapter:
http://books.google.com/books?id=Mu...a&dq=rindler 9.6&pg=PA183#v=onepage&q&f=false

 

[starthaus]

Thanks, this is helpful. I am looking at the whole chapter:
http://books.google.com/books?id=Mu...a&dq=rindler 9.6&pg=PA183#v=onepage&q&f=false

Excellent, we can have a common base of discussion.
Could you please give a clear answer to questions 1 thru 5?

 

[yuiop]

First things first.
1. Do you agree that post 1 of the OP is a mess?

Post 1 is just a collection of statements and assumptions requesting input from other members of this forum. Whether it is a mess or not, depends on whether the statements are true or not and that remains to be determined.

2. That the results are put in by hand, none of them is derived?

I never claimed that #1 was a formal derivation. (See accusation 1.)

3. That the only correct formula is the one for proper acceleration a_0=\frac{GM}{r^2}(1-\frac{GM}{rc^2})^{-1/2}?.

We agree on the equation for proper acceleration. I disgree about your claim that the rest of the equations are wrong.

5. Since kev did not derive the above for proper acceleration until very late under my guidance, it isn't clear if he din't simply lucked out in the OP.

If the statement's in #1 are correct, it does not really matter how I obtained them, but for the record they are substantially based on the mathpages relativity website which is usually fairly reliable and rigorous.

Now, to the justification of the derivation based on the strong field:

6. Rindler gives the justification for this approach in chapter 9.6.
7. Rindler uses it for deriving proper acceleration for rotation in 9.7.
8. Rindler uses it for deriving proper acceleration in a gravitational field in 11.2.

From the above it is clear that Rindler has not directly derived the coordinate gravitational acceleration and so the incorrect equation you have obtained for coordinate acceleration, is your own hash up rather than Rindler's.

9. What troubled me was something different, the fact that he uses the approach by equating the strong field metric

ds^2=e^{2 \Phi/c^2} dt^2-...

with the Newtonian approximation for the weak field ds^2=(1-\frac{2m}{r})dt^2-.... I wrote to him about this (in conjunction to criticizing his circular derivation of the equations of accelerated motion in 3.7). He (Rindler) got very defensive and said (textually) that he prefers simpler proofs to the more rigorous ones, so I let the issue drop.

Allow me to clear up some of your misconceptions. The metric ds^2=(1-\frac{2m}{r})dt^2-... is not an approximation. It is the exact Schwarzschild solution (with units of c=1). The weak field approximation is the assumption that the gravitational potential is given by \Phi = GM/r (same as the Newtonian potential) and strong field approximation is the assumption that the potential is given by \Phi = (1/2)c^2 \log(1-2GM/(rc^2))

The trouble with the field approximations, is that while it fairly clear what the limitations of the weak field are, it is not clear how strong the field has to be before the strong field approximation breaks down (as Dalespam has pointed out).

Here is another misconception of yours:

That's too bad, if you end up buying Rindler, you will find out that, contrary to your post 1 (and to your incorrect claims above),

\vec{f} = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-1}

is indeed the coordinate force per unit mass. You can multiply by m_0 all by yourself. Besides, you have now the results derived rather than put in by hand as you did in post 1.

You seem to think that coordinate force is obtained by multiplying the coordinate acceleration by the rest mass. That is simply wrong in both SR and GR. It is very easy to demonstrate that your assumption is false in SR.

...
Both fields are required (see point 9 , above).

This is just all wrong. See my comments above. It is the strong field approximation of potential substituted into the Schwarzschild metric. It is not a combination of weak and strong field approximations. Does Rindler really give the exact equation you have quoted above for coordinate acceleration or is it just a botched extrapolation of what he said, by you?

P.S. I think it is time to end the feud and if you agree to stop being insulting, demeaning, evasive, inflamatory and confrontational, then I will agree to be civil with you too.

 

[starthaus]

Post 1 is just a collection of statements and assumptions requesting input from other members of this forum. Whether it is a mess or not, depends on whether the statements are true or not and that remains to be determined.

I never claimed that #1 was a formal derivation. (See accusation 1.)

No, it is just a collection of errors put in by hand. That much we have established.

We agree on the equation for proper acceleration. I disgree about your claim that the rest of the equations are wrong.

How about the blunder:

a_0=a\gamma^3 ?

If the statement's in #1 are correct,

They aren't, see bove.

it does not really matter how I obtained them,

Sure it does, physics is about deriving results. You are just putting in results by hand. Sometimes you hit but most often you miss.

 

[starthaus]

Allow me to clear up some of your misconceptions. The metric ds^2=(1-\frac{2m}{r})dt^2-... is not an approximation. It is the exact Schwarzschild solution (with units of c=1). The weak field approximation is the assumption that the gravitational potential is given by \Phi = GM/r (same as the Newtonian potential) and strong field approximation is the assumption that the potential is given by \Phi = (1/2)c^2 \log(1-2GM/(rc^2))

LOL.

ds^2=e^{2\Phi/c^2}dt^2-...

is approximated in the weak field by:

ds^2=(1+\frac{2\Phi}{c^2})dt^2-...

(simple Taylor expansion)

Now, substitute \Phi=-mc^2/r above.

 

[yuiop]

ds^2=e^{2\Phi/c^2}dt^2-...

is approximated in the weak field by:

ds^2=(1+\frac{2\Phi}{c^2})dt^2-...

(simple Taylor expansion)

Now, substitute \Phi=-mc^2/r above.

I concede that

ds^2= (1+\frac{2\Phi}{c^2})dt^2 - (1+\frac{2\Phi}{c^2})^{-1}dr^2 - ...

approximates to the Newtonian gravitational field when (1-GM/r) is approximately equal to unity and when \Phi=-GMc^2/r is substituted. It is also independent of the value or units used for c.

I was thrown off course by this contradictory post https://www.physicsforums.com/showpost.php?p=2730381&postcount=81 by you in another thread:

Start with the Schwarzschild solution in the weak field approximation:

(cd\tau)^2=(1+\frac{2\Phi}{c^2})(cdt)^2+(1+\frac{2\Phi}{c^2})^{-1}(dr)^2+...
...
...
At the Earth surface :

\Phi_1=-\frac{GM}{R}

The result of the statements you make in the above quote is that the weak field approximation is:

(cd\tau)^2=(1-\frac{2GM}{Rc^2})(cdt)^2+(1-\frac{2GM}{Rc^2})^{-1}(dr)^2+...

which is not correct.

At least the equation for the weak field you give in this thread does approximate to the Newtonian one, as shown here:

d\tau^2= (1-\frac{2GM}{r})dt^2 - (1+\frac{2GM}{r})^{-1}dr^2

Assume radial motion only and assume dr/dt is small so that
d\tau \approx dt

dt^2 = (1-\frac{2GM}{r})dt^2 - (1+\frac{2GM}{r})^{-1}dr^2

\frac{dr}{dt} = \sqrt{ \frac{4G^2M^2}{r^2}-\frac{2GM}{r} }

\frac{d^2r}{dt^2} = \frac{d(dr/dt)}{dt} = \frac{d(dr/dt)}{dr}\frac{dr}{dt}= \frac{GM}{r^2}(1-2GM/r) - \frac{2G^2M^2}{r^3}

In the weak field (1-2GM/r) is approximately unity so:

\frac{d^2r}{dt^2} = \frac{GM}{r^2}(1-2GM/r) - \frac{2G^2M^2}{r^3} \approx \frac{GM}{r^2} - \frac{2G^2M^2}{r^3} \approx \frac{GM}{r^2} (1-2GM/r) \approx \frac{GM}{r^2}

...
How about the blunder:

a_0=a\gamma^3 ?

That is not a mistake. I still maintain the equations in #1 are correct. I can admit and correct my mistakes as I have done in this post. I learn and move on. You should try doing the same. It makes for much better progress all round.

Anyway, the weak field approximation is not important to this thread. We are more interested in where the fields are much stronger.

However, this post does establish that your derivation uses the strong field potential approximation substituted in the Schwarzschild metric (See post #95 https://www.physicsforums.com/showpost.php?p=2731106&postcount=95) and not a combination of the strong and weak field approximations as you claim.

 

[starthaus]

I concede that

ds^2= (1+\frac{2\Phi}{c^2})dt^2 - (1+\frac{2\Phi}{c^2})^{-1}dr^2 - ...

approximates to the Newtonian gravitational field when (1-GM/r) is approximately equal to unity and when \Phi=-GMc^2/r is substituted. It is also independent of the value or units used for c.

I was thrown off course by this contradictory post https://www.physicsforums.com/showpost.php?p=2730381&postcount=81 by you in another thread:


The result of the statements you make in the above quote is that the weak field approximation is:

(cd\tau)^2=(1-\frac{2GM}{Rc^2})(cdt)^2+(1-\frac{2GM}{Rc^2})^{-1}(dr)^2+...

which is not correct.

THat was another thread. That was a typo. That has no bearing on the final result, a result that to this day you still do not understand since you don't know that d\tau_1/d\tau_2 is not the ratio of frquencies. You really need to read on this subject, I can recommend pages 346-7 in Moller's "Theory of Relativity".

For the time being, stick to the subject.

Assume radial motion only and assume dr/dt is small so that
d\tau \approx dt

But this is a horrible hack that is not true. You know very well the value of the raitio d\tau/dt=\sqrt{1-2m/r}

That is not a mistake. I still maintain the equations in #1 are correct. I can admit and correct my mistakes as I have done in this post. I learn and move on. You should try doing the same. It makes for much better progress all round.

LOL. You are plugging in formulas that don't apply because you don't know how they were derived

 

[yuiop]

...
Assume radial motion only and assume dr/dt is small so that
d\tau \approx dt

But this is a horrible hack that is not true. You know very well the value of the raitio d\tau/dt=\sqrt{1-2m/r}

I know that, but approximation solutions always involve "horrible hacks" and should be used with caution. If you do not like hacks, then you should probably avoid derivations based on the weak or strong field approximations. They start with failure built in.

There is no notion of any distinction between proper time and coordinate time in Newtonian equations and from that point of view all Newtonian equations are a horrible hack as you put it.

The formula you base you derivation on, of

\vec{F}=-grad\Phi

is true in Newtonian physics and happens to also be true for proper acceleration and force in relativity, but it is not true for coordinate acceleration without modification.

 

[yuiop]

This is a question for Dalespam.

I know you are an expert in 4 vectors and have already obtained a correct solution for the proper acceleration. Would you be able to obtain a solution for the coordinate gravitational acceleration for us, using those methods? It would be great help to this thread.

 

[yuiop]

Why don' you two guys do the following: You must arrive to Newtons law of Gravitation in the weak field also. Suppose two spherical masses of 1 and 2 kg at a distance of 1m.For example made of iron with density 7.9 g/cm^3.
Write on the left hand side :
m1 = 2 kg
m2 = 1 kg
r1 = 1m
G = 6,67 *10^-11*m^3/(kg*s^2)

F = G* m1*m2/r^2

F = 1.334*10^-10 N

Now write down on the right hand side every single step you need to calculate to get this single number in General relativity including conversions for G = c = 1 etc. So this would settle this discussion (hopefully!)

Hi pepeherborn and welcome to PF . Sorry, didn't mean to ignore you, but we were caught up in a rather heated argument. (You might have noticed LOL). Unfortunately the solution is not as simple as you suggest and there are many ways of interpreting mass and distance (r) in General Relativity and that is probably why we often appear to be in conflict here. For simplicity it is also probably best to use to use one mass that is significantly larger than the other. If we ever come to an agreement in this thread, we may be able to answer you question in your terms.

 

[starthaus]

...
Assume radial motion only and assume dr/dt is small so that
d\tau \approx dtI know that, but approximation solutions always involve "horrible hacks" and should be used with caution. If you do not like hacks, then you should probably avoid derivations based on the weak or strong field approximations. They start with failure built in.

Physics and numerology are different disciplines. What you are doing isn't physics. I suggest that you invest in some good books , I recommended quite a few.(Rindler,Moller)

There is no notion of any distinction between proper time and coordinate time in Newtonian equations and from that point of view all Newtonian equations are a horrible hack as you put it.

While true this is not a valid argument for hacking GR. There are perfectly correct solutions to the problem that don't involve hacks. You have been given at least two in this thread, I don't see the point in your continuing to produce hacks.

The formula you base you derivation on, of

\vec{F}=-grad\Phi

is true in Newtonian physics and happens to also be true for proper acceleration and force in relativity, but it is not true for coordinate acceleration without modification.

Are you reading these things in any book or are you simply making them up?

 

[starthaus]

Assume radial motion only and assume dr/dt is small so that
d\tau \approx dt

dt^2 = (1-\frac{2GM}{r})dt^2 - (1+\frac{2GM}{r})^{-1}dr^2

\frac{dr}{dt} = \sqrt{ \frac{4G^2M^2}{r^2}-\frac{2GM}{r} }

\frac{d^2r}{dt^2} = \frac{d(dr/dt)}{dt} = \frac{d(dr/dt)}{dr}\frac{dr}{dt}= \frac{GM}{r^2}(1-2GM/r) - \frac{2G^2M^2}{r^3}

In the weak field (1-2GM/r) is approximately unity so:

\frac{d^2r}{dt^2} = \frac{GM}{r^2}(1-2GM/r) - \frac{2G^2M^2}{r^3} \approx \frac{GM}{r^2} - \frac{2G^2M^2}{r^3} \approx \frac{GM}{r^2} (1-2GM/r) \approx \frac{GM}{r^2}

If the above were correct (it isn't) and since we have already established that :

a_0= \frac{GM}{r^2}(1-2GM/(rc^2))^{-1/2}

it becomes clearly apparent that you can't have

a_0=a\gamma^3

So, is the your coordinate acceleration expression false or is the relationship between proper and coordinate acceleration false?
Hint: they are both false.

 

[yuiop]

If the above were correct (it isn't) and since we have already established that :

a_0= \frac{GM}{r^2}(1-2GM/(rc^2))^{-1/2}

it becomes clearly apparent that you can't have

a_0=a\gamma^3

So, is the your coordinate acceleration expression false or is the relationship between proper and coordinate acceleration false?
Hint: they are both false.

Wrong again. As I said a few posts before, if you start a derivation from an approximation, you have failure built in from the start, if you try to come to any general conclusions outside the scope of the approximation.


This is the (long awaited) correct way to derive the coordinate acceleration:

(It is the substance of the derivation I sent to Dalespam and DrGreg, but tidied up and debugged. Thanks are due to Dalespam for looking over it and picking up some of the original typos. )

Derivation of the coordinate acceleration of in Schwarzschild spacetime.

Starting with the radial Schwarzschild metric:

ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2

\alpha=1-\frac{2M}{r}

Solve for dr/dt:

\frac{dr}{dt} \;=\; \sqrt{\alpha^2 - \alpha\left( \frac{ds}{dt}\right )^2}

From the Lagragian and the equations of motion we have the conserved quantity k defined as:

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

(See #2 by Starthaus https://www.physicsforums.com/showthread.php?p=2710570#post2710570 )

(Note that the equations of motion are only valid for a particle in freefall.)

This value (k/\alpha) is substituted for dt/ds in the equation above it.

\frac{dr}{dt} \;=\; \sqrt{\alpha^2 - \frac{\alpha^3}{k^2}}

Now dr/dt is differentiated again:

\frac{d^2r}{dt^2} \;=\; \frac{d(dr/dt)}{dr}\frac{dr}{dt}\;=\; -\frac{M}{r^2}\left(\frac{3\alpha^2}{k^2}-2\alpha \right )

For a particle falling from infinity, k has the value 1, but for a particle at its apogee, k = \sqrt{\alpha} .

(See https://www.physicsforums.com/showpost.php?p=2714572&postcount=4 for how the value of k is obtained).

Substituting k=\sqrt{\alpha} gives the coordinate acceleration of a particle at its apogee at r as:

\frac{d^2r}{dt^2} \;=\;-\frac{M}{r^2}\alpha \;=\; -\frac{M}{r^2}\left(1-\frac{2M}{r}\right )

Note that the particle is stationary, in the sense that it is at the peak of its trajectory and is in freefall.

If a value of k=1 is used, then the coordinate acceleration of a particle falling from infinity is obtained.

The derivation of the proper acceleration is given in the next post.

 

[yuiop]

Derivation of the proper acceleration of a particle in Schwarzschild spacetime.

Starting with the radial Schwarzschild metric:

ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2

\alpha=1-\frac{2M}{r}

Solve the metric for dr/ds:

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

Substitute k/\alpha for dt/ds: (See previous post).

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

Differentiate again with respect to s:

\frac{d^2r}{ds^2} = \frac{d(dr/ds)}{dr} \frac{dr}{ds} = -\frac{M}{r^2}

For a particle in freefall at r, but stationary or nearly stationary (i.e at the apogee of its trajectory):

\frac{dr'}{dr} = \frac{1}{\sqrt{1-2M/r}}

and the proper acceleration is:

\frac{d^2r'}{ds^2} = \frac{d(dr)}{ds^2} \frac{dr'}{dr} = \frac{d^2r}{ds} \frac{1}{\sqrt{1-2M/r}} = -\frac{M}{r^2}\frac{1}{ \sqrt{1-2M/r}}

(Note that proper acceleration is independent of the k parameter and therefore the proper acceleration of a free falling particle is indepenent of its velocity.)

This is the initial acceleration of a particle released from radial coordinate r as measured by a local observer at r and is equal in magnitude to the proper acceleration of a particle at rest at r, as measured by an accelerometer.

 

[starthaus]

Now dr/dt is differentiated again:

\frac{d^2r}{dt^2} \;=\; \frac{d(dr/dt))}{dr}\frac{dr}{dt}\;=\; -\frac{m}{r^2}\left(\frac{3\alpha^2}{k^2}-2\alpha \right )

For a particle falling from infinity, k has the value 1, but for a particle at rest at a given radial coordinate, k = \sqrt{\alpha} .

(See https://www.physicsforums.com/showpost.php?p=2714572&postcount=4 for how the value of k is obtained).

Substituting k=\sqrt{\alpha} gives the coordinate acceleration of a particle at its apogee at r as:

Ummm, no. You found the calculations "ready made" by a very good guy (prof. Kevin Brown).
But he goofed badly on his page, one of the rare instances that he made a very serious mistake.

Here is the problem : when you (or more exactly he) differentiated wrt r he did it as if k is a constant, i.e. it does not depend of r.
Unfortunately he later "forgets" about k being a constant when he makes k=\sqrt{\alpha}=\sqrt{1-2m/r} dependent on r. Even the very best make mistakes and this is a very bad one. So, prof. Brown's proof is invalid.

 

[starthaus]

Substitute k/\alpha for dt/ds: (See previous post).

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

Nope, coupled with k=\sqrt{\alpha} (see your previous post) that would make

\frac{dr}{ds} = 0

This is really bad.

Differentiate again with respect to ds:

\frac{d^2r}{ds^2} = \frac{d(dr/ds)}{dr} \frac{dr}{ds} = -\frac{m}{r^2}

Nope, same mistake as before, you can't forget that k is a function of r and differentiate as if k is a constant. (it isn't).

I think I have Kevin's email somewhere, I used to email him suggestions how to improve his book.

 

[Al68]

Nope, coupled with k=\sqrt{\alpha} (see your previous post) that would make

\frac{dr}{ds} = 0

This is really bad.

Why is that bad? Shouldn't \frac{dr}{ds} = 0 be true for a particle at apogee?

Not trying to butt in, just trying to follow along.

 

[yuiop]

Nope, coupled with k=\sqrt{\alpha} (see your previous post) that would make

\frac{dr}{ds} = 0

This is really bad.

Why is that bad? Shouldn't \frac{dr}{ds} = 0 be true for a particle at apogee?

Not trying to butt in, just trying to follow along.

You are quite right Al. I think the objection that Starthaus is making is that this implies the next step is differentiating zero with respect to s. The clever part of Prof Brown's derivation is that by using the constant k he can get around this and carry out the differentiation. If Starthaus was as good at calculus as he makes out he is, he would realize there is no problem with the step taken by K. Brown. You only have to read Prof Brown's website to realize he very good at calculus and physics.

Here is another way of looking at it.

x = \sqrt{\frac{1}{2}-\left(1-\frac{2M}{r}\right)}

\frac{dx}{dr} = \frac{M}{r^2\sqrt{2M/r-1/2}}

The above is a perfectly valid derivative, even though for a certain value of r, x=0.

The above can rephrased using k^2=1/2 and \alpha = (1-2M/r) as:

x = \sqrt{k^2-\alpha}

and as long as it understood that k is a constant and \alpha is a function of the variable r, there is no problem differentiating it.

 

[starthaus]

You are quite right Al. I think the objection that Starthaus is making is that this implies the next step is differentiating zero with respect to s.

It is much worse than that. Face the music , kev, prof. Kevin Brown's proof is fatally flawed. See post #137. Time for you to go fishing for a different proof on the web.

 

[yuiop]

It is much worse than that. Face the music , kev, prof. Kevin Brown's proof is fatally flawed. See post #137. Time for you to go fishing for another proof on the web.

I don't need to. I have shown in the last post how your main objection is invalid.

I can address your other objections too. There is nothing wrong K. Brown's derivation.

Time for you start quoting from textbooks that you actually understand. The equation you have given for coordinate acceleration is not actually given by Rindler in his book and is just a hash up by you based on a extrapolated misunderstanding of what Rindler actually says .

 

[starthaus]

The clever part of Prof Brown's derivation is that by using the constant k he can get around this and carry out the differentiation.

No, kev, you can't do that , this is basic calculus. You admitted earlier on that you are weak on this subject, you really need to learn it if you want to learn physics. You can't differentiate as if k is a constant when k is really a function of r. Prof. Kevin Brown goofed, what is your excuse?

 

[yuiop]

No, kev, you can't do that , this is basic calculus. You admitted earlier on that you are weak on this subject, you really need to learn it if you want tolearn physics. You can't differentiate as if k is aconstant when k is really a function of r. Prof. Kevin Brown goofed, what is your excuse?

It was you that introduced the Lagrangian and the equations of motion, but it seems you really do not understand the physics behind them. k is a constant for a free falling particle. Read your own post #2 of this thread quoted below:

This is very wrong. You need to start with the metric:


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

\alpha=1-\frac{2m}{r}

From this you construct the Lagrangian:

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

From the above Lagrangian, you get immediately the equations of motion:

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

r^2 \frac{d\phi}{ds}=h

whre h,k are constants.

A classic case of being "hung by your own petard" I think. TIme to bow out graciously and admit you are wrong and apologise to K.Brown.

When a particle is in freefall, \alpha changes in such a way that exactly cancels out the change in the ratio dt/ds so that k remains constant for any value of r. k is what Prof Brown calls an affine parameter. Read up on Killing vectors.

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

 

[starthaus]

It was you that introduced the Lagrangian and the equations of motion, but it seems you really do not understand the physics behind them. k is a constant for a free falling particle. Read your own post here:
When a particle is in freefall, \alpha changes in such a way that exactly cancels out the change in the ratio dt/ds so that k remains constant for any value of r. k is what Prof Brown calls an affine parameter. Read up on Killing vectors.

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

No , kev, you really don't understand calculus,

The Euler-Lagrange equation is:

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

This means that :

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

where k is NOT a function of s. But, as Kevin Brown does in his failed proof, k can be a function of both t and r. In his case, k=\sqrt{\alpha}=\sqrt{1-2m/r}.

So, he cannot differentiate wrt r as if k were not a function of r. Because it is.

 

[yuiop]

No , kev, you really don't understand calculus,

The Euler-Lagrange equation is:

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

This means that :

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

where k is NOT a function of s. But, as Kevin Brown does in his failed proof, k can be a function of both t and r. In his case, k=\sqrt{\alpha}=\sqrt{1-2m/r}.

So, he cannot differentiate wrt r as if k were not a function of r. Because it is.

k=\sqrt{\alpha}=\sqrt{1-2m/r}

is really:

k=\sqrt{1-2m/r_a}

where r_a is a constant which is equal to the radial coordinate of the particle at its apogee. As the particle falls the value of \sqrt{1-2m/r_a} does not change. It is only at the apogee that r = r_a and the two values can be used interchangeably and cancel out. It is only at the apogee that you can say \sqrt{1-2m/r_a} = \sqrt{1-2m/r} = \sqrt{\alpha}

Why did you say in #2 that k is a constant and now you are vehemently arguing that it is not? Do you suffer from multiple personality disorder? Either you are wrong now or you were wrong in #2. Do you retract your statements in #2? You probably shouldn't. I can find plenty of references that show k is a constant. (Not just K.Brown).

 

[starthaus]

k=\sqrt{\alpha}=\sqrt{1-2m/r}

is really:

k=\sqrt{\alpha}=\sqrt{1-2m/r_a}

What do you mean "is really"? You are changing your tune from post to post.

 

[starthaus]

\alpha=1-\frac{2M}{r}

Substituting k=\sqrt{\alpha}

So, you need to make up your mind, what is the value for k?

 

[yuiop]

k=\sqrt{\alpha}=\sqrt{1-2m/r}

is really:

k= \sqrt{\alpha} =\sqrt{1-2m/r_a}

What do you mean "is really"? You are changing your tune from post to post.

\alpha=1-\frac{2M}{r}


Substituting k=\sqrt{\alpha} gives the coordinate acceleration of a particle at its apogee at r as:

So, you need to make up your mind, what is the value for k?

I though I made it clear in the last post that:

k=\sqrt{\alpha}=\sqrt{1-2m/r}

is interchangeable (synonymous) with

k=\sqrt{\alpha}=\sqrt{1-2m/r_a}

at the apogee, where r=r_a

It is a bit tricky. You have to pay attention

 

[starthaus]

I though I made it clear in the last post that:

k=\sqrt{\alpha}=\sqrt{1-2m/r}

is interchangeable (synonymous) with

k=\sqrt{\alpha}=\sqrt{1-2m/r_a}

at the apogee, where r=r_a

It is a bit tricky. You have to pay attention

..and its value at infinity is 1. So, varies from 1 to \sqrt{1-2m/r_a} as a function of r k=\sqrt{1-2m/r}. What does all this have to do with your inability to perform a derivative of a function with respect to the r variable?