Gravitational force and acceleration in General Relativity.

[starthaus]

Well, at least we agree on that.

If you want to back up this claim then you are more than welcome to do so and I will be more than glad to point out once again that you have derived coordinate acceleration instead of proper acceleration.

Sure, we'll go over it again. Should be fun, once we are done, you an also take up your claim with Moller.

 

[yuiop]

It took a painful 150 posts to guide him through a proper derivation.

It took 150 posts to figure out that you were talking about some form of coordinate acceleration when everyone else was talking about proper acceleration.

Your derivation concluded that proper centrifugal acceleration is zero and in another place you claimed proper centrifugal acceleration is defined as a_0 = a\gamma and in another place you got a_0 = a. You never did arrive at the conclusion that everyone else arrived at, that proper centrifugal acceleration is a_0 = a\gamma^2. So can you leave out the condescending tone?

 

[Dale]

Sure, we'll go over it again. Should be fun, once we are done, you an also take up your claim with Moller.

I just had a kind of flash-back from 2nd grade. "I'll tell my Moller on you".

 

[starthaus]

I just had a kind of flash-back from 2nd grade. "I'll tell my Moller on you".

No, no :lol:. You'll have to tell Moller, not me :lol:

 

[starthaus]

It took 150 posts to figure out that you were talking about some form of coordinate acceleration when everyone else was talking about proper acceleration.

Your derivation concluded that proper centrifugal acceleration is zero and in another place you claimed proper centrifugal acceleration is defined as a_0 = a\gamma and in another place you got a_0 = a. You never did arrive at the conclusion that everyone else arrived at, that proper centrifugal acceleration is a_0 = a\gamma^2. So can you leave out the condescending tone?

You need to learn how to stop making false claims.

 

[Dale]

You need to learn how to stop making false claims.

And the part of The Pot will be played by starthaus.

 

[starthaus]

And the part of The Pot will be played by starthaus.

So, "science advisor" job description includes "casting director"? :lol:

 

[Dale]

Does that seem about right?

My results for proper acceleration agree with yours.

So, using the convention that timelike intervals squared are positive, in Schwarzschild coordinates given by:
(t,r,\theta,\phi)

The line element is:
ds^2=\left(1-\frac{R}{r}\right) c^2 dt^2-\left(1-\frac{R}{r}\right)^{-1}dr^2-r^2 d\theta^2- r^2 \sin ^2(\theta ) d\phi^2
where
R=\frac{2GM}{c^2}

And the metric tensor is:
\mathbf g = \left(<br /> \begin{array}{cccc}<br /> c^2 \left(1-\frac{R}{r}\right) &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; -\left(1-\frac{R}{r}\right)^{-1} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -r^2 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; -r^2 \sin ^2(\theta )<br /> \end{array}<br /> \right)

In this Schwarzschild metric, the Christoffel symbols
\left(<br /> \begin{array}{cccc}<br /> \left\{\Gamma _{t t}^t,\Gamma _{t r}^t,\Gamma _{t \theta }^t,\Gamma _{t\phi<br /> }^t\right\} &amp; \left\{\Gamma _{r t}^t,\Gamma _{r r}^t,\Gamma _{r \theta }^t,\Gamma<br /> _{r \phi }^t\right\} &amp; \left\{\Gamma _{\theta t}^t,\Gamma _{\theta r}^t,\Gamma<br /> _{\theta \theta }^t,\Gamma _{\theta \phi }^t\right\} &amp; \left\{\Gamma _{\phi<br /> t}^t,\Gamma _{\phi r}^t,\Gamma _{\phi \theta }^t,\Gamma _{\phi \phi<br /> }^t\right\} \\<br /> \left\{\Gamma _{t t}^r,\Gamma _{t r}^r,\Gamma _{t \theta }^r,\Gamma _{t \phi<br /> }^r\right\} &amp; \left\{\Gamma _{r t}^r,\Gamma _{r r}^r,\Gamma _{r \theta }^r,\Gamma<br /> _{r \phi }^r\right\} &amp; \left\{\Gamma _{\theta t}^r,\Gamma _{\theta r}^r,\Gamma<br /> _{\theta \theta }^r,\Gamma _{\theta \phi }^r\right\} &amp; \left\{\Gamma _{\phi<br /> t}^r,\Gamma _{\phi r}^r,\Gamma _{\phi \theta }^r,\Gamma _{\phi \phi<br /> }^r\right\} \\<br /> \left\{\Gamma _{t t}^{\theta },\Gamma _{t r}^{\theta },\Gamma _{t \theta }^{\theta<br /> },\Gamma _{t \phi }^{\theta }\right\} &amp; \left\{\Gamma _{r t}^{\theta },\Gamma<br /> _{r r}^{\theta },\Gamma _{r \theta }^{\theta },\Gamma _{r \phi }^{\theta }\right\}<br /> &amp; \left\{\Gamma _{\theta t}^{\theta },\Gamma _{\theta r}^{\theta },\Gamma<br /> _{\theta \theta }^{\theta },\Gamma _{\theta \phi }^{\theta }\right\} &amp;<br /> \left\{\Gamma _{\phi t}^{\theta },\Gamma _{\phi r}^{\theta },\Gamma _{\phi<br /> \theta }^{\theta },\Gamma _{\phi \phi }^{\theta }\right\} \\<br /> \left\{\Gamma _{t t}^{\phi },\Gamma _{t r}^{\phi },\Gamma _{t \theta }^{\phi<br /> },\Gamma _{t \phi }^{\phi }\right\} &amp; \left\{\Gamma _{r t}^{\phi },\Gamma<br /> _{r r}^{\phi },\Gamma _{r \theta }^{\phi },\Gamma _{r \phi }^{\phi }\right\} &amp;<br /> \left\{\Gamma _{\theta t}^{\phi },\Gamma _{\theta r}^{\phi },\Gamma _{\theta<br /> \theta }^{\phi },\Gamma _{\theta \phi }^{\phi }\right\} &amp; \left\{\Gamma _{\phi<br /> t}^{\phi },\Gamma _{\phi r}^{\phi },\Gamma _{\phi \theta }^{\phi },\Gamma<br /> _{\phi \phi }^{\phi }\right\}<br /> \end{array}<br /> \right)

are given by:
\left(<br /> \begin{array}{cccc}<br /> \left\{0,\frac{R}{2 r^2-2 r R},0,0\right\} &amp; \left\{\frac{R}{2 r^2-2 r<br /> R},0,0,0\right\} &amp; \{0,0,0,0\} &amp; \{0,0,0,0\} \\<br /> \left\{\frac{c^2 (r-R) R}{2 r^3},0,0,0\right\} &amp; \left\{0,-\frac{R}{2 r^2-2 r<br /> R},0,0\right\} &amp; \{0,0,R-r,0\} &amp; \left\{0,0,0,(R-r) \sin ^2(\theta )\right\} \\<br /> \{0,0,0,0\} &amp; \left\{0,0,\frac{1}{r},0\right\} &amp; \left\{0,\frac{1}{r},0,0\right\} &amp;<br /> \{0,0,0,-\cos (\theta ) \sin (\theta )\} \\<br /> \{0,0,0,0\} &amp; \left\{0,0,0,\frac{1}{r}\right\} &amp; \{0,0,0,\cot (\theta )\} &amp;<br /> \left\{0,\frac{1}{r},\cot (\theta ),0\right\}<br /> \end{array}<br /> \right)

So, without loss of generality the worldline of a particle at rest in these coordinates is given by:
\mathbf X = (t,r_0,0,0)

Then we can derive the four-velocity as follows:
\mathbf U = \frac{d \mathbf X}{d \tau} = c \frac{d \mathbf X}{ds} = c \frac{d \mathbf X}{dt} \frac{dt}{ds} = c \; (1,0,0,0) \; \frac{1}{\sqrt{c^2 \left(1-\frac{R}{r}\right)}} = \left(\left(1-\frac{R}{r}\right)^{-1/2},0,0,0\right)

And we can verify that the norm of the four-velocity is equal to:
||\mathbf U||^2=U_{\mu} U^{\mu}= g_{\mu\nu} U^{\nu} U^{\mu} = c^2

Now we can derive the four-acceleration as follows:
A^{\mu}=\frac{DU^{\mu}}{d\tau}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}

\frac{d \mathbf U}{d\tau}= c\frac{d \mathbf U}{ds}= c\frac{d \mathbf U}{dt}\frac{dt}{ds}= c \; (0,0,0,0) \; \frac{dt}{ds}=(0,0,0,0)

There is only one non-zero component of:
\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma^{r}_{tt}U^{t}U^{t}=\frac{c^2 (r-R) R}{2 r^3}

So, substituting back in we obtain the four-acceleration in the Schwarzschild coordinates:
\mathbf A = \left(0,\frac{c^2 R}{2 r^2},0,0\right) = \left(0,\frac{G M}{r^2},0,0\right)

Finally, the norm of the four-acceleration, which is equal to the magnitude of the proper acceleration, is given by:
-||\mathbf A||^2= -A_{\mu} A^{\mu}= -g_{\mu\nu} A^{\nu} A^{\mu} = \frac{c^4 R^2}{4 r^4 (1-R/r)} = \left( \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} \right)^2

So this also agrees with your previous result. As expected, the proper acceleration for a particle at rest in the Schwarzschild coordinates goes to infinity as r goes to R and is imaginary for r<R.

 

[starthaus]

My results for proper acceleration agree with yours.

So, using the convention that timelike intervals squared are positive, in Schwarzschild coordinates given by:
(t,r,\theta,\phi)

The line element is:
ds^2=\left(1-\frac{R}{r}\right) c^2 dt^2-\left(1-\frac{R}{r}\right)^{-1}dr^2-r^2 d\theta^2- r^2 \sin ^2(\theta ) d\phi^2
where
R=\frac{2GM}{c^2}

And the metric tensor is:
\mathbf g = \left(<br /> \begin{array}{cccc}<br /> c^2 \left(1-\frac{R}{r}\right) &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; -\left(1-\frac{R}{r}\right)^{-1} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -r^2 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; -r^2 \sin ^2(\theta )<br /> \end{array}<br /> \right)

In this Schwarzschild metric, the Christoffel symbols
\left(<br /> \begin{array}{cccc}<br /> \left\{\Gamma _{t t}^t,\Gamma _{t r}^t,\Gamma _{t \theta }^t,\Gamma _{t\phi<br /> }^t\right\} &amp; \left\{\Gamma _{r t}^t,\Gamma _{r r}^t,\Gamma _{r \theta }^t,\Gamma<br /> _{r \phi }^t\right\} &amp; \left\{\Gamma _{\theta t}^t,\Gamma _{\theta r}^t,\Gamma<br /> _{\theta \theta }^t,\Gamma _{\theta \phi }^t\right\} &amp; \left\{\Gamma _{\phi<br /> t}^t,\Gamma _{\phi r}^t,\Gamma _{\phi \theta }^t,\Gamma _{\phi \phi<br /> }^t\right\} \\<br /> \left\{\Gamma _{t t}^r,\Gamma _{t r}^r,\Gamma _{t \theta }^r,\Gamma _{t \phi<br /> }^r\right\} &amp; \left\{\Gamma _{r t}^r,\Gamma _{r r}^r,\Gamma _{r \theta }^r,\Gamma<br /> _{r \phi }^r\right\} &amp; \left\{\Gamma _{\theta t}^r,\Gamma _{\theta r}^r,\Gamma<br /> _{\theta \theta }^r,\Gamma _{\theta \phi }^r\right\} &amp; \left\{\Gamma _{\phi<br /> t}^r,\Gamma _{\phi r}^r,\Gamma _{\phi \theta }^r,\Gamma _{\phi \phi<br /> }^r\right\} \\<br /> \left\{\Gamma _{t t}^{\theta },\Gamma _{t r}^{\theta },\Gamma _{t \theta }^{\theta<br /> },\Gamma _{t \phi }^{\theta }\right\} &amp; \left\{\Gamma _{r t}^{\theta },\Gamma<br /> _{r r}^{\theta },\Gamma _{r \theta }^{\theta },\Gamma _{r \phi }^{\theta }\right\}<br /> &amp; \left\{\Gamma _{\theta t}^{\theta },\Gamma _{\theta r}^{\theta },\Gamma<br /> _{\theta \theta }^{\theta },\Gamma _{\theta \phi }^{\theta }\right\} &amp;<br /> \left\{\Gamma _{\phi t}^{\theta },\Gamma _{\phi r}^{\theta },\Gamma _{\phi<br /> \theta }^{\theta },\Gamma _{\phi \phi }^{\theta }\right\} \\<br /> \left\{\Gamma _{t t}^{\phi },\Gamma _{t r}^{\phi },\Gamma _{t \theta }^{\phi<br /> },\Gamma _{t \phi }^{\phi }\right\} &amp; \left\{\Gamma _{r t}^{\phi },\Gamma<br /> _{r r}^{\phi },\Gamma _{r \theta }^{\phi },\Gamma _{r \phi }^{\phi }\right\} &amp;<br /> \left\{\Gamma _{\theta t}^{\phi },\Gamma _{\theta r}^{\phi },\Gamma _{\theta<br /> \theta }^{\phi },\Gamma _{\theta \phi }^{\phi }\right\} &amp; \left\{\Gamma _{\phi<br /> t}^{\phi },\Gamma _{\phi r}^{\phi },\Gamma _{\phi \theta }^{\phi },\Gamma<br /> _{\phi \phi }^{\phi }\right\}<br /> \end{array}<br /> \right)

are given by:
\left(<br /> \begin{array}{cccc}<br /> \left\{0,\frac{R}{2 r^2-2 r R},0,0\right\} &amp; \left\{\frac{R}{2 r^2-2 r<br /> R},0,0,0\right\} &amp; \{0,0,0,0\} &amp; \{0,0,0,0\} \\<br /> \left\{\frac{c^2 (r-R) R}{2 r^3},0,0,0\right\} &amp; \left\{0,-\frac{R}{2 r^2-2 r<br /> R},0,0\right\} &amp; \{0,0,R-r,0\} &amp; \left\{0,0,0,(R-r) \sin ^2(\theta )\right\} \\<br /> \{0,0,0,0\} &amp; \left\{0,0,\frac{1}{r},0\right\} &amp; \left\{0,\frac{1}{r},0,0\right\} &amp;<br /> \{0,0,0,-\cos (\theta ) \sin (\theta )\} \\<br /> \{0,0,0,0\} &amp; \left\{0,0,0,\frac{1}{r}\right\} &amp; \{0,0,0,\cot (\theta )\} &amp;<br /> \left\{0,\frac{1}{r},\cot (\theta ),0\right\}<br /> \end{array}<br /> \right)

So, without loss of generality the worldline of a particle at rest in these coordinates is given by:
\mathbf X = (t,r_0,0,0)

Then we can derive the four-velocity as follows:
\mathbf U = \frac{d \mathbf X}{d \tau} = c \frac{d \mathbf X}{ds} = c \frac{d \mathbf X}{dt} \frac{dt}{ds} = c \; (1,0,0,0) \; \frac{1}{\sqrt{c^2 \left(1-\frac{R}{r}\right)}} = \left(\left(1-\frac{R}{r}\right)^{-1/2},0,0,0\right)

And we can verify that the norm of the four-velocity is equal to:
||\mathbf U||^2=U_{\mu} U^{\mu}= g_{\mu\nu} U^{\nu} U^{\mu} = c^2

Now we can derive the four-acceleration as follows:
A^{\mu}=\frac{DU^{\mu}}{d\tau}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}

\frac{d \mathbf U}{d\tau}= c\frac{d \mathbf U}{ds}= c\frac{d \mathbf U}{dt}\frac{dt}{ds}= c \; (0,0,0,0) \; \frac{dt}{ds}=(0,0,0,0)

There is only one non-zero component of:
\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma^{r}_{tt}U^{t}U^{t}=\frac{c^2 (r-R) R}{2 r^3}

So, substituting back in we obtain the four-acceleration in the Schwarzschild coordinates:
\mathbf A = \left(0,\frac{c^2 R}{2 r^2},0,0\right) = \left(0,\frac{G M}{r^2},0,0\right)

Finally, the norm of the four-acceleration, which is equal to the magnitude of the proper acceleration, is given by:
-||\mathbf A||^2= -A_{\mu} A^{\mu}= -g_{\mu\nu} A^{\nu} A^{\mu} = \frac{c^4 R^2}{4 r^4 (1-R/r)} = \left( \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} \right)^2

So this also agrees with your previous result. As expected, the proper acceleration for a particle at rest in the Schwarzschild coordinates goes to infinity as r goes to R and is imaginary for r<R.

Excellent, you can now use the same exact source and you will get the proper centripetal acceleration as a_0=r\omega^2. See bottom of the paragraph. Exactly what I have been showing over several posts.

 

[Dale]

What are you talking about? There is no centripetal acceleration, this is a particle which is stationary in the Schwarzschild coordinates, not stationary in a rotating reference frame.

If you want me to derive the proper acceleration for a particle which is moving in uniform circular motion in the Schwarzschild coordinates then I can certainly do that.

 

[starthaus]

What are you talking about? There is no centripetal acceleration, this is a particle which is stationary in the Schwarzschild coordinates, not stationary in a rotating reference frame.

.

Why don't you read the paragraph I linked for you? All the way, to the end.

If you want me to derive the proper acceleration for a particle which is moving in uniform circular motion in the Schwarzschild coordinates then I can certainly do that

That would not be necessary, it is already done (correctly). See the link.

 

[Dale]

I did read it, I just don't see how it relates to kev's question which is concerning a stationary particle in the Schwarzschild spacetime, not a rotating particle.

 

[starthaus]

I did read it, I just don't see how it relates to kev's question which is concerning a stationary particle in the Schwarzschild spacetime, not a rotating particle.

The same method produces the correct proper centripetal accelleration, i.e a_0=r\omega^2.

 

[Dale]

The correct centripetal acceleration for a stationary particle in the Schwarzschild metric is 0.

 

[starthaus]

The correct centripetal acceleration for a stationary particle in the Schwarzschild metric is 0.

Try clicking on your own link, ( "centripetal acceleration" ) , in your above post.

 

[Dale]

That isn't my link, it is an automatic link to the PF library. It is not really relevant to kev's question since it is for a particle moving in flat spacetime instead of a stationary particle in a curved spacetime. In any case even in the limit R->0 (flat spacetime) for a stationary particle in the Schwarzschild metric \omega = 0 implies centripetal acceleration equals 0.

 

[starthaus]

That isn't my link, it is an automatic link to the PF library. It is not really relevant since it is for a flat spacetime. In any case even in the limit R->0 (flat spacetime) for a stationary particle in the Schwarzschild metric \omega = 0 implies centripetal acceleration equals 0.

OBVIOUSLY...but NOT \gamma^2r\omega^2. This is the point of what I have been telling you for days. There is no \gamma in the correct derivation.

 

[Dale]

OBVIOUSLY...but NOT \gamma^2r\omega^2. This is the point of what I have been telling you for days. There is no \gamma in the correct derivation.

I think you are talking about the other thread where we were dealing with rotational motion in flat spacetime. None of your last 5 posts have any relevance whatsoever to this thread where we are dealing with stationary particles in a curved Schwarzschild spacetime. Certainly my derivation here never used the variables \omega or \gamma.

 

[starthaus]

I think you are talking about the other thread where we were dealing with rotational motion in flat spacetime. None of your last 5 posts have any relevance whatsoever to this thread where we are dealing with stationary particles in a curved Schwarzschild spacetime. Certainly my derivation here never used the variables \omega or \gamma.

I simply pointed out that the same formalism, based on covariant derivatives produces the answer r\omega^2 for proper acceeration. You introduced the rotational motion earlier on in this thread. So, my post is an answer to your earlier post. As such, it is quite relevant.

 

[Dale]

The discussion about a rotating particle in flat spacetime is not relevant to this discussion of a stationary particle in a curved spacetime. I will point out that I used exactly the same covariant derivative based formalism in both threads and worked it out clearly step by step in both the flat-rotating and curved-stationary cases. There is no error in either derivation and every step is clearly provided for scrutiny. DrGreg already explained in the other thread how my derivation and the Wikipedia result are in agreement for the case of a rotating particle in flat spacetime.

So, the pertinent question for this thread is: Do you find any error in my derivation of the proper acceleration for a stationary particle in the curved Schwarzschild spacetime?

 

[starthaus]

So, the pertinent question for this thread is: Do you find any error in my derivation of the proper acceleration for a stationary particle in the curved Schwarzschild spacetime?

Your derivation is correct, I never contested that.
So is my derivation for proper acceleration in a rotating frame.

 

[yuiop]

Finally, the norm of the four-acceleration, which is equal to the magnitude of the proper acceleration, is given by:
-||\mathbf A||^2= -A_{\mu} A^{\mu}= -g_{\mu\nu} A^{\nu} A^{\mu} = \frac{c^4 R^2}{4 r^4 (1-R/r)} = \left( \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} \right)^2

So this also agrees with your previous result. As expected, the proper acceleration for a particle at rest in the Schwarzschild coordinates goes to infinity as r goes to R and is imaginary for r<R.

Thanks Dale, impressive as always. I envy your skill with four vectors. I was wondering if you can produce a transformation matrix for the Schwarzschild spacetime, perhaps in its own thread? I have never seen one and so I imagine it is non-trivial.

I also imagine it will analogous to the Lorentz transformation:

\begin{bmatrix}<br /> c t&#039; \\ x&#039; \\ y&#039; \\ z&#039;<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> \gamma &amp; -\beta \gamma &amp; 0 &amp; 0\\<br /> -\beta \gamma &amp; \gamma &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 1 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 1\\<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> c\,t \\ x \\ y \\ z<br /> \end{bmatrix}

where in this case things have been kept relatively simple by limiting the boost to the x direction. Even nicer would be a Schwarzschild transformation matrix for (dt, dr, d\theta, d\phi) to ((dt&#039;, dr&#039;, d\theta&#039;, d\phi&#039;).

 

[yuiop]

Your derivation is correct, I never contested that.
So is my derivation for proper acceleration in a rotating frame.

Dalespam's result agrees with my statement in #1 and yet you say his result is correct and mine is wrong

Can we just clarify something? Do you agree that the results I posted in #1 are correct but you simply don't like the fact that I have not shown how they are derived?

 

[starthaus]

Dalespam's result agrees with my statement in #1 and yet you say his result is correct and mine is wrong

Can we just clarify something? Do you agree that the results I posted in #1 are correct but you simply don't like the fact that I have not shown how they are derived?

DaleS derived his result, you didn't.

 

[yuiop]

...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
...

... we can get immediately:

\frac{dt}{ds}=\frac{k}{1-\frac{2m}{r}}

From the above, we can get the relationship between proper and coordinate speed:

\frac{dr}{ds}=\frac{dr}{dt}\frac{dt}{ds}

Differentiate one more time and you will obtain the correct relationship between proper and coordinate acceleration. ...

You claim you have done all the hard work, but the information you have given is standard fare and can be found in many references including online. The difficult part is going from the equations of motion to the proper acceleration of a stationary particle. It is also obvious that are you are on the wrong tack because the time dilation factor dt/ds that you are using is the time dilation of a particle falling from infinity.

It can be shown that the local velocity of a particle falling from infinity dr'/dt' or dr/ds is \sqrt{(2M/r)} (where ds is the proper time of the particle and primed quantities are the measurements according to a stationary observer at r) and this falling velocity contributes a factor of 1/\sqrt{(1-2M/r)} to the total time dilation of the particle. For a stationary particle this velocity contribution to the time dilation has to be factored out and the time dilation of the stationary particle is simply 1/\sqrt{(1-2M/r)} and not 1/(1-2M/r).

Next you have to show how to obtain the relationship between dr' and dr and it not obvious how you are going to obtain that from the equations of motion that you have given.

Anyway, let's complete your "derivation" and see where that gets us.

...

\frac{dr}{ds}=\frac{dr}{dt}\frac{dt}{ds}

Differentiate one more time and you will obtain the correct relationship between proper and coordinate acceleration. ...

In order to differentiate dr/ds we need to know its value. This can be obtained from the Lagrangian (with d\phi set to zero).

L = \frac{1}{2} = \frac{1}{2} \left( \alpha \left(\frac{dt}{ds}\right)^2 - \frac{1}{\alpha}\left(\frac{dr}{ds}\right)^2 \right)

(See section 13.1 of http://people.maths.ox.ac.uk/~nwoodh/gr/gr03.pdf )

Solve for dr/ds:

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

Now your value for dt/ds is 1/\alpha so this value is substituted into obtain:

\frac{dr}{ds} \; = \; \sqrt{ 1 - \alpha} \; = \; \sqrt{1-\left(1-\frac{2M}{r}\right)} \; = \; \sqrt{\frac{2M}{r}}

The second derivative of dr/ds is:

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

Your "proper" result for the acceleration of a stationary particle is not correct and the reason your result differs from my result is that you are considering the motion of a particle with significant velocity while we are consdering a stationary or nearly stationary particle and also because you have not taken into the account the transformation between dr' and dr.

 

[yuiop]

DaleS derived his result, you didn't.

So if I say E = mc^2, then that is "wrong" if I do not show how I derived the result ?

It follows that your result is also wrong, because you have to yet to show how you have derived your result from the equations of motion and as Luke said, it is doubtful that you are able to do it.

 

[yuiop]

Can we just clarify something? Do you agree that the results I posted in #1 are correct but you simply don't like the fact that I have not shown how they are derived?

DaleS derived his result, you didn't.

Ok I take that to mean that you agree the results posted in #1 are correct and you retract your statement in #2 that they are wrong.

 

[starthaus]

So if I say E = mc^2, then that is "wrong" if I do not show how I derived the result ?

You mean E^2 - (\vec{p}c)^2=(mc^2)^2 right?

 

[starthaus]

Ok I take that to mean that you agree the results posted in #1 are correct and you retract your statement in #2 that they are wrong.

Physics is not copying and pasting together stuff you glean off the internet.

 

[yuiop]

You mean E^2 - (\vec{p}c)^2=(mc^2)^2 right?

I mean:

E = mc^2 = \frac{m_0 c^2}{\sqrt{1-v^2/c^2}} = \sqrt{(m_0c^2)^2 +(\vec{p}c)^2}

By your "logic" E^2 - (\vec{p}c)^2=(mc^2)^2 is "wrong" because you have not derived it.