Gravitational force and acceleration in General Relativity.

[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:
$$\mathbf g = \left( \begin{array}{cccc} c^2 \left(1-\frac{R}{r}\right) & 0 & 0 & 0 \\ 0 & -\left(1-\frac{R}{r}\right)^{-1} & 0 & 0 \\ 0 & 0 & -r^2 & 0 \\ 0 & 0 & 0 & -r^2 \sin ^2(\theta ) \end{array} \right)$$

In this Schwarzschild metric, the Christoffel symbols
$$\left( \begin{array}{cccc} \left\{\Gamma _{t t}^t,\Gamma _{t r}^t,\Gamma _{t \theta }^t,\Gamma _{t\phi }^t\right\} & \left\{\Gamma _{r t}^t,\Gamma _{r r}^t,\Gamma _{r \theta }^t,\Gamma _{r \phi }^t\right\} & \left\{\Gamma _{\theta t}^t,\Gamma _{\theta r}^t,\Gamma _{\theta \theta }^t,\Gamma _{\theta \phi }^t\right\} & \left\{\Gamma _{\phi t}^t,\Gamma _{\phi r}^t,\Gamma _{\phi \theta }^t,\Gamma _{\phi \phi }^t\right\} \\ \left\{\Gamma _{t t}^r,\Gamma _{t r}^r,\Gamma _{t \theta }^r,\Gamma _{t \phi }^r\right\} & \left\{\Gamma _{r t}^r,\Gamma _{r r}^r,\Gamma _{r \theta }^r,\Gamma _{r \phi }^r\right\} & \left\{\Gamma _{\theta t}^r,\Gamma _{\theta r}^r,\Gamma _{\theta \theta }^r,\Gamma _{\theta \phi }^r\right\} & \left\{\Gamma _{\phi t}^r,\Gamma _{\phi r}^r,\Gamma _{\phi \theta }^r,\Gamma _{\phi \phi }^r\right\} \\ \left\{\Gamma _{t t}^{\theta },\Gamma _{t r}^{\theta },\Gamma _{t \theta }^{\theta },\Gamma _{t \phi }^{\theta }\right\} & \left\{\Gamma _{r t}^{\theta },\Gamma _{r r}^{\theta },\Gamma _{r \theta }^{\theta },\Gamma _{r \phi }^{\theta }\right\} & \left\{\Gamma _{\theta t}^{\theta },\Gamma _{\theta r}^{\theta },\Gamma _{\theta \theta }^{\theta },\Gamma _{\theta \phi }^{\theta }\right\} & \left\{\Gamma _{\phi t}^{\theta },\Gamma _{\phi r}^{\theta },\Gamma _{\phi \theta }^{\theta },\Gamma _{\phi \phi }^{\theta }\right\} \\ \left\{\Gamma _{t t}^{\phi },\Gamma _{t r}^{\phi },\Gamma _{t \theta }^{\phi },\Gamma _{t \phi }^{\phi }\right\} & \left\{\Gamma _{r t}^{\phi },\Gamma _{r r}^{\phi },\Gamma _{r \theta }^{\phi },\Gamma _{r \phi }^{\phi }\right\} & \left\{\Gamma _{\theta t}^{\phi },\Gamma _{\theta r}^{\phi },\Gamma _{\theta \theta }^{\phi },\Gamma _{\theta \phi }^{\phi }\right\} & \left\{\Gamma _{\phi t}^{\phi },\Gamma _{\phi r}^{\phi },\Gamma _{\phi \theta }^{\phi },\Gamma _{\phi \phi }^{\phi }\right\} \end{array} \right)$$

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

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:
$$\mathbf g = \left( \begin{array}{cccc} c^2 \left(1-\frac{R}{r}\right) & 0 & 0 & 0 \\ 0 & -\left(1-\frac{R}{r}\right)^{-1} & 0 & 0 \\ 0 & 0 & -r^2 & 0 \\ 0 & 0 & 0 & -r^2 \sin ^2(\theta ) \end{array} \right)$$

In this Schwarzschild metric, the Christoffel symbols
$$\left( \begin{array}{cccc} \left\{\Gamma _{t t}^t,\Gamma _{t r}^t,\Gamma _{t \theta }^t,\Gamma _{t\phi }^t\right\} & \left\{\Gamma _{r t}^t,\Gamma _{r r}^t,\Gamma _{r \theta }^t,\Gamma _{r \phi }^t\right\} & \left\{\Gamma _{\theta t}^t,\Gamma _{\theta r}^t,\Gamma _{\theta \theta }^t,\Gamma _{\theta \phi }^t\right\} & \left\{\Gamma _{\phi t}^t,\Gamma _{\phi r}^t,\Gamma _{\phi \theta }^t,\Gamma _{\phi \phi }^t\right\} \\ \left\{\Gamma _{t t}^r,\Gamma _{t r}^r,\Gamma _{t \theta }^r,\Gamma _{t \phi }^r\right\} & \left\{\Gamma _{r t}^r,\Gamma _{r r}^r,\Gamma _{r \theta }^r,\Gamma _{r \phi }^r\right\} & \left\{\Gamma _{\theta t}^r,\Gamma _{\theta r}^r,\Gamma _{\theta \theta }^r,\Gamma _{\theta \phi }^r\right\} & \left\{\Gamma _{\phi t}^r,\Gamma _{\phi r}^r,\Gamma _{\phi \theta }^r,\Gamma _{\phi \phi }^r\right\} \\ \left\{\Gamma _{t t}^{\theta },\Gamma _{t r}^{\theta },\Gamma _{t \theta }^{\theta },\Gamma _{t \phi }^{\theta }\right\} & \left\{\Gamma _{r t}^{\theta },\Gamma _{r r}^{\theta },\Gamma _{r \theta }^{\theta },\Gamma _{r \phi }^{\theta }\right\} & \left\{\Gamma _{\theta t}^{\theta },\Gamma _{\theta r}^{\theta },\Gamma _{\theta \theta }^{\theta },\Gamma _{\theta \phi }^{\theta }\right\} & \left\{\Gamma _{\phi t}^{\theta },\Gamma _{\phi r}^{\theta },\Gamma _{\phi \theta }^{\theta },\Gamma _{\phi \phi }^{\theta }\right\} \\ \left\{\Gamma _{t t}^{\phi },\Gamma _{t r}^{\phi },\Gamma _{t \theta }^{\phi },\Gamma _{t \phi }^{\phi }\right\} & \left\{\Gamma _{r t}^{\phi },\Gamma _{r r}^{\phi },\Gamma _{r \theta }^{\phi },\Gamma _{r \phi }^{\phi }\right\} & \left\{\Gamma _{\theta t}^{\phi },\Gamma _{\theta r}^{\phi },\Gamma _{\theta \theta }^{\phi },\Gamma _{\theta \phi }^{\phi }\right\} & \left\{\Gamma _{\phi t}^{\phi },\Gamma _{\phi r}^{\phi },\Gamma _{\phi \theta }^{\phi },\Gamma _{\phi \phi }^{\phi }\right\} \end{array} \right)$$

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

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:

$$\begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma & -\beta \gamma & 0 & 0\\ -\beta \gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix}$$

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 ($(dt', dr', d\theta', d\phi'$).

 

[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.

 

[yuiop]

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

If your definition of physics is showing how you derive your conclusions, then you are not doing physics in this thread, because you have yet to shown how you derived your results from the standard Schwarzschild equations of motion.

Several times you have stated that your final results differ from the equations I gave in #1. Now you have begrudgingly admitted that the equations in #1 are correct that means the results you have obtained are wrong.

You also stated the results obtained by Moller differ from the results obtained by Dalespam. Now that you have admitted Dalespam's results are correct, you are in fact saying the results given by Moller are wrong. Are you going to write to Moller and explain to him where he is going wrong?

 

[yuiop]

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

I am keen to learn. Post your derivation and I will see what I can learn from it.

 

[starthaus]

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).

There is no reason to set $d\phi=0$. You loose one set of equations of motion for no reason whatsoever.

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

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

Solve for dr/ds:

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

What do you mean "solve for dr/ds"? There is no equation.
Physics is not a collection of hacks.

 

[starthaus]

Now convert to ds to dt to get the coordinate result:

$$\frac{d^2r}{dt^2} = \frac{d^2r}{ds^2}\left(\frac{ds}{dt}\right)^2 \;$$
.

Why on Earth would anyone do such an elementary calculus mistake? Do you plan to learn differentiation any time soon?:lol:

 

[yuiop]

Why on Earth would anyone do such an elementary calculus mistake? ...

Yep, I slipped at at the last step. This is how it should done.

Assuming initial conditions of a stationary particle at infinity that radially freefalls:

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

Solve for dr/dt:

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

Differentiate dr/dt with respect to t:

$$\frac{d^2r}{dt^2} \; = \; \frac{d(dr/dt)}{dr} * \frac{dr}{dt} \;=\; \left(\frac{2m}{r^2}\sqrt{\frac{2M}{r}} -\frac{m}{r^2} \frac{\alpha}{\sqrt{2M/r}}\right) \alpha \sqrt{\frac{2M}{r}} \; = \; \frac{-m}{r^2} \left(1-\frac{2M}{r}\right) \left(1-\frac{6M}{r}\right)$$

This is the correct equation for the Schwarzschild coordinate acceleration of a radially free falling particle, initially at rest at infinity.

Now it your turn to stop being a coward and show your derivation from the equations of motion for the proper/ coordinate acceleration of a stationary particle in Schwarzschild coordinates. My guess is that we will never see it, because you realize by now that your derivation was wrong.

 

[starthaus]

Yep, I slipped at at the last step. This is how it should done.

Assuming initial conditions of a stationary particle at infinity that radially freefalls:

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

What gives you this idea? What is the connection to the Lagrangian you put up before?

Solve for dr/dt:

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

Differentiate dr/dt with respect to t:

$$\frac{d^2r}{dt^2} \; = \; \frac{d(dr/dt)}{dr} * \frac{dr}{dt} \;=\; \left(\frac{2m}{r^2}\sqrt{\frac{2M}{r}} -\frac{m}{r^2} \frac{\alpha}{\sqrt{2M/r}}\right) \alpha \sqrt{\frac{2M}{r}} \; = \; \frac{-m}{r^2} \left(1-\frac{2M}{r}\right) \left(1-\frac{6M}{r}\right)$$

This is the correct equation for the Schwarzschild coordinate acceleration of a radially free falling particle, initially at rest at infinity.

No, it isn't.

Now it your turn to stop being a coward and show your derivation from the equations of motion for the proper/ coordinate acceleration of a stationary particle in Schwarzschild coordinates. My guess is that we will never see it, because you realize by now that your derivation was wrong.

The equations of motion are all the way back in post #2. For someone who makes so many mistakes, you are quite abusive.

 

[starthaus]

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

What gives you the bright idea that you can set the Lagrangian to a number?

 

[yuiop]

Differentiate dr/dt with respect to t:

$$\frac{d^2r}{dt^2} \; = \; \frac{d(dr/dt)}{dr} * \frac{dr}{dt} \;=\; \left(\frac{2m}{r^2}\sqrt{\frac{2M}{r}} -\frac{m}{r^2} \frac{\alpha}{\sqrt{2M/r}}\right) \alpha \sqrt{\frac{2M}{r}} \; = \; \frac{-m}{r^2} \left(1-\frac{2M}{r}\right) \left(1-\frac{6M}{r}\right)$$

This is the correct equation for the Schwarzschild coordinate acceleration of a radially free falling particle, initially at rest at infinity.

No, it isn't.

That exact equation is given here http://www.mathpages.com/rr/s6-07/6-07.htm

Do you think that reference is wrong? What do you think the equation should be?

The equations of motion are all the way back in post #2. For someone who makes so many mistakes, you are quite abusive.

You gave the equations of motion which is effectively giving nothing because they are standard and quoted in hundreds of online references. However you claim to be able to teach me how to derive the proper and coordinate acceleration of a stationary particle from those equations of motion, which would be impressive if you could actually do it but so far you have come up with nothing.

What gives you the bright idea that you can set the Lagrangian to a number?

I got the idea from here: http://people.maths.ox.ac.uk/~nwoodh/gr/gr03.pdf section 13.1

L = constant. ... Since s is proper time, $g_{ab}\dot{x}^{a} \dot{x}^{b} = 1$, and therefore the third conservation law is L = 1/2 .

That suggests to me that the Lagrangian is a constant.

Anyway, I can do it a different way and obtain the same result.

Using units of G=c=1 the Schwarzschild metric is;

$$ds^2 = (1-2M/r)dt^2 - (1-2m/r)^{-1} dr^2 - r^2 d\phi^2$$

Divide both sides by ds^2.

$$\frac{ds^2}{ds^2} = (1-2M/r) \frac{dt^2}{ds^2} - (1-2m/r)^{-1} \frac{dr^2}{ds^2} - r^2 \frac{d\phi^2}{ds^2}$$

$$1 = (1-2M/r) \frac{dt^2}{ds^2} - (1-2m/r)^{-1} \frac{dr^2}{ds^2} - r^2 \frac{d\phi^2}{ds^2}$$

 

[starthaus]

That exact equation is given here http://www.mathpages.com/rr/s6-07/6-07.htm

Do you think that reference is wrong? What do you think the equation should be?

The correct result has already been derived eons ago by using covariant derivatives. Look up Dalespam's post.

You gave the equations of motion which is effectively giving nothing

You just asked for the equations of motion, I told you that I derived them at post #2. You got exactly what you asked for.

because they are standard and quated in hundreds of online references. However you claim to be able to teach me how to derive the proper and coordinate acceleration of a stationary particle from those equations of motion, which would be impressive if you could actually do it but so far you have come up with nothing.

I can do it in three lines.

I got the idea from here: http://people.maths.ox.ac.uk/~nwoodh/gr/gr03.pdf section 13.1

You obviously know nothing about Lagrangian mechanics.

That suggests to me that the Lagrangian is a constant.

Yep, you don't.

Anyway, I can do it a different way and obtain the same result.

Using units of G=c=1 the Schwarzschild metric is;

$$ds^2 = (1-2M/r)dt^2 - (1-2m/r)^{-1} dr^2 - r^2 d\phi^2$$

$$\frac{ds^2}{ds^2} = (1-2M/r) \frac{dt^2}{ds^2} - (1-2m/r)^{-1} \frac{dr^2}{ds^2} - r^2 \frac{d\phi^2}{ds^2}$$

$$1 = (1-2M/r) \frac{dt^2}{ds^2} - (1-2m/r)^{-1} \frac{dr^2}{ds^2} - r^2 \frac{d\phi^2}{ds^2}$$

So, what do you do next? We already know the correct result, ir is $-\frac{m}{r^2\sqrt{1-2m/r}}$
It can be derived either through covariant derivatives or , more directly, through using the metric.

 

[yuiop]

You just asked for the equations of motion, I told you that I derived them at post #2. You got exactly what you asked for.

You know that is not true, as I have asked you many times to derive the proper and coordinate acceleration of a stationary particle from the equations of motion, as you claim to be able to do.

I can do it in three lines.

Lets see it then!


For your convenience, here is the start of your derivation;

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.


From the first equation, 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.