# Gravitational force and acceleration in General Relativity.

P: 3,967
 Quote by starthaus ...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: $$L=\alpha (\frac{dt}{ds})^2-\frac{1}{\alpha}(\frac{dr}{ds})^2-r^2(\frac{d\phi}{ds})^2$$ ...
 Quote by starthaus ... 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, lets complete your "derivation" and see where that gets us.
 Quote by starthaus ... $$\frac{dr}{ds}=\frac{dr}{dt}\frac{dt}{ds}$$ 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 in to 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.
P: 3,967
 Quote by starthaus 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.
P: 3,967
Quote by kev
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?

 Quote by starthaus 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.
P: 1,568
 Quote by kev 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?
P: 1,568
 Quote by kev 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.
P: 3,967
 Quote by starthaus 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.
P: 3,967
 Quote by starthaus 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?
P: 3,967
 Quote by starthaus 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.
P: 1,568
 Quote by kev 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.
P: 1,568
 Quote by kev 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:
P: 3,967
 Quote by starthaus 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 realise by now that your derivation was wrong.
P: 1,568
 Quote by kev 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 realise 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.
P: 1,568
 Quote by kev $$L = \frac{1}{2} = \frac{1}{2} \left( \alpha \left(\frac{dt}{ds}\right)^2 - \frac{1}{\alpha}\left(\frac{dr}{ds}\right)^2 \right)$$
What gives you the bright idea that you can set the Lagrangian to a number?
P: 3,967
Quote by kev
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.

 Quote by starthaus 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?

 Quote by starthaus 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.

 Quote by starthaus 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.

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

$$1 = (1-2M/r) \frac{dt^2}{ds^2} - (1-2m/r)^{-1} \frac{dr^2}{ds^2} - r^2 \frac{d\phi^2}{ds^2}$$
P: 1,568
 Quote by kev 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.
P: 3,967
 Quote by starthaus 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.

 Quote by starthaus I can do it in three lines.
Lets see it then!

 Quote by starthaus 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.
P: 1,568
 Quote by kev 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. Lets see it then! For your convenience, here is the start of your derivation;
The first two equations are your own personal hacks, I never wrote them.

Let me give you the correct start that solves the problem in three lines:

$$\vec{F}=-grad\Phi$$

where $$\Phi$$ is the potential that you can read straight off the metric. You have two more lines to write, see if you can do it all by yourself, I have already shown you how to do that for centripetal acceleration, you only need to copy the method.
P: 3,967
Quote by kev
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?

 Quote by starthaus The correct result has already been derived eons ago by using covariant derivatives. Look up Dalespam's post.
The equation given in the reference is for the coordinate acceleration of a free falling particle. It is not wrong. It differs from Dalespam's result, because his result is for the proper acceleration of a stationary particle, which is more relevent to this thread. Both equations are right in their own contexts.

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