Gravitational force and acceleration in General Relativity.

[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.

 

[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.

 

[starthaus]

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.

 

[yuiop]

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.

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 relevant to this thread. Both equations are right in their own contexts.

 

[yuiop]

The first two equations are your own personal hacks, I never wrote them.

They are direct copy and pastes, word for word from your posts https://www.physicsforums.com/showpost.php?p=2710570&postcount=2" of this thread.

 

[Tomsk]

Noob question here: why is the lagrangian
L = \frac{1}{2}g_{ab}\dot{x}^a\dot{x}^b
? Shouldn't it be
L = \sqrt{g_{ab}\dot{x}^a\dot{x}^b}
since it's the path length we want to maximize?

I tried it that way and got different equations of motion. I'm not sure what to do with them now though! I think I prefer the derivation posted by DaleSpam a couple of pages ago.

 

[yuiop]

I think I prefer the derivation posted by DaleSpam a couple of pages ago.

Just curious. Prefer it to what? So far Dalespam is the only person to give a derivation of the acceleration of a stationary particle, in this thread.

 

[Tomsk]

Just curious. Prefer it to what? So far Dalespam is the only person to give a derivation of the acceleration of a stationary particle, in this thread.

To getting it from the lagrangian. I nearly did it but I was just saying the one with 4-vectors looks much easier.

 

[yuiop]

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.

I see you now want to move the goal posts. Back in #2 and #4 you started a derivation using the Lagrangian and "left it as an exercise" for me to complete. That normally implies that you know how complete the derivation starting with the Lagrangian and the equations of motion, but despite many requests you have been unable to do so and now you want to do it using covariant derivatives (already done by Dalespam) or using potentials. Now if you want to do it doing using potentials that is fine and I would be glad to see it because I have never aproached the problem from that angle, but for future reference, if you set people "exercises" you should be able to complete the exercise yourself.

 

[Tomsk]

I managed the lagrangian derivation, the answer I got was
\frac{d^2 r}{dt^2} = -\frac{m}{r^2}\alpha\left( 1 - \frac{3}{\alpha^2} \left( \frac{dr}{dt}\right)^2\right)
I think that agrees with the other results in this thread, depending on the initial velocity. Does that seem right to anyone? I am not quite sure how to interpret it though. This isn't a proper acceleration is it? So it only applies to free falling particles. But is it equal and opposite to the proper acceleration required to stay a fixed distance from the black hole as well?

 

[yuiop]

I managed the lagrangian derivation, the answer I got was
\frac{d^2 r}{dt^2} = -\frac{m}{r^2}\alpha\left( 1 - \frac{3}{\alpha^2} \left( \frac{dr}{dt}\right)^2\right)
I think that agrees with the other results in this thread, depending on the initial velocity. Does that seem right to anyone? I am not quite sure how to interpret it though. This isn't a proper acceleration is it? So it only applies to free falling particles. But is it equal and opposite to the proper acceleration required to stay a fixed distance from the black hole as well?

I too completed the "lagrangian derivation" and sent copies of my derivation to Dalespam and DrGreg to check over for me. I will post the derivation here in a day or two, but I thought I would first give Starthaus the opportunity to show that he is not full of "it" and show he can complete the derivation he started and left for us to complete as "an exercise".

 

[starthaus]

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^2{dt}{ds})-\frac{1}{\alpha}(\frac^2{dr}{ds})-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.

The correct Lagrangian (see above) is derived directly from the metric.
From the Lagrangian, you get the equation of motion (see above).

There is a third Euler-Lagrange equation, that I have left for last:

2\frac{d}{ds} (1/\alpha \frac{dr}{ds})-[\frac{2m}{r^2}(\frac{dt}{ds})^2-(\frac{dr}{ds})^2 \frac{d}{dr}(1/\alpha)-2r(\frac{d\phi}{ds})^2]=0

The above, for \frac{dr}{ds}=0 produces:

\omega^2=\frac{m}{r^3}

i.e. Kepler third law.

For a complete derivation of acceleration in rotating frames, see the other thread.

On the other hand, if the motion is radial, i.e. \frac{d\phi}{ds}=0 the equation of motion becomes:

2\frac{d}{ds} (1/\alpha \frac{dr}{ds})-[\frac{2m}{r^2}(\frac{dt}{ds})^2-(\frac{dr}{ds})^2 \frac{d}{dr}(1/\alpha)]=0

 

[yuiop]

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

You mean the correct result obtained by Dalespam that agrees with the result I gave in #1 right?

...

On the other hand, if the motion is radial, i.e. \frac{d\phi}{ds}=0 the equation of motion becomes:

2\frac{d}{ds} (1/\alpha \frac{dr}{ds})-[\frac{2m}{r^2}(\frac{dt}{ds})^2-(\frac{dr}{ds})^2 \frac{d}{dr}(1/\alpha)]=0

Your derivation is incomplete. You have not shown that you can apply the equations you have mindlessly copied from a textbook to a physical situation. You should be able to answer the questions posed in the OP and give answers in Newtons for force or meters/second^2 for acceleration.

... we should be be able to answer simple questions like "what does General Relativity predict that the weight of a 1kg mass on the surface of a very massive gravitational body to be?" or "what is the initial acceleration of 1kg mass when released from a short distance above a very massive gravitational body?"

I can answer those questions unambiguously. Can you?

 

[starthaus]

You mean the correct result obtained by Dalespam that agrees with the result I gave in #1 right?

Physics is not putting in results by hand. You need to learn how to derive them, kev. For this, you need to learn.

Your derivation is incomplete.

I am trying to teach you how to derive results (instead of putting them in by hand, as you have been doing), so I left the easy part for you, as an exercise. You still have no clue how to derive the equations of motion from the Lagrangian, eh?

 

[yuiop]

Your derivation is incomplete. You have not shown that you can apply the equations you have mindlessly copied from a textbook to a physical situation. You should be able to answer the questions posed in the OP and give answers in Newtons for force or meters/second^2 for acceleration. I can answer those questions unambiguously. Can you?

Physics is not putting in results by hand. You need to learn how to derive them, kev. For this, you need to learn.

I am trying to teach you how to derive results (instead of putting them in by hand, as you have been doing), so I left the easy part for you, as an exercise. ...

You are missing the point. As I said in post 79, I have already derived the equations and Dalespam and DrGreg hold copies of that derivation, as proof that I have done so. I have given you over a week to show you can complete your own "exercises" and prove yourself worthy of being "my teacher" and not just some guy trying to look clever by posting exercises from a textbook that he has no idea how to complete. So far you have not completed any of the exercises or derivations you have posted in any thread, that are not already completed in the textbooks you quote from. I am still waiting and so is Tomsk who would like to see the derivation.

 

[starthaus]

You are missing the point. As I said in post 79, I have already derived the equations and Dalespam and DrGreg hold copies of that derivation, as proof that I have done so. I have given you over a week to show you can complete your own "exercises" and prove yourself worthy of being "my teacher" and not just some guy trying to look clever by posting exercises from a textbook that he has no idea how to complete. So far you have not completed any of the exercises or derivations you have posted in any thread,

LOL

that are not already completed in the textbooks you quote from. I am still waiting and so is Tomsk who would like to see the derivation.

I don't think he's waiting on anything, he knows what to do.

 

[yuiop]

Starthaus: If you don't want want to post the derivation because you are worried that I will miss the learning opportunity of deriving for myself, then you have no need to worry because I have availed myself of the opportunity and done just that. You now no longer have any excuses not to publish the completion of the exercise you set and the I suspect the reason you are stalling, is that you do not know how to to complete your own exercise.

P.S. Tomsk clearly expressed an interest in #78.

 

[starthaus]

Starthaus: If you do want want to post the derivation because you are worried that I will miss the learning opportunity

No, I am not worried at all. It is clear that you don't know what to do with the Lagrangian.

 

[yuiop]

No, I am not worried at all. It is clear that you don't know what to do with the Lagrangian.

I have done the derivation from the initial information (What you call "the hard work" and what I call quoting from a textbook.) that you provided on the first page of this thread. It seems you want to move the goal posts again. All I am asking you to do is complete ANY of the exercises/derivations that have started and show you can end up with the equations given in #1.

 

[starthaus]

I have done the derivation from the initial information (What you call "the hard work" and what I call quoting from a textbook.)

No, you haven't. That would require knowledge of the Euler-Lagrange formalism of which you have none. Otherwise, you would have derived the equations of motion from the Lagrangian I derived for you long,long ago.

 

[yuiop]

No, you haven't. That would require knowledge of the Euler-Lagrange formalism of which you have none.

Yes I have and Dalespam and DrGreg have copies of my derivation to prove it. If you publish your derivation and it is different to mine I will learn from the different approach. If you show your derivation (if you are in fact able to do it) I will post my derivation and if it is wrong wrong or flawed, I will learn from that too. You claim your purpose here is to educate me, but so far I have seen no sign of you trying to be genuinely helpful and your posts are just confrontational and inflamatory as possible. Does the word "troll" mean anything to you?

 

[starthaus]

You claim your purpose here is to educate me,

Education doesn't mean doing all the homework for you, you need to learn how to do it yourself.

 

[starthaus]

your posts are just confrontational and inflamatory as possible. Does the word "troll" mean anything to you?

You mean like this:

Your derivation is incomplete. You have not shown that you can apply the equations you have mindlessly copied from a textbook to a physical situation.

:-)

 

[yuiop]

Education doesn't mean doing all the homework for you, you need to learn how to do it yourself.

I keep telling you I have already done "the homework" and two people on this forum have seen it. I am asking you to show you have the right to be a self proclaimed guru/teacher on this forum by showing you can complete your own homework. We can all say "the completion of the derivation is left as an exercise for the reader". This is your opportunity to show that you have the guru status you are trying to portray and are not just full of hot air.

 

[starthaus]

I keep telling you I have already done "the homework" and two people on this forum have seen it.

You mean, you finally cleaned up the mess from post 1? Good for you.

 

[starthaus]

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.

Ok, I see that you don't want to make the effort to learn the Euler-Lagrange formalism, so we can use the simple method I showed for reading the potential straight off the metric.

Here is the second line:

e^{2\Phi/c^2}=1-\frac{2m}{r}

You have one line to write in order to find the correct expression of the force.

 

[yuiop]

You mean, you finally cleaned up the mess from post 1? Good for you.

There you go, being all confrontational and aggressive again without looking into where and why we differ and seeing if in fact our results are in agreement and just expressed in different ways.It does not help that it has taken over 80 posts to discover what you think the results should be.

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.

Ok, I see that you don't want to make the effort to learn the Euler-Lagrange formalism, so we can use the simple method I showed for reading the potential straight off the metric.

Here is the second line:

e^{2\Phi/c^2}=1-\frac{2m}{r}

You have one line to write in order to find the correct expression of the force.

OK, straight up, I am not familiar with the formalism you are using, as I do not come across it very often. If you care to state the exact titles and ISBN's of the Gron and Rindler books (or the Amazon links), I will order them and at least then we will be using the same references and notation and perhaps we might understand each other better.

Now for your derivation.

The strong field aproximation metric is:

ds^2= \left(e^{\frac{2\Phi}{c^2}}\right)c^2dt^2 - \left(e^{\frac{2\Phi}{c^2 }}\right)^{-1}dr^2 - r^2(d \theta^2 - \sin^2 \theta d \phi^2)

and the Regular Schwarzschild metric is:

ds^2= \left(1-\frac{2GM}{rc^2}}\right)c^2dt^2 - \left(1-\frac{2GM}{rc^2 }}\right)^{-1}dr^2 - r^2(d \theta^2 - \sin^2 \theta d \phi^2)

From the above two forms of the metric it is easy to make the identity:

e^{2\Phi/c^2}=\left(1-\frac{2GM}{rc^2}\right)

Solve the above for \phi:

\phi = \frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right)

Now you state:

\vec{F}=-grad\Phi

(as does Wikipedia here http://en.wikipedia.org/wiki/Force#Potential_energy)

so:

\vec{F}= -grad\Phi = -grad\left(\frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right)\right)

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

Now your result for coordinate force does not agree with the equations I got for force in #1 quoted below:

The proper force acting on the stationary test mass on the surface of the gravitational body (i.e its weight as measured by a set of weighing scales on the surface) is:

F_0 = \frac{GMm_0}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} = m_0a_0

The Schwarzschild coordinate force as measured by an observer at infinity is:

F = \frac{GMm}{r^2}

Now before you jump on your "Starthaus is right and Kev is wrong" high horse I think we should explore why we get different results and see if we are saying the same thing in different ways. You should bear in mind that several posters have stated that the equations I gave in #1 agree with equations given by textbooks that they have. I am sure your derivation is also from a textbook, so it likely that both results are right but stated in different ways.

The first thing to observe is that your equation for force fails on dimensional analysis because it does not have the units of force. It would appear your equation has units of acceleration, but even then it does not agree with the equation I gave for coordinate acceleration in #1:

...the initial coordinate acceleration of a test mass released at r is (according to the observer at infinity) is:

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

I think when you fix your equation so that it has the correct units of force, rather than units of acceleration and identify who makes the measurements, then we might find some agreement.

 

[starthaus]

There you go, being all confrontational and aggressive again without looking into where and why we differ and seeing if in fact our results are in agreement and just expressed in different ways.It does not help that it has taken over 80 posts to discover what you think the results should be.
OK, straight up, I am not familiar with the formalism you are using, as I do not come across it very often. If you care to state the exact titles and ISBN's of the Gron and Rindler books (or the Amazon links), I will order them and at least then we will be using the same references and notation and perhaps we might understand each other better.

Now for your derivation.

The strong field aproximation metric is:

ds^2= \left(e^{\frac{2\Phi}{c^2}}\right)c^2dt^2 - \left(e^{\frac{2\Phi}{c^2 }}\right)^{-1}dr^2 - r^2(d \theta^2 - \sin^2 \theta d \phi^2)

and the Regular Schwarzschild metric is:

ds^2= \left(1-\frac{2GM}{rc^2}}\right)c^2dt^2 - \left(1-\frac{2GM}{rc^2 }}\right)^{-1}dr^2 - r^2(d \theta^2 - \sin^2 \theta d \phi^2)

From the above two forms of the metric it is easy to make the identity:

e^{2\Phi/c^2}=\left(1-\frac{2GM}{rc^2}\right)

Solve the above for \phi:

\phi = \frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right)

Now you state:

\vec{F}=-grad\Phi

(as does Wikipedia here http://en.wikipedia.org/wiki/Force#Potential_energy)

so:

\vec{F}= -grad\Phi = -grad\left(\frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right)\right)

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

Good, you finally did the computations. It is not "my result", it is the "correct result". Now, you need to think a little how you calculated the gradient, this will explain why you got the result you got.

Now your result for coordinate force does not agree with equations I got for force in #1 quoted below.

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. I did not mislead you, I guided you to discovering the correct results.

Now, if you could perform one more calculation (exactly one line) you would also get the correct expression for the proper force.

 

[starthaus]

The Schwarzschild coordinate force as measured by an observer at infinity is:

F = \frac{GMm}{r^2}

Nope, I just guided you through the correct derivation, results put in by hand are not physics. Besides, they are most often likely to be wrong.

 

[yuiop]

\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. I did not mislead you, I guided you to discovering the correct results.

Now, if you could perform one more calculation (exactly one line) you would also get the correct expression for the proper force.

I am afraid I have no idea of how to obtain the proper force from your coordinate force equation, as I am unable to relate to the physical meaning of your equation. It differs by orders of magnitude from my equations (which others in this forum say are correct). If you state what you and Rindler claim the proper force to be, (and how it is obtained) I will try and figure out why our equations differ.

 

[starthaus]

It differs by orders of magnitude from my equations (which others in this forum say are correct).

They aren't. The one that you posted in post 1 is wrong, the one that you derived following my hints is correct. You are just a few steps away from getting the correct results.

If you state what you and Rindler claim the proper force to be, (and how it is obtained) I will try and figure out why our equations differ.

I gave you a hint. You need to look at how you calculated the gradient, grad \Phi for \Phi = \frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right). How did you do it?

 

[yuiop]

I gave you a hint. You need to look at how you calculated the gradient, grad \Phi for \Phi = \frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right). How did you do it?

Force is the differential of potential so I took the differential with respect to r. Put another way, the differential of a curve is the gradiant of the curve. How does that help?