Gravitational force and acceleration in General Relativity.

[yuiop]

I was pondering this thread https://www.physicsforums.com/showthread.php?t=401713" when it occurred to me that 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?".

Not being able to find a clear, simple definitive answer to that question on the internet, I decided to do some "back of the envelope" calculations and this is what I came up with (using a combination of hunches, intuition and guesses):

Assuming Schwarzschild geometry (uncharged non-rotating body) and defining the gravitational gamma factor as:

\gamma = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}

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)

The initial acceleration of the test mass when released at r, as measured by a local observer at r is:

a_0 = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} = a \gamma^3

This is equal to the proper acceleration of the test mass when it is stationary in the gravitational field at r, as measured by an accelerometer.

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}

Note that:

a_0 = a\gamma ^3

and:

F_0 = F = m_0 a_0

just as in the linear analogue in Special Relativity for longitudinal acceleration and force.

Does that seem about right? Can anyone find any solutions in "the literature"?

P.S. I have not shown my working or referenced any sources that I based my calculations on, but I can if anyone is interested.

 

[starthaus]

I was pondering this thread https://www.physicsforums.com/showthread.php?t=401713" when it occurred to me that 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?".

Not being able to find a clear, simple definitive answer to that question on the internet, I decided to do some "back of the envelope" calculations and this is what I came up with (using a combination of hunches, intuition and guesses):

Assuming Schwarzschild geometry (uncharged non-rotating body) and defining the gravitational gamma factor as:

\gamma = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}

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)

This is very wrong. You need to start with the metric:ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2

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

From this you construct the Lagrangian:

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

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

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

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

whre h,k are constants.

 

[yuiop]

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


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

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

From this you construct the Lagrangian:

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

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

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

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

whre h,k are constants.

and what do you conclude that the coordinate and proper acceleration should be?

 

[starthaus]

and what do you conclude that the coordinate and proper acceleration should be?

I derived for you the equations of motion, I left this as an exercise for you to calculate.

 

[yuiop]

I derived for you the equations of motion, I left this as an exercise for you to calculate.

I have shown my conclusions, so why don't you show your conclusions and then we will see where we differ? That is of course assuming you know how to calculate the proper and coordinate acceleration from the Lagrangian.

You may not have noticed, but my original post is about particles that are stationary or very nearly stationary (eg a free falling particle near its apogee). I do not know why you always need to ovecomplicate things by introducing things that were not in the original post like horizontal/orbital motion. To get to the crux of the matter, you need to simplify things as much as possible, so set d\phi = 0 and forget about it.

Remember I said in the OP that I am looking for a clear simple answer. If you have not got one, you might as well stay out of this thread. There are plenty of general solutions that cover every possible permutation of variables, but I am trying to cut out the crap. I suggest you do the same.

Remember Schwarzschild found the first correct solution to the EFEs by assuming charge = 0 and angular momentum of the gravitational body = 0. That is the way to go. Take the simplest case first.

 

[Ich]

Not being able to find a clear, simple definitive answer to that question on the internet...

You should have asked your https://www.physicsforums.com/showthread.php?p=1807379#post1807379".

 

[starthaus]

That is of course assuming you know how to calculate the proper and coordinate acceleration from the Lagrangian.

You don't calculate it from the Lagrangian, you calculate it from the equations of motion. It is a simple exercise in calculus, this is why I left it for you. I did all the heavy lifting.

You may not have noticed, but my original post is about particles that are stationary or very nearly stationary (eg a free falling particle near its apogee). I do not know why you always need to ovecomplicate things by introducing things that were not in the original post like horizontal/orbital motion. To get to the crux of the matter, you need to simplify things as much as possible, so set d\phi = 0 and forget about it.

It is not correct to "set d\phi = 0 and forget about it. "The simple reason is that \phi is variable

Remember I said in the OP that I am looking for a clear simple answer. If you have not got one, you might as well stay out of this thread. There are plenty of general solutions that cover every possible permutation of variables, but I am trying to cut out the crap. I suggest you do the same.

You are becoming abusive every time you are shown wrong.

 

[starthaus]

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. You can do the same exercise for the angular coordinate , \phi.

 

[DrGreg]

Prof Nick Woodhouse of Oxford University shows http://people.maths.ox.ac.uk/~nwoodh/gr/index.html (Section 12.1) that the "acceleration due to gravity", i.e. the proper acceleration of a hovering object, is

\frac{M}{r^2\sqrt{1 - 2M/r}}​

(in units where G=c=1) although the method he uses requires you to be familiar with covariant differentiation.

 

[LukeD]

We did this calculation in my GR class last semester. It took me a bit of time to find my book, but as I suspected (the answer is given on p.261 of Hartle with other relevant formulas scattered throughout the chapters), every formula that you wrote is exactly correct.

You are becoming abusive every time you are shown wrong.

Spoken with the certainty of someone who didn't bother to check if there were any differences between what you two wrote! There aren't of course. You're both absolutely correct. (You may have missed where kev said that he wasn't reproducing his work)
By the way, the Lagrangian is independent of angular momentum, so by conservation of angular momentum (your h), we can in fact just set \mtext{d}\phi = 0. This is because if we start on a geodesic with \frac{\mtext{d}\phi}{\mtext{d}\tau} = 0, this remains true for all time. It is a boundary condition on our motion that imposes no extra forces.

 

[starthaus]

By the way, the Lagrangian is independent of angular momentum, so by conservation of angular momentum, we can in fact just set \mtext{d}\phi = 0.

If you do that, you won't get the second equation of motion:

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

 

[starthaus]

Spoken with the certainty of someone who didn't bother to check if there were any differences between what you two wrote!

I checked, the two solutions produce different answers. As to rudeness, look at his tone.

(You may have missed where kev said that he wasn't reproducing his work)

I didn't miss anything.

 

[yuiop]

You should have asked your https://www.physicsforums.com/showthread.php?p=1807379#post1807379".

LOL, your right. It seems I end up back where I started every couple of years. Also, I was hoping for a more authorative second opinion than myself! At the time of that old thread I felt the conclusions were inconclusive, but looking back at the old thread all the information is there and I simply had not made sufficient connections to totally convince myself I was on the right track. I think I have a clearer picture now (hopefully).

One thing that always confused me was this statement from http://www.mathpages.com/rr/s6-04/6-04.htm :

At the apogee r = R where dr/dt = 0 the quantity in the square brackets is unity and this equation reduces to

\frac{d^2r}{d\tau^2} = \frac{GM}{r^2} \; \; \; (6)

This is a measure of the acceleration of a static test particle at the radial parameter r.

I always took that to mean that the proper acceleration of the static test particle is given by (6) but with hindsight I now see that they do not claim that and the correct equation is given later in http://www.mathpages.com/rr/s7-03/7-03.htm :

However, this acceleration is expressed in terms of the Schwarzschild radial parameter r, whereas the hovering observer’s radial distance r' must be scaled by the “gravitational boost” factor, i.e., we have dr' = dr/(1-2m/r)^(1/2). Substituting this expression for dr into the above formula gives the proper local acceleration of a stationary observer

\frac{d^2r'}{d\tau^2} \;\; = \;\; \frac{GM}{r^2} \frac{1}{\sqrt{1-2GM/r}}

This value of acceleration corresponds to the amount of rocket thrust an observer would need in order to hold position, and we see that it goes to infinity as r goes to 2m.

One thing I would quibble over is that the above equation proves conclusively that the proper acceleration becomes infinite at the event horizon. The acceleration equation can be written in a more general way as :

a \;\; = \;\; \frac{GM}{r^2} \frac{(1-2GM/r)}{(1-2GM/r_o)^{3/2}}

where r_o is the location of the observer and r is the location of the stationary particle. Setting r_o = \infty gives the coordinate acceleration and setting r_o = r gives the proper acceleration. When r = 2GM the proper acceleration is then:

a_0 \;\; = \;\; \frac{GM}{r^2} \frac{0}{0}

which is indeterminate.

 

[starthaus]

When r = 2GM/r the proper acceleration is then:

a_0 \;\; = \;\; \frac{GM}{r^2} \frac{0}{0}

which is indeterminate.

I think you meant r=2GM, not r=2GM/r.
The more important thing is that it is easy to prove that physically possible solutions for the equations of motion I derived exist only for r>3GM (or, in my notation , r>3m), so the above is a non-issue. You can never get infinite or indeterminate proper acceleration.

 

[yuiop]

I checked, the two solutions produce different answers. As to rudeness, look at his tone.

The tone you detect is my irritation and frustration at your habit of saying my conclusions are wrong without saying why they are wrong or what your conclusions are.

You always expect me to draw your conclusions for you "as an exercise". That would end up in "circular criticism". If I draw your conclusions for you and then say your conclusions are wrong I would infact be criticising my own conclusions. For me to draw your conclusions for you, requires me to read your mind and that is very difficult at the best of times and even more so over the internet when you may be thousands of miles away and I can not see your facial expressions or body language.

So why not state what YOUR conclusions are and maybe we can have a sensible conversation?

Here is an attempt to read your mind:

From the first equation, we can get immediately:

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

You think the above is the relationship between the coordinate time and the proper time of a stationary particle. Your exercise is to show why the above is not applicable to a stationary particle. For a hint, see my last post.

If that is not what you are thinking, then like I said I am not a mind reader. Why not simply state what you are thinking?

 

[yuiop]

I think you meant r=2GM, not r=2GM/r.
The more important thing is that it is easy to prove that physically possible solutions for the equations of motion I derived exist only for r>3GM (or, in my notation , r>3m), so the above is a non-issue. You can never get infinite or indeterminate proper acceleration.

Your right about the typo. I should have wrote r = 2GM and have now corrected it that post. Thanks. I usually write r = 2GM/c^2 but we are now using units of c=1.

The equations of motion are valid for r>2m but there are no circular orbits between r=2m and r=3m.

 

[starthaus]

The tone you detect is my irritation and frustration at your habit of saying my conclusions are wrong without saying why they are wrong or what your conclusions are.

I showed you why your solution was wrong, you can't "fiddle" or "guess" the correct solutions, you need to derive them from base principles. I have done it in the past, every time I have pointed out your errors.

 

[yuiop]

I showed you why your solution was wrong. I have done it in the past, every time I have pointed out your errors.

If you are referring to this thread https://www.physicsforums.com/showthread.php?t=397403&page=11 then it clear that atyy, George Jones and Dalespam showed you were wrong and no one supported your argument. If fact Dalespam gave a lengthy, rigorous and very impressive demonstration that you were wrong starting at #165 of that thread and ending at #169.

Actually most of the stuff you say is not incorrect in itself, but for some reason you always seem to be talking about a different subject to the subject of thread and what everyone else is talking about. You are doing that in this thread too.

 

[yuiop]

I showed you why your solution was wrong, you can't "fiddle" or "guess" the correct solutions,

You can if you are lucky and it looks like I was.

Actually it wasn't luck. I just refer to conclusions that are rigorously derived by other people that know what they are doing.

 

[starthaus]

If you are referring to this thread https://www.physicsforums.com/showthread.php?t=397403&page=11 then it clear that atyy, George Jones and Dalespam showed you were wrong and no one supported your argument. If fact Dalespam gave a lengthy, rigorous and very impressive demonstration that you were wrong starting at #165 of that thread and ending at #169.

I knew you were going to go there. I don't know why you would do that since there were so many calculus errors that you made in that thread. As to the proof produced by Dalespam, I repeatedly pointed out the error in his approach so, in the end, we agreed to disagree. If you want to discuss more on the subject of transformation of centripetal acceleration in SR, we can do it in the appropriate thread.

 

[LukeD]

One thing that always confused me was this statement from http://www.mathpages.com/rr/s6-04/6-04.htm :

At the apogee r = R where dr/dt = 0 the quantity in the square brackets is unity and this equation reduces to

<br /> \frac{d^2r}{d\tau^2} = \frac{GM}{r^2} \; \; \; (6)<br />

This is a measure of the acceleration of a static test particle at the radial parameter r.I always took that to mean that the proper acceleration of the static test particle is given by (6) but...

Right, this is the radial acceleration of our test particle as measured against the coordinate r. But this isn't the acceleration measured by our test particle. It's what we'd get if we parallel transported the acceleration vector all the way to infinity and then asked the observer at infinity how much acceleration he would measure if he had that acceleration vector.

To get the acceleration measured by our observer at r, you need to measure the acceleration vector in the orthonormal basis used by an observer at r. So the observed acceleration is
\frac{d^2 x^\alpha}{d\tau^2} \cdot \hat e^r
Where |\hat e^r|^2=1 and \frac{d x^a}{d\tau} \cdot \hat e^r = 0. e^r is directed toward larger r value.

It is very important to remember that formula for the dot product involves the metric.
Because our observer is at constant r, just like our observer at infinity, we can choose \hat e^r to be in the same direction as the observer @ infinity's radial vector, but
\hat e^r has a different length
(remember, a length of a vector is only defined at a point in spacetime. If we put both of these vectors at the same point, then e^r is bigger by a factor of (1-2M/R)^(-1/2))

 

[yuiop]

I knew you were going to go there. I don't know why you would do that since there were so many calculus errors that you made in that thread. As to the proof produced by Dalespam, I repeatedly pointed out the error in his approach so, in the end, we agreed to disagree. If you want to discuss more on the subject of transformation of centripetal acceleration in SR, we can do it in the appropriate thread.

In that thread it took over 150 posts and a lot of bickering to play the "guess what Starthaus is thinking" game when you could have simply stated what you were thinking in one or two posts and now it seems you want to play that game again in this thread.

So far, in our new "guess what Starthaus is thinking" game, I have guessed in #15 that you are thinking that the relationship between the proper time and the coordinate time of a stationary particle is given by:

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

and you have not denied it. Right there is a mistake in your thinking.

 

[Dale]

As to the proof produced by Dalespam, I repeatedly pointed out the error in his approach so, in the end, we agreed to disagree.

We certainly can go back and do this again if you wish. My derivation and my results are correct, the "error" you pointed out comes straight from the reference (Gron) that you agreed to use, and your error lies in not knowing what proper acceleration is.

 

[LukeD]

I checked, the two solutions produce different answers. As to rudeness, look at his tone.

If you got different answers by the 2 methods, then you were doing something wrong. I've checked kev's original answer with a few different sources as well as with my own work, and he made no mistakes at all. I also checked your formulas (the few that you wrote), and they were going in the right direction. You never got very far with them though, and your later comments suggest to me that you probably don't know the correct way to proceed.

In particular, you mention dr/ds as being the proper speed, but this is incorrect. The observer at r has to deal with length contraction in addition to the time dilation that you handled. He measures a dr'/ds, not dr/ds.
Working with 4-vectors instead of the individual coordinates leads to a much cleaner treatment that allows you to transform between observers' measurements easily.

Attacking someone and insisting that they are wrong just because they didn't show their work (which he said he would if you asked) usually doesn't get you a nice response...

Anyway, the question's been cleared up..

 

[starthaus]

In that thread it took over 150 posts and a lot of bickering

It is not my problem that you had so much trouble with getting a few basic derivatives correct <shrug>. So many errors in your math makes it so painful teaching you.

 

[starthaus]

We certainly can go back and do this again if you wish. My derivation and my results are correct, the "error" you pointed out comes straight out of the published literature of two different sources,

I can add a very nice derivation from Moller that supports my point and contradicts yours to the one from Rindler.

and your error lies in not knowing what proper acceleration is.

Nothing productive will come out of this.

 

[starthaus]

If you got different answers by the 2 methods, then you were doing something wrong. I've checked kev's original answer with a few different sources as well as with my own work, and he made no mistakes at all. I also checked your formulas (the few that you wrote), and they were going in the right direction. You never got very far with them though, and your later comments suggest to me that you probably don't know the correct way to proceed.

In particular, you mention dr/ds as being the proper speed, but this is incorrect. The observer at r has to deal with length contraction in addition to the time dilation that you handled. He measures a dr'/ds, not dr/ds.

How do you "manufacture" dr' out of dr? In a rigorous way, not by putting in a scaling factor by hand?

Working with 4-vectors instead of the individual coordinates leads to a much cleaner treatment that allows you to transform between observers' measurements easily.

Agreed but all you have are the equations of motion that I derived correctly.

Attacking someone and insisting that they are wrong just because they didn't show their work (which he said he would if you asked) usually doesn't get you a nice response...

There is no work to be shown, the different expressions were put in by hand. You do not know but this is a repeat of another thread on the same exact subject where kev did the same thing, put in the results by hand. It took a painful 150 posts to guide him through a proper derivation.

 

[yuiop]

It is not my problem that you had so much trouble with getting a few basic derivatives correct <shrug>. So many errors in your math makes it so painful teaching you.

I admitted in that thread that I am not a mathematician and my calculus is lousy. Maybe you get a kick out of getting me to say that again and again. Your calculus might be good, but you are lousy at understanding the physical interpretation of the equations.

We did this calculation in my GR class last semester. It took me a bit of time to find my book, but as I suspected (the answer is given on p.261 of Hartle with other relevant formulas scattered throughout the chapters), every formula that you wrote is exactly correct.

Thanks Luke

 

[starthaus]

I admitted in that thread that I am not a mathematician and my calculus is lousy. Maybe you get a kick out of getting me to say that again and again. Your calculus might be good, but you are lousy at understanding the physical interpretation of the equations.

Physics is about deriving results, not putting them in by hand . For that, you need to know math. As to my physical interpretation, it comes through the ability to derive results rather than cobbling them from websites.

 

[Dale]

Nothing productive will come out of this.

Well, at least we agree on that.

I can add a very nice derivation from Moller that supports my point and contradicts yours to the one from Rindler.

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 a coordinate acceleration instead of the proper acceleration.

 

[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?