# Gravitational force and acceleration in General Relativity.

#### yuiop

Yeah, took only about 200 posts to convince you of your error.
Actually you have spent over 300 posts since post #2 moving the goal posts from where they were set in #1. This is known as derailing the thread and is against the rules of this forum. However I was also interested in the more general result so I let it go, but you seem to have a knack for turning what could be said in in 30 posts into 300 posts by being secretive, confrontational and evasive.

Having said that, I have learnt a lot in thread and I think others that have the stamina to read through all the background noise of the bickering ( like Geigerclick ) have found it instructive too. You too have learnt some things (or you should of) such as that the Schwarzschild coordinate acceleration is not greater than proper acceleration and that a free falling particle does not have non-zero proper acceleration.

#### starthaus

Actually you have spent over 300 posts since post #2 moving the goal posts from where they were set in #1. This is known as derailing the thread and is against the rules of this forum. However I was also interested in the more general result so I let it go, but you seem to have a knack for turning what could be said in in 30 posts into 300 posts by being secretive, confrontational and evasive.
No, I told you from post 2 that you didn't know what you were talking about claiming that $$a_0=a\gamma^3$$

#### yuiop

No, I told you from post 2 that you didn't know what you were talking about claiming that $$a_0=a\gamma^3$$
It was valid in the context of #1 but I have conceded that it is not valid in the more general case (outside the context of the OP).

#### starthaus

It was valid in the context of #1 but I have conceded that it is not valid in the more general case (outside the context of the OP).
Yes, it took you up to post 324 or so and even after I showed you the calculus, you were still arguing.
Now, the formula for coordinate acceleration in post 1 is also incorrect, I showed you how to derive it some time ago. Agreed?

#### yuiop

Now, the formula for coordinate acceleration in post 1 is also incorrect, I showed you how to derive it some time ago. Agreed?
I was not aware there was still any dispute about that. I still stand by what I said in #1 in the context that it was given in i.e:

... 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 the coordinate acceleration of a free falling particle when it is at its apogee at r. Rofle2, myself and DaleSpam all agreed it is correct and Dalespam did a thorough derivation using four vectors and Rofle2 did a thorough job of showing how it agrees with the equations given by Kevin Brown and also demonstrated that the derivation given in mathpages by Prof. Brown are pefectly valid (something you disputed). We have even shown you how your own equations agree with coordinate acceleration equation given in #1. I do not see what there is left to dispute.

#### starthaus

I was not aware there was still any dispute about that. I still stand by what I said in #1 in the context that it was given in i.e:
This is the coordinate acceleration of a free falling particle when it is at its apogee at r. Rofle2, myself and DaleSpam all agreed it is correct and Dalespam did a thorough derivation using four vectors
No, Dalespam and I have worked on deriving the proper (not coordinate) acceleration at apogee by using covariant derivatives. I used the same method as one of the ways to calculate proper acceleration in rotating frames.

and Rofle2 did a thorough job of showing how it agrees with the equations given by Kevin Brown and also demonstrated that the derivation given in mathpages by Prof. Brown are pefectly valid (something you disputed). We have even shown you how your own equations agree with coordinate acceleration equation given in #1. I do not see what there is left to dispute.
For the limited context yes. For the general case , no.
This is the problem with cutting and pasting from websites instead of deriving.
Well, at least now you know how to derive the general formula for coordinate acceleration, correct?
You also know not to try to get the proper acceleration by multiplying the coordinate acceleration by $$\gamma^3$$, correct?

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#### yuiop

No, Dalespam and I have worked on deriving the proper (not coordinate) acceleration at apogee by using covariant derivatives. I used the same method as one of the ways to calculate proper acceleration in rotating frames.
This is not true. DaleSpam calculated the coordinate acceleration of a momentarily stationary particle at r, in post #155 quoted here:

OK, but the truth is that I am an expert in Mathematica , not 4-vectors. I will use the metric, Christoffel symbols, etc. given https://www.physicsforums.com/showpost.php?p=2712746&postcount=38". The worldline of a particle moving only in the radial direction in these coordinates is given by:
$$\mathbf X = (t,r,0,0)$$

Where r is now a function of t.

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

Where the ' denotes differentiation wrt the coordinate t.

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 it gets ugly. We can derive the four-acceleration as follows:
$$A^{\mu}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda} =$$
$$\left(\frac{c^2 r' \left(c^2 R (R-r)^2+2 r^3 (r-R) r''-3 r^2 R \left(r'\right)^2\right)}{2 \left(c^2 (R-r)^2-r^2 \left(r'\right)^2\right)^2},\frac{c^4 (R-r)^2 \left(c^2 R (R-r)^2+2 r^3 (r-R) r''-3 r^2 R \left(r'\right)^2\right)}{2 \left(c^2 r (R-r)^2-r^3 \left(r'\right)^2\right)^2},0,0\right)$$

So, we can set all of these equal to 0 to get the geodesic equation describing an object in free-fall along a radial trajectory. Doing so and solving for r'' gives
$$r''=\frac{c^2 R (R-r)^2-3 r^2 R \left(r'\right)^2}{2 r^3 (R-r)}$$

For a momentarily stationary particle r'=0 so
$$r''=\frac{GM}{r^2}\left(\frac{2GM}{c^2r}-1 \right)$$

where I have substituted back in for R.
....
DaleSpams formula agrees with the one I gave in #1 and not with the formula you gave in #96 quoted here:

... 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. ...
Are you still defending this formula that you incorrectly extrapolated from a Rindler text?

We have already established that Rindler never directly gave any equation for the coordinate acceleration.

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#### starthaus

This is not true. DaleSpam calculated the coordinate acceleration of a momentarily stationary particle at r, in post #155 quoted here:
I was refering to post 38 (the derivation of proper acceleration at apogee)

$$a_0=-\frac{m/r_0^2}{\sqrt{1-2m/r_0}}$$

Are you still defending this formula that you incorrectly extrapolated from a Rindler text?
It was a good attempt at teaching you how to derive (instead of copying) formulas. It worked fine for teaching you how to derive the proper acceleration, right?
. I have derived the complete formula for coordinate acceleration, it is in the blog, you even quoted it a few times.

$$a=\frac{m}{r^2}(1-2m/r)(2-3\frac{1-2m/r}{1-2m/r_0})$$

At apogee it reduces to the formula derived via covariant differentiation in post 155. The covariant method cannot calculate the general formula for coordinate acceleration, the lagrangian method I developed does. Are we on the same page now?

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#### yuiop

That is proper acceleration, you seem not to know what covariant differentiation produces.
You are wrong. DaleSpan is answering the question "What is the coordinate accelertion?" https://www.physicsforums.com/showpost.php?p=2734621&postcount=155"as you will see if you read it again.

The derivation of proper time is given by DaleSpam in #38 quoted here:
Finally, the norm of the four-acceleration, which is equal to the magnitude of the proper acceleration, is given by:
$$-||\mathbf A||^2= -A_{\mu} A^{\mu}= -g_{\mu\nu} A^{\nu} A^{\mu} = \frac{c^4 R^2}{4 r^4 (1-R/r)} = \left( \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} \right)^2$$
and this agrees with the equation for proper acceleration given in #1.

I have derived the complete formula for coordinate acceleration, it is in the blog, you even quoted it a few times.

$$a=\frac{m}{r^2}(1-2m/r)(2-3\frac{1-2m/r}{1-2m/r_0})$$
Several people have already shown you that if a particle is at apogee and $r= r_o$ then your equation reduces to

$$a=\frac{m}{r^2}(1-2m/r)$$

in total agreement with #1 and yet you say the formual in #1 is wrong??

Your blog formula does not agree with the coordinate acceleration formula you claimed you derived from the Rindler text using potentials in #96.

Do you now agree that your derivation is #96 is wrong?

Clearly either your blog formula or your #96 formula is wrong, as they do not agree.

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#### starthaus

...which you "derived" by multiplying the coordinate acceleration by $$\gamma^3$$

Duh, if you make $$r=r_0$$ in the general formula, of course you get the correct formula for coordinate acceleration at apogee. The point is that you only got the coordinate acceleration at apogee correct in post 1.

Your blog formula does not agree with the coordinate acceleration formula you claimed you derived from the Rindler text using potentials in #96.
Of course it doesn't. That was an attempt to give you a dumbed down derivation since you couldn't follow the derivation based on lagrangian mechanics. So, I gave you a dumbed down version (that gives the correct answer in the case of proper acceleration) since it was clear that trying to teach you lagrangian mechanics was hopeless. How many times do you need to have this explained to you?

#### yuiop

...
Duh, if you make $$r=r_0$$ in the general formula, of course you get the correct formula for coordinate acceleration at apogee. The point is that you only got the coordinate acceleration at apogee correct in post 1.
Fine that's settled. You have spent 300 odd posts trying to say post #1 is wrong, but now you concede that post #1 is correct in the narrow context that it was given in.

#### starthaus

Fine that's settled. You have spent 300 odd posts trying to say post #1 is wrong, but now you concede that post #1 is correct in the narrow context that it was given in.
No, it isn't. The only correct thing is that it took up to post 324 for you to understand why you can't use the $$a_0=a\gamma^3$$ hack.

#### Altabeh

No, Dalespam and I have worked on deriving the proper (not coordinate) acceleration at apogee by using covariant derivatives. I used the same method as one of the ways to calculate proper acceleration in rotating frames.

For the limited context yes. For the general case , no.
This is the problem with cutting and pasting from websites instead of deriving.
Well, at least now you know how to derive the general formula for coordinate acceleration, correct?
You also know not to try to get the proper acceleration by multiplying the coordinate acceleration by $$\gamma^3$$, correct?
And you also know how to use covariant derivatives to define "proper acceleration" in the context of GR from now on, correct?

In the page 39 of the http://books.google.com/books?id=slA4CFfIpHAC&lpg=PA39&ots=eGVWMXPMag&dq=acceleration in free fall vanishes&pg=PA39#v=onepage&q&f=false" "General relativity and the Einstein equations" by Yvonne Choquet-Bruhat, we can see that author says:

"...It is always possible to choose local coordinates such that Christoffel symbols vanish at that point; gravity and relative accelerations are then, at that point, exactly balanced. It is even possible to choose local coordinates along one given geodesic*: astronauts have made popular the fact that in free-fall one feels neither acceleration nor gravity; in a small enough neighborhood of a geodesic the relative accelerations of objects in free-fall are approximately zero. "

This is strongly confirmed in some other book, Papapetrou's Lectures on GR, p. 57, as he puts it this way:

"In a Riemannian space we can always make the Christoffel symbols vanish on a
given curve by an appropriate coordinate transformation. According to the principle
of equivalence this property of the Riemannian space should mean physically that the
sum of the inertial and the gravitational acceleration** can be made equal to zero on the
given curve. This is really the case, as we see at once if we consider a freely falling lift
or a freely moving non-rotating artificial satellite: An observer inside the satellite,
using the frame which is connected rigidly with the satellite - a comoving frame -,
will observe neither inertial non gravitational accelerations."

By all of this stuff we are to back up the idea that proper acceleration is zero along any geodesic and what you're taking as a cheap guess at the riddle "what is proper acceleration!?" leads to nowhere but to where you can possibly get stuck in the mud of nonsense beliefs. Mathematically, these quotes are hinting at the fact that in the geodesic equation,

$$\frac{d^2x^a}{ds^2}+\Gamma^{a}_{bc}\frac{dx^b}{ds}\frac{dx^c}{ds}=0$$

the first term $$\frac{d^2x^a}{ds^2}$$ represents the total acceleration due to both gravity and inertial forces, not as that nonsense in your article stands for "the proper acceleration". It is apparent that if $$\Gamma^{a}_{bc}=0$$, then this total acceleration vanishes, leading to the obvious fact that we don't have any kind of proper acceleration felt by the freely falling object because now the first term is simply the proper acceleration of freely falling object as in Minkowski spacetime. So applying a Fermi-like transformation or using Riemann normal coordinates to make the Christoffel symbols vanish is just a clue on the reduction of the total proper acceleration to the famous term $$d^2x^a/ds^2=0$$ in SR. Without this transformation, the two terms in geodesic equation have to cancel out each other in order for the proper acceleration to vanish.

-------------------------------------------------------------------
*This kind of coordinate system is called "Fermi normal coordinates" that gives the coordinate conditions by which the metric tensor is to be equivalent to $$\eta_{\mu\nu}$$ along an entire geodesic. In fact, this coordinate system would make the local-flatness happen to exist along a geodesic instead of forcing us to believe the old cliche of local-flatness occurring only exactly at some given point and approximately in the vicinity of it (Riemann normal coordinates). For a complete discussion of this along any time-like geodesic see, for example, An advanced course in GR by E. Poisson, p. 14.

** Earlier on the page 56, Papapetrou makes it fully clear what role Christoffel symbols are exactly playing in the geodesic equations:

"More exactly, the Christoffel symbols will now describe, according to the principle of equivalence, the sum of the inertial and the gravitational acceleration."

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#### starthaus

And you also know how to use covariant derivatives to define "proper acceleration" in the context of GR from now on, correct?
What do you mean "from now on"? Are you trolling? Drop the attitude and the colored text and try to stick to physics.

#### Altabeh

What do you mean "from now on"? Are you trolling? Drop the attitude and the colored text and try to stick to physics.
Nonsense! You're the one who has been trolling for like 2 straight weeks and now that is time for the "teacher" to confess to what I can confidently call a huge error about a 100% physical stuff!! Don't speak of physics in case you're making nonsense! LOL

#### starthaus

Nonsense! You're the one who has been trolling for like 2 straight weeks and now that is time for the "teacher" to confess to what I can confidently call a huge error about a 100% physical stuff!! Don't speak of physics in case you're making nonsense! LOL
So, you are trolling. Fine, I'll ignore you.

#### Geigerclick

I was refering to post 38 (the derivation of proper acceleration at apogee)

$$a_0=-\frac{m/r_0^2}{\sqrt{1-2m/r_0}}$$

Wow, the "teacher" now is making nonsense again. What about you copying Schwartzchild metric and all Euler-Lagrange equations from textbooks? Trying to seal off the letter of another mistake would not erase the wrong equations wrapped in.

I have derived the complete formula for coordinate acceleration, it is in the blog, you even quoted it a few times.

Did you really get this "huge" achievement on your own, "teacher"? Writing a bunch of basic equations based on an even more basic calculus doesn't have anything worthy of debate in like 23 pages!

AB
Now that is a pretty good point. I don't see how your (starthaus') position has not become a bit mad. Maybe this thread needs to take a break. There is some great material here, especially for a novice such as myself, but this devolving into a pissing match is not doing anyone favors. Listen to Altabeh, he seems like the calmest fellow in the room so to speak.

A

#### Al68

Now, the formula for coordinate acceleration in post 1 is also incorrect, I showed you how to derive it some time ago. Agreed?
Still claiming post 1 is wrong about coordinate acceleration? Oh, wait:
The point is that you only got the coordinate acceleration at apogee correct in post 1.
So, post 1 was correct about coordinate acceleration? I just reread post 1 and it contained no equation for coordinate acceleration other than specifically at apogee.

So, I'll ask again: Do you still thing post 1 is incorrect in any way?

#### starthaus

Still claiming post 1 is wrong about coordinate acceleration? Oh, wait:
So, post 1 was correct about coordinate acceleration? I just reread post 1 and it contained no equation for coordinate acceleration other than specifically at apogee.
-Yet, the equation is invalid for any other point but the apogee.
-Everywhere else another equation applies.
-Yet, the relationship $$a_0=a\gamma^3$$ has also been shown to be false., so it can't be used (as kev attempted) to derive proper acceleration from coordinate acceleration

So, I'll ask again: Do you still thing post 1 is incorrect in any way?
Can you read above? Of course, given your continued trolling in this thread, I don't expect you to admit that post 1 has been shown invalid.

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A

#### Al68

-Yet, the equation is invalid for any other point but the apogee.
-Everywhere else another equation applies.
-Yet, the relationship $$a_0=a\gamma^3$$ has also been shown to be false., so it can't be used (as kev attempted) to derive proper acceleration from coordinate acceleration

Can you read above? Of course, given your continued trolling in this thread, I don't expect you to admit that post 1 has been shown invalid.
Again, you are saying things that don't show post 1 to be incorrect, then claiming post 1 to be invalid.

Are you under the impression that an equation is invalid unless its derivation is properly shown?

Are you under the impression that an equation must apply to cases other than the one specified in order to be valid?

Is there some other reason for your incoherent and bizarre statements?

#### starthaus

Again, you are saying things that don't show post 1 to be incorrect, then claiming post 1 to be invalid.

Are you under the impression that an equation is invalid unless its derivation is properly shown?

Are you under the impression that an equation must apply to cases other than the one specified in order to be valid?

Is there some other reason for your incoherent and bizarre statements?
What in "$$a_0=a\gamma^3$$ is false and any derivation based on a false premise is invalid" don't you understand?

A

#### Al68

What in "$$a_0=a\gamma^3$$ is false and any derivation based on a false premise is invalid" don't you understand?
You have already admitted that that equation was true for the specific case referred to in post 1, and post 1 contained no derivation based on it, anyway.

In fact, post 1 contains no derivations, so claiming post 1 is wrong because some derivation is invalid is just illogical nonsense.

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