Gravitational force and acceleration in General Relativity.

[starthaus]

So, now you're saying that a particle in free-fall does not follow a timelike geodesic?

This is not what I am saying. Do you have some comprehension problem?

 

[Rolfe2]

The point I made in the previous post that all methods obtain the same expression for proper acceleration at the apogee.

No, you're confused. The expression -m/r^2 [1/sqrt(1-2m/r)] does not represent the proper acceleration of a free-falling particle at the apogee, it represents the proper acceleration of a _stationary_ particle at the Schwarzschild radial coordinate r. If you're talking about the proper acceleration of a particle on a free radial trajectory, that is obviously zero, regardless of whether it is at the apogee or at any other point on its trajectory.

 

[starthaus]

As I cautioned you several posts ago, there are infinitely many possible coordinate systems that can be defined on a manifold, whereas there is a unique proper acceleration for a stationary particle in a gravitational field, so you ought to be asking yourself how a definition of proper acceleration in terms of [arbitrary] coordinates can yield a unique coordinate-independent result. The obvious answer is that, regardless of what coordinates we choose, we operate on them with the corresponding metric coefficients (different for each choice of coordinates), and those coefficients essentially convert the arbitrary coordinate measures into the unique proper measures, which are the basis of the proper acceleration.

I think that you are are clearly confused on the subject. Of course you will get different results if you define proper acceleration using different mathematical expressions. If you define it as :

$$a_0=\frac{d^2r}{d\tau^2}$$

or $$a_0=\frac{d^2\rho}{d\tau^2}$$

and wiki , defines it as:

$$\frac{d}{dt}(\frac{dr}{d\tau})$$

you will be getting different expressions. You think not?
Now, if you dropped the condescending tone you adopted and maintained from your very first post, maybe we could have a more constructive discussion. How about it?

 

[Rolfe2]

This is not what I am saying.

Yes it is. Your previous message said (and I quote verbatim) "along a <b>timelike</b> geodesic the proper acceleration is zero." But you have also insisted that the proper acceleration is NOT zero for a free-falling particle, and you've even given an expression for what you think the proper acceleration of such a particle is along its [geodesic] trajectory. So the only conclusion one can draw from your statements is that you believe a particle in free-fall does not follow a <b>timelike</b> geodesic. If this is not what you mean, then your statements are self-contradictory.

 

[starthaus]

If you're talking about the proper acceleration of a particle on a free radial trajectory, that is obviously zero,

...provided that the norm of the tangent vector to the trajectory is constant. This is not the case for the radial field produced by the Earth.
Another way of looking at it is:

$$a=-grad(\Phi)$$ where $$\Phi$$ is not constant, it is actually a function of $$r$$.

 

[Rolfe2]

I think that you are are clearly confused on the subject. Of course you will get different results if you define proper acceleration using different mathematical expressions.

Again, you're totally mistaken. The issue here is not whether we can dream up alternative definitions for the term "proper acceleration". There is only one standard accepted definition of that term. The point is that you totally mis-understand what that term means, as shown by your belief that a particle in free-fall is subject to non-zero proper acceleration. You arrived at this belief because you mistakenly think that proper acceleration is a coordinate dependent quantity, so I explained to you why it is actually based on the proper measures of space and time. You're welcome.

 

[starthaus]

Again, you're totally mistaken. The issue here is not whether we can dream up alternative definitions for the term "proper acceleration". There is only one standard accepted definition of that term.

Yet, the one that you claim is contradicted by the definition in wiki.

The point is that you totally mis-understand what that term means, as shown by your belief that a particle in free-fall is subject to non-zero proper acceleration.

The acceleration is zero only if the norm of the tangent vector to the geodesic is constant. This is not the case for the radial, non-uniform field produced by the Earth. Think about it:

$$a=-grad(\Phi)$$. If $$\Phi$$ is constant, the proper acceleration is indeed zero, yet for the case of the field produced by the Earth, $$\Phi$$ is definitely not constant.

 

[Rolfe2]

...provided that the norm of the tangent vector to the trajectory is constant. This is not the case for the radial field produced by the Earth.

You're trying to apply the formalism of special relativity to physics in a gravitational field. That doesn't work.

Another way of looking at it is:

$$a=-grad(\Phi)$$ where $$\Phi$$ is not constant, it is actually a function of $$r$$.

Nope. Once again, you're confusing coordinate acceleration with proper acceleration. The proper acceleration of a particle has a definite physical meaning, in that it represents the amount of acceleration that would be "felt" or measured by a co-moving accelerometer. If you agree that an accelerometer in freefall measures no acceleration (which is true by definition of freefall), then you must agree that the proper acceleration of a particle in freefall is zero (again, by definition of "proper acceleration"). Try re-reading the previous explanations that have been provided to you, or maybe consult a good introductory textbook on general relativity.

 

[DrGreg]

I think that you are are clearly confused on the subject. Of course you will get different results if you define proper acceleration using different mathematical expressions. If you define it as :

$$a_0=\frac{d^2r}{d\tau^2}$$

or


$$a_0=\frac{d^2\rho}{d\tau^2}$$

and wiki , defines it as:

$$\frac{d}{dt}(\frac{dr}{d\tau})$$

you will be getting different expressions. You think not?
Now, if you dropped the condescending tone you adopted and maintained from your very first post, maybe we could have a more constructive discussion. How about it?

Wikipedia is not a reliable source.

First of all, the Wikipedia article proper acceleration (today as I write this) gives the almost-correct definition in the first paragraph as "acceleration relative to a free-fall, or inertial, path", although it fails to mention the concept of "co-moving" which is also a necessary part of the definition. The definition as acceleration relative to a co-moving free-falling observer (in locally Minkowski coordinates) is the standard definition you will find in all reputable textbooks, and it's a pretty trivial consequence that a free-falling particle has zero proper acceleration (in all circumstances). See, for example, Rindler (2006), Relativity: Special, General, and Cosmological, p. 53.

The Wikipedia article goes on to claim that in the second paragraph that "for unidirectional motion, proper acceleration is the rate of change of proper velocity with respect to coordinate time". That's true only relative to an inertial coordinate system. It can't possibly be true relative to an accelerating coordinate system in which the particle being measured is at rest.

So the Wikipedia article is misleading, to say the least. I'll probably have a go at rewording it slightly to avoid this error.

 

[Rolfe2]

Yet, the one that you claim is contradicted by the definition in wiki.

Again, you're mistaken. Here is what Wikipedia says about proper acceleration (please note that this is exactly what everyone here has been telling you):

"In relativity theory, proper acceleration is the physical acceleration (i.e., measurable acceleration as by an accelerometer) experienced by an object. It is acceleration relative to a free-fall, or inertial, path. It is opposed to the coordinate acceleration, which is dependent on choice of coordinate systems and thus upon choice of observers."

Now, Wikipedia also contains an article on "four-acceleration", and here is what it says

Wiki: Four-acceleration
"In special relativity, four-acceleration is a four-vector and is defined as the change in four-velocity over the particle's proper time..."

Do you see the words "special relativity"? As I've mentioned to you multiple times, you are trying to apply the formalism of special relativity to curved spacetime, and it doesn't work, because in curved spacetime the metric coefficients do not have their Minkowskian values. The formulas you are trying to apply to a gravitational field are simply not applicable without replacing the Minkowskian metric coefficients of special relativity with the general metric coefficients of general relativity. This has the effect of accounting for the curvature of spacetime that corresponds to the gravitational field. Again, I urge you to consult with an introductory text on general relativity.

 

[starthaus]

Nope. Once again, you're confusing coordinate acceleration with proper acceleration.

Rindler (11.15) p.230 clearly disagrees with what you are saying. See his derivation for proper acceleration.

Try re-reading the previous explanations that have been provided to you, or maybe consult a good introductory textbook on general relativity.

You know where you can put your condescending tone. I asked you to drop it but you continue with this.

 

[Rolfe2]

The Wikipedia article goes on to claim that in the second paragraph that "for unidirectional motion, proper acceleration is the rate of change of proper velocity with respect to coordinate time". That's true only relative to an inertial coordinate system. It can't possibly be true relative to an accelerating coordinate system in which the particle being measured is at rest.

Exactly. The wiki article makes statements that apply in the context of special relativity, but that are not valid in the presence of gravitational fields. In some places, such as the article on four-acceleration, it specifically says it is talking about special relativity, but some readers might fail to see the significance.

 

[starthaus]

Wikipedia is not a reliable source.

First of all, the Wikipedia article proper acceleration (today as I write this) gives the almost-correct definition in the first paragraph as "acceleration relative to a free-fall, or inertial, path", although it fails to mention the concept of "co-moving" which is also a necessary part of the definition. The definition as acceleration relative to a co-moving free-falling observer (in locally Minkowski coordinates) is the standard definition you will find in all reputable textbooks, and it's a pretty trivial consequence that a free-falling particle has zero proper acceleration (in all circumstances). See, for example, Rindler (2006), Relativity: Special, General, and Cosmological, p. 53.

Yes, $$\frac{d\phi}{d\tau}$$ is constant ($$\phi$$ is rapidity). This seems to argue for using the definition $$c\frac{d\phi}{d\tau}$$ for proper acceleration. This leads to a formula dependent on the first and second derivatives of $$r$$ wrt $$\tau$$:

$$a_p=\frac{d^2r}{d\tau^2}/\sqrt{1+(dr/d\tau)^2}$$

The Wikipedia article goes on to claim that in the second paragraph that "for unidirectional motion, proper acceleration is the rate of change of proper velocity with respect to coordinate time". That's true only relative to an inertial coordinate system. It can't possibly be true relative to an accelerating coordinate system in which the particle being measured is at rest.

So the Wikipedia article is misleading, to say the least. I'll probably have a go at rewording it slightly to avoid this error.

Thank you, this is a reasonable , level-headed , non-condescending post.
Where does this leave us in terms of the dispute, i.e. what is a reasonable expression for proper acceleration in GR that is consistent with the SR definition?

 

[espen180]

Thank you, this is a reasonable , level-headed , non-condescending post.
Where does this leave us in terms of the dispute, i.e. what is a reasonable expression for proper acceleration in GR that is consistent with the SR definition?

Whatever it turns out to be, I'm pretty sure it won't predict

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

for a freefalling particle at apogee.

 

[DrGreg]

Thank you, this is a reasonable , level-headed , non-condescending post.
Where does this leave us in terms of the dispute, i.e. what is a reasonable expression for proper acceleration in GR that is consistent with the SR definition?

See post #9 in this thread!

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.

That's essentially the same method that some others have mentioned in this thread, using the facts that

1. proper acceleration is the magnitude of 4-acceleration
2. 4-acceleration is the covariant derivative of 4-velocity with respect to proper time
3. 4-velocity is the derivative of worldline-coordinates with respect to proper time

That's not the only method of doing it, but as the result is published in a reputable textbook -- Nick Woodhouse's General Relativity (p.99) based on the website given above -- (and lots of other books too), we can be pretty confident it's correct -- any method that gets a different answer must be defective.

 

[DrGreg]

Whatever it turns out to be, I'm pretty sure it won't predict

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

for a freefalling particle at apogee.

No, a free-falling particle always has a proper acceleration of zero, apogee or not. That expression is the proper acceleration of a particle at rest in the coordinate system.

 

[starthaus]

See post #9 in this thread! That's essentially the same method that some others have mentioned in this thread, using the facts that

1. 
   
   Yes, this is the same result as the one obtained in post 38 (using the covariant derivatives) and the same result as the one obtained using the gradient of the gravitational field (I forget the post) and the same result that I get in my blog through variational mechanics, a particle hovering at $$r=r_0$$ experiences a proper acceleration:
   
   $$a_0=-\frac{m}{r_0^2}/\sqrt{1-2m/r_0}$$
   
   The disagreement is about what happens when the particle starts falling in the non-uniform gravitational field.

 

[espen180]

The disagreement is about what happens when the particle starts falling in the non-uniform gravitational field.

Why is there a disagreement? Proper acceleration for a freefalling particle is zero per definition.

 

[Passionflower]

Why is there a disagreement? Proper acceleration for a freefalling particle is zero per definition.

Well that is a little strong, real objects after all are spatially extended.

 

[Al68]

Yes, this is the same result as the one obtained in post 38 (using the covariant derivatives) and the same result as the one obtained using the gradient of the gravitational field (I forget the post) and the same result that I get in my blog through variational mechanics, a particle hovering at $$r=r_0$$ experiences a proper acceleration:

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

The disagreement is about what happens when the particle starts falling in the non-uniform gravitational field.

According to post 1 of this thread:

It's initial coordinate acceleration relative to an observer hovering at $$r_0$$ is numerically equal to $$a_0$$, since that's how (in reverse) proper acceleration is defined.

It's initial coordinate acceleration relative to an observer at infinity is $$\frac{m}{r^2}(1-2m/r)$$.

Do you now agree with the acceleration equations in post 1?

 

[Altabeh]

See post #9 in this thread! That's essentially the same method that some others have mentioned in this thread, using the facts that

1. proper acceleration is the magnitude of 4-acceleration
2. 4-acceleration is the covariant derivative of 4-velocity with respect to proper time
3. 4-velocity is the derivative of worldline-coordinates with respect to proper time

Very well-worded and clear definitions that I agree with, as well. Maybe it's now too late to get involved in this catchy brawl but since the thread goes as fast as it can, I've lost almost the subject of battle but definitely I don't think starthaus has been telling beyond these, has he?

AB

 

[Al68]

You know where you can put your condescending tone. I asked you to drop it but you continue with this.

So, you're complaining about someone else's condescending tone? Seriously? After this:

It's not my fault that you don't understand basic calculus.
Is that you can't read math or are you just trolling?
Do you have some comprehension problem?
I think that you are are clearly confused on the subject.
Hey, thanks for playing anyway, better luck next time.
What in the definition : "proper acceleration is the derivative of proper speed wrt coordinate tiime" did you not understand?

I'm way too lazy to quote a significant fraction of your rude and condescending remarks, so that's just a small sampling.

It's obvious who the most condescending person in this thread is, by far.

 

[starthaus]

Very well-worded and clear definitions that I agree with, as well. Maybe it's now too late to get involved in this catchy brawl but since the thread goes as fast as it can, I've lost almost the subject of battle but definitely I don't think starthaus has been telling beyond these, has he?

AB

You are right, I haven't. I simply showed a few methods of deriving the expressions, rather than putting them in by hand, mindlessly. The most interesting approach is the one explained in my blog, using variational mechanics. The approach seems even more powerful (and certainly has a richer physics content) than even the approach based on covariant derivatives.

 

[Altabeh]

Rindler (11.15) p.230 clearly disagrees with what you are saying. See his derivation for proper acceleration.

Where in that page does he claim that the equation (11.15) belongs to proper acceleration of a particle following a time-like geodesic or whatever you might think of other than saying that it's just the field strength?

AB

 

[Altabeh]

In the second post of this thread, starthaus brings up the first of theories "kev's wrong because I'm right" by notifying us (!) of the following error (!) that he probably finds uninteresting just because kev is the poster:

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.

This is in response to the very correct formula,

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

which shows the initial coordinate acceleration of a rest particle released at the distance r according to an observer at infinity. Well if I'm not wrong about realizing what you seem to be pointing at in kev's post to be incorrect, I can very beautifully mathematically prove that the above formula is correct. I know that you're a knowledgeable guy but sometimes I feel you really get gutless to confess to what I call "the ordinary mistakes" that every physicist can make multiple times in his long-lived career. At least I'm trying to hint at the moot point everyone else here has been leaving out: If you've been trying to make kev learn from you, why don't I teach you something that you as an expert in calculus and everything probably don't know? If you're ready, I'm going to start debating about your numerous mistakes here!

I'm not going to be spineless when I find myself getting corrected as in a thread I really found the whole view of mine on the issue of "proper vs coordinate" wrong due to a debate with JesseM. You see that we confess to the wrongness or correctness of our mistakes as every normal physicist of Einsteinian level does.

AB

 

[Altabeh]

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:
$$\mathbf A = \left(0,\frac{c^2 R}{2 r^2},0,0\right) = \left(0,\frac{G M}{r^2},0,0\right)$$

Would someone please make it clear for me why the second term in

$$\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}$$

disappears in the end?

AB

 

[starthaus]

Would someone please make it clear for me why the second term in

$$\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}$$

disappears in the end?

AB

It doesn't "disappear", $$\mathbf X=(t,r_0,0,0)$$ with $$r_0$$ fixed, so all you are left with is the term in $$U^tU^t$$

 

[Altabeh]

In fact, I checked starthaus's derivation and it is flawless and mine is a sort of similar approach but definitely doesn't result in an expression different from his for the coordinate acceleration. All I nag about is the mistakes he makes when facing with the definition of proper acceleration in GR as he clearly does this in his article "General Euler-Lagrange Solution for Calculating Coordinate Acceleration" where he says the proper acceleration is $$\frac{d^2r}{ds^2}$$ while it's obviously not. As is clear, the proper acceleration here is to be defined by

$$\frac{Dr}{Ds}=\frac{d^2r}{ds^2}+\Gamma^{r}_{\mu\nu}U^{\mu}U^{\nu}.$$

(This was the idea that I came to when getting involved in a debate with JesseM.)

If we can settle these issues, I'll go on.

AB

 

[Altabeh]

It doesn't "disappear", $$\mathbf X=(t,r_0,0,0)$$ with $$r_0$$ fixed, so all you are left with is the term in $$U^tU^t$$

No, I was asking this question that if I want to get from

$$\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma ^{r}_{tt}U^{t}U^{t}=\frac{c^2 (r-R) R}{2 r^3}$$

to

$$\mathbf A = \left(0,\frac{c^2 R}{2 r^2},0,0\right) = \left(0,\frac{G M}{r^2},0,0\right)$$

it is required to have the term $$\frac{-c^2R^2}{2 r^3}$$ disappear. How does it occur?

AB

 

[starthaus]

No, I was asking this question that if I want to get from

$$\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma ^{r}_{tt}U^{t}U^{t}=\frac{c^2 (r-R) R}{2 r^3}$$

to

$$\mathbf A = \left(0,\frac{c^2 R}{2 r^2},0,0\right) = \left(0,\frac{G M}{r^2},0,0\right)$$

it is required to have the term $$\frac{-c^2R^2}{2 r^3}$$ disappear. How does it occur?

AB

It doesn't either. $$R=\frac{2GM}{rc^2}$$

 

[starthaus]

As is clear, the proper velocity here is to be defined by

$$\frac{Dr}{Ds}=\frac{dr}{ds}+\Gamma^{r}_{\mu\nu}U^{\mu}U^{\nu}.$$

You sure about that? :-)

 

[Altabeh]

It doesn't either. $$R=\frac{2GM}{rc^2}$$

Look, I know all of this. The problem that I'm fed up with is how the second expression is derived from the first.

AB

 

[Altabeh]

You sure about that? :-)

I am, aren't you?]

Edit: All typos and stuff were edited. You see, I intentionally can make an error and you come and point it out to me somehow and I then stand corrected. What happened? Do you know more than rest of us, then? LOL

Anyways, I hope I'll know today whether the question in the post 305 is right!

AB

 

[starthaus]

In fact, I checked starthaus's derivation and it is flawless

1. So, why are you wasting time?

and mine is a sort of similar approach but definitely doesn't result in an expression different from his for the coordinate acceleration.

2. I must have missed your derivation, where is it?

All I nag about is the mistakes he makes when facing with the definition of proper acceleration in GR as he clearly does this in his article "General Euler-Lagrange Solution for Calculating Coordinate Acceleration" where he says the proper acceleration is $$\frac{d^2r}{ds^2}$$ while it's obviously not.

Well, it wasn't that obvious to me when I started developing the alternate approach since I was trying to avoid covariant derivatives. Otherwise the approach would not be original, would it? Besides, I was interested in getting an alternate definition for proper acceleration. To date, none of the definitions (see my exchange with Rolfe2) has been satisfactory. This may mean that the variational mechanics method is good for deriving coordinate acceleration (something that the covariant derivative method can't do) and the covariant derivative method is good at deriving the proper acceleration for the hovering particle only. So, we need both methods.

As is clear, the proper acceleration here is to be defined by

$$\frac{Dr}{Ds}=\frac{d^2r}{ds^2}+\Gamma^{r}_{\mu\nu}U^{\mu}U^{\nu}.$$

(This was the idea that I came to when getting involved in a debate with JesseM.)

If we can settle these issues, I'll go on.

AB

3. You editted your post from inadvertently trying to correct my definition of proper "speed" to trying to correct my definition of proper "acceleration". Do you think anyone is missing your moving the goalposts between posts?
4. BTW , your new definition (of proper acceleration) is still wrong, though you copied from post 38.

 

[yuiop]

... a particle hovering at $$r=r_0$$ experiences a proper acceleration:

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

The disagreement is about what happens when the particle starts falling in the non-uniform gravitational field.

Good to see you have finally come to the conclusion that the equation I gave in #1 for proper acceleration (based mainly on information from the mathpages website) is correct in the context that it was given in. The claim you started in #2 that the equations I gave in #1 are wrong, has been shown by the vast majority of posters in this thread (i.e. everyone but you), to be invalid.

You have expressed an interest to extend the equations beyond the context of #1, to the more general situation. In order to do that, we should first establish a "baseline" of what we do agreee on. A good place to start, would be for you to openly concede that the equations in #1 are correct in their context.