Gravitational force and acceleration in General Relativity.

[yuiop]

Why does this nonsense about a_0=a\gamma^3 has so much fascination for you?

Do you agree that in the colinear case in SR, the relationship between the proper acceleration of a particle as measured in the MCIF (S') and the acceleration in inertial reference frame S is a' =a\gamma^3 where \gamma = 1/\sqrt{(1-v^2/c^2)} and v is the relative velocity between inertial reference frames S and S'?

Do you agree that a' as measured in the MCIF is equivalent in magnitude to the proper acceleration of the particle? i.e a' = a_0

Do you agree that a_0 = d^2x'/dt'^2 and not a_0 = d^2x/dt'^2\;\;? (i.e. the local or proper acceleration is measured using local rulers and clocks and not coordinate distance)

 

[Rolfe2]

I gave you three references that disagree with what you are saying, how about you gave me a couple of references that agree with what you are saying.

Again, the references that have been cited all agree with what I'm saying, and they all disagree with what you are saying. The x, y, z, and t coordinates appearing in the four-vector expression for proper acceleration are local co-moving inertial coordinates, which signifies that x,y,z are the proper space coordinates (by definition). If you think those coordinates represent something else (Schwarzschild coordinates?? Starthaus Normal Coordinates??) then feel free to say so.

 

[starthaus]

Do you agree that in the colinear case in SR, the relationship between the proper acceleration of a particle as measured in the MCIF (S') and the acceleration in inertial reference frame S is a' =a\gamma^3 where \gamma = 1/\sqrt{(1-v^2/c^2)} and v is the relative velocity between inertial reference frames S and S'?

Sure, I told you this several times in this thread.
The reason is that in SR \gamma=1/\sqrt{1-(v/c)^2}

Can you take the above in really prove that a_0=a\gamma^3

Once you do that, try the same thing with \gamma=1/\sqrt{1-2m/r}

Do you agree that a' as measured in the MCIF is equivalent in magnitude to the proper acceleration of the particle? i.e a' = a_0

Do you agree that a_0 = d^2x'/dt'^2 and not a_0 = d^2x/dt'^2\;\;? (i.e. the local or proper acceleration is measured using local rulers and clocks and not coordinate distance)

This has nothing to do with the disproof to your claims I just posted. Simple algebra says you're wrong.

 

[yuiop]

But you know that the above can't be true. For a particle dropped from r_0:

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

\frac{d^2r}{d\tau^2}=-\frac{m}{r^2}

(Even if we accepted that the above might be of the form:

\frac{d^2r}{d\tau^2}=-\frac{m}{r^2 \sqrt{1-2m/r}}

it still isn't true)

For the stationary observers on the surface, the local clock rate (dt') is not the same as the proper time of the particle (dtau) when the particle has significant velocity relative to the observers.

 

[starthaus]

Again, the references that have been cited all agree with what I'm saying, and they all disagree with what you are saying. The x, y, z, and t coordinates appearing in the four-vector expression for proper acceleration are local co-moving inertial coordinates, which signifies that x,y,z are the proper space coordinates (by definition). If you think those coordinates represent something else (Schwarzschild coordinates?? Starthaus Normal Coordinates??) then feel free to say so.

So, you have no references. Thank you.

 

[starthaus]

For the stationary observers on the surface, the local clock rate (dt') is not the same as the proper time of the particle (dtau) when the particle has significant velocity relative to the observers.

True but this is a total non-sequitur, the first expression is the coordinate acceleration and the second one is the proper acceleration and they are obviously not in the ratio a_0/a=\gamma^3.

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

2a. a_0=\frac{d^2r}{d\tau^2}=-\frac{m}{r^2}

(Even if we accepted that the above might be of the form:

2b. a_0=\frac{d^2\rho}{d\tau^2}=-\frac{m}{r^2 \sqrt{1-2m/r}}

it still isn't true, there is no way you can ignore the term 3\frac{1-2m/r}{1-2m/r_0}-2}

 

[Al68]

But you know that the above can't be true. For a particle dropped from r_0:

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

a_0=\frac{d^2r}{d\tau^2}=-\frac{m}{r^2}

(Even if we accepted that the above might be of the form:

a_0=\frac{d^2r}{d\tau^2}=-\frac{m}{r^2 \sqrt{1-2m/r}}

it still isn't true)

Why does this nonsense about a_0=a\gamma^3 has so much fascination for you?

A simple rearrangement of the first and last equations above shows that:

\frac{d^2r}{d\tau^2}=\frac{d^2r}{dt^2}(1/\sqrt{1-2m/r})^3 when r=r_0

 

[starthaus]

A simple rearrangement of those two equations shows that:

\frac{d^2r}{d\tau^2}=\frac{d^2r}{dt^2}(1/\sqrt{1-2m/r})^3 when r=r_0

Yes, I can see you can perform simple substitutions. Yet, the above is not true in general. It isn't true for any other value of r. Don't let that trouble you.

 

[Al68]

A simple rearrangement of the first and last equations above shows that:

\frac{d^2r}{d\tau^2}=\frac{d^2r}{dt^2}(1/\sqrt{1-2m/r})^3 when r=r_0

Yes, I can see you can perform simple substitutions. Yet, the above is not true in general. It isn't true for any other value of r. Don't let that trouble you.

Has anyone in this thread claimed that a_0=a\gamma^3 was true in any case other than when r=r_0 as specified in post 1?

If anyone neglected to specify "when r=r_0 ", I'm sure it's because (almost) everyone in this thread was talking about the case specified in post 1, where it was made perfectly clear that a referred to "the initial coordinate acceleration of a test mass released at r".

Was it unclear that "the initial coordinate acceleration of a test mass released at r" means r=r_0 ?

 

[Rolfe2]

So, you have no references. Thank you.

You mis-read my message. I provided the references you requested, and I also explained your errors. You're welcome.

 

[starthaus]

The proper acceleration is actually the second derivative of what you might call the proper radial coordinate rho with respect to the proper time, and it is given by -m/r^2 [1/sqrt(1-2m/r)].

I don't think so. For the correct definition, see http://wapedia.mobi/en/Proper_acceleration .
Using the correct definition, proper acceleration for the general case is:

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

 

[Rolfe2]

I don't think so. For the correct definition, see http://wapedia.mobi/en/Proper_acceleration .
Using the correct definition, proper acceleration for the general case is:

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

Since you're unable or unwilling to follow rational explanations, and will only accept things on "authority", please check page 152 of Wald (for just one example), where he says "It is easy to check that static observers in the Schwarzschild spacetime must undergo a proper acceleration (in order to "stand still" in the "gravitational field") given by a = (1-2M/r)^-1/2 M/r^2..." This of course is perfectly consistent with the correct definition of proper acceleration, as has been explained to you already.

 

[starthaus]

Since you're unable or unwilling to follow rational explanations, and will only accept things on "authority", please check page 152 of Wald (for just one example), where he says "It is easy to check that static observers in the Schwarzschild spacetime must undergo a proper acceleration (in order to "stand still" in the "gravitational field") given by a = (1-2M/r)^-1/2 M/r^2..." This of course is perfectly consistent with the correct definition of proper acceleration, as has been explained to you already.

You gave a bogus definition of proper acceleration. So I pointed you towards the correct one (derivative of proper speed wrt coordinate time).

PS; if you made a little effort you would have recovered the formula in Wald by making r=r_0 in my formula . I gave you the general formula, derived from scratch, not gleaned from a book. <shrug>

Hey, thanks for playing anyway, better luck next time.

 

[Shackleford]

Hey, who's right? It seems like that would be easy to determine on here. lol.

 

[Al68]

I don't think so. For the correct definition, see http://wapedia.mobi/en/Proper_acceleration .
Using the correct definition, proper acceleration for the general case is:

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

What do you mean by "general case"? Doesn't proper acceleration by definition mean the acceleration relative to a local (r=r_0) inertial path?

 

[starthaus]

What do you mean by "general case"? Doesn't proper acceleration by definition mean the acceleration relative to a local (r=r_0) inertial path?

If you drop a particle from r_0 what will be its acceleration at r?
If a particle hovers at r_0 what is its acceleration?

 

[Rolfe2]

You gave a bogus definition of proper acceleration. So I pointed you towards the correct one (derivative of proper speed wrt coordinate time).

Nope. See the definition of proper acceleration in any reputable reference, such as page 67 of Rindler's Essential Relativity. Again, it is the second derivative of the local co-moving inertial coordinates (including the proper space coordinates) with respect to proper time. This isn't controversial, it is simply what the term "proper acceleration" means and has always meant.

PS; if you made a little effort you would have recovered the formula in Wald by making r=r_0 in my formula .

First you insisted that the proper acceleration is -m/r^2, and you insisted that I was wrong for giving the correct proper acceleration of a stationary particle at a radial position r, along with the derivation. Then you switch to a different expression, one which you proudly announce reduces to my formula (the one you had been insisting was wrong for the past several posts) for a stationary particle, which is what my formula was stated to be in the first place. Weird.

I gave you the general formula, derived from scratch, not gleaned from a book.

You refused to accept the correct derivation provided to you, back when you were insisting the proper acceleration was -m/r^2, and you demanded that I provide a reference for the correct expression. Now when I provide a reference, you snidely acuse me of "gleaning it from a book". Very odd.

 

[Al68]

What do you mean by "general case"? Doesn't proper acceleration by definition mean the acceleration relative to a local (r=r_0) inertial path?

If you drop a particle from r_0 what will be its acceleration at r?
If a particle hovers at r_0 what is its acceleration?

LOL. The first question refers to coordinate acceleration. The second question refers to proper acceleration (r=r_0).

So, again, what do you mean by proper acceleration for the "general case"?

 

[Cyosis]

So, what value does GR predict for a_0 ? You have two choices:

1. \frac{m}{r^2} (my derivation based on lagrangian mechanics and K.Brown's)

2. \frac{m}{r^2\sqrt{1-2m/r}} ?

The above is definitely at odds with this.

U=(dx/d\tau,dy,d\tau,dz/d\tau,d(ct)/d\tau)

(see also Rindler, p.99, Moller p.288))IF what you were saying were true, the coordinate acceleration a would not show up in the definition of the four-vector A, nor would we be able to calculate proper acceleration a_0 from the conditon A=(a_0,0) for u=0

The expression you obtained for the four-acceleration, using your convention, is A=(-m/r^2,0,0,0). Its magnitude given by

\sqrt{-g_{\mu \nu}A^\mu A^\nu}=\sqrt{-g_{00}A^0 A^0}=\frac{m}{r^2\sqrt{1-\frac{2m}{r}}}<br />.

 

[starthaus]

LOL. The first question refers to coordinate acceleration. The second question refers to proper acceleration (r=r_0).

So, again, what do you mean by proper acceleration for the "general case"?

You are trolling again. What is the proper acceleration of a particle dropped from r_0 when the particle arrives at location r? The observer "rides" on the particle.

 

[starthaus]

First you insisted that the proper acceleration is -m/r^2, and you insisted that I was wrong for giving the correct proper acceleration of a stationary particle at a radial position r, along with the derivation. Then you switch to a different expression, one which you proudly announce reduces to my formula (the one you had been insisting was wrong for the past several posts) for a stationary particle, which is what my formula was stated to be in the first place. Weird.

The formulas are all derivatives of distance wrt time. Depending on what you choose for expressing time and distance, you get different values for the derivatives. What in the definition : "proper acceleration is the derivative of proper speed wrt coordinate tiime" did you not understand? You have the http://wapedia.mobi/en/Proper_acceleration all you need is to do the calculations<shrug>

 

[starthaus]

The expression you obtained for the four-acceleration, using your convention, is A=(-m/r^2,0,0,0). Its magnitude given by

\sqrt{-g_{\mu \nu}A^\mu A^\nu}=\sqrt{-g_{00}A^0 A^0}=\frac{m}{r^2\sqrt{1-\frac{2m}{r}}}<br />.

Yes, for a complete derivation see post 38. It uses the derivative of the coordinate distance wrt proper time as definition for four-speed:\mathbf X = (t,r_0,0,0)

(Note that there is no such thing as "proper radial coordinate" in the definition of \mathbf X,
\mathbf X is simply the four vector defined by the Schwarzschild coordinates (t,r,\theta,\phi))
and

\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)

(Note that the factor \frac{1}{\sqrt{1-R/r}} arises simply from taking the derivative of \mathbf X wrt proper time \tau)

and the derivative of four-speed wrt coordinate time as four-acceleration (see Rindler, p99):

\mathbf A=\frac{d \mathbf U}{d\tau}

Choosing \mathbf X = (t,r,0,0) one gets yet a different set of results since \mathbf U now depends on \frac{dr}{dt}. These are the type of results I have obtained by using the lagrangian method in my blog. Of course, one can get the same results through covariant derivatives starting from \mathbf U={\frac{1}{\sqrt{1-R/r}}(1,\frac{dr}{dt},0,0).

Clearly, what was calculated in post 38 uses a different definition than the one given in http://wapedia.mobi/en/Proper_acceleration reference. Obviously, you get different results starting from different definitions. Using the wiki definition, one gets the expression I posted in post 261: a_0=-\frac{m}{r^2}\frac{\sqrt{1-2m/r_0}}{1-2m/r}
I think what gets Rolfe2 all twisted in his knickers is that either definition reduces to the same expression for r=r_0. Yet, they are obviously different for all other values of r.

Yet, a different (and probably the best) definition for proper acceleration is given in the attachment "Accelerated Motion in SR, Part II", where the proper acceleration is defined as:

a_p=c\frac{d\phi}{d\tau}
where:
sinh(\phi)=\frac{1}{c}\frac{dx}{d\tau}

x=coordinate distance, \tau=proper time, \phi=rapidity

 

[Rolfe2]

Yes, for a complete derivation see post 38.

Notice that this is the same result about which, in your previous message, you said "I don't think so", and you presented a different expression. But now you say "Yes" to the expression that you previously denied. Later on you admit that your proposed alternative came from an alternative definition of proper acceleration - a subject which you still clearly don't understand - see below.

It uses the derivative of the coordinate distance wrt proper time as definition for four-speed...

Excuse me for saying so, but what you're saying simply makes no sense. 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.

To help you see this, try this little exercise: You talk about how X is just based on the Schwarzschild coordinates, but you ought to be able to get the very same answer for proper acceleration using any other system of coordinates, right? Think about why you get the same answer, regardless of what coordinate system you choose, if (as you claim) the result is based on the coordinate measures of distance rather than the proper distances. The answer is simply that it works for any system of coordinates because the corresponding metric coefficients for any other coordinates are just such as to make the invariant proper measures come out the same. This is the physical basis of the derivation.

Remember, there is nothing magical about Schwarzschild coordinates. So you are obviously wrong to say, unequivocally that "X" represents the Schwarzschild coordinates. And likewise you are wrong to say that the definition of proper acceleration is not based on the proper measures of spatial distance as well as proper time. You're failing to grasp what the symbols (including the metric coefficients) represent physically. Fundamentally, proper acceleration is defined in terms of the essentially unique local co-moving inertial coordinates, which correspond to the proper time and proper spatial lengths. The fact that we can express these proper measures in terms of arbitrary coordinates along with the corresponding metric coefficients is trivially obvious, but does not imply that proper acceleration is a coordinate-dependent quantity.

 

[yuiop]

I don't think so. For the correct definition, see http://wapedia.mobi/en/Proper_acceleration .
Using the correct definition, proper acceleration for the general case is:

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

Section 4.3 of the reference you quote http://wapedia.mobi/en/Proper_acceleration?t=4.#4 . specifically states:

The magnitude of the above proper acceleration 4-vector, namely

a = \sqrt{1/(1-GM/r)}\; GM/r^2,

is however precisely what we want i.e. the upward frame-invariant proper acceleration needed to counteract the downward geometric acceleration felt by dwellers on the surface of a planet.

which agrees with what everyone else is saying the proper acceleration is. It also agrees with what Cyosis says here:

The expression you obtained for the four-acceleration, using your convention, is A=(-m/r^2,0,0,0). Its magnitude given by

\sqrt{-g_{\mu \nu}A^\mu A^\nu}=\sqrt{-g_{00}A^0 A^0}=\frac{m}{r^2\sqrt{1-\frac{2m}{r}}}<br />.

If you drop a particle from r_0 what will be its acceleration at r?

Its proper acceleration is zero.

If a particle hovers at r_0 what is its acceleration?

Its proper acceleration is \frac{m}{r^2\sqrt{1-\frac{2m}{r}}}

Your equation:

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

does not work, because the proper acceleration of a free falling particle is zero. (Attach an accelerometer to a free falling particle and see what it reads.)

 

[Al68]

What is the proper acceleration of a particle dropped from r_0 when the particle arrives at location r? The observer "rides" on the particle.

Are you joking? Are you seriously asking me what an accelerometer would read on a free falling object? I obviously agree with kev's answer: Zero.

It's no wonder this thread has dragged on this long.

 

[starthaus]

To help you see this, try this little exercise: You talk about how X is just based on the Schwarzschild coordinates, but you ought to be able to get the very same answer for proper acceleration using any other system of coordinates, right?

Make r=r_0 in a_0=-\frac{m}{r^2}\frac{\sqrt{1-2m/r_0}}{1-2m/r}.
The point I made in the previous post that all methods obtain the same expression for proper acceleration at the apogee. (i.e. a_0=-\frac{m}{r_0^2}\frac{1}{\sqrt{1-2m/r_0}}).
The difference is that the lagrangian method obtains the general expressions for both coordinate and proper acceleration for arbitrary r. The result at apogee is simply a particularization of the general formula for r=r_0

 

[starthaus]

, because the proper acceleration of a free falling particle is zero.

...in a uniform gravitational field, yes. Not in a radial, variable field. This is what this thread is all about, a radial, variable field, no?
The math used for deriving the proper and coordinate acceleration in the radial field does not support your claim above. If you think otherwise try deriving the formula for the proper acceleration of a particle dropped from r=r_0 in a variable field as a function of the radial coordinate r. Show that your obtain a_0=0 for any r

 

[espen180]

...in a uniform gravitational field, yes. Not in a radial, variable field. This is what this thread is all about, a radial, variable field, no?
The math used for deriving the proper and coordinate acceleration in the radial field does not support your claim above. If you think otherwise try deriving the formula for the proper acceleration of a particle dropped from r=r_0 in a variable field as a function of the radial coordinate r. Show that your obtain a_0=0 for any r

I thought proper acceleration was always zero for particles following a geodesic. Are you saying a particle in free fall does not?

 

[starthaus]

I thought proper acceleration was always zero for particles following a geodesic. Are you saying a particle in free fall does not?

Wait until kev answers the question. I'll give you a hint: "not along a geodesic", but "along a timelike geodesic" the proper acceleration is zero. Now, you need to think about what a timelike geodesic means.

 

[Rolfe2]

Wait until kev answers the question. I'll give you a hint: "not along a geodesic", but "along a timelike geodesic" the proper acceleration is zero. Now, you need to think about what a timelike geodesic means.

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

 

[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[/color] (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[/color] (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?