Gravitational force and acceleration in General Relativity.

[Al68]

It obviously depends on whether $$a_0$$ is defined as $$\frac{d^2r}{d\tau^2}$$ or $$\frac{d^2r'}{d\tau^2}$$.

LOL. You need to pick the one that matches the measurement. This is not up to debate based on definition.

If you want me to pick between your choices, I would be glad to if you specify which definition of $$a_0$$ you want me to base it on.

Otherwise, I will choose a third option allowed by GR:

3. $$a_0=2\pi r$$, where $$a_0$$ is defined as the circumference of the moon.

 

[Rolfe2]

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)

No, as always, the references you've cited confirm what I said and refute what you are saying. The x, y, and z coordinates appearing in those expressions are local co-moving inertial coordinates, which in the context of a gravitational field are the proper distance coordinates.

IF what you were saying were true, the ccordinate 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 $$u=0$$

What you typed there is gibberish. You need to learn the meanings of various systems of coordinates, both in flat spacetime, and in the presence of a gravitational field. Again, the proper acceleration is, by definition, expressed in terms of the local co-moving inertial coordinates. You don't appear to understand what that means, so I'd suggest you start by going back to basics and trying to learn the meanings of coordinate systems in general relativity.

 

[yuiop]

There is no :

$$a_0=a\gamma^3$$

Imagine we have some observers on the surface of a massive body of Schwarzschild radius r. They have a set of ideal clocks and a set of vertical rulers that are calibrated using radar devices, so that the average vertical speed of light is c by their measurements. A test particle is dropped from higher up. (It does not have to be from infinity.)

The acceleration of the particle as it passes the surface observers is measured by the Schwarzschild observer at infinity using Schwarzschild radial coordinates and Schwarzschild coordinate time. Let's call this the coordinate acceleration of the particle $$a = (d^2r/dt^2)$$.

The observers on the shell measure the acceleration using their local measuring devices. Let's call this the local acceleration $$a' = (d^2r'/dt' ^2)$$

The ratio between $$(d^2r/dt^2)$$ and $$(d^2r'/dt' ^2)$$ will always be:

$$a' = a\gamma^3$$

where $\gamma = 1/\sqrt{(1-2M/r)}$ and r is the radius of the massive body, to first order and for a sufficiently local measurement, for a particle dropped from any height greater than r. (as far I can tell)

 

[starthaus]

No, as always, the references you've cited confirm what I said and refute what you are saying. The x, y, and z coordinates appearing in those expressions are local co-moving inertial coordinates, which in the context of a gravitational field are the proper distance coordinates.

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.

What you typed there is gibberish. You need to learn the meanings of various systems of coordinates, both in flat spacetime, and in the presence of a gravitational field. Again, the proper acceleration is, by definition, expressed in terms of the local co-moving inertial coordinates. You don't appear to understand what that means, so I'd suggest you start by going back to basics and trying to learn the meanings of coordinate systems in general relativity.

How about you get off your high horse?

 

[starthaus]

Imagine we have some observers on the surface of a massive body of Schwarzschild radius r. They have a set of ideal clocks and a set of vertical rulers that are calibrated using radar devices, so that the average vertical speed of light is c by their measurements. A test particle is dropped from higher up. (It does not have to be from infinity.)

The acceleration of the particle as it passes the surface observers is measured by the Schwarzschild observer at infinity using Schwarzschild radial coordinates and Schwarzschild coordinate time. Let's call this the coordinate acceleration of the particle $$a = (d^2r/dt^2)$$.

The observers on the shell measure the acceleration using their local measuring devices. Let's call this the local acceleration $$a' = (d^2r'/dt' ^2)$$

The ratio between $$(d^2r/dt^2)$$ and $$(d^2r'/dt' ^2)$$ will always be:

$$a' = a\gamma^3$$

where $\gamma = 1/\sqrt{(1-2M/r)}$ and r is the radius of the massive body, to first order and for a sufficiently local measurement, for a particle dropped from any height greater than r. (as far I can tell)

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?

 

[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

$$\sqrt{-g_{\mu \nu}A^\mu A^\nu}=\sqrt{-g_{00}A^0 A^0}=\frac{m}{r^2\sqrt{1-\frac{2m}{r}}}$$.

 

[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

$$\sqrt{-g_{\mu \nu}A^\mu A^\nu}=\sqrt{-g_{00}A^0 A^0}=\frac{m}{r^2\sqrt{1-\frac{2m}{r}}}$$.

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

$$\sqrt{-g_{\mu \nu}A^\mu A^\nu}=\sqrt{-g_{00}A^0 A^0}=\frac{m}{r^2\sqrt{1-\frac{2m}{r}}}$$.

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. :uhh:

 

[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?