Gravitational force and acceleration in General Relativity.

[Al68]

I can't help but point out the obvious logical fallacy here, and persisting in your posts. The fallacy is claiming that a result is wrong because you disagree with the way it was derived.

For example, consider this simple derivation:

Given: x = 16/64.

Canceling the 6 on the top and bottom, we obtain: x = 1/4.

Claiming that this result is incorrect would be a logical fallacy.

It's not my fault that you don't understand basic calculus. Based on your logic someone copying a correct result and producing a bogus derivation has produced valuable work instead of junk.

Another example of a post that makes it clear that you either didn't read or didn't comprehend the post you responded to.

I said it would be a logical fallacy to refer to the result as wrong, not to call its derivation "bogus" or "not valuable".

 

[yuiop]

I explained to you repeatedly that the derivation you copied from K. Brown is wrong. I explained why you can't do the derivatives wrt $$r$$ while considering $$k$$ constant wrt $$r$$.

I have spent enough time on this nonsense.

The strength with which you proclaim that k is not a constant with respect to r has been bothering me and started to sow seeds of doubt, so I have decided to look into it a bit more.

When I google "conserved energy Schwarzschild" I find plenty of references that say

$$\left(1-\frac{2M}{r}\right) \frac{dt}{ds}$$

is the "conserved energy" of a falling particle. This is almost universely agreed and to me, "conserved" means not changing as the particle falls.

Your derivation depends on k not changing with with the proper time of the falling particle. Now let us say we drop a particle from a height R (where capital R is a constant as apposed to the variable r in lower case). At some lower radius the proper time (s) of the particle has increased and k holds the same value because it is independent of the proper time. Now it is obvious that at the lower radius the variable r has changed and so has the coordinate time (t), so if k is independent of s, then it is also independent of t and r.

That suggests to me that your main objection to the derivation by K.Brown on mathpages http://www.mathpages.com/rr/s6-04/6-04.htm" is not valid.

 

[starthaus]

The strength with which you proclaim that k is not a constant with respect to r has been bothering me and started to sow seeds of doubt, so I have decided to look into it a bit more.

When I google "conserved energy Schwarzschild" I find plenty of references that say

$$\left(1-\frac{2M}{r}\right) \frac{dt}{ds}$$

is the "conserved energy" of a falling particle. This is almost universely agreed and to me, "conserved" means not changing as the particle falls.

Your derivation depends on k not changing with with the proper time of the falling particle. Now let us say we drop a particle from a height R (where capital R is a constant as apposed to the variable r in lower case). At some lower radius the proper time (s) of the particle has increased and k holds the same value because it is independent of the proper time.

OK.

Now it is obvious that at the lower radius the variable r has changed and so has the coordinate time (t), so if k is independent of s, then it is also independent of t and r.

Can you prove this ? By using math?
How does your claim jibe with $$k=\sqrt{1-2m/r}$$ used in K.Brown's proof that you are using?

That suggests to me that your main objection to the derivation by K.Brown on mathpages http://www.mathpages.com/rr/s6-04/6-04.htm" is not valid.

Sure it is, K. Brown uses $$k=\sqrt{\alpha}=\sqrt{1-2m/r}$$ and you use it since you copied his solution. So, K.Brown's solution is invalid.

On the other hand, $$\frac{dk}{ds}=0$$ represents the second Euler-Lagrange equation, so, k does not depend on s. This is what I use throughout my derivation.

 

[Rolfe2]

How does your claim jibe with $$k=\sqrt{1-2m/r}$$ used in K.Brown's proof that you are using?

I think you're mis-informed. The derivation on mathpages does not use k = sqrt(1-2m/r). The quantity k is introduced as a constant of integration, and it is shown (for bound paths) to equal sqrt(1-2m/R) where R is the apogee. This is a constant, not a variable.

Sure it is, K. Brown uses $$k=\sqrt{\alpha}=\sqrt{1-2m/r}$$ and you use it since you copied his solution. So, K.Brown's solution is invalid.

I think you have your facts wrong. In the mathpages derivation k is always a constant of integration, equal to sqrt(1-2m/R) for bound paths, where R is the radial coordinate of the apogee.

 

[starthaus]

I think you're mis-informed. The derivation on mathpages does not use k = sqrt(1-2m/r). The quantity k is introduced as a constant of integration, and it is shown (for bound paths) to equal sqrt(1-2m/R) where R is the apogee. This is a constant, not a variable.
I think you have your facts wrong. In the mathpages derivation k is always a constant of integration, equal to sqrt(1-2m/R) for bound paths, where R is the radial coordinate of the apogee.

You can check it out here.
According to Brown, k is a parameter and it is constant wrt the proper time, $$\tau$$. It takes the value $$\sqrt{1-2m/R}$$ at the apogee. It takes the value of 1 (obviously) at infinity. You can see the diagram of trajectories function of k as a parameter near the bottom of the page.
If you want to see a correct derivation, you can find one in the first attachment in my blog.
The interesting conclusion that you can draw drom both Brown's derivation and mine is that:

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

so, post 1 is in error. This is what we spent close to 200 posts to prove.

 

[Al68]

You can check it out here.
According to Brown, k is a parameter and it is constant wrt the proper time, $$\tau$$. It takes the value $$\sqrt{1-2m/R}$$ at the apogee. It takes the value of 1 (obviously) at infinity. You can see the diagram of trajectories function of k as a parameter near the bottom of the page.
If you want to see a correct derivation, you can find one in the first attachment in my blog.
The interesting conclusion that you can draw drom both Brown's derivation and mine is that:

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

so, post 1 is in error. This is what we spent close to 200 posts to prove.

Can you reference the part of post 1 you are claiming to be in error?

 

[Rolfe2]

You can check it out here.

The link you provided confirms what I said, and refutes what you are saying. The link says "where kappa [the reciprocal of k] is a constant parameter of the given trajectory..." Do you see the word 'constant'? It also refers back to a previous section, where k is introduced as a constant of integration (note again the word 'constant'), which is shown to equal the constant value sqrt(1-2m/R) for a given bound trajectory, where R is the radial coordinate of the apogee.

According to Brown, k is a parameter...

Again, k is identified not just as a "parameter", but as a "constant parameter" of a given trajectory. Obviously for other trajectories this parameter has other values, but for any given trajectory it is constant.

...and it is constant wrt the proper time, $$\tau$$.

For any given trajectory, it is constant, period. It is a function only of the radial coordinate R of the apogee and the mass m of the gravitating body, both of which are constant for any given trajectory. It is not a function of anything else, i.e., it is not a function of t our tau or r or anything else. Again the value of k is sqrt(1-2m/R). You can tell directly from this expression what k is a function of, and what it is not a function of. Look at the symbols in the expression sqrt(1-2m/R). (By the way, the symbols "sqrt" just represent the square root, they do not signify the product of s, q, r, and t. Hopefully this isn't what has confused you.)

The interesting conclusion that you can draw drom both Brown's derivation and mine is that:

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

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

so, post 1 is in error. This is what we spent close to 200 posts to prove.

Your comment here seems confused. If you've managed to confirm [after several false starts] some interesting results from mathpages, then congratulations, but your previous comments indicated that you believed the derivations of those results were invalid, because you believed that k was a function of r in those derivations. Hopefully you now realize that k is explicitly a constant in those derivations. There was no need to spend 200 posts to realize this. (Actually, I see that in the 6th or 7th post of this thread, Ich had already pointed out that this thread is a replica of a thread that Kev initiated 2 years ago, in which he asked the very same question, and the very same things were painstakingly explained to him in great detail. The word "troll" comes to mind...)

 

[starthaus]

Can you reference the part of post 1 you are claiming to be in error?

can you compare the formulas for :

-proper acelleration
-coordinbate acceleration
-relationship between proper and coordinate acceleration

in post 1 vs. post 215?

Is that you can't read math or are you just trolling?

 

[Al68]

Can you reference the part of post 1 you are claiming to be in error?

can you compare the formulas for :

-proper acelleration
-coordinbate acceleration
-realtionship between proper and coordinate acceleration

in post 1 vs. post 215?

I'll gladly observe forum rules and clarify/substantiate any claim I make. I made no claims relevant to your request.

You did however claim that post 1 was in error.

Again, can you reference the part of post 1 you are claiming to be in error?

 

[starthaus]

I'll gladly observe forum rules and clarify/substantiate any claim I make. I made no claims relevant to your request.

You did however claim that post 1 was in error.

Again, can you reference the part of post 1 you are claiming to be in error?

You are trolling. I will answer nevertheless:

Proper acceleration at the apogee (or at the drop point $r=r_0$)

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

Coordinate acceleration:$$a=-\frac{m}{r_0^2}(1-2m/r_0)$$

There is no :

$$a_0=a\gamma^3$$

(The above expression can be derived from base principles for the particular case when $$\gamma=1/\sqrt{1-(v/c)^2}$$ . You can try proving that, it might be a more useful activity than trolling.)

For the general case:

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

The above are derived in a rigorous way in the first attachment in my blog.

 

[Al68]

Proper acceleration at the apogee (or at the drop point $r=r_0$)

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

Coordinate acceleration:$$a=-\frac{m}{r_0^2}(1-2m/r_0)$$

There is no :

$$a_0=a\gamma^3$$

Now, we're getting somewhere. So $$a_0=a\gamma^2$$ at apogee?

And was this:

...the only correct formula is the one for proper acceleration $a_0=\frac{GM}{r^2}(1-\frac{GM}{rc^2})^{-1/2}$?.

a typo?

That post gave me the impression that you agreed with kev on the equation for proper acceleration in post 1 but disagreed on coordinate acceleration. Is it really vice versa?

 

[starthaus]

Now, we're getting somewhere. So $$a_0=a\gamma^2$$ at ]

apogee?

What do you think? the point is that post 1 is wrong on the issue.

 

[Rolfe2]

Proper acceleration at the apogee (or at the drop point $r=r_0$)

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

No, that's not the proper acceleration. What you've written there is the second derivative of the Schwarzschild radial coordinate r with respect to the proper time tau. 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)].

 

[Al68]

Now, we're getting somewhere. So $$a_0=a\gamma^2$$ at apogee?

What do you think? the point is that post 1 is wrong on the issue.

This is the type of non-answer that has led to hundreds of fruitless posts in this thread.

So, did it really just take over 200 posts to realize that the real issue is that you are using $$a_0=\frac{d^2r}{d\tau^2}$$ while kev was using $$a_0=\frac{d^2r'}{d\tau^2}$$, where r' is proper distance?

 

[yuiop]

It's not my fault that you don't understand basic calculus. Based on your logic someone copying a correct result and producing a bogus derivation has produced valuable work instead of junk.

This is just unwarranted abuse directed at Al68. Attacking the person rather than their argument is against the forum rules. It seems that anyone who disagrees with your opinion becomes a target for a personal attack from you.

 

[starthaus]

This is the type of non-answer that has led to hundreds of fruitless posts in this thread.

So, did it really just take over 200 posts to realize that the real issue is that you are using $$a_0=\frac{d^2r}{d\tau^2}$$ while kev was using $$a_0=\frac{d^2r'}{d\tau^2}$$, where r' is proper distance?

Since kev invariably puts in expressions by hand without defining the variables and with no derivation, there was no way to find out what he was talking about.

 

[starthaus]

No, that's not the proper acceleration. What you've written there is the second derivative of the Schwarzschild radial coordinate r with respect to the proper time tau. 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)].

Interesting. So, you are saying that:

$$a_0=\frac{d^2\rho}{d\tau^2}=\frac{d}{d\tau}(\frac{\d\rho}{d\tau})=\frac{d}{d\tau}(\frac{d\rho}{dr}\frac{dr}{d\tau})= \frac{d\rho}{dr}\frac{d^2r}{d\tau^2}=-\frac{m}{r^2}\frac{1}{\sqrt{1-2m/r}}$$

I was under the impression that proper acceleration is related to the four-acceleration $$\vec{A}=\frac{d^2\vec{R}}{d\tau^2}$$ where $$R$$ is the coordinate , not proper distance (see here).
Has the definition of proper acceleration changed recently?

 

[espen180]

Since kev invariably puts in expressions by hand without defining the variables, there was no way to find out what he was talking about.

From what I have seen, kev has been sticking to the same definitions for his variables and constants since the thread was started, and they have all benn defined many times.

However, the issue in this thread seems to be your inability to give a straightforward answer to any question.

 

[Al68]

This is the type of non-answer that has led to hundreds of fruitless posts in this thread.

So, did it really just take over 200 posts to realize that the real issue is that you are using $$a_0=\frac{d^2r}{d\tau^2}$$ while kev was using $$a_0=\frac{d^2r'}{d\tau^2}$$, where r' is proper distance?

Since kev invariably puts in expressions by hand without defining the variables, there was no way to find out what he was talking about.

I merely assumed that kev was using $$a_0=\frac{d^2r'}{d\tau^2}$$, where r' is proper distance, because that's the most commonly used definition of proper acceleration.

 

[anuraaggenius]

can anyone tell me of what is this photon made of? how is it possible for the photon to exist without mass. i want the explanantin. iknow it is massless but why?

 

[anuraaggenius]

This is just unwarrented abuse directed at Al68. Attacking the person rather than their argument is against the forum rules. It seems that anyone who disagrees with your opinion becomes a target for a personal attack from you.

this isn't the way a science student should reply. no one is born genius neither can they say that they know everything.be simple and sober. then only can u learn science.

 

[Al68]

Interesting. So, you are saying that:

$$a_0=\frac{d^2\rho}{d\tau^2}=\frac{d}{d\tau}(\frac{\d\rho}{d\tau})=\frac{d}{d\tau}(\frac{d\rho}{dr}\frac{dr}{d\tau})= \frac{d\rho}{dr}\frac{d^2r}{d\tau^2}=-\frac{m}{r^2}\frac{1}{\sqrt{1-2m/r}}$$

Now you know that can't be right, starthaus, because that would make $$a_0=a\gamma^3$$ at apogee.

 

[starthaus]

Now you know that can't be right, because that would make $$a_0=a\gamma^3$$.

You are trolling again.

 

[Al68]

You are trolling again.

Yes, I made a sarcastic post. But I hardly think you are in a position to complain about anyone else's trolling in this thread.

 

[starthaus]

I merely assumed that kev was using $$a_0=\frac{d^2r'}{d\tau^2}$$, where r' is proper distance, because that's the most commonly used definition of proper acceleration.

Hmm, I was under the impression that proper acceleration is related to the four-acceleration $$\vec{A}=\frac{d^2\vec{R}}{d\tau^2}$$ where $$R$$ is clearly the coordinate , not proper ,distance (see here). Has the definition of proper acceleration changed lately? Dalespam and I went over this issue in another thread.

 

[Al68]

Hmm, I was under the impression that proper acceleration is related to the four-acceleration $$\vec{A}=\frac{d^2\vec{R}}{d\tau^2}$$ where $$R$$ is clearly the coordinate , not proper ,distance (see here). Has the definition of proper acceleration changed lately? Dalespam and I went over this issue in another thread.

The issue of how different people define proper acceleration is a semantical issue, not worthy of argument.

What's relevant here is that in post 1, kev clearly defined $$a_0$$ as the proper acceleration measured by an accelerometer by a local stationary observer.

 

[yuiop]

At some lower radius the proper time (s) of the particle has increased and k holds the same value because it is independent of the proper time. Now it is obvious that at the lower radius the variable r has changed and so has the coordinate time (t), so if k is independent of s, then it is also independent of t and r.

Can you prove this ? By using math?

It seems so obvious, I am not sure why a proof is needed. I can not prove it any more than I can rigously prove 2+1=3. It seems obvious to any normal person, but actually proving it formally and rigorously would probably take a mathematician a whole book to do.

The best I can do is demonstrate that using k as a constant that is indepent of r produces reasonable results and using k as a function of r produces incorrect results.

Let us initially assume the value of k is $$\sqrt{1-2M/R}$$ where M and R are constants of a given trajectory and R is the apogee.

This gives:

$$\frac{dt}{ds} = \frac{\sqrt{1-2M/R}}{(1-2M/r)}$$

where r is a variable of the trajectory. Now if the location of the particle (r) coincides with the apogee (R) this gives

$$\frac{dt}{ds} = \frac{1}{\sqrt{1-2M/r}}$$

in this specific instance and is what we would expect for the difference between coordinate time and proper time for a stationary particle at R.

If the apogee of the particle is at infinity the equation reduces to:

$$\frac{dt}{ds} = \frac{1}{(1-2M/r)}$$

which is also what we would expect the time dilation ratio to be for a particle falling from infinity to r and takes both the gravitational and velocity time dilation components into account. This result has been obtained in many other threads.

Now let us assume k is a function of (r) such that $$k=\sqrt{1-2M/r}$$ where r is a variable.

This gives:

$$\frac{dt}{ds} = \frac{\sqrt{1-2M/r}}{(1-2M/r)} = \frac{1}{\sqrt{1-2M/r}}$$

This is an incorrect result because it says the time dilation ratio is independent of the height of the apogee and of the instantaneous velocity of the falling particle and so is not generally true.

It is easy to see that time dilation ratio of the falling particle does depend on the velocity of the falling particle by taking the radial Schwarzschild solution:

$$ds^2 = \alpha dt^2 - \alpha^{-1} dr^2$$

and solving for ds/dt:

$$\frac{ds}{dt} = \sqrt{\alpha - \frac{1}{\alpha} \frac{dr^2}{dt^2}}$$

From the above it obvious that ds/dt is only independent from dr/dt when $dr^2/(\alpha dt ^2)= 0$ which is not generally true.

That should be enough to convince any reasonable person that k is not a function of the variable r.

Treating k as constant is consistent with what we know about the Schwarzschild metric.

If the constant k is inserted into the Schwarzschild metric it is easy to obtain the falling velocity in terms of proper time as:

$$\frac{dr}{ds} = \sqrt{k^2 -\alpha} = \sqrt{(1-2M/R) -(1-2M/r)} = \sqrt{2M/r - 2M/R}$$

This allows us to calculate the terminal velocity of a falling particle at r when released from a height R. Setting the apogee R to infinite gives $$dr/ds = \sqrt{2M/r}$$ which is the Newtonian escape velocity at r and is related to the conversion of potential energy to kinetic energy. I guess a formal proof of the constancy of k would be based on energy considerations. I leave that to a better person.

 

[starthaus]

The issue of how different people define proper acceleration is a semantical issue, not worthy of argument.

But it is very worthy. You need to be able to choose a set of rules and stick to it.

What's relevant here is that in post 1, kev clearly defined $$a_0$$ as the proper acceleration measured by an accelerometer by a local stationary observer.

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

1. $$\frac{m}{r_0^2}$$ (my derivation based on lagrangian mechanics and K.Brown's)

2. $$\frac{m}{r_0^2\sqrt{1-2m/r_0}}$$?

 

[yuiop]

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

so, post 1 is in error. This is what we spent close to 200 posts to prove.

When the particle is at apogee and $$r=r_0$$ then the equation resolves to:

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

or:

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

(the two equations are synonymous at that instant).

This agrees with the equation given for coordinate acceleration in post #1 for the specific instance of a particle at its apogee and with the equation beautifully derived by Dalespam for the same specific instance.

 

[yuiop]

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

1. $$\frac{m}{r_0^2}$$ (my derivation based on lagrangian mechanics and K.Brown's)

2. $$\frac{m}{r_0^2\sqrt{1-2m/r_0}}$$?

K.Brown does not call equation (1.) the "proper acceleration. He calls it "the second derivative of r with respect to the proper time t of a radially moving particle" which is exactly what it is. Interesting that you are now using K.Brown in your defence when very recently you were calling him a blundering fool.

No, that's not the proper acceleration. What you've written there is the second derivative of the Schwarzschild radial coordinate r with respect to the proper time tau. 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)].

Agree.

This is the type of non-answer that has led to hundreds of fruitless posts in this thread.

So, did it really just take over 200 posts to realize that the real issue is that you are using $$a_0=\frac{d^2r}{d\tau^2}$$ while kev was using $$a_0=\frac{d^2r'}{d\tau^2}$$, where r' is proper distance?

Agree.

 

[starthaus]

K.Brown does not call equation (1.) the "proper acceleration. He calls it "the second derivative of r with respect to the proper time t of a radially moving particle" which is exactly what it is. Interesting that you are now using K.Brown in your defence when very recently you were calling him a blundering fool.

I am not "using" him as any defense. I am just pointing out that his solution coincides with mine. How about answering the question I asked Al68 instead of trolling?

 

[Al68]

The issue of how different people define proper acceleration is a semantical issue, not worthy of argument.

But it is very worthy.

It's not worthy of my time and attention. If someone chose to define $$a_0$$ as the circumference of the moon for the purpose of a post, I would accept their choice and move on instead of trying to convince him to use a different symbol. And I wouldn't try to convince him that $$a_0=\frac{d^2r}{d\tau^2}$$ instead of $$a_0=2\pi r$$.

What's relevant here is that in post 1, kev clearly defined $$a_0$$ as the proper acceleration measured by an accelerometer by a local stationary observer.

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

1. $$\frac{m}{r_0^2}$$ (my derivation based on lagrangian mechanics and K.Brown's)

2. $$\frac{m}{r_0^2\sqrt{1-2m/r_0}}$$?

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

 

[Rolfe2]

Hmm, I was under the impression that proper acceleration is related to the four-acceleration $$\vec{A}=\frac{d^2\vec{R}}{d\tau^2}$$ where $$R$$ is clearly the coordinate , not proper ,distance (see here). Has the definition of proper acceleration changed lately? Dalespam and I went over this issue in another thread.

No, the definition of proper acceleration hasn't changed. The space-time coordinates comprising the four-vector R are (and always have been) local co-moving inertial coordinates, which implies that the spatial components are the "proper" space coordinates.

By the way, it isn't a good idea to talk about "the coordinate" variables, because there are infinitely many coordinate systems, and even the so-called "proper" variables are equal to coordinate variables for some suitable choice of coordinates. To avoid confusion, it's best to just say exactly what you mean (assuming you know what you mean).

 

[starthaus]

No, the definition of proper acceleration hasn't changed. The space-time coordinates comprising the four-vector R are (and always have been) local co-moving inertial coordinates, which implies that the spatial components are the "proper" space coordinates.

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 $$u=0$$

 

[starthaus]

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. Which of the two matches "what a comoving observer dynamometer will measure"? This is not up to debate based on definition.