Is there a difference between normal and anti-gravitating geodesics?

[George Jones]

I, too, feel that the fact that Lorentz boosts have different actions under the different transformation laws is the key to what's happening.

Regards,
George

 

[vanesch]

BTW, equation (26), is this a typo ?

g_tt underscore is -1/gamma, and gamma = 1 - 2 M / r

-1/2 d/dr (-1/gamma) = MINUS m/r^2 in my book

(according to mathematica, it is -m/(-2m+r)^2, which reduces to -m/r^2 in the case of r>>m.

If this is a sign error, then, eh, your a-particle falls down...

 

[vanesch]

Hi again,

The new paper by Sabine made it possible for me to find out how one can deduce the quantities from the metric representation in a given coordinate system, so I programmed this in a few mathematica notebooks.

In hossi1.nb, I calculated the E, tau, g_underscore and the gamma_underscore components in the Schwarzschild metric. Apart from the previously mentionned typos, I find agreement with her calculations in appendix 1.

In hossi2.nb, I did the same for the FRW metric. I find the same coefficients, but with an index permuted systematically, in appendix B. I wonder if it is me or Sabine.

In hossi3.nb, as a joke, I put in the Minkowski metric, and of course I find 0 everywhere.

The next step will be to find the expression for the a-particle geodesic.

 

[vanesch]

Ok, I'm stuck with equation (13). Of course I can work out the coefficients (using the previously calculated expressions), but then you end up with a funny system of "differential equations" which do not make sense. I didn't realize this immediately, but equation 13 does not, after all seem to be an equation of a world line, or curve or what so ever. A normal equation of a world line would have COORDINATES as functions of lambda (so that the coefficients, expressed in coordinates, become also functions of the unknowns and that we have a genuine set of differential equations of coordinates as a function of lambda). Its solution would then give you the 4 unknowns as a function of lambda, and hence, trace out a world line on the manifold. But apparently the x-underscore quantities are NOT coordinates. Then equation (13) doesn't make much sense to me as defining a world line, and doesn't even make much sense as an equation. How do the coefficients now depend on the unknowns x-underscore ?
Is it possible to write down a genuine equation of a world line so that we can KNOW what we are talking about all the time ?
Or do we have to bluntly substitute the derivatives of x-underscore to lambda, by the expression given in the text underneath it ? But how does the second derivative act upon this ?

 

[hossi]

Ok, I'm stuck with equation (13). Of course I can work out the coefficients (using the previously calculated expressions), but then you end up with a funny system of "differential equations" which do not make sense. I didn't realize this immediately, but equation 13 does not, after all seem to be an equation of a world line, or curve or what so ever. A normal equation of a world line would have COORDINATES as functions of lambda (so that the coefficients, expressed in coordinates, become also functions of the unknowns and that we have a genuine set of differential equations of coordinates as a function of lambda). Its solution would then give you the 4 unknowns as a function of lambda, and hence, trace out a world line on the manifold. But apparently the x-underscore quantities are NOT coordinates. Then equation (13) doesn't make much sense to me as defining a world line, and doesn't even make much sense as an equation. How do the coefficients now depend on the unknowns x-underscore ?
Is it possible to write down a genuine equation of a world line so that we can KNOW what we are talking about all the time ?
Or do we have to bluntly substitute the derivatives of x-underscore to lambda, by the expression given in the text underneath it ? But how does the second derivative act upon this ?

Dear vanesh,

I can't avoid being flattered by your attention. Or is it vengeance of the GR-defenders?

First, I apologize for not having much time to be around at PF these weeks.

Second, the question about Eq (13) should be answered in the three points listed directly below this equation. Alternatively, you can rewrite the equation into an equation for the tangential vector (this is in the first paper, and I didn't repeat it in the 2nd, maybe I should have). The x-underscores are coordinates - as explained in the footnote on this side. The underscore is just a notation to remind you that you are currently investigating the world-line of an a-grav. particle. Well, you can drop the underscore if you just keep in mind what question you are currently investigating. There are no space-time coordinates that belong to the underlined g. The basis in the underlined TM's is not a basis of partial derivatives of any kind.

Third, I unfortunately can't check on the index or its permutations right now but will do so asap. (Have no mathematica license here). Do we have the same definition of angles in the metric (I sometimes mix up phi and theta).

Forth, I apologize for the confusion with the Lorentz boosts. From reading the above, it occurred to me that the motivation is very sloppy. The product in the group is meant to be the one that preserves the structure, so the relation between the representations is \eta^-1 G^T \eta = G^{-1}, which enters the definition of \tau and was the actual form that I used, but it turns out that \eta doesn't change the following calculation (the minus drops out). (As a sideremark I would have been surprised if there had been an essential difference for euklidean spacetime regarding the Lagrangian). The 'new' particle transforms under Lorentz boosts as the usual one does. I should make that clearer in an updated version, thanks very much for pointing it out.



B.

 

[vanesch]

I can't avoid being flattered by your attention. Or is it vengeance of the GR-defenders?

There are not many "personal feelings" mixed into this, so there's no vengence. It's just that it is probably more productive to do explicit calculations than repeating my argument for the 25th time.
In fact, since your second paper, it is clear now - I think - that all the quantities can be expressed as a function of the metric representation (which doesn't necessarily mean that they have geometrical existence). I want to find out if the final a-geodesics depend upon the choosen coordinate system in which they are calculated, whether (which comes down to the same) there's some arbitrary choice to be made at a certain point, or whether they indicate that they fall down like normal geodesics (maybe different from normal geodesics, but only on the level of tidal effects). These are the three options that I keep for possible, which you will of course dispute.
So one of both of us is in for a surprise and instead of arguing endlessly over it, it is probably much easier to just do the entire calculation.

Second, the question about Eq (13) should be answered in the three points listed directly below this equation.

Yes, but I don't know what to do with the derivative to lambda in the first term. Do I FIRST substitute d (x-underscore)^(alpha-underscore)/d lambda by the (inverse of) the last equation on p 7 and THEN I apply the derivative to lambda, or vice versa ?
Because both are of course not equivalent: the derivative to lambda will act upon the elements in the tau matrix of course in the first case and not in the second.

In other words, if I consider the t-underscore^alpha in the last equation on p 7 (in the second point of the algorithm) to be the normal tangent vector to the curve we want to obtain (so, dx-underscore^alpha/dlambda) then tau is function of these coordinates (at different values of lambda, we are at different points on the manifold, and hence tau has a different expression at this other point).

This is why I don't understand the "first instruction": integrate equation (13) once. You cannot integrate equation (13) as it stands, because the tau and gamma coefficients are expressed as a function of the coordinates on the manifold, while the unknowns are not those coordinates, so the coefficients are not functions of the unknowns - hence it is not a differential equation in the usual way of speaking.
I stuck on it when writing it out in mathematica.

Alternatively, you can rewrite the equation into an equation for the tangential vector (this is in the first paper, and I didn't repeat it in the 2nd, maybe I should have). The x-underscores are coordinates - as explained in the footnote on this side. The underscore is just a notation to remind you that you are currently investigating the world-line of an a-grav. particle. Well, you can drop the underscore if you just keep in mind what question you are currently investigating. There are no space-time coordinates that belong to the underlined g. The basis in the underlined TM's is not a basis of partial derivatives of any kind.

I prefer that approach. I'll look it up again.EDIT: could it be that there is a typo in eq. (19) of your first paper ? The last alpha is a lower one, and I think it should be an upper one.

Third, I unfortunately can't check on the index or its permutations right now but will do so asap. (Have no mathematica license here). Do we have the same definition of angles in the metric (I sometimes mix up phi and theta).

If it is just to read the notebooks, you can always download mathreader freely from Wolfram's site.

 

[vanesch]

Progress...

Hello Sabine and others,

Ok, I think that I finally managed to finish the calculation. I completed the calculation for the Schwarzschild metric in the attached notebook hossi1a.nb. I added comments so that the calculation should be readable.

I started from equation (39) in the first paper. However, there were two modifications necessary. You say that you can substitute equation (19), but tau is not in the right from there (it is the opposite tau that appears in (39)). So I hope I introduced the correct "covariant derivative" of this other tau.
The other point was the covariant derivative of the tangent vector to the curve, which needs to be re-written in order to be able to use the derivative towards the affine parameter of the curve. These substitutions are commented in the notebook.

In the end, I do find that, for an a-geodesic which starts out at rest (tangent vector = (1,0,0,0)), the derivative of the tangent vector corresponds to an upward acceleration with a radial derivative of +M/r^2, as you announce.

Could you check whether this calculational scheme is correct ?

 

[vanesch]

Conclusion ...

Finally, I applied the same calculation sheet to a Rindler coordinate system. It is described on p 173 of MTW, but the (T,X,Y,Z) coordinate system is the coordinate system of an observer which is accelerated in the PLUS Z direction wrt an inertial frame, in flat space with an acceleration + GG.

So, if the "a-geodesics" are world lines, the calculation in this coordinate system should give me an acceleration of - GG (as does a normal geodesic), because in flat space, a-geodesics are the same as normal geodesics.

Well, at the end of the calculation, I find an acceleration IN THE PLUS Z DIRECTION.

So this clearly shows that the "a-geodesic" is dependent on the frame in which it is worked out, as I was claiming all along.

It is the famous particle that "accelerates away" from you when you accelerate towards it, and the "a-geodesics" are not geometrical world lines.

Unless there's a mistake in the calculation of course...

Your Honor, I rest my case.

 

[vanesch]

for completeness...

Just for completeness, I did also the calculation for the FRW metric. I also found the error which gave the difference with the paper: the function a[t] was used, and I summed over an index a.
This is corrected, and now there is agreement with the coefficients in appendix B of the second paper.

 

[hossi]

Finally, I applied the same calculation sheet to a Rindler coordinate system. It is described on p 173 of MTW, but the (T,X,Y,Z) coordinate system is the coordinate system of an observer which is accelerated in the PLUS Z direction wrt an inertial frame, in flat space with an acceleration + GG.

So, if the "a-geodesics" are world lines, the calculation in this coordinate system should give me an acceleration of - GG (as does a normal geodesic), because in flat space, a-geodesics are the same as normal geodesics.

Well, at the end of the calculation, I find an acceleration IN THE PLUS Z DIRECTION.

So this clearly shows that the "a-geodesic" is dependent on the frame in which it is worked out, as I was claiming all along.

It is the famous particle that "accelerates away" from you when you accelerate towards it, and the "a-geodesics" are not geometrical world lines.

Unless there's a mistake in the calculation of course...

Your Honor, I rest my case.

Hi vanesh,

could you please explain what exactly you have done? I don't have MTW here, so I can not look up the reference. If you could refer to

http://en.wikipedia.org/wiki/Rindler_coordinates"

that would be more useful. You have taken Rindler coordinates in flat space. And computed a geodesic in this space? And then you have computed the anti-geodesic? I don't really get what acceleration you are talking about. Both have no acceleration. The Rindler coordinates belong to an observer that is accelerated but that is not a geodesic. You know all that, just that from your description it is not clear to me what you actually have computed.

B.

 

[vanesch]

Hi vanesh,

could you please explain what exactly you have done? I don't have MTW here, so I can not look up the reference. If you could refer to

http://en.wikipedia.org/wiki/Rindler_coordinates"

that would be more useful. You have taken Rindler coordinates in flat space. And computed a geodesic in this space? And then you have computed the anti-geodesic? I don't really get what acceleration you are talking about. Both have no acceleration. The Rindler coordinates belong to an observer that is accelerated but that is not a geodesic. You know all that, just that from your description it is not clear to me what you actually have computed.

B.

Hi Sabine,

Mm, maybe these coordinates are not called Rindler coordinates, I don't know.

In MTW, they take on the following form. We are in flat space, and consider x0,x1,x2,x3 an inertial frame (Minkowski coordinates).

Then the thing that I called Rindler coordinates, are given by the following transformation, and are called \xi0, \x1...

The transformation is given by:
$$x^0 = (1/g + \xi^1) sinh(g \xi^0)$$
$$x^1 = (1/g + \xi^1) cosh(g \xi^0)$$
$$x^2 = \xi^2$$
$$x^3 = \xi^3$$

Now, I took the liberty to flip the 1 and the 3 (to use z instead of x).
The associated metric is given by:
$$ds^2 = \eta_{\mu,\nu}dx^{\mu}dx^{\nu} = -(1+g\xi^1)^2 (d\xi^0)^2 + (d\xi^1)^2 + (d\xi^2)^2 + (d\xi^3)^2$$

But the important point is that the $$\xi^{\mu}$$ coordinates are "the coordinates relative to the accelerated observer" and this "accelerated observer" is accelerated with acceleration g wrt the inertial frame in the x1 direction.
In other words, this is the rocket frame.

If you doubt about this, put a point "at rest" in the \xi frame, so that \xi = (T,X,0,0). For small g \xi0, you have a Newtonian approximation, and you find that
x0 ~ T
x1 ~ g^(-1) + X + g/2 T^2

In other words, up to a fixed translation 1/g, you find that x1(t) = X + g/2 t^2, which proves that a point at the "origin" in the T,X,Y,Z frame accelerates with acceleration g in the +x1 direction in the Minkowski frame.

(and then I preferred to work with Z instead of with X).

Now, if you want to know in detail what I did in the calculation, the best you can do is to check the mathematica notebook (using the free Mathreader from wolfram) ; I provided comments.

In short, I wrote out all quantities in equation (39) of your first paper as a function of the metric (as you do, in your examples), and I calculate the different elements of d t^a / d lambda as a function of the metric and initial values of t^a. I take the initial values to be (1,0,0,0) (a particle at rest).

I succeeded in reproducing your "upward falling" particle in the Schwarzschild coordinates that way, where I found a positive radial acceleration of M/r^2 (well, to be precise, of M(r - 2M)/r^3 ) when I gave as initial condition t^a = (1,0,0,0) = particle at rest.
I also reproduced similar calculations for the FRW metric. For instance, a particle at rest in FRW, stays at rest, compatible with your ci = 0 initial condition in equation (38). I didn't check other cases, but you're free to do so, the equations are printed out in the worksheet hossi2a.nb.I then applied the identical worksheet to these "Rindler" coordinates (well, to the \xi^a, which I called T,X,Y,Z, and the acceleration was in +Z, not in +X). I found, solving for your equation (39) in the same way, a POSITIVE acceleration in the Z direction (that is, the t^3 component had a POSITIVE derivative equal to g + g^2 Z). Now, a normal particle undergoes in such an accelerated frame of course a "negative" acceleration, that is, the derivative of the Z component of t should be something like - g.
But as this is nothing else but a coordinate transformation, and if we would have worked in normal Minkowski coordinates in this space, you would have found no acceleration for both a normal particle and an a-particle and both world lines would be identical, this means that the "world line" is dependent on the coordinate system where you work this out.

 

[vanesch]

The transformation is given by:
$$x^0 = (1/g + \xi^1) sinh(g \xi^0)$$
$$x^1 = (1/g + \xi^1) cosh(g \xi^0)$$
$$x^2 = \xi^2$$
$$x^3 = \xi^3$$

When I look at the Wiki entry, I think that it is about the same transformation, except for a translation \xi^1 + 1/g --> \xi^1' and a normalisation g \xi^0 --> \xi^0'

 

[vanesch]

In fact, I just added the anti-geodesic equation in the radial case for the FRW metric (hossi2b.nb).

I find the equation d t^t/d lambda = a(t) a'(t) (t^r)^2
and the other t are constant.

This doesn't correspond with your (36) and (37), but those are expressed with these underscore quantities, while I give you the tangent vector along the line (according to your paper1 equation (39)).

I find hence that the time dilatation gets the opposite sign, than for the normal geodesic (your equation (34) in paper2) which, I guess, is what you are saying with the red/blue shift flip.

I admit being a bit puzzled by the absence of any acceleration in the r direction.

 

[hossi]

Hi vanesh,

I am sorry cause I can't use the mathematica stuff. I can open the files but can't execute them. So I am poking around in the dark. I read in your text (hossi4a) So, as said, we'll need the "covariant derivative" of the second form of tau. This flips the signs of the two terms (at least if I understood how things are done). Does this refer to Eqs (18) and (19) of the first paper? And if so, why do you change both signs? The signs change when an index is pulled up or down, only one index changes from (18) to (19). It's the 'normal' index, thus only first sign changes, which belongs to the 'normal' connection. Does that help? And yes thanks, indeed there is a typo in Eq (19), the last alpha should be up, not down.

B.

 

[hossi]

Yes, but I don't know what to do with the derivative to lambda in the first term. Do I FIRST substitute d (x-underscore)^(alpha-underscore)/d lambda by the (inverse of) the last equation on p 7 and THEN I apply the derivative to lambda, or vice versa ?
Because both are of course not equivalent: the derivative to lambda will act upon the elements in the tau matrix of course in the first case and not in the second.

Hi vanesh, I just understood the problem. First convert the index, then take the derivation. The other option belongs to the usual geodesics, just with the tangential vector expressed in the basis of TM, which does't change the curve. It becomes clearer from the equation for t, or from the defining equation resp. But you have figured that out by now anyway, as I see from your above posts.

B.

 

[vanesch]

Hi vanesh,

I am sorry cause I can't use the mathematica stuff. I can open the files but can't execute them. So I am poking around in the dark. I read in your text (hossi4a) So, as said, we'll need the "covariant derivative" of the second form of tau. This flips the signs of the two terms (at least if I understood how things are done). Does this refer to Eqs (18) and (19) of the first paper? And if so, why do you change both signs?

When looking at your equation (39) first paper, I take it that this $$t^{\nu}$$ is the equation of the tangent vector of the world line of the a-particle, right ? (I drop the underscores here)
So $$t^{\nu}$$ is simply $$d x^{\nu}(\lambda) / d\lambda$$ where $$x^{\nu}(\lambda)$$ is the coordinate description of the world line on the manifold of this a-particle, right ?

Ok, in order to write this equation fully out in known quantities, I needed two things. The first thing I needed was to find, in the last term, the expression for the covariant derivative of tau. Now, you say, use equation (19), but equation (19) gives you the covariant derivative of the OTHER tau (upper and lower interchanged). So I had to find out myself what was the expression for THIS tau.

I wrote that it is:
$$D_{\nu} (\tau_{\kappa})^{\alpha-u} = \partial_{\nu} (\tau_{\kappa})^{\alpha-u} - (\Gamma^{\beta})_{\nu \kappa} (\tau_{\beta})^{\alpha-u} + (\Gamma^{\alpha-u})_{\nu \beta-u}<br /> (\tau_{\kappa})^{\beta-u}$$

(I use D for del)

as an equivalent for (19) for the term I needed. Could you confirm/correct this ? (I use -u for "underscore")

Next, the second point I needed, was that in the expression (39), there's a (normal) covariant derivative of t, but as t is only defined as a function of lambda, I needed to rewrite this.
I wrote $$d t^{\alpha}/d\lambda + t^{\nu} (\Gamma^{\alpha})_{b \nu}t^b$$
for the first term, given that the tangent vector is a normal vector.
(I replaced the covariant derivative by the coordinate derivative + connection term, and then contracting with t^nu, I replaced $$t^{\nu}\partial_{\nu}t^{\alpha}$$ by $$d t^{\alpha}/d\lambda$$.

I hope that's correct too. These were the only places where I needed to take some "initiative" because this was not written down explicitly.

After all that, I have my equations for the derivatives of t to lambda, as a function of those t themselves, and coordinate expressions.

 

[vanesch]

Hi vanesh,

I am sorry cause I can't use the mathematica stuff. I can open the files but can't execute them.

Yes, with the reader you can only view them, but as I executed them in Mathematica, you can read the results of the commands...

Although it is a bit clumsy, the expressions should speak for themselves if you are used to Mathematica. (and with a bit more effort and guess work, even if you are not).

I'll try to explain the gist of the approach: I use two kinds of objects: matrices (nested lists with depth 2 - standard convention in Mathematica), and functions.

On the first line, g, for instance, is a matrix. To address elements of a matrix, one writes g[[ a, b]] where a and b are the number of the row and the column, respectively (from 1 to 4).

On the next line, I print it out in "matrixform".

Same for eta: it's a matrix.

Now, E1 is your famous E matrix (but E exists in mathematica). Now, I think there's a bug there (or, better, a limitation), in that I'm not sure that Sqrt takes the MATRIX square root, should check it. But for diagonal matrices as we are using here, that doesn't matter.

So E1 is a matrix too.

tau1 is calculated from E1 (Inverse is the instruction for matrix inversion, Transpose for transposition, should be evident), and printed out.

I print out the result of tau1 in "matrix form". Tau1 is the matrix of the form of tau as in your equation (22) second paper.

gunder is the inverse matrix of g (and hence also the matrix of g^{a,b}).
I print it out.

tau2 is the other form of tau, which is the inverse transposed of tau1. It is also a matrix, and printed out.

And then things get a bit more involved.

First, var is a list of 4 symbols, which represent the 4 coordinates (in which the elements of all these matrices are expressed). I need this in order to be able to translate something like partial_3 (g_{1,1}):
I take g_{1,1}, which is, in mathematica, g[[1,1]], and then I have to compute its derivative wrt the third variable. The symbol of the third variable is var[[3]], so this derivative becomes:
D[g[[1,1]] , var[[3]] ]
(D is the derivative function in mathematica: D[f(x,y),x] is \partial_x f )

Mathematica doesn't know the Einstein summation convention, so I have to explicitly sum over, eh, sums:

Sum[expression,{i,1,4}] will be the sum of expression, 4 times, where i takes on respectively the values 1,2,3, and 4.

gammaunder[nu_,lambda_,kappa_] is NOT a matrix, but a function (of 3 numbers, all within range 1 to 4). That is, if you type gammaunder[1,3,3], it will return you an expression, supposed to be the Christoffel symbols of the kind you give in appendix A.

In its description, one has to imagine that at the moment of calling, the nu, lambda and kappa take on the numerical values given by the caller.

I first need to sum over the index alpha for the entire expression and will need to further sum in certain terms over k and aa.

With what I've explained, you should be able to verify that this implements your expression (10) of the second paper.
Proof of the pudding is the eating:

I next construct the list of all the possible calls to the function gammaunder, in the big list gams. I do this by looping over the three call parameters, from 1 to 4.
Then I print out the result, together with the corresponding call parameters:
So the first line is 1 1 1 0, which means that gamma^t_{t,t} = 0
(1 = first variable, = t).
As such I reconstruct the list you give in appendix A (equation (83)).
You can check it, we are in agreement, which is also a proof that this approach works well.

Next, I apply the same technique to calculate the normal Christoffel symbols, and I print out the list.

The function deltau1 is made in a similar way. It is the implementation of equation (19) of your first paper. And then I realized I didn't need it.

deltau2 is again build up in a similar way. It is the implementation of the covariant derivative of tau2, as in my previous message. It is the equivalent of equation (19), but for this form of tau which appears in the last term of equation (39), first paper.

Next step, I define a list of 4 functions, which is going to represent the t in equation (39): t1[lambda], t2[lambda], t3[lambda] and t4[lambda].
These are abstract functions in mathematica, it only says that they are functions of lambda (will be nice to take the derivative to lambda).

speedequation[alpha_] is a function that will write out explicitly equation (39) for a given numerical value of alpha (1 to 4), in the form described in my previous message.

speed[[nu]] stands for t^{nu}. For instance, speed[[3]] is t3[lambda].

I make the list of the 4 equations (by calling speedequation[ww] for ww 1,2, 3 and 4), which gives me the set of 4 equations, and this list of equations is called speedeqs.

In it, we have the expression of the derivatives of t1, t2,... to lambda, the functions t1, t2... themselves, and expressions as a function of the coordinates.

I next use a replacement rule in which I replace the functions t1, t2, t3 and t4 themselves (but not their derivatives) by 1, 0, 0, and 0.
This comes down to setting d x^0 / dlambda = 1, and d x^1/d lambda ... = 0, in other words, a particle which is initially, as a function of lambda, at rest in the coordinate frame. The d x^0 / dlambda = 1 makes that the rate of change of x^0 (the frame time coordinate) equals the curve parameter lambda change (so initially, the curve is parametrised in coordinate time).

As such, the derivatives of t1, t2 and t3 represent the "accelerations" in the given coordinate system (the second derivatives to time = curve parameter of the coordinate values).

 

[vanesch]

general solution

I calculated the general solution for a diagonal metric, and I also calculated it according to a second method, starting from equation (13) (or 38 in the first paper). Indeed, the t^(a-underscore) quantities are functions of the t^a quantities via the tau matrix, so it is sufficient to substitute them (taking into account the derivative of the tau matrix itself) into equation (13) to obtain an equation for the components of the normal tangent vector t^a.

The reassuring thing is that this gives us the same equations as through the derivation from equation (39), so both are consistent. This gives us a good check on our calculation.

All this is done in hossi5.nb. We see that in the end, the equation of motion for an initially stationary a-particle is relatively simple (equations obtained through both the methods):

(shift from indices 1,2,3,4 to 0,1,2,3)

if we had ds^2 = g_00 dt^2 + g_11 dx^2 + g_22 dy^2 + g_33 dz^2

(note the positive sign of g_00: minus sign to be included in the definition of g_00)

we obtain for the equations of the a-geodesic, starting from (1,0,0,0) for the tangent vector:
2 g_00 t0' = - \partial_0 g_00
2 g_11 t1' = - \partial_1 g_00
2 g_22 t2' = - \partial_2 g_00
2 g_33 t3' = - \partial_3 g_00

For comparison, we applied the same reasoning to the same metric for the normal geodesic equation in general relativity. The calculation can be found in geodesic1.nb

We find the same equations up to a sign flip:
2 g_00 t0' = - \partial_0 g_00
2 g_11 t1' = + \partial_1 g_00
2 g_22 t2' = + \partial_2 g_00
2 g_33 t3' = + \partial_3 g_00

So this shows us that the a-particle "geodesic" acceleration has a flipped sign for the 3 space components as compared to the normal geodesic acceleration. Which was indeed what we found, for the Rindler case, for instance, proving in much more general terms, the coordinate system dependency of these "a-geodesics".