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

[twin]

Hi Sabine,

I think there cannot be any difference between geodesics for "ordinary" and "anti-gravitating" particles for the following reasons:

In (3) you define the (ordinary) tangent bundle as a bundle of "column vectors" with diffeomorphisms operationg from the left, and the (ordinary) cotangent bundle as bundle of "row vectors" with diffeomorphisms operating from the right. Whether you define your bundles as row or column vectors is of course only a matter of convention.
The underlined bundles defined in (4) are nothing else but the cotengent bundle and the tangent bundle, but with the opposite conventions: The underlined TM is the cotangent bundle, considered as bundle of column vectors, and the underlined TM* is the tangent bundle, considered as bundle of row vectors. Therefore, your anti-gravitating particles essentially live in the cotangent bundle.
Your map \tau, which identifies the ordinary and underlined bundles, is nothing else but the canonical identification of tangent and cotangent bundle provided by the metric (and therefore it comes as no surprise that the composite map defined in (7), which identifies the two tangent bundles, does not depend on the metric and is merely given by a transposition).
You go on and define the covariant derivative for the underlined bundles. However, since you use no extra data apart from the metric, this must result in the ordinary covariant derivative in the cotangent bundle (modulo transpositions perhabs).
The geodesics for this covariant derivative are nothing else than the ordinary geodesics.


Twin

 

[Careful]

**
your objection was the toughest one. It took me some while to figure out the reason for my confusion. The point is, your comment is right, but you have asked the wrong question.

The essence of the approach I have proposed is that there are two ways a field can behave under coordinate trafos. The one is mapped to the other by \tau. Whether you like that or not, it is handy. I personally don't like it, but I couldn't come up with anything better. Note that the elements of both spaces don't transform alike. **

I was already way past that issue.

** How could they possibly have the same properties under parrallel transport?! They can't (unless spacetime is globally flat.) **

Their properties under parallel transport map to each other under \tau.

** You see that most easily by looking at the covariant derivative. For a field A in TM (or higher products), you have two ways to introduce a covariant derivative. The one is the usual \nabla A. The other one is \tau \nabla \tau^{-1} A. (You have to distribute indices according to what A is.) The latter means that you charge-conjugate your field, transport it, then push it back to the manifold. Since \nabla \tau does not vanish, the result of both is not identical. **

Well, two comments. First, you say that A is in TM, so I cannot apply \tau^{-1} on it. Second, I presume you just mean that I should consider \nabla_{\beta} g_{\gamma \kappa} A^{\kappa} instead of \nabla{\beta} A^{\kappa}. But that does not change zip, since the metric goes through the covariant derivative.


**
What you have to consider instead is taking the derivatives \tau \nabla \tau^{-1} (which have the same commutation behaviour as the usual ones). **

But that does not make sense : \tau is supposed to be a map from TM to \underline{TM} so, it should map a basis in TM to a basis in \underline{TM}. The \underline{\partial} you wrote down here is the \partial in the operational sense ON \underline{TM}. But that is meaningless since the latter is not an intrinsic operation.

Look, it is really very simple. Given the metric g, there is only one prescription to go to the connection which does not require extra fields INDEPENDENT of it and that is the Christoffel prescription. For example: the introduction of spinors requires *extra structure*, that is a spinor bundle attached to a preferred tetrad.

Cheers,

Careful

 

[Careful]

Hi Sabine,

I think there cannot be any difference between geodesics for "ordinary" and "anti-gravitating" particles for the following reasons:

In (3) you define the (ordinary) tangent bundle as a bundle of "column vectors" with diffeomorphisms operationg from the left, and the (ordinary) cotangent bundle as bundle of "row vectors" with diffeomorphisms operating from the right. Whether you define your bundles as row or column vectors is of course only a matter of convention.
The underlined bundles defined in (4) are nothing else but the cotengent bundle and the tangent bundle, but with the opposite conventions: The underlined TM is the cotangent bundle, considered as bundle of column vectors, and the underlined TM* is the tangent bundle, considered as bundle of row vectors. Therefore, your anti-gravitating particles essentially live in the cotangent bundle.
Your map \tau, which identifies the ordinary and underlined bundles, is nothing else but the canonical identification of tangent and cotangent bundle provided by the metric (and therefore it comes as no surprise that the composite map defined in (7), which identifies the two tangent bundles, does not depend on the metric and is merely given by a transposition).
You go on and define the covariant derivative for the underlined bundles. However, since you use no extra data apart from the metric, this must result in the ordinary covariant derivative in the cotangent bundle (modulo transpositions perhabs).
The geodesics for this covariant derivative are nothing else than the ordinary geodesics.


Twin

Exactly, that is the ``non-conventional isomorphism´´ I was talking about.

Cheers,

Careful

 

[vanesch]

Both free-falling frames are not identical, but for both seperately, the eq. principle holds. That's why I speak of a relaxation of the equivalence principle, not of a generalization. You are of course completely right that in a gravitational field, there is no elevator that could make BOTH particles at once not notice it. That is exactly the point. Both particles will be indistinguishable if and only if spacetime is globally flat.

It depends on what you mean by "globally". If you mean by that, the *entire* manifold, ok, but then there's nothing that remains of the equivalence principle (we now have something like an ether theory).
But if you mean by "globally" in a finite patch of spacetime, instead of a point, then I don't follow you.

If you want to apply that to your accelerated observer in (globally) flat
space you have to compare the one particle to acceleration a, the other one to acceleration -a. That is, you can not map both to the same Rindler-observer, since the ratio of intertial to gravitational mass is +1 for the one, and -1 for the other.

?? So for a GIVEN Rindler metric, does the a-particle fall as a normal one, or not ?

If your answer is yes (which means that in the local Lorentzian metric over the patch - which is possible, it is a flat space - both describe a uniform motion. But then I have the problem that we have THE SAME METRIC for an observer at the surface of the earth, so same metric -> same equations -> same geodesic, and hence your a-particle falls just as well down as a normal particle.

Now, if your answer is no, it means that in the local Lorentzian metric over the patch, we DO NOT have uniform motion of the a-particle, and this in flat space, which is against what you postulated ? So I take it that this is impossible.

The answer to your question is of course yes, the motion of the a-particle can be derived from the local metric (over a patch of spacetime). Just that this is not exactly the easiest way to do it.

Difficult or not, if this is *in principle* possible, I don't see how this can lead to an "up falling" equation of motion at the surface of the earth.

In the case your answer was "yes" to the first discussion, I'd think that the SAME metric must give rise to the SAME equation of motion, no ? If this is the only thing from which it is derived. And there, it was "downfalling" in a Rindler set of coordinates.
Otherwise we have the funny situation that, in flat space, the particle, in a Rindler metric, falls down, and at the surface of the earth, in just as flat a space, in the Rindler metric, the particle falls up.

Because at the surface of the eath, there is (almost) no difference between this situation, and the free, flat space situation.

How can you obtain a totally different equation of motion, starting from exactly the same inputs (the metric over a local patch) ?

The "metric" of the accelerated observer in flat space you look at

ds^2 = -(1+aZ)^2 dT^2 + dX^2 + dY^2 + dZ^2,

Is not in space-time coordinates (see e.g. MTW 6.18), therefore you can't just take the inverse and get the geodesics.

That's a coordinate system as any other, no ?
It is in fact the coordinate system you would normally obtain "in a rocket".

For small a.z, you can expand this, and this comes down to the Newtonian transformation:

T = t
X = x - a/2 t^2
Y = y
Z = z

and this coordinate system, and its metric, is just as "real" as the Lorentzian coordinate system in flat spacetime, right ?

 

[hossi]

In (3) you define the (ordinary) tangent bundle as a bundle of "column vectors" with diffeomorphisms operationg from the left, and the (ordinary) cotangent bundle as bundle of "row vectors" with diffeomorphisms operating from the right. Whether you define your bundles as row or column vectors is of course only a matter of convention.
The underlined bundles defined in (4) are nothing else but the cotengent bundle and the tangent bundle, but with the opposite conventions: The underlined TM is the cotangent bundle, considered as bundle of column vectors, and the underlined TM* is the tangent bundle, considered as bundle of row vectors. Therefore, your anti-gravitating particles essentially live in the cotangent bundle.

In globally flat space it is a matter of convention (when I say globally, I mean on the whole manifold). In a curved-space you usually don't go from the tangential space to the co-tangential space by changing "row vectors" into "column vectors". Instead, you use the metric. Therefore, the space TM^* is not identical to TM. You can eighter use the metric to go from TM to TM^*, or you use \tau to go from TM to TM^* .

You see from the first some equations, that the elements of both spaces do not transform identically. You get an element of TM to look like that of TM^*, by applying the transposition. You call it operating from the left or from the right: but this is not how, in GR, you usally identify tangential and co-tangential space.

You go on and define the covariant derivative for the underlined bundles. However, since you use no extra data apart from the metric, this must result in the ordinary covariant derivative in the cotangent bundle (modulo transpositions perhabs).
The geodesics for this covariant derivative are nothing else than the ordinary geodesics.

They are identical if and only if the covariant derivative on \tau vanishes. In general, this will not be the case. E.g. in terms of the tetrads (Eq 10), \tau is (EE^T){-1}, which simplifies if E is diagonal to E^-2, the covariant derivative on which usually doesn't vanish. The covariant derivative in the cotangent bundle is the same as always. You can of course formulate the geodesic equation for tangential or co-tangential vectors. It looks different, but is the same curve, the reason for which is essentially that \nabla g_munu vanishes.



B.

 

[hossi]

Their properties under parallel transport map to each other under \tau.

Well, two comments. First, you say that A is in TM, so I cannot apply \tau^{-1} on it. Second, I presume you just mean that I should consider \nabla_{\beta} g_{\gamma \kappa} A^{\kappa} instead of \nabla{\beta} A^{\kappa}. But that does not change zip, since the metric goes through the covariant derivative.

I am sorry, I might have mixed up \tau with \tau^-1 . Let us define TM: A -> \tau A \in TM. Okay, then I agree, it should have been \tau^-1 \nabla \tau acting on A in TM. The metric goes through the covariant derivative but \tau doesnt. That's the reason why you can of course 'relate' the properties of elements in TM to those in TM under parallel transport, but the resulting curves are not identical (see Eq (39).

But that does not make sense : \tau is supposed to be a map from TM to \underline{TM} so, it should map a basis in TM to a basis in \underline{TM}. The \underline{\partial} you wrote down here is the \partial in the operational sense ON \underline{TM}. But that is meaningless since the latter is not an intrinsic operation.

It is meaningless, or is it \partial in the operational sense? I didn't use the basis in \underline TM as an operator, that's why I explicitly wrote it's a notation.

Look, it is really very simple. Given the metric g, there is only one prescription to go to the connection which does not require extra fields INDEPENDENT of it and that is the Christoffel prescription. For example: the introduction of spinors requires *extra structure*, that is a spinor bundle attached to a preferred tetrad.

Look, is it really very simple? The prescription to go the Christoffelsymbols is unique. But it matters what field you transport whether the Christoffelsymbols are the symbols to use or not. The extra structure is the properties of elements in TM to react to general coordinate transformations. If I remember that correctly, you can express the connexion for spinors in terms of the usual Christoffels by using the generators of the approriate spinor representation. In which sense is this more independent of the usual Christoffels than in my scenario?



B.

 

[hossi]

It depends on what you mean by "globally". If you mean by that, the *entire* manifold, ok, but then there's nothing that remains of the equivalence principle (we now have something like an ether theory).
But if you mean by "globally" in a finite patch of spacetime, instead of a point, then I don't follow you.

By globally I always mean on the entire manifold. I guess we have some disagreement on the equivalence principle. As I understand it it says: the local effects of gravity in curved space are identical to those that an uniformly accelerated observer in globally flat space experiences. The uniformly accelerated observer in globally flat space is your Rindler observer. Now you have two of them, with positive or negative acceleration. The point is to relate the local properties in curved space-time (General Relativity) to those of Special Relativity (which does not require dealing with curved spaces, otherwise you haven't won anything).

Since space is globally flat for the accelerated observer, you can of course then say, well, then it's the same everywhere and I only look at a local piece again. That I think, is what you are doing (?). But that doesn't change the fact that the space for the Rindler observer is globally flat, and, in particular, if this observer is uniformly accelerated, he doesn't move on a geodesic.

?? So for a GIVEN Rindler metric, does the a-particle fall as a normal one, or not ?

If your answer is yes[...]

Now, if your answer is no[...]

So I take it that this is impossible.

The Rindler space is flat and therefore the particle, no matter which, does not 'fall' at all.

Otherwise we have the funny situation that, in flat space, the particle, in a Rindler metric, falls down, and at the surface of the earth, in just as flat a space, in the Rindler metric, the particle falls up.

Because at the surface of the eath, there is (almost) no difference between this situation, and the free, flat space situation.

How can you obtain a totally different equation of motion, starting from exactly the same inputs (the metric over a local patch) ?

You are repeating your mistake. I have the same metric over a local patch. In the one case, I attribute it (using the equivalence principle) to an Rindler observer (I am trying to use your reasoning) with positive acceleration, in the other case with negative acceleration. The prescription you need to 'simulate' gravitational effects in flat space is different for both particles.

That's a coordinate system as any other, no ?
It is in fact the coordinate system you would normally obtain "in a rocket".

It's not a global coordinate system, therefore you can't just take the coefficients in ds^2 and say, it's a metric of the manifold. Look, if spacetime is flat, you can accelerate your observer as you like, it better stays flat.

You can of course use local coordinates like the ones you use, but you have to be careful what conclusions you draw from the coefficients, that was all I wanted to say.

B.

 

[Careful]

** I am sorry, I might have mixed up \tau with \tau^-1 . Let us define TM: A -> \tau A \in TM. Okay, then I agree, it should have been \tau^-1 \nabla \tau acting on A in TM. The metric goes through the covariant derivative but \tau doesnt. **

But, \tau is constructed trough transposition of indices and/or application of g^{\alpha \beta} so \tau goes through the covariant derivative! If you contest this, give us a detailed calculation as well as your definitions.
Another reason why both geodesics are equal is that there exists no nonvanishing (1,2) tensor which is constructed from the metric and first derivatives alone. In more detail: you claim that we have to look at

A \tau^(-1) \nabla \tau A = A \nabla A + A \tau^{-1} ( \nabla \tau ) A

so \tau^{-1} ( \nabla \tau ) must be a (1,2) tensor depending upon g and \partial g only. Hence it must be zero.



**
Look, is it really very simple? The prescription to go the Christoffelsymbols is unique. But it matters what field you transport whether the Christoffelsymbols are the symbols to use or not. The extra structure is the properties of elements in TM to react to general coordinate transformations. **

But the elements in TM react identically to those in \underline{TM}^* and vice versa ! In a spinor bundle you DO add something, for example you ENLARGE the gauge group (= Lorentz group) by going over to the universal cover. Here, there is no enlarged gauge group whatsoever.




**If I remember that correctly, you can express the connexion for spinors in terms of the usual Christoffels by using the generators of the approriate spinor representation. In which sense is this more independent of the usual Christoffels than in my scenario? **

This gauge connection uses more than simply a local property of the metric !

 

[hossi]

But, \tau is constructed trough transposition of indices and/or application of g^{\alpha \beta} so \tau goes through the covariant derivative! If you contest this, give us a detailed calculation as well as your definitions.

Definitions are in the paper. Detailed calculations are to follow - I am still waiting for feedback from some friends/colleagues. For now, take the above example for a diagonal metric, then \tau is essentially E^-2 and \tau^-1 \nabla \tau is -2 E^-1 \nabla E.

Another reason why both geodesics are equal is that there exists no nonvanishing (1,2) tensor which is constructed from the metric and first derivatives alone. In more detail: you claim that we have to look at

A \tau^(-1) \nabla \tau A = A \nabla A + A \tau^{-1} ( \nabla \tau ) A

so \tau^{-1} ( \nabla \tau ) must be a (1,2) tensor depending upon g and \partial g only. Hence it must be zero.

Why do you assume \tau is a function of g? In certain cases you can express elements of \tau as functions of elements of g - as you have also done (how I conclude from your post above), but that does not mean \tau is a function of g in general. You can write it as a function of E and E^T, but the ^T is something you don't usually do in GR, so I am not sure how your argument goes with it.

But the elements in TM react identically to those in \underline{TM}^* and vice versa ! In a spinor bundle you DO add something, for example you ENLARGE the gauge group (= Lorentz group) by going over to the universal cover. Here, there is no enlarged gauge group whatsoever.

No, it is an additional discrete symmetry. If you want to get that formally right, you probably had some direct product of twice the gauge groups. The elements in TM do not react identically to those in \underline TM^* to general diffeomorphism.

This gauge connection uses more than simply a local property of the metric !

Right! It also uses the transformation behaviour of the fields, which differs whether it's a vector or spinor - even which type of (Weyl) spinor for that matter.



B.

 

[Rade]

...are we talking anti-matter (conjugation of electic charge), or are we talking anti-gravitating-matter (negative gravitational mass), or both? ... what is the model you study, and could you give me some references?

It is my understanding that we talk about both, anti-matter clusters with conjugation of electric charge, and also having negative gravitational mass. For details of what is known please see this link:http://d2800693.u47.phoenixrising-web.net/downloads/antimttr.pdf
Additional information is available at this web page:http://www.brightsenmodel.phoenixrising-web.net/Download.html

 

[vanesch]

By globally I always mean on the entire manifold. I guess we have some disagreement on the equivalence principle. As I understand it it says: the local effects of gravity in curved space are identical to those that an uniformly accelerated observer in globally flat space experiences.

Yes, which comes down to saying that the effects of gravity (the kinematical effects: we're only concerned with test particles here, not with experiments where the dynamics of gravity enters into account (where the test masses MODIFY the metric or something else) are solely dependent on the metric on a local, finite patch around where we try to establish them.

The uniformly accelerated observer in globally flat space is your Rindler observer. Now you have two of them, with positive or negative acceleration.

Eh, no. I only have one of them: it is my right to consider one, or the other and that choice is set up by the experimental situation. I mean: in a rocket, there are not TWO but only ONE coordinate system which is "fixed to the rocket", and in this system of coordinates, the acceleration is well-defined: it is defined by the behavior of NORMAL particles (of which the experimental existence doesn't need any explanation).
With these normal particles, in this rocket frame (X,Y,Z,T) we can establish the local metric, and it is the Rindler metric, and the sign of the acceleration in this metric is unambiguously established. In fact, it is established by "letting the particles fall" in this rocket frame. They will "fall down" (in the rocket frame = Rindler metric) in the sense that their equation of motion will be something of the kind: Z(T) = Z0 + V0 T - a/2 T^2 (Newtonian approximation). These are the geodesics of the Rindler metric. These same geodesics (world lines) are of course straight lines because when we transform this to the Lorentz frame, we have free particles in flat space. Now, this transformation between the Lorentz frame and the rocket frame is a transformation of coordinates on the manifold, and one doesn't have any choice there in the sign of the acceleration.
And if a-particles 1) describe a world line and 2) follows uniform motion in the Lorentz frame, then its world lines, are geodesics and hence will remain geodesics of the Rindler metric in the X,Y,Z,T coordinate system.

If you insist that in the coordinate frame XYZ,T (fixed to the rocket, so experimentally entirely unambiguous), the equation of motion of an a-particle is not a geodesic Z(T) = Z0 + V0 T - a/2 T^2, but rather
Z(T) = Z0 + V0 T PLUS a/2 T^2, then the only possibilities are:
a) that this equation of motion does not describe a world line but is observer-dependent OR
b) that the equation of motion in the Lorentz frame is NOT uniform motion.

The reason is of course that a world line cannot be a geodesic of a metric in one coordinate system and NOT be a geodesic in another coordinate system (a geodesic is a concept that is coordinate-system independent).

Now, both claims are denied by you, so this cannot be true.

Hence, the only possible motion for an a-particle, in the rocket frame, is given by Z(T) = Z0 + V0 T - a/2 T^2 (= a geodesic of the metric), which can be called "falling down in the rocket".

Since space is globally flat for the accelerated observer, you can of course then say, well, then it's the same everywhere and I only look at a local piece again. That I think, is what you are doing (?). But that doesn't change the fact that the space for the Rindler observer is globally flat, and, in particular, if this observer is uniformly accelerated, he doesn't move on a geodesic.

Yes, but an *observer* doesn't have to move on a geodesic! An "observer" is a coordinate system, meaning: mapping 4 numbers on a patch of spacetime. That's the whole idea of coordinate systems on a manifold.

The Rindler space is flat and therefore the particle, no matter which, does not 'fall' at all.

Well, it "falls" of course from the POV of the coordinate system of the rocket (with the Rindler metric), where the equation of motion is:
Z(T) = Z0 + V0 T - a/2 T^2 (in the Newtonian limit)
which is the equation of motion of a "falling" particle.

You are repeating your mistake. I have the same metric over a local patch. In the one case, I attribute it (using the equivalence principle) to an Rindler observer (I am trying to use your reasoning) with positive acceleration, in the other case with negative acceleration.

You can't. In a GIVEN coordinate system, (such as the one in a rocket), the acceleration is unambiguously defined. You cannot assign TWO accelerations to it. The unambiguously defined acceleration defines the sign of the acceleration in the metric expressed in this ONE AND UNIQUE coordinate system, which is nothing else but a Rindler metric. There's no way to assign TWO DIFFERENT accelerations to the same coordinate system, because the transformation between this coordinate system (of the rocket) and a given Lorentz frame is UNIQUELY DEFINED. It is a *coordinate transformation*. X,Y,Z and T are FUNCTIONS of (x,y,z,t) in a unique way.

The prescription you need to 'simulate' gravitational effects in flat space is different for both particles.

But as I outlined above, there is only ONE way to be compatible with:
a) a-particles describe world lines (coordinate-independent)
b) a-particles describe geodesics in the Lorentz frame.

If the world lines are geodesics of the metric in ONE frame, they are geodesics in ALL frames, because "geodesic" and "world line" are geometric concepts which are coordinate independent.

It's not a global coordinate system, therefore you can't just take the coefficients in ds^2 and say, it's a metric of the manifold. Look, if spacetime is flat, you can accelerate your observer as you like, it better stays flat.

Of course. The Rindler metric is a flat metric, but it is not Lorentzian. It isn't "global" because some piece of spacetime is missing, but it surely is "global enough" to define the metric over a finite local patch. (and not just in one point). By the equivalence principle, on which we both seem to agree, this should be good enough to determine uniquely the equation of motion in it.

You can of course use local coordinates like the ones you use, but you have to be careful what conclusions you draw from the coefficients, that was all I wanted to say.

But "local coordinates over a finite patch of spacetime" is all you have in GR. You have to be able to deduce everything from within local coordinates and the metric expressed in it. It's the essence of the equivalence principle. And what I'm saying is that these local coordinates are THE SAME for an observer on the surface of the earth, and in an accelerated rocket (up to tiny tidal effects).
If these are the same, then the equation of motion that you can derive of it, should be the same. And you claim it doesn't.

 

[Careful]

**Definitions are in the paper. Detailed calculations are to follow - I am still waiting for feedback from some friends/colleagues. For now, take the above example for a diagonal metric, then \tau is essentially E^-2 and \tau^-1 \nabla \tau is -2 E^-1 \nabla E. **

This is damn amazing I have worked out the definitions in your paper and came to the conclusions mentioned previously ! You first publish (miracles happen every day still), then you claim that you did not check things yet but at the same time you are telling to everyone who is kindly trying to inform you that the miracle does not happen that it somehow HAS to occur. As far as I can guess your notion of transpose : E_\mu E^T_\nu = g_\mu \nu (I remind you : I asked you to clarify this issue in the beginning), so there you have it.

**
Why do you assume \tau is a function of g? In certain cases you can express elements of \tau as functions of elements of g - as you have also done (how I conclude from your post above), but that does not mean \tau is a function of g in general. You can write it as a function of E and E^T, but the ^T is something you don't usually do in GR, so I am not sure how your argument goes with it. **

I can do the ^T in GR as much as I want to without changing anything (as twin correctly asserted). And if \tau does not depend upon g and its derivatives, then you introduce an extra field (see later) - eather, here we come.

**
No, it is an additional discrete symmetry. **

That is what you think ! But we don't see anything, and you are repeating the same issues we already know for 4 pages on this thread now.


**Right! It also uses the transformation behaviour of the fields, which differs whether it's a vector or spinor - even which type of (Weyl) spinor for that matter.
**

Correct, and hence it uses the extra (not uniquely fixed by the metric) structure (a preferred tetrad), and it has an enlarged gauge group SL(2,C) - exactly what I said and you denied.

I made in the beginning the comment that you do not need the tetrad in your construction, and you *agreed*. You still claim that everything can be done using local properties of the metric.

So, for a change I would like to have some scientific arguments, not just your beliefs about what it should be. We are not doing quantum gravity here, this topic should be simple enough.

Careful

 

[josh1]

Pair creation (and annihilation) in the antigravity sector, which by the uncertainty principle must occur everywhere, would provide a nonsensical mechanism by which black hole mass decreases but charge increases.

 

[hossi]

Dear Careful,

I sense you are getting angry at me for reasons I can't quite follow. I am sorry if this topic annoys you, I assure you I don't try to be stupid on purpose.

This is damn amazing I have worked out the definitions in your paper and came to the conclusions mentioned previously ! You first publish (miracles happen every day still), then you claim that you did not check things yet but at the same time you are telling to everyone who is kindly trying to inform you that the miracle does not happen that it somehow HAS to occur. As far as I can guess your notion of transpose : E_\mu E^T_\nu = g_\mu \nu (I remind you : I asked you to clarify this issue in the beginning), so there you have it.

I pointed out that your conclusions are right, but you have investigated the wrong question. I did not come across the question you had because I took another way, how I tried to explain above. I don't know what you mean with my (?) notion of transpose, the E's are the usual tetrads, i.e. E^i_\nu E^j_\mu \eta_ij = g_\mu\nu and there is no ^T involved here.

(The paper actually took a long time to be published and went through a quite lengthy review process. I am the first to admit that this still does not mean it's content is without doubt. However, I would say, science lives from controversy.)

I can do the ^T in GR as much as I want to without changing anything (as twin correctly asserted). And if \tau does not depend upon g and its derivatives, then you introduce an extra field (see later) - eather, here we come.

Then you are also changing the transformation from G to G^T. The second point actually is a very good point. I don't think \tau has any additional degrees of freedom, at least in my version of the model it hasn't (exept maybe some additional gauge-freedom like rotations or so that wouldn't really change anything). However, maybe it ought to be dynamical (credits for this idea don't go to me, and I am still not sure about it). I have no idea why this is an eather, it certainly has no fixed background. There are plenty of examples where additiona fields are coupled to the SM that change the dynamics without violating GR or SR.

That is what you think ! But we don't see anything, and you are repeating the same issues we already know for 4 pages on this thread now.

Well, I could say the same thing.

Correct, and hence it uses the extra (not uniquely fixed by the metric) structure (a preferred tetrad), and it has an enlarged gauge group SL(2,C) - exactly what I said and you denied.

I made in the beginning the comment that you do not need the tetrad in your construction, and you *agreed*. You still claim that everything can be done using local properties of the metric.

Hmm. Let me think. I certainly use the tetrad to make my computations. Now do I really need it? How do I find out? I admit, I am actually not sure about it I found it handy to use. Anyway, I don't deny that a spinor transforms under a different representation than a vector does.

So, for a change I would like to have some scientific arguments, not just your beliefs about what it should be. We are not doing quantum gravity here, this topic should be simple enough.

You are aware that you are being very insulting? Try the following: take a vector field V and apply some general coordinate transformation to it. You get V' = ... Now take some field-with-one-index W and transform it under the inverse and transponed of the above matrix. Tell me where the field with this property appears in GR, what it means and how it behaves.



B.

 

[hossi]

Pair creation (and annihilation) in the antigravity sector, which by the uncertainty principle must occur everywhere, would provide a nonsensical mechanism by which black hole mass decreases but charge increases.

Pair creation (and annihilation) in the antigravity sector does not occur at the black hole horizon, because the anti-g particles don't experience a horizon at this surface. (Well, they get repelled by the thing, how could they get trapped?) Nevertheless, good question,

B.

 

[hossi]

It is my understanding that we talk about both, anti-matter clusters with conjugation of electric charge, and also having negative gravitational mass. For details of what is known please see this link:http://d2800693.u47.phoenixrising-web.net/downloads/antimttr.pdf
Additional information is available at this web page:http://www.brightsenmodel.phoenixrising-web.net/Download.html

Hi Rade,

okay, I see what you are talking about. Naturally, I came across some of these ideas. However, I have a serious problem with the idea that anti-matter is also anti-gravitating. How do you make sure the vacuum is stable and what do you do with loop-contributions in, say QED. I know that the actual MEASUREMENT of the mass has not been done with very high precision, but if anti-graviating particles exist, I think they shouldn't be able to annihilate with usual matter.

B.

 

[josh1]

Hi Hossi,

Is there an antigravity analog of a black hole?

 

[hossi]

Dear vanesh,

you are still making the same mistakes. You are not using the equivalence principle as it's use is intended.

Eh, no. I only have one of them: it is my right to consider one, or the other and that choice is set up by the experimental situation. I mean: in a rocket, there are not TWO but only ONE coordinate system which is "fixed to the rocket", and in this system of coordinates, the acceleration is well-defined: it is defined by the behavior of NORMAL particles (of which the experimental existence doesn't need any explanation).

This is completely right. If you are in your rocket and describe the acceleration you experience it's the inertial mass that plays a role and it's the same for both particles. The difference between usual and anti-g particle lies in the comparison to gravitational acceleration, which in the one case has to go in the same direction (upward acceleration equals positively gravitating mass on the bottom) in the other case in the opposide direction (upward acceleration equals positively gravitating mass on the ceiling).

Now, this transformation between the Lorentz frame and the rocket frame is a transformation of coordinates on the manifold, and one doesn't have any choice there in the sign of the acceleration.

No, of course not, the acceleration is a parameter you have to compare to the graviational acceleration. If both are identical you have GR (thats exactly what the eq. principle says). If they are identical only up to a sign, you get what I say.

You don't get the point, for the Rindler observer both particles ARE identical. There is no gravitational effect, only acceleration, they both do the same thing. The difference is how you translate the particle's behavior to it's reaction to the gravitational pull.

Of course. The Rindler metric is a flat metric, but it is not Lorentzian.

I have no idea what you are trying to say.

Yes, but an *observer* doesn't have to move on a geodesic! An "observer" is a coordinate system, meaning: mapping 4 numbers on a patch of spacetime. That's the whole idea of coordinate systems on a manifold.).

It's not and an observer is not a coordinate system. A coordinate system has to cover the whole manifold, whereas you have a collection of local maps, and locally these are always flat. You have to patch the local maps toghether to get an atlas of the manifold. Then you get what I call a coordinate system, that is an appropriate base of 1-forms dx^\nu whose coefficients are the metric tensor and actually say something about the properties of the space.

B.

 

[hossi]

Hi Hossi,

Is there an antigravity analog of a black hole?

Hi Josh ,

never thought I would be happy about a comment of yours - the old guys are really stubborn, sigh. That's quite an interesting point, I have thought about that a bit, but not really come to any conclusions. The first question is of course how the anti-grav. matter behaves, like, how does it clump and structure? And, is there enough stuff around of it to actually make such a thing? Also: would it be stable? (There was some paper about the issue recently, will try to find the reference).

Unfortunately, that does not change the fact that these anti-g black holes would be unobservable I mean, they would be even more black than black holes. Ultra-black holes so to say.



B.

 

[Careful]

**
I pointed out that your conclusions are right, but you have investigated the wrong question. I did not come across the question you had because I took another way, how I tried to explain above. I don't know what you mean with my (?) notion of transpose, the E's are the usual tetrads, i.e. E^i_\nu E^j_\mu \eta_ij = g_\mu\nu and there is no ^T involved here.
**

There are no wrong questions, there are only inaccurate answers.
So define **rigorously** your sacred T and the controversy is over (I asked this since my first post).
You are not answering my objections why the two geodesics are identical. In this case, I presume you mean by E^T_(a \mu) = \eta_ab E^b_\mu .

** I don't think \tau has any additional degrees of freedom, at least in my version of the model it hasn't (exept maybe some additional gauge-freedom like rotations or so that wouldn't really change anything). However, maybe it ought to be dynamical (credits for this idea don't go to me, and I am still not sure about it). I have no idea why this is an eather, it certainly has no fixed background. There are plenty of examples where additiona fields are coupled to the SM that change the dynamics without violating GR or SR. **

So, here you say you don't know whether you add additional degrees of freedom or not ?? Hello you were claiming since the beginning nothing was added.


**You are aware that you are being very insulting? Try the following: take a vector field V and apply some general coordinate transformation to it. You get V' = ... Now take some field-with-one-index W and transform it under the inverse and transponed of the above matrix. Tell me where the field with this property appears in GR, what it means and how it behaves.
**

Twin has already *explicitely* mentioned that you only have to take the transpose of the cotangent basis to achieve that (and I acknowledged that I did the same thing before) - it is as simple as that.

Tell us EXACTLY - using MATHEMATICS only - what you mean. In each message you are trying to tell us what it *could* be - tell us what it IS. So, you see, I am (with good reason) insulted by your messages since on the contrary to you, I am giving mathematical arguments, using real definitions, why I think it does not work and you do not respond to that at all. For example : you previously said you did not need the tetrad and now you claim you don't know ?

Furthermore, I must tell you that Vanesch' interpretation of the equivalence principle is entirely correct - as any relativist (and that includes myself) will acknowledge.

Careful

 

[josh1]

Hi again Hossi,

The additional covariant derivatives you define reflect the main idea of your paper which is conjecturing couplings that mix degrees of freedom transforming contravariantly and covariantly under lorentz and gauge transformations and vice versa. Correct?

 

[hossi]

Hi Careful,

So, you see, I am (with good reason) insulted by your messages since on the contrary to you, I am giving mathematical arguments,

I certainly don't mean to insult your intelligence. I try to follow your arguments. I would appreciate it if you tried to follow mine.

There are no wrong questions, there are only inaccurate answers.

If you are aiming to examine something, you have to ask the right question. I was wondering why your result doesn't agree with mine: because you have examined another question than I have (see above).

So define **rigorously** your sacred T and the controversy is over (I asked this since my first post).
You are not answering my objections why the two geodesics are identical. In this case, I presume you mean by E^T_(a \mu) = \eta_ab E^b_\mu .

T is a transposition. That is a well-defined procedure to apply to G (the coordinate transformation), since G is nothing but a matrix. If you think about it for a while you will find that you can't write this procedure using standard index gymnastics in GR. That is the reason why I introduced \tau. It does this for you. \eta_ab E^b_\mu - the internal indices are raised and lowered by \eta, i.e. = E_a\mu .

So, here you say you don't know whether you add additional degrees of freedom or not ?? Hello you were claiming since the beginning nothing was added.

I said in my model I haven't added additional degrees of freedom. That does not mean it's not possible to make \tau dynamic. Before you frown at my again, let me add that I don't particularly like it, but I don't see why it should not be possible to make such an extension.

Twin has already *explicitely* mentioned that you only have to take the transpose of the cotangent basis to achieve that (and I acknowledged that I did the same thing before) - it is as simple as that.

Good. Then take it.

Tell us EXACTLY - using MATHEMATICS only - what you mean. In each message you are trying to tell us what it *could* be - tell us what it IS.

Maybe you are not listening closely enough.

For example : you previously said you did not need the tetrad and now you claim you don't know ?

Actually, it was you who said that I don't need the tetrad. I was stupid enough to agree on that without really thinking about it. That should teach me to keep my mouth shut. I have never been a fast thinker, and I am definately not suited for online discussions, blogs and the modern world. Yes, I admit, I don't know, it will take me time (that I don't have) to come up with an answer.

Furthermore, I must tell you that Vanesch' interpretation of the equivalence principle is entirely correct - as any relativist (and that includes myself) will acknowledge.

Okay, maybe I misinterpret what he interprets. I thought he takes an uniformly accelerated observer, say with acceleration a (not specifying any global structure on the manifold), claims that the equivalence principle states that this looks like gravity with acceleration a=g. And thus, every particle needs to have a=g. So far so good.

This is the same as saying that inertial mass equals gravitational mass. That's why it's important that the accelerated observer is in flat space. Otherwise you haven't made the connection beetween special and general relativity, there would be already gravitational effects present and you wouldn't gain anything.

However, from a theoretical point of view nothing hinders you to say, a uniform acceleration belongs to a gravitational acceleration with g=-a. That is inertial mass = - gravitational mass. In this regard, you might want to look at

"[URL Mass in General Relativity
H.Bondi, Rev. Mod. Phys. 29, 423-428 (1957).[/URL]

(if you have no access, I can send you a copy)

Just that we have never seen such a particle, and it's not considered to be a good choice. If you make the one choice for all particles, or the other choice for all particles doesn't matter much for the application of the equivalence principle.

What vanesh does now is to assumes that the acceleration of both types of particles (inertial mass is responsible for this) has to belong to the same gravitational acceleration (gravitational mass is responsible for this). Then of course both behave the same way. But that's because he has already assumed they do.

This is exactly the 'relaxation' of the equivalence principle I mean. You give up that the ratio intertial/gravitational mass = 1, then you can map the accelerated observer either to a down- or up-falling frame in the gravitational field.

Now he says I can not map ONE particle in the rocket in two different ways to the curved space. The point is that the mapping implies the equivalence principle, alias, whether a/g = + or -1. What I was trying to say is that one better thinks about TWO rockets, one with +a one with -a, instead of mapping ONE rocket differently.

To summarize: vanesh says the equivalence principle fixes the ratio of gravitational/intertial mass. True in the usual case. I say, relax it up to a sign. vanesh says it's not possible and proves this by applying the equivalence principle with the ratio fixed to plus one. Small wonder he is confused.



B.

 

[hossi]

Hi again Hossi,

The additional covariant derivatives you define reflect the main idea of your paper which is conjecturing couplings that mix degrees of freedom transforming contravariantly and covariantly under lorentz and gauge transformations and vice versa. Correct?

I am not conjecturing couplings, and I don't mix co- and contravariant degrees of freedom (not sure actually, what you mean with 'mix'). I think you could state it better like this:

The additional covariant derivatives I define reflect the main idea that the local transformation behaviour can be extended in two different ways to a global transformation behaviour, thus requiring a covariant derivative that transforms accordingly.



B.

 

[josh1]

Okay hossi,

The footnote on page 10 of your paper reads "… an anti-gravitating scalar field is possible since for a scalar field the covariant and the anti-covariant derivative are identical. Nevertheless, the arising equations of motions are second order and therefore distinguish between the gravitational or anti-gravitational nature of the field whose nature is encoded in the full action.”

Maybe you meant that despite the fact that for a scalar field the covariant and the anti-covariant derivative are identical, an anti-gravitating scalar field is possible since the equations of motions are second order and therefore distinguish between the gravitational or anti-gravitational nature of the field whose nature is encoded in the full action.

What is it in the full action that would allow nature to make the distinction?

 

[hossi]

The footnote on page 10 of your paper reads

Uhm, not sure which version of the paper you are talking about. In the most recent one there is no footnote on page 10.

Maybe you meant that

despite the fact that for a scalar field the covariant and the anti-covariant derivative are identical, an anti-gravitating scalar field is possible since the equations of motions are second order and therefore distinguish between the gravitational or anti-gravitational nature of the field whose nature is encoded in the full action.

Smart josh.

There is a reason why the footnote is gone. Here is the point I stumbled over: the action for a scalar field is not unique. You can eigher take (\nabla_\nu \phi) (\nabla^\nu \phi) or - \phi (\nabla^\nu \nabla_nu \phi). The first does not change when you replace \phi with some anti-gravitating pendent (which doesn't make very much sense to me), the second does. Either way, I wasn't sure what to say about this and decided to stay on the save side and better not say anything until I have come to some conclusions. Should you come to any, let me know. Fortunately, there are not so many scalar fields in the SM.

B.

 

[josh1]

Maybe you meant that despite the fact that for a scalar field the covariant and the anti-covariant derivative are identical...

For other people paying attention to this thread, by "scalar field" is just meant a field which carries no internal charges.

Uhm, not sure which version of the paper you are talking about. In the most recent one there is no footnote on page 10.

That may explain some of my earlier confusion, but I'm fairly confident there's no natural way to deal with scalars short of arguing they don't exist.

 

[Careful]

Ok, last try (and then I give up):

**
I certainly don't mean to insult your intelligence. I try to follow your arguments. I would appreciate it if you tried to follow mine. **

Ok, imagine yourself : Josh, Vanesch and I are three PhD's and we all get your paper wrong (according to you). My arguments are first year tensorial calculus, Vanesch' are introductory course GR - so it should be clear what we try to point out.


**T is a transposition. That is a well-defined procedure to apply to G (the coordinate transformation), since G is nothing but a matrix. If you think about it for a while you will find that you can't write this procedure using standard index gymnastics in GR. That is the reason why I introduced \tau. It does this for you. \eta_ab E^b_\mu - the internal indices are raised and lowered by \eta, i.e. = E_a\mu . **

Ok, I shall present in ultimate detail what I mean. Take a point p of the manifold M, and local charts (x^\mu), (y^\nu) defined on an open neighborhood O of p. Then y^\mu(x^\alpha) and x^\mu(y^\alpha) are assumed to be of class C^2. Now, define the matrix G^{\mu}_{\nu} = \partial y^{\mu} / \partial x^{\nu}, then you know that any contravariant vector V = V^{\mu} \partial / ( \partial x^{\mu} ) transforms as : V^{ \mu} -> G^{\mu}_{\nu} V^{\nu}, or as twin says : V'(y^{\mu}(x)) = G(y,x) V(x) where V and V' are considered to be column vectors with respect to the bases \partial / ( \partial x^{\mu}) and \partial / ( \partial y^{\nu} ) respectively. So here you have G as a matrix written out with respect to two coordinate bases. Now, take the transpose of this, then the above rule becomes : V'^T = V^T G^T which is exactly the defining relation of \underline{TM}^*. We can do the same with the one forms : V'_{\mu} = V_{\nu} (G^{-1})^{\nu}_{\mu}. Hence V'_{\mu} is a row vector and V' = VG^{-1}. Again, take the transpose : V'T = (G^{-1})^T V and this is exactly the defining relation of \underline{TM}. So, if I were to take this definition of transpose, then nothing happens. At first, I thougt : this cannot be since you were alluding to G^T = G^{-1} for a Lorentz transformation which is not true in the defining representation, UNLESS you mean by ^T the involution with respect to the Minkowskian scalar product in which case you use \eta to lower and raise indices. But that would be in contradiction with you claim that ^T is metric independent : so we have two conflicting statements here. Now, if you take the first notion of transpose, then nothing happens.

**
Actually, it was you who said that I don't need the tetrad. I was stupid enough to agree on that without really thinking about it. **

That is not only stupid but deceit. Don't you see that the whole discussion is turning around this very important point whether you need extra structure besides the local expression of the metric or not ?? So, either you do need a PREFERRED tetrad (or other field) or not. If you do, then there is no discussion since then we can all come up with new anti-gravity theories within a few hours.

**
That should teach me to keep my mouth shut. I have never been a fast thinker, and I am definately not suited for online discussions, blogs and the modern world. Yes, I admit, I don't know, it will take me time (that I don't have) to come up with an answer.
**

Euhhh ?? We are still dealing with an attempt to discuss YOUR paper beneath the level of abstract and you claim you have no time to understand your own work in some more depth ??


So, stick now for a moment with the first above thing, say where we are wrong and educate us.

Careful

 

[vanesch]

You don't get the point, for the Rindler observer both particles ARE identical. There is no gravitational effect, only acceleration, they both do the same thing. The difference is how you translate the particle's behavior to it's reaction to the gravitational pull.

Ok, so I take it that for a Rindler observer, the a-particle "falls down" in the same way as a normal particle. Because in flat space, both follow geodesics of the metric (which is, in an appropriate Minkowskian coordinate system, a uniform straight motion, and only "looks" like falling down in the rocket because of the coordinate transformation).
Mind you that nowhere MASS enters into this consideration. What enters into consideration is the TRANSFORMATION between the two coordinate systems (Minkowskian and Rindler), and this coordinate transformation is unique for a given setup. You can flip the sign of mass or anything, if we know the world line in ONE system, then it simply TRANSFORMS (by substitution!) into the description of the same world line in the other, independent of any sign of mass or anything. The "geodesics" only come about because this is the POSTULATED movement in a flat spacetime in Minkowski coordinates (well, in Minkowski coordinates, it is postulated to be uniform straight motion - something that only makes sense in a coordinate system of course), and as this corresponds with geodesics in the Minkowski frame, it corresponds to geodesics (coordinate-independent geometrical concept) in all coordinate systems.

It's not and an observer is not a coordinate system. A coordinate system has to cover the whole manifold, whereas you have a collection of local maps, and locally these are always flat.

First of all a coordinate system has not to cover the entire manifold ; for many manifolds this is even not possible (hence the concept of an ATLAS in differential geometry). A coordinate system has to cover an open domain around a point of a manifold. That's what I meant with a local, finite patch.
You seem to think that coordinate systems can always be made to look flat, locally, but that is not true. If space is curved, from a "local coordinate system over a patch" you can very well derive (over the same patch), the Rieman tensor, and it will be non-zero if that space is not flat. Of course in a SINGLE POINT you can have a Minkowski metric, but ITS DERIVATIVES will not be zero (which will give rise to the non-zero Rieman tensor).
But the fun thing about the surface of the Earth is that the Rieman tensor is ALMOST zero, so the space is ALMOST flat (and I already explained a few times how to keep the same g, and smaller and smaller Rieman tensor, by going to a bigger and bigger planet/black hole). So physically, concerning curved spacetime, over this local patch, there's not much difference (and the difference can be made arbitrarily small) between this situation, and our flat space with our Rindler coordinates. So I don't see how your a-particle can fall DOWN in a Rindler system, and fall UP in an identical setting (if things only depend - as you acknowledged - on the local metric over the patch).

You have to patch the local maps toghether to get an atlas of the manifold. Then you get what I call a coordinate system, that is an appropriate base of 1-forms dx^\nu whose coefficients are the metric tensor and actually say something about the properties of the space.

Well, first of all, concerning what happens in this local patch, it should not depend on what happens on a far-away patch. But then, nothing stops you to complete my local coordinate system (map of the atlas) into an entire atlas of the manifold ! If I have such a local map, you can complete it ALWAYS into an atlas. And that won't change then any conclusions I draw from my local map of the atlas, right ?
Especially as you confirmed that (= equivalence principle) everything depends only on the local (over finite open domain = local finite patch) metric (Rieman tensor if you want ; both are equivalent under the assumption of no torsion).

Again, I think you think that I'm confusing the fact that one can always choose IN A POINT a LOB that I erroneously think that space is flat there. That's of course not true - I know that.
But it is not what I'm thinking. I'm saying that the RIEMAN TENSOR at the surface of the Earth is small, and that in a local patch, the metric can be made Minkowskian (or very close) OVER AN ENTIRE PATCH (and not just at a point). As such, I *DO* can claim that locally, space is (almost) flat - not just in a point, but over a finite open domain. The "almost" is the tidal effects, which are small compared to the 1g.
So I was trying to fish out if you wanted to use the small deviation from true flatness as something to derive the 1 g -> -1 g flip for the a-particle (which would be difficult, given that I can make this deviation as small as I want). I COULD accept SMALL deviations of a geodesic from a world line (small "different" tidal effects or something). But I cannot consider this 1 g -> -1 g flip, because the flip does NOT occur in flat space (Rindler...) and it is according to you occurring in an ALMOST IDENTICAL situation (surface of the earth).

 

[Careful]

Vanesch, you might replace lorentzian by minkowskian since that lead to some confusion before.

 

[vanesch]

Vanesch, you might replace lorentzian by minkowskian since that lead to some confusion before.

Yes, you're right. I have tendency to confuse the vocabulary of the geometrical concepts with the vocabulary of their coordinate representations (because I'm more used to thinking in the latter).

I edited my previous post to take into account this indeed confusing abuse of language and to correct it.

 

[josh1]

Hi hossi,

About uncharged fields, the only thing I can think of is charging them.

 

[hossi]

Ok, last try (and then I give up)
Ok, imagine yourself : Josh, Vanesch and I are three PhD's and we all get your paper wrong (according to you). My arguments are first year tensorial calculus, Vanesch' are introductory course GR - so it should be clear what we try to point out.

It is clear to me what you try to point out. I try to point out where the introductory GR does not suffice any more, but you have missed it repeatedly.

and this is exactly the defining relation of \underline{TM}. So, if I were to take this definition of transpose, then nothing happens.

Thanks for you summary, you got that correctly. You have mapped an element of TM to one of TM. They can be identified with each other. You call that: nothing happens. I call that: gravitational charge conjugation.

**
Actually, it was you who said that I don't need the tetrad. I was stupid enough to agree on that without really thinking about it. **

That is not only stupid but deceit. Don't you see that the whole discussion is turning around this very important point whether you need extra structure besides the local expression of the metric or not ?? So, either you do need a PREFERRED tetrad (or other field) or not. If you do, then there is no discussion since then we can all come up with new anti-gravity theories within a few hours.

Interesting, you know, I had to look up the word 'deceit' in a dictionary. I will repeat for you again what I said before, and I apologize for being stupid: I used the tetrad. But I don't know whether I HAVE to use it. No, I actually didn't see the importance of whether I use it or not. I don't know what you mean with preferred tetrad, but I invite you to come up with anti-gravity theories, just try to make them self-consistent. I assure you, it's not easy. People will yell at you for being stupid.

**
That should teach me to keep my mouth shut. I have never been a fast thinker, and I am definately not suited for online discussions, blogs and the modern world. Yes, I admit, I don't know, it will take me time (that I don't have) to come up with an answer.
**

Euhhh ?? We are still dealing with an attempt to discuss YOUR paper beneath the level of abstract and you claim you have no time to understand your own work in some more depth ??

Correct. Huge problem. I wish I had all the time of the world to think about anti-gravity, but as you might suspect from this discussion, it's not really a topic that my collegues appreciate. Meaning, I have about 5 other research topics going on, and I begin to wonder why I spend the little time I have here with repeating my words.

So, stick now for a moment with the first above thing, say where we are wrong and educate us.

So far, you did not go wrong. You made the identification between elements of TM and TM. Now you have both particles (fields, types of vectors, whatever), and can ask for their derivative. The one is \nabla, the other one is \tau \nabla \tau^-1. Since elements of TM react differently to general diffeomorphism, their derivative is modified. Just go ahead and see where it leads you.

B.

 

[Careful]

**. Since elements of TM react differently to general diffeomorphism, their derivative is modified. Just go ahead and see where it leads you. **

Have done that one week ago, it leads me to the usual geodescis. How to see that ? Put B \in \underline{TM} equal to \tau(A), A \in TM then B satisfies the anti-geodesic equation if and only if A does satisfy the geodesic one. Although both are different vectors in different spaces, they lead to the same curve on the manifold.

Cheers,

Careful

 

[hossi]

Put B \in \underline{TM} equal to \tau(A), A \in TM then B satisfies the anti-geodesic equation if and only if A does satisfy the geodesic one. Although both are different vectors in different spaces, they lead to the same curve on the manifold.

Excuse me for being slow. How do you know that this implies the curve is the same?

 

[Careful]

Excuse me for being slow. How do you know that this implies the curve is the same?

A curve is a map from R to M, the *only* way to extract `infinitesimal´information from it is to take the derivative with respect to its parameter. This automatically gives you an element in TM, not in \underline{TM}. In general fibre bundle theory, you can parallel transport a `vector´ over a GIVEN curve, which defines the corresponding holonomy (with respect to the associated connection). In order to generate equations for the curve however, you can only refer to TM, \underline{TM} is entirely abundant. So, although you associate exotic `vectors´ to the anti-gravitating particles, the geodesics are identical.

 

[hossi]

A curve is a map from R to M, the *only* way to extract `infinitesimal´information from it is to take the derivative with respect to its parameter. This automatically gives you an element in TM, not in \underline{TM}. In general fibre bundle theory, you can parallel transport a `vector´ over a GIVEN curve, which defines the corresponding holonomy (with respect to the associated connection). In order to generate equations for the curve however, you can only refer to TM, \underline{TM} is entirely abundant. So, although you associate exotic `vectors´ to the anti-gravitating particles, the geodesics are identical.

Now we are getting somewhere The tangential vector of a curve is an element of TM, it always is, you are absolutely right with this. However, the prescription how you get the curve, or, define the equations of motion, is a different question. And it is subject of the interaction the field undergoes with it's sourrounding. The usual prescription is to parallel transport the tangential vector, i.e. it refers to \nabla acting on what you might interpret as the momentum. The defining equation for the anti-g particle is accordingly with respect to \tau^-1 \nabla \tau (see Eq (36)). Don't mix up the two points: in both cases the tangential vector is a tangential vector. But in the one case it is parallel transported, in the usual sense, in the other case it isn't (unless spacetime is globally flat).

This is the reason why t in Eq(36) is underlined: not because it is an element of TM, but because it belongs to the curve Eq.(36) which is the world line of a particle whose kinetical momentum is not equal the gravitational one. Having solved for t, you can pull it back and forth from TM to TM using \tau, that doesn't change a thing, as you correctly point out.



B.

 

[vanesch]

The defining equation for the anti-g particle is accordingly with respect to \tau^-1 \nabla \tau (see Eq (36)). Don't mix up the two points: in both cases the tangential vector is a tangential vector. But in the one case it is parallel transported, in the usual sense, in the other case it isn't (unless spacetime is globally flat).

So I expect that if you could express how the USUAL tangential vector is transported along the curve as a function of the metric, without using the underscore spaces and the tau, that this would lead us somewhere.
(it would then finally be the equation of motion, expressed locally, as a function of the metric)

Because then I'd also see how you can find, in a coordinate system at the surface of the earth, as a function of the metric there, that this curve would be described by:

Z(T) = Z0 + v0 T + g/2 T^2

and not as:
Z(T) = Z0 + v0 T - g/2 T^2

as do the usual geodesics (in Newtonian approximation), with the Z-axis up.

 

[Careful]

**Now we are getting somewhere **

We were there one week ago.

Since my objections did not change, I see only one solution for this. You write out explicitely in ordinary coordinates x^{\mu}, using the standard metric only, the equation of motion for the anti-particle (on TM !). This is possible since you have an explicit prescription for \tau which does not require any tetrad (yeh that's right), from TM \to underline{TM} it is simply v^{\underline{\alpha}} = g_{\alpha \beta} v^{\beta}. Typing this in should only take 10 minutes, and I assume you have worked it out already. Then we, the ones who refuse to see the shining light, can see :
(a) wether it has the correct behavior under coordinate transformations
(b) it is really different from standard geodesic

Cheers,

Careful

 

[hossi]

So I expect that if you could express how the USUAL tangential vector is transported along the curve as a function of the metric, without using the underscore spaces and the tau, that this would lead us somewhere.

See Eq. (39) together with Eq. (10).

Since my objections did not change, I see only one solution for this. You write out explicitely in ordinary coordinates x^{\mu}, using the standard metric only, the equation of motion for the anti-particle (on TM !). This is possible since you have an explicit prescription for \tau which does not require any tetrad (yeh that's right), from TM \to underline{TM} it is simply v^{\underline{\alpha}} = g_{\alpha \beta} v^{\beta}. Typing this in should only take 10 minutes, and I assume you have worked it out already. Then we, the ones who refuse to see the shining light, can see :
(a) wether it has the correct behavior under coordinate transformations
(b) it is really different from standard geodesic

First, the coordinates are always the "ordinary" coordinates. You are completely right, it is of course possible to explicitly obtain the equation of motion, since the necessary expressions are already in the paper, one just has to apply them (thats why I wrote the paper). Your prescription to go from TM to \underline TM does not take into account the transformation of the basis elements. Indeed, I have worked out the equations of motion for some cases I found interesting. I apologize that these concrete examples were not printed in the published version, but PLB has a rather strict page limit. I am not posting any results online before the second paper is on the arxiv -- should be out in some weeks (hopefully). I invite you to beat me at it.



B.

 

[Careful]

**First, the coordinates are always the "ordinary" coordinates. You are completely right, it is of course possible to explicitly obtain the equation of motion, since the necessary expressions are already in the paper, one just has to apply them (thats why I wrote the paper). Your prescription to go from TM to \underline TM does not take into account the transformation of the basis elements. Indeed, I have worked out the equations of motion for some cases I found interesting. I apologize that these concrete examples were not printed in the published version, but PLB has a rather strict page limit. I am not posting any results online before the second paper is on the arxiv -- should be out in some weeks (hopefully). I invite you to beat me at it. **

Euuh , first you say that it is easy to explicitely obtain them from the paper and then you pretend like it is a crucial result in a new paper you cannot share yet. And sure, my expressions do take into account the transformation of the basis elements. For some incomprehensible reason, I am quite confident nobody will try to steal this new, magic geodesic formula of yours, so go ahead. Me, Vanesch and Twin already explained you why we even don't have to make calculations to know that what you say cannot be unless you entirely give up covariance or introduce a preferred frame.

 

[hossi]

Euuh , first you say that it is easy to explicitely obtain them from the paper and then you pretend like it is a crucial result in a new paper you cannot share yet.

You are probably right, Careful, I just have to read through this thread to figure that it's apparently hard to get the idea of my paper, so maybe it's not easy at all.

And sure, my expressions do take into account the transformation of the basis elements.

Then compare their transformation with Eq. (4) in my paper and see if it matches.

For some incomprehensible reason, I am quite confident [...] already explained you why we even don't have to make calculations to know that what you say cannot be [...]

Glad to hear you are so confident. Thanks for the discussion, I appreciate you took your time to think about my work,

B.

 

[Careful]

**
Then compare their transformation with Eq. (4) in my paper and see if it matches. **

Of course it does, you already agreed that my transformation formula were correct

**
Glad to hear you are so confident. Thanks for the discussion, I appreciate you took your time to think about my work,
**

But give us this formula in TM, and silence all critisicm - that is the standard practice in science you know.

 

[vanesch]

I propose we simply wait for Sabine's new paper.
Although I would certainly be surprised to see this equation for a a-geodesic expressed explicitly in a coordinate system, using only the metric and its derivatives, and corresponding to a geometrical object, such that such an a-particle falls *up* for an observer fixed on the surface of the earth, if she has a way I can surely understand that she wants to deposit it first on at least the arxiv and not on a discussion forum.
As Careful said, I'm convinced it simply cannot be done, but then I can be wrong of course - which would probably mean I'd have to revise entirely what I thought I understood of GR. Could happen.

I think I'll refrain from this discussion until I finally see this result, because it seems like everybody camps on its positions and the arguments are just turning in circles. I certainly don't want this discussion to turn personal, and the only way for doing this is to have explicit results and calculations to discuss, so I'll wait for that. It is what is usually to be done when confronted with an extraordinary claim which seems to go against established truths: explicit proof. If this is what Sabine is writing up, then I won't bother her any further until she has done this.

 

[vanesch]

See Eq. (39) together with Eq. (10).

I'll make exception to what I just said because it is an explicit claim, even though you leave the burden upon me to do the substitutions.

For (10) on the earth, I take the LOB to be the (essentially Minkowskian) basis of the coordinate system in a falling elevator. This makes E equal to the unity matrix and hence tau too in (10), over a patch.

Next I fill this in (39). As in this frame, the covariant derivative is essentially the coordinate derivative, the second term is 0 (covariant derivative of constant unity matrix tau), and hence I find the first term, which means that tau-underscore is parallel transported in the elevator frame. In other words, your curve is a geodesic in the usual sense and your particle is in uniform motion in the elevator, which means it falls DOWN for a surface based observer.

 

[hossi]

I certainly don't want this discussion to turn personal

I appreciate that.

For (10) on the earth, I take the LOB to be the (essentially Minkowskian) basis of the coordinate system in a falling elevator. This makes E equal to the unity matrix and hence tau too in (10), over a patch.

You are still mixing up LOB with locally free falling frames (including the local sorrounding) and coordinate systems. E converts the coordinate system into the LOB and thus transforms the metric into the Minkowskian one. If \tau is globally identical to the unit matrix this means spacetime is flat. If you go to a locally free falling frame (this does not cover the whole manifold) then you have to decide what 'free falling' means, and you have gone around in a circle.

Concerning some of the above discussion, it seems to me that we have a different notion of coordinate system. I was referring to coordinate system as an atlas on the whole manifold, whereas you also consider a local patch to be a coordinate system.



B.

 

[vanesch]

You are still mixing up LOB with locally free falling frames (including the local sorrounding) and coordinate systems. E converts the coordinate system into the LOB and thus transforms the metric into the Minkowskian one. If \tau is globally identical to the unit matrix this means spacetime is flat.

I know you think I mix this up, but I don't.
A LOB is defined in a POINT, but a coordinate patch is defined over an open domain. Now, I gathered from what you said, that the motion of an a-particle is derivable from the metric over an open domain. This is the essence of the equivalence principle: that you can derive motion from the metric over an open domain. If not, you're in deep trouble (I'll come to that).
Now, starting from a LOB in point P, one can always extend it to a coordinate system over an open domain D around P, but then of course the dx, dy... in this coordinate system will not form a LOB in other points beside the point P where we started (in other words, the coordinate representation of the metric tensor in this basis will not be the Minkowski form (-1 1 1 1) diagonal, except in this single point P).
But the point I'm arguing, from the beginning, is this:
If you construct your LOB at the surface of the earth, and you extend this into a coordinate patch over an open domain (say, a few meters and seconds), then you have "falling elevator coordinates" over this patch, and IN THIS ENTIRE PATCH, not just in one event, the coordinate representation of the metric tensor is essentially Minkowskian. The deviation is very tiny from the Minkowskian metric. (it is zero of course in the chosen point, but it REMAINS very small away from the point).
So YES, spacetime is flat at the surface of the Earth (or almost so), over a patch which is by far big enough to find out if a particle falls "up" or falls "down" (a few meters and a few seconds are enough), and in our constructed coordinate system over a patch, it is entirely Minkowskian. So it should follow "uniform straight motion" in this coordinate system over a patch.

If you go to a locally free falling frame (this does not cover the whole manifold) then you have to decide what 'free falling' means, and you have gone around in a circle.

"Locally free falling frame" is a coordinate map over an open domain in which the metric takes on a Minkowski form diag(-1,1,1,1). Now, if spacetime is strongly curved, this only works in one point. However, if spacetime is flat there, this works over the entire patch. This is what happens on the surface of the earth.
The definition of "free falling" is clear up to a Lorentz transformation. There is no ambiguity on a manifold over which a metric is defined, what it means, to be free falling. It is an entirely clear notion. I have even difficulties understanding what you want to say, because there is absolutely no "decision" to be taken.

Concerning some of the above discussion, it seems to me that we have a different notion of coordinate system. I was referring to coordinate system as an atlas on the whole manifold, whereas you also consider a local patch to be a coordinate system.

I realized that, but I answered that issue: indeed, for me, a local patch (an open domain D of the manifold which maps 1-1 onto an open domain of R^4) is a coordinate system, let's call it a map. I said that if I give you ONE MAP, then you can always build an entire atlas containing that map.

So saying that, because I only define a coordinate map (X,Y,Z,T) over a domain D, that dX, dY, dZ and dT are not 1-forms is not correct: they ARE 1-forms (and their definition is explicit over the part of the cotangent bundle that corresponds to D, but can be extended to the entire manifold - just pick an atlas of your choice that contains my map).

In any case, this shouldn't matter, because the world line of the a-particle was to be dependent only ON THE METRIC OVER THE LOCAL PATCH. It is hard to see how you are going to introduce a dependency on the *global topology* of the manifold, no ? Because that would mean that the world line of your a-particle here on Earth will depend crucially on the extact configuration of black holes somewhere near Andromeda... OR you will need an extra field. In both cases, the equivalence principle is dearly harmed.

How do you handle otherwise the case of the motion of an a-particle when, say, 3 black holes are in a complicated dynamics somewhere far away, if your local piece of world line does NOT depend only on the LOCAL metric over the patch of spacetime of the laboratory ? If your local motion depends on the global structure of the manifold, you're in deep trouble, so I assumed (also because you said so) that the world line of an a-particle is derivable from the metric over an open domain around the a-particle.

So IF your a-particle world line only depends on the metric over a local open domain D, then this metric IS entirely expressible as a function of the 1-forms of my local coordinate map (in my falling elevator). In this expression, it DOES take (to a very good approximation) the form -dT^2 + dX^2 + dY^2 + dZ^2 over D (and will take on of course a deviating form in the extensions of this coordinate map into an entire atlas, when complementing D into the entire manifold, but we agreed that it should only depend on the form of the LOCAL metric over D, so my Minkowski expression is all I should care about, because that's its form over D).
Now, when the metric takes on the form -dT^2 + dX^2 + dY^2 + dZ^2 over an open domain, I know 2 things:
1) space is essentially flat over D
2) the coordinate system T,X,Y,Z is an inertial frame a la Lorentz.

As such we are in an identical situation as with a patch in a globally flat spacetime. And then its world line is "uniform motion" in this coordinate map (over D), which coincides with "geodesic for the metric" and hence which should correspond to a particle falling down for an observer fixed at the Earth surface.

Now, I only see one way out for you (if no other field is introduced), which is to say that the world line of an a-particle does not only depend on the LOCAL metric, but on the global metric. First of all, this is of course a violation of a basic idea in relativity (things should depend only on LOCAL stuff, no action-at-a-distance).
But it would essentially mean that your theory has no predictivity, because NOBODY KNOWS the global structure of spacetime. It would mean that the world line of your a-particle depends on the precise structure of the big bang ; on whether or not we are in "one bubble" of an inflationary universe or in just one of many ; on whether the universe is open or closed... All this to determine the motion, over a few meters, of an a-particle, in a lab on earth, in entirely Newtonian conditions.

Now, to come back to your initial remark:

You are still mixing up LOB with locally free falling frames (including the local sorrounding) and coordinate systems. E converts the coordinate system into the LOB and thus transforms the metric into the Minkowskian one. If \tau is globally identical to the unit matrix this means spacetime is flat.

When I consider my coordinate system my initial "falling elevator" map, extended to the entire manifold in an atlas of your choice with maps of your choice outside of D (but making a smooth transition to my falling elevator map on its border), then E IS the unity matrix over D (and deviates from it outside of D). tau is then the unity matrix over D, and deviates from it outside of D.
I guess it is because this is a very "skewed" coordinate system which doesn't look like the nice symmetrical coordinate system you have in, say, Schwarzschild coordinates, that you didn't consider it, and that you "calibrated" your tau in some other point. But this means that you introduced a PREFERENTIAL COORDINATE SYSTEM and that your world line is entirely dependent on HOW AND WHERE you decided to calibrate your tau.
And if not, then I am free to pick the "initial" coordinate system of my choice, which is the one I just described. In that case, as I said, E is the unity matrix (because over D, my initial coordinate system IS Minkowskian entirely over D, so it doesn't need any transformation in a LOB), hence tau too.

 

[hossi]

I know you think I mix this up, but I don't.
A LOB is defined in a POINT, but a coordinate patch is defined over an open domain.

Right.

I will try it again: you start in a LOB with internal Minkowskian metric, extend it to a local patch (as you say "a few meters and a few seconds") to what you call the elevator frame. The elevator frame is the frame of the free falling particle. In requires a definition. The definition is that the connection coefficients vanish. The connection coefficients depend on the transformation properties of the particle. Hence, the definition for the free falling frame of the gravitating are not identical to the definition of the free falling frame for the anti-gravitating particle.

Both can be derived from the metric though.

The definition of "free falling" is clear up to a Lorentz transformation. There is no ambiguity on a manifold over which a metric is defined, what it means, to be free falling. It is an entirely clear notion. I have even difficulties understanding what you want to say, because there is absolutely no "decision" to be taken.

Free falling means the geodesic equation takes the form it has in flat space. One point is not enough to define that, it requires a local sourrounding. That means it also requires the knowledge of how to transport quantities in this local sourrounding. The derivative acting on the quantities, and the question whether it looks like that in flat space, depend on the transformation behaviour of the quantity to be transported.

That is the point you are missing.

I absolutely don't understand your argument with the "local" and "global" metric. It is sufficient to know the metric in a local sourrounding from where the particle is to derive the transport laws.

I think what you are trying to say with 'calibrating' the \tau is that it's covariant derivative does not vanish. I.e. if it has a value at one point and I transport it around, then it changes. If I tranport it around in a circle in space, then it might end up differently than how it started. That is what happens in curved spaces. I don't see why this means I have introduced a preferred frame.



B.

 

[vanesch]

I will try it again: you start in a LOB with internal Minkowskian metric, extend it to a local patch (as you say "a few meters and a few seconds") to what you call the elevator frame. The elevator frame is the frame of the free falling particle. In requires a definition. The definition is that the connection coefficients vanish. The connection coefficients depend on the transformation properties of the particle. Hence, the definition for the free falling frame of the gravitating are not identical to the definition of the free falling frame for the anti-gravitating particle.

You don't need a particle for that, you know. I take it that the metric is given, right ? If the metric is given, I can construct, by parallel transporting my initial tetrad in a specified way, a coordinate map. Now of course this parallel transport is the "usual" parallel transport which is associated with the metric. But I don't see what's wrong with that: it's what a metric is for ! But further, exactly how I construct my coordinate map shouldn't matter to you: I GIVE you the representation of the metric in it, so that's all you should know about it: how the metric tensor is expressed in it.
So, again: from my initial tetrad which is a LOB wrt the metric, I extend this into a "coordinate grid" over a local patch, by (arbitrarily, true) choosing a specific procedure (first x times e1 then y times e2 then...). The arbitrariness resides in the order in which I build up the grid, because in curved spacetime, the point of arrival will depend on the choosen order.
But the less spacetime is curved, the less there is arbitrariness in this.
This comes in fact about the specific way in which the observer in the falling elevator frame sets up his coordinate frame.
And the point is, that in a falling elevator, over a few meters and a few seconds, there is not much ambiguity because space is essentially flat. In other words, given the metric at the surface of the earth, the construction of the elevator frame is essentially unambiguous (as is obvious for any engineer who has made an elevator!).

Now, you can of course say that for my parallel transport, I should use your "alternative connection coefficients", but this is not correct. I can use whatever procedure is valid to construct a coordinate map, and I prefer to use "normal" parallel transport, which is perfectly well defined. To a metric is associated a well-defined connection, which is the one of the standard procedure. I only have to establish that my coordinate map over the local domain D EXISTS and is well-defined. I don't have to justify its particular choice.

And again, this has a priori nothing to do with "the choice of a particle". It happens to be the case that normal particles follow geodesics of the metric, but that's an extra postulate. Geodesics STILL are geodesics, even if normal particles or other particles wouldn't follow them: they are geometical objects associated with the metric, so these geodesics are well defined if the metric is given.
And, again, all this doesn't matter: I construct my coordinate map the way I want. I only have to show that it exists, and then give you the metric expressed in it: then you know all there is to know about the metric over this domain.
If you claim (as you do) that you can construct the equation of motion from the metric and the metric only over D, then you have now all you need: the expression of the metric tensor in the coordinate frame over D.
So exactly HOW I constructed the map doesn't matter in the end (whether I used geodesics, or other constructions): what matters is what is the coordinate expression of the metric tensor. It turns out to be diag(-1,1,1,1) over the local patch.

Both can be derived from the metric though.

That's what I don't believe.

Let's be clear about what this means:
it means, when I give you the coordinate expression of the metric tensor over a domain D, that you can express the equation of motion (which gives me the piece of worldline in D) of the a-particle, ONLY USING THE COORDINATE EXPRESSION OF THE METRIC TENSOR OVER D.

Free falling means the geodesic equation takes the form it has in flat space. One point is not enough to define that, it requires a local sourrounding. That means it also requires the knowledge of how to transport quantities in this local sourrounding. The derivative acting on the quantities, and the question whether it looks like that in flat space, depend on the transformation behaviour of the quantity to be transported.

Yes, but you claim these to be expressible as a function of the metric over D.
Mind you, I have repeated often that you might very well find OTHER curves (call them a-geodesics) which are derivable from the metric. But look: we have a given metric. From this metric, I derive (in an unambiguous way) a local coordinate system using the NORMAL transport over a patch. By doing so, I assert that this coordinate map exists over D, and that's all I have to do. In that patch, I express the metric in this local coordinate system and it corresponds to a minkowski tensor. You may not "agree" that this is a LOB, but it is a perfectly well defined procedure to give you the metric over the patch, so you should be able to take this as ALL THE INFORMATION YOU HAVE ABOUT THE METRIC over this patch.
STARTING FROM THIS, you claim that you can construct your a-geodesics. I tell you: do it, and explain me how this is different from the IDENTICAL situation in a patch of free flat space.

That is the point you are missing.

I absolutely don't understand your argument with the "local" and "global" metric. It is sufficient to know the metric in a local sourrounding from where the particle is to derive the transport laws.

Well, then: do it! In my given coordinate map (of which I told you how I constructed it, but that shouldn't matter: this only served to show you that it EXISTS as a coordinate map over D), the metric tensor is in Minkowski form diag(-1,1,1,1). You now HAVE a description of the metric tensor in a coordinate map of my choice over an open domain. You claim that you can derive the a-geodesics in this frame from the metric over this open domain, and from this metric only.
What are the a-geodesics, expressed in this coordinate system ?

(and how do you know that this metric tensor is to be treated differently if original in a falling elevator frame, or in free flat space, because their expressions are identical and you're only supposed to use the expressions of the metric - as you claim them to be derivable from the metric over D)

 

[Careful]

Euuuh, I thought this conversation was closed unless our Newton lover, hossi, gave us the explicit formula using the metric g_{\mu \nu} and partial derivatives thereof only.

By the way, it is not only possible to put the christoffels to zero in a point, but on a *worldline* by good choice of coordinate system. So, I am still waiting for
(a) a general prescription for the connection
(b) the full geometry = matter dynamics

 

[George Jones]

By the way, it is not only possible to put the christoffels to zero in a point, but on a *worldline* by good choice of coordinate system.

While it is possible to choose a frame (tetrad) with respect to which the Christoffels vanish along a geodesic, it is not possible, in general, to choose a coordinate system with respect to which the Christoffels vanish along a geodesic. See the bottom of page 331/top of page 332 in Misner, Thorne, and Wheeler.

Regards,
George