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

[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

 

[Careful]

** 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.

**

Can you give an explicit counter example (sorry, don't have misner and thorne here) ? I will look up the paper where this result was claimed to be true (obviously the statement you make is weaker and therefore true).

 

[George Jones]

Can you give an explicit counter example (sorry, don't have misner and thorne here) ?

I will have to think about this.

I will look up the paper where this result was claimed to be true (obviously the statement you make is weaker and therefore true).

I would like to see the paper.

Regards,
George

 

[Careful]

I will have to think about this.



I would like to see the paper.

Regards,
George

Ok, give me a day or two, it came to my mind now that when we took it from the arxiv we knew the result was in contradiction to the statement in MTW (which does not contain a counter example neither a proof if I remember well), a friend of mine checked it mathematically and told me it was ok (I did not pay further attention to it). So, I will ask him tomorrow.

As an aside, you could perhaps give your impression about Hossi's anti gravity ideas. It would be useful to have some comments from some of the participants of the GR forum here.

Cheers,

Careful

 

[hossi]

Hi vanesh,

I think we have eventually come to the reason for our misunderstanding.

You don't need a particle for that, you know. [...]

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, [...]

And again, this has a priori nothing to do with "the choice of a particle". of the metric - as you claim them to be derivable from the metric over D)

I apologize, I used "particle" as a synonym for "field with transformation properties under general coordinate transformations". The whole idea of my paper is that the additional particles (fields) have a different transformation behavior. If you prefer to use 'normal' parallel transport, then you are just describing 'normal' particles. I don't doubt that. But that's not the point. The point is whether it is self-consistently possible to have particles with a different transformation behaviour.

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.

Well, as Careful has pointed out, it takes the tetrad to get \tau and so it takes the tetrad also to get the connection. (I think he is right, but I can't prove whether it's really necessary or not to use the tetrad.)

But otherwise, yes, that's what I say. You find the equation in the paper, believe it or not.



B.

PS @ George Jones: thanks for the clarification!

 

[vanesch]

I think we have eventually come to the reason for our misunderstanding.

I'm affraid we're not home yet...

The whole idea of my paper is that the additional particles (fields) have a different transformation behavior. If you prefer to use 'normal' parallel transport, then you are just describing 'normal' particles. I don't doubt that. But that's not the point. The point is whether it is self-consistently possible to have particles with a different transformation behaviour.

I think I understood that, and I think that the problem still remains. Because, as I tried to point out, I only used this "normal transport" to set up a coordinate frame, but I tried to point out too that the specific procedure for constructing a coordinate frame shouldn't matter (the procedure only indicates that the coordinate frame EXISTS, but once it is there, where it came from doesn't matter). So once I have constructed my coordinate frame, it doesn't matter HOW I did so, and I can express the metric tensor in this coordinate frame. I'm not committing, in this way, to any "choice of particle", and I'm not pretending that the only curves that make sense in this way, are the geodesics of the metric. I'm JUST constructing a coordinate frame, and express the metric tensor in it. So whatever strange "transformation properties" certain entities might have, this should be expressible in this coordinate frame, if the claim is that this is derivable from, and only from, the local metric over domain D. If the transformation properties give rise to curves different from the geodesics, there's no problem with that, but it should STILL be expressible in this coordinate frame, ONLY using the expression of the metric tensor, if that's the ONLY thing on which we are depending (which is the essence of the content of the equivalence principle).
In other words, the equation of the world line of any "thing" (no matter what are its transformation properties) in this coordinate map should be expressible as a function of the coordinate representation of the metric tensor over D. The solution doesn't necessarily have to be a geodesic. But it needs to be expressible by JUST using the coordinate representation of the local metric.

Let's summarize the reasoning I think you wanted to expose:
You take a manifold with a metric on it, and this metric gives us of course "normal" geodesics. But you seem to introduce another kind of connection on top of this, which gives rise to OTHER curves (I call them a-geodesics) on the manifold, which are such that they "parallel transport" the underscore quantities. So far, so good. This "other connection" should however, still be expressible as a function of the metric on D.
But then you claim that, in the case of the surface of the earth, these a-geodesics satisfy the following properties:
1) they correspond to particles falling "up" (with about an upward acceleration of 1g)
2) in free flat space, the a-geodesics are identical with normal geodesics (that is to say: in the associated orthogonal coordinate frame, they correspond to uniform straight motion).
3) the a-geodesics are determined by the metric over a local domain D.

I don't think that it is possible to satisfy these 3 conditions together, for the following reason, which is always the same:
at the surface of the earth, space IS essentially flat, and the corresponding orthogonal coordinate frame is the one of the falling elevator. (there's no discussion about this: it is in a FALLING elevator that the metric takes on the Minkowski form, not in a "rising" elevator, because the metric itself is a 2-tensor and *its* transformation properties ARE well-defined, no matter what OTHER things you might introduce with OTHER transformation properties - this is NOT committing myself to any 'particles'). But now we DO have a coordinate system over D, in which the metric takes on the Minkowski form (and hence can be considered flat over D).

As such,by 2) the a-geodesics should correspond to normal geodesics, at least in a local patch where the space can be considered flat enough ; and by 3) this patch is enough to determine the local piece of a-geodesic (which corresponds to "falling down"), hence in contradiction with 1).

Now of course from the moment you allow for another field, specified independently of the original metric, over the patch, to be used too in the transformation properties, and hence in the equation of an a-geodesic, then there's of course no contradiction anymore. But then it is NOT purely expressible as a function of the metric, and the metric alone, over D. (and to me, the equivalence principle is dead). This seems to be what you are saying now:

Well, as Careful has pointed out, it takes the tetrad to get \tau and so it takes the tetrad also to get the connection. (I think he is right, but I can't prove whether it's really necessary or not to use the tetrad.)

Ok, that's what I said from the beginning: I suspected this tau to be related to an extra field (take 4 vector fields, which each determine one leg of the vierbein you need in each point to fix tau). So you introduce arbitrary extra fields *which are not derivable from the local metric over D*. But if an extra field is allowed for, then it is not difficult to introduce anti-gravity, or simultaneity or whatever. This is completely killing off the equivalence principle. For instance, you can introduce an extra scalar field over the manifold, and call it "absolute time". You now have an ether theory.

But otherwise, yes, that's what I say. You find the equation in the paper, believe it or not.

Yes, yes, the equation is there IF YOU KNOW WHAT TAU IS. But you claimed it was to be written PURELY from the metric, so I wanted you to express tau from the metric.

Now, if tau is determined by the choice of a vierbein in each point (or even in a single point), and this tau determines now your extra connection, aren't we now just inventing a complicated scheme for just introducing a SECOND metric on the manifold, and aren't we now not just back to the initial case where we have a funny manifold over which two metrics are defined ?

In any case, there's something arbitrary to be fixed (vierbein, tau, second metric...). Your a-geodesics will then crucially depend on this arbitrary choice. In that case, you can of course do what you want, that's not a big surprise. I call that "killing entirely the equivalence principle".
Because nothing stops me now from introducing, say, ANOTHER vierbein, which will fix a second tau, and introduce a THIRD set of connection coefficients, which will in turn fix "b-geodesics", and introduce YET ANOTHER vierbein, with a third tau, and "c-geodesics"... no ?

It seems to me that we now simply have two different metric manifolds which have been "identified" in an arbitrary way.

 

[Careful]

Hi P,

**
It seems to me that we now simply have two different metric manifolds which have been "identified" in an arbitrary way. **

As you know, I fully agree (modulo equivalent formulations of your statement ) with your position here. Trying to make such scheme dynamical will bring along some further problems as I mentioned to you before.

Cheers,

C

 

[vanesch]

at the surface of the earth, space IS essentially flat, and the corresponding orthogonal coordinate frame is the one of the falling elevator. (there's no discussion about this: it is in a FALLING elevator that the metric takes on the Minkowski form, not in a "rising" elevator, because the metric itself is a 2-tensor and *its* transformation properties ARE well-defined, no matter what OTHER things you might introduce with OTHER transformation properties - this is NOT committing myself to any 'particles'). But now we DO have a coordinate system over D, in which the metric takes on the Minkowski form (and hence can be considered flat over D).

I would like to follow up on this, because it is the crucial point on which we seem to miss each other (from my PoV, it seems that I cannot make you see what I'm trying to say, and I'm getting desperate at it, because it seems to be such an elementary point).

At a certain point in the discussion, you said that the coordinate frame I "should" choose to describe an a-particle (a-geodesic, a-something) was an "up-falling" elevator frame (say, a rocket accelerating up with acceleration 1 g wrt the Earth surface). I didn't want to consider that because it didn't fit in my explanation, but of course we can do that. If I go to a coordinate frame attached to such a rocket (we could do it in detail, but I take it now that you can accept that this WILL give rise to a coordinate map over a patch D around the rocket which is a smooth mapping between D and R^4), then my metric does NOT take on the Minkowski form in this coordinate system (but rather a Rindler form with an acceleration of 2 g).

You can say that this is the right frame in which your alternative connection will parallel transport your alternative quantities along curves of constant coordinates, but that doesn't change the fact that the METRIC tensor isn't in Minkowski form, because the transformation of the metric tensor is well-defined.
Now, it is of course tempting to introduce a SECOND METRIC which will generate this alternative parallel transport. But that's indeed what it is: a second metric, which DOES take on the Minkowksi form in this rocket coordinate frame. It is now up to you to say whether this second "metric" is going to transform as a 2-tensor or as something else between coordinate changes. However, what is clear is that this "second metric" is not derivable from the original one without the extra input which simply comes down in STATING that in this frame, this second metric takes on the Minkowski form (or stating that it is here that tau = 1, or making the choice of the tetrad from which the tau=1 is derived).

And then my question is: why in THIS frame ? Why not, in a rocket frame that shoots off, say, not vertically, but horizontally to the East with 1 g with respect to the Earth surface ? We could say now that it is in THIS frame that the "second metric" takes on its "Minkowski form" and that it is in THIS frame that the coordinate lines are "a-geodesics". Or a rocket frame that shoots off horizontally to the north. We could say that THIS is the frame in which a-parallel transport is along the coordinate lines. Or we could say that it is in a frame fixed to the surface of the Earth that this happens. Or in a rocket frame that doesn't shoot off with 1g, but rather with 7g. Or...

In these examples, a-particles would "fall to the east" or would "fall to the north", or would "float at the surface". See, it is totally arbitrary to fix it as "falling up" (or "to the East", or "to the North", or "float at the surface"). The a-geodesic is entirely dependent on the choice of how we "fix" this tau=1 condition.

 

[George Jones]

Ok, give me a day or two, it came to my mind now that when we took it from the arxiv we knew the result was in contradiction to the statement in MTW (which does not contain a counter example neither a proof if I remember well), a friend of mine checked it mathematically and told me it was ok (I did not pay further attention to it).

Misner, Thorne, and Wheeler refer to local Lorentz coordinate systems, so maybe there exist coordinate systems that are not local Lorentz coordinate system, and that have vanishing Christoffels along a worldline.

As an aside, you could perhaps give your impression about Hossi's anti gravity ideas.

I have been following this thread superficially, but not in great detail. Below I offer some thoughts that are pure specultion, like the ideas that one might float at tea or coffee, and so might be quite wrong. While doing this, I shall try to perch precariously on the fence.

Like Patrick and you, I, too, have questions about which bundle and which metric.

I think it might be appropriate to consider the bundles of frames (4 linearly independent vectors) and tetrads (4 orthonormal vectors)

Drat, didn't finish ... got to run. I'll finish this post via editing in 2 or 3 hours.

Regards,
George

 

[Careful]

**Misner, Thorne, and Wheeler refer to local Lorentz coordinate systems, so maybe there exist coordinate systems that are not local Lorentz coordinate system, and that have vanishing Christoffels along a worldline.**

**

Here is the paper math.DG/0304157 (check for two other papers of the same author). It is quite easy to show that for any non self intersecting curve \gamma, there exists a local basis (on a neighborhood of \gamma) such that the connection coefficients vanish on \gamma. Now, for any torsion free connection, this implies that such basis must be holonomic on the curve. The author further claims (but I did not check that) the stronger result that there exists a coordinate system in a neighborhood of \gamma such that the associated connection coefficients vanish on \gamma. Have to go now.

EDIT : an older version of the same paper does - strangely enough - not make this claim. I will check the book of Kobayashi on this. In any case, if the stronger claim were not true, it would be interesting to have a counterexample (anyway if someone else knows this, please go ahead).

Cheers,

Careful

 

[hossi]

In these examples, a-particles would "fall to the east" or would "fall to the north", or would "float at the surface". See, it is totally arbitrary to fix it as "falling up" (or "to the East", or "to the North", or "float at the surface"). The a-geodesic is entirely dependent on the choice of how we "fix" this tau=1 condition.

vanesh, don't you see that I could use exactly the same argument to tell you that the equivalence principle is arbitrary? Why do you fix on a downfalling elevator? Why not one falling "to the East" or "to the North"? The reason is that you do not need to specify this when you use the equivalence principle. There is only one geodesic for the usual particles - because they have a fixed and well defined transformation behaviour under general diffeomorphism. You can always rotate your internal frame (the tetrads), it is fixed only up to Lorentz-transformations. These include rotaions, but that doesn't change the geodesic.

The new particles also have a well defined transformation behaviour, which uniquely determines the curve they move on. The \tau is not fixed to the identity in the local frame just because I want so, but because that is required for the transformation of both particles to be related as in Eq.(3) and (4).

Now, it is of course tempting to introduce a SECOND METRIC which will generate this alternative parallel transport. But that's indeed what it is: a second metric, which DOES take on the Minkowksi form in this rocket coordinate frame.

You don't want any second metric to take Minkowskian form, you want the connection coefficients (of the new particles) to vanish.



B.

 

[George Jones]

I think it might be appropriate to consider the bundles of frames (4 linearly independent vectors) and tetrads (4 orthonormal vectors). The bundle of frames is a principal bundle with structure group GL(4), and a connection on this bundle determines connections on all tensor bundles. A connection on the frame bundle is itself determined by a connection on the tetrad bundle, which has the Lorentz group as structure group. In non-bundle language this is just the requirement that the connection be metric-compatible, i.e, that the metric be covariantly constant.

So, the action of Lorentz transformations is important, even for the frame bundle, which has general GL(4) transformations. This might mean that what Careful writes in post #6 is important. Different actions of Lorentz group on tetrads can be defined by G and (G^-1)^T, where G is a Lorentz transformation, which leads to similar, but different(?), tetrad bundles, since an action of the structure group is include implicitly in the definition of a principal bundle. Maybe these different actions lead to different connect, with one connection being compatible with the standard metric, and other connection being compatible with a different "derived" metric.

Note that, if for some frame, G is a spatial rotation, then G = (G^-1)^T, and if G is a pure boost, (G^-1)^T is a boost in the *opposite* direction.

These thoughts are just speculations, and I have made little attempt to be accurate, or even correct, and I have no idea "what it al means".

Regards,
George

 

[vanesch]

vanesh, don't you see that I could use exactly the same argument to tell you that the equivalence principle is arbitrary? Why do you fix on a downfalling elevator?

Because this is the frame in which the metric takes on the Minkowski form. Now, if you are going to discuss WHY the metric takes on this form here, that is relatively simply answered: because the metric is a 2-tensor. Something else might be another object, with other transformation properties, but the metric, in order to be a metric, is a 2-tensor.

I DO NOT NEED ANY CONNECTION TO FIND THE TRANSFORMATION PROPERTIES OF A 2-TENSOR.

If you give me a 2-tensor in a coordinate system A, and you give me the relationship between coordinate system A and coordinate system B - all over a domain D of course - (these are a tupel of 4 real functions from the R^4 domain for A to the R^4 domain of B), then I know how to write out the metric tensor in coordinate system B. I repeat: I DO NOT NEED ANY CONNECTION FOR THAT.

So, if you give me, for instance, a Schwarzschild metric with its associated set of coordinates (R, theta, phi, T), and I apply the transformation (which I wrote out in detail somewhere in this thread) to a coordinate patch which would correspond to a falling elevator around the point theta = phi = 0, R = radius of the earth, and T=0, then simply by writing out the functions that map (R, theta, phi, T) onto (x,y,z,t) over this domain, I know how to transform the 2-tensor which we call "metric", and it happens to take the form (-1,1,1,1) to a high degree of accuracy (small tidal effects still present). At no point, I USE a connection ; the only thing I use is the Jacobian of the mapping (R, theta, phi, T) onto (x,y,z,t), which is not open to any arbitrariness.

Of course, if you're now going to claim that the metric is not a 2-tensor, then I have to say I don't know what to answer...

 

[vanesch]

The \tau is not fixed to the identity in the local frame just because I want so, but because that is required for the transformation of both particles to be related as in Eq.(3) and (4).

I never clearly understood those equations (that's why I argued starting from your conclusions, and not from your derivation).
General covariance, to me, means that "physical objects have to live on the tangent/cotangent bundle (or powers thereof)". I never understood this as them having to satisfy certain ACTIVE transformations, but rather, them being geometrical objects living on the tangent / cotangent bundle, their COORDINATE REPRESENTATIONS change according to certain rules when we change the coordinate mapping to another one. But the "geometrical object" remains the same. (this is sometimes called passive transformations, as transformations of the coordinate representation, and not of the geometrical object).
So the only way I can understand equation (3) is not by G being a mapping from the bundle onto itself, but as a mapping between two coordinate representations of the SAME geometrical object (in other words, mappings between R^4^n and R^4^n, in normal speak: tensor transformations).

Now, you want to introduce an extra class of potentially physical objects, which do not live on the tangent/cotangent bundle, but on these underscore bundles. I have no problem with that, a priori. In (4), I interpret that as a transformation rule for "a-tensor" representations under a coordinate transformation G of one and the same geometrical object (and not, as an active mapping of a geometrical object onto another one).

Now, in this view (the only one I can make sense of as considering it related to general covariance), where G is thus a CHANGE OF COORDINATE REPRESENTATION, equation (5) tells me something about how the relationship between the tangent/cotangent bundle and the a-tangent and a-cotangent bundle CHANGES when I go to another coordinate representation. This relationship is tau.
In this view, G is simply the jacobian of the transformation between the two coordinate patches, and tau is a 4x4 matrix, which tells us how to go from the coordinate expression of an element of TM in the TM basis associated with the coordinate system at hand, onto the coordinate expression of an element of a-TM. In other words, given tau, it fixes the basis in a-TM when we have the coordinate basis in TM.

But we never fixed tau WITHIN a coordinate frame. You only tell us how it is CHANGING when I go to another coordinate system. So, somehow, I can CHOOSE a specific coordinate system, and PUT TAU EQUAL TO THE UNITY MATRIX in each point of the manifold in this coordinate system. This comes down in saying that the (coordinate system induced) basis in TM is now identified with my basis in a-TM.
But the funny thing is that if I were to do that in ANOTHER coordinate system, I would find ANOTHER relationship between TM and a-TM.
In fact, the specific choice of tau, in the relationship between TM and a-TM, is quite analoguous in the specific choice of relating TM to TM*. I could pick a given coordinate system, and tell you that in this system, I identify the basis of TM with the one of TM*. But if I did it in another coordinate system, my mapping between TM and TM* would be different. This mapping (depending on where I "calibrate" it), between TM and TM*, is what people normally call a metric, and "picking the coordinate system where we identify TM with TM*" comes down to defining the metric (in fact, setting the metric equal to the Minkowski form in said coordinate system).
Your equation (5) would then simply be the equivalent for the transformation of the coordinate representation of a metric in one coordinate system to another (it would take on the form or the transformation rule of the representation of a 2-tensor).

You do something very analoguous between TM and a-TM: you introduce a mapping tau between TM and a-TM, and your equation (5) is the equivalent of the "tensor transformation" when we change coordinate representation. But in the same way as the rule of transformation of a 2-tensor doesn't FIX the 2-tensor, and still leaves entirely open the SPECIFICATION of the metric, in the same way, your equation (5) doesn't fix, at all, the content of tau (it only specifies how its representation should transform between different coordinate representations).

So I don't see any difference in principle between defining the metric g, which fixes the map between TM and TM*, and defining the a-metric tau, which fixes the map between TM and a-TM.

In the same way as g is a 2-tensor field, tau is an extra field over the manifold (this time fixing the relationship between TM and a-TM and not between TM and TM*).

So tau needs to be specified. It has not much to to, a priori, with g. It is a second kind of "metric" (although tau is not a 2-tensor, but something that transforms differently, according to (5)).

tau cannot a priori be deduced from the metric (in the same way as the metric cannot be deduced ab initio!). The specification of tau will determine what are those a-geodesics.

The danger is, of course, that by working in a preferred coordinate system one puts accidentally tau equal to 1. It is what I think you do when you say that a-particles fall upward on the surface of the earth. In the same way as one can accidentally introduce a metric by identifying TM and TM* in a preferred coordinate system.

So up to here, tau and g are independent quantities.

However, above equation 10, you seem to say that you want to couple tau to the metric: you seem to postulate that tau must take on the unity matrix form in a coordinate system where the metric takes on the minkowski form.

As such, you DO couple tau to g, but you have to understand that this is an extra requirement which you now impose, and which fixes tau from the metric.

Well, if this is the case, in my falling elevator frame, the metric DOES take on the minkowski form (because of the transformation property of the metric 2-tensor), so I take it that I can apply your rule and set tau = 1 here.

And we're back home now: tau = 1 in the falling elevator frame, TM and a-TM coordinate representations are identical (tau being equal to 1), and hence a-geodesics are geodesics, and my a-particle falls down with the elevator.

In my rocket-going-upward frame, the metric does NOT take on the Minkowski form, hence I don't know what tau is (it's only specification being that it takes on the form 1 when the metric takes on the Minkowski form). But assuming that the entire mapping (g from TM to TM*, tau from TM to a-TM...) is geometrical, the a-geodesics which were coincident with geodesics in the falling elevator frame are geometrical objects independent of any coordinate representation. So I take it that if all your transformation rules are correct, in my upgoing rocket frame (where the metric has a Rindler form with acceleration 2 g), the a-geodesics are STILL geodesics. And in this coordinate system, they don't take on the form of uniform, straight motion, as an upfalling particle worldline would.

If you insist on this, you CAN do so, but this time you'll have to fix tau in a different way (and NOT having tau = 1 in the falling elevator frame where the metric took on the Minkowski form, so dropping your link between the metric and tau which you specified above equation 10). This is what I called this arbitrary fixing (to the north, to the east...). Tau is now an independent field.

But you'll have to choose. If you fix tau starting from the metric, then your a-geodesic IS expressible purely from the metric, and your a-particle falls down along a geodesic on the surface of the earth.
If you prefer your a-particle to "fall up" you can do this, but this is by uncoupling tau from the metric, and by fixing this yourself using a specific choice of tau. As such, tau is an extra field over the manifold (which takes on aspects of a second metric, which is independently fixed).

There is even another mystery which remains, to me, and that is how you find g_underscore. In equation (11), you seem to put the coordinate representation of g-underscore equal to the inverse matrix of the coordinate representation of g, but I don't see where this comes from. Why can't I, say, fix g_underscore to be diag(-1,1,1,1) in an arbitrary basis (in other words, totally arbitrary) ?
This would then be a genuine new a-metric between a-TM and a-TM*

 

[marcus]

new paper by Sabine

just posted today:
http://arxiv.org/abs/gr-qc/0605083
Cosmological Consequences of Anti-gravitation
S. Hossenfelder
Comments: are welcome

"The dynamics of a universe with an anti-gravitating contribution to the matter content is examined. The modified Friedmann equations are derived, and it is shown that anti-gravitating radiation is the slowest component to dilute when the universe expands. Assuming an interaction between both kinds of matter which becomes important at Planckian densities, it is found that the universe undergoes a periodic cycle of contraction and expansion. Furthermore, the possibility of energy loss in our universe through separation of both types of matter is discussed."

 

[Careful]

just posted today:
http://arxiv.org/abs/gr-qc/0605083
Cosmological Consequences of Anti-gravitation
S. Hossenfelder

Simply watch formula (20) till (24) and you will see that she uses the preferred anholonomic static frame non-dynamically (moreover, there is a typo in (21)). In the friedmann robertson walker case, she picks out the preferred (by symmetry) harmonic frame. That is : in the construction of \tau she manifestly breaks local lorentz invariance (and no prescription for \tau in terms of the metric is given as far as I see), hello Galileian mechanics.

Cheers,

Careful

 

[vanesch]

Simply watch formula (20) till (24) and you will see that she uses the preferred anholonomic static frame non-dynamically (moreover, there is a typo in (21)). That is : in the construction of \tau she manifestly breaks local lorentz invariance (and no prescription for \tau in terms of the metric is given as far as I see), hello Galileian mechanics.

I don't think it is not only \tau but also fixing g-underscore which is frame dependent. Indeed, that's what I thought she did: she decides to go from g to g-underscore (by taking the inverse matrix of the representation of g to represent g-underscore) in the Schwarzschild frame. If she would have done it in a falling elevator frame, (where the metric is essentially eta), she'd have arrived at another g-underscore, and hence other anti-geodesics.
We are here again at finding the "new geodesics" corresponding to the "metric" which is given by the inverse matrix of the coordinate representation of the normal metric g. But this operation results in "geodesics" which are dependent on the choice of the frame in which you do this inversion.
If you do it in a frame in which the metric is \eta, then the inverse is also \eta, and the geodesics coincide with the "new" geodesics. If you do it in a frame which suffers an acceleration g, then you'll find new geodesics which will have acceleration g in that frame (while normal geodesics have acceleration -g of course).

So we ARE again back to our particle which is accelerating away from us depending on our own acceleration. As the "Schwarzschild" observer, at the surface of the earth, sees an acceleration of g in the + z direction, our "new geodesic" in this frame will also have an acceleration of 1 g in + z. And as the acceleration in the falling elevator is 0, it will give a "new geodesic" there which is also 0.

So let's repeat the exercise in (20)-(24), in the "falling elevator frame". After long discussion with Sabine, she finally granted that this coordinate frame exists. It can be found, starting from the Schwarzschild coordinates, by the transformation I sketched in post number 32 in this thread.
Now, we have:
ds^2 = -dT^2 + dX^2 + dY^ + dZ^2 for (20)

Going to the local orthonormal basis gives us that E is the unit matrix (21) - because we ARE already in an essentially orthonormal basis.
As such, \tau^\nu_\nu is diag(1,1,1,1) (22), and so is
\tau_{\nu,\nu} (23)
Hence, g_underscore is diag(-1,1,1,1) and all the new connection symbols vanish in this frame.

Finding the geodesic will lead us to a uniform straight motion, which is NOT the same world line than was the case in the derivation in the paper. Hence, this derivation is coordinate-system dependent.

There's nothing wrong with that of course: one may arbitrary select a coordinate system, which comes down to arbitrary select the "metric" g_underscore. But this choice fixes a second metric, and it is an arbitrary choice.

But it was useful that Sabine wrote this down explicitly, so that we could see the calculations explicitly, instead of having to guess them.

 

[vanesch]

My remark can be stated differently. If we take it that \tau is essentially the metric inversed, and the gamma-underscore connection coefficients are expressed starting from the inverse matrix of the metric g, as given by equation 10, then the curve given by equation (13) is dependent on the coordinate frame in which the calculation is performed ; in other words, it is not a world line on the manifold (or it is a world line on the manifold, if it is specified in which preferred frame the calculation has to be performed).

This could probably be established in detail if all the necessary substitutions as a function of the local metric would be performed (the thing I have been asking for ages now), but the simple example in the Schwarzschild coordinate frame and in the falling elevator frame giving different results, proves this already.

 

[hossi]

Simply watch formula (20) till (24) and you will see that she uses the preferred anholonomic static frame non-dynamically (moreover, there is a typo in (21)). In the friedmann robertson walker case, she picks out the preferred (by symmetry) harmonic frame. That is : in the construction of \tau she manifestly breaks local lorentz invariance (and no prescription for \tau in terms of the metric is given as far as I see), hello Galileian mechanics.

Cheers,

Careful

Hi Careful,

thanks for pointing out the typo. Indeed, to explicitly calculate a curve, I pick a coordinate system. Why do you think the construction of \tau breaks covariance? It is constructed from the tetrads. Do these break covariance? Yes, in a certain coordinate system, \tau will take a certain form.



B.

 

[vanesch]

thanks for pointing out the typo. Indeed, to explicitly calculate a curve, I pick a coordinate system. Why do you think the construction of \tau breaks covariance? It is constructed from the tetrads. Do these break covariance? Yes, in a certain coordinate system, \tau will take a certain form.

The problem doesn't lie in tau by itself. It is indeed correct to define a transformation which transforms the coordinate basis in an orthogonal basis (but which one :-). The trouble comes when you fix g_underscore and the connection coefficients which follow from this. As long as these are just definitions, there's nothing wrong with it, but the problem is that when you construct a "geodesic" with it, that this geodesic is now dependent on the choice of frame in which you've done this.

The reason is that a world line is a map from R onto the manifold x(lambda). You can construct now an equation, in a coordinate system, using all the quantities you defined, to give you an equation for x(lambda), this does not necessarily result in a world line, in that when you apply the same mechanical rules to do this in another coordinate system, you'll find an equation which gives you y(lambda'), and the mapping between the two coordinate systems will not map x(lambda) onto y(lambda').
This is EXACTLY what we've done. You've constructed your radial world line starting from a Schwarzschild coordinate frame, and using your tau, and g-underscore, and the underscore connection coefficients and all that, and you came to the conclusion that the world line described an up-falling particle in the Schwarzschild frame.
I did the same in the falling elevator frame (linked to the Schwarzschild coordinates by a specific transformation, see post 32), and I applied the same rules for obtaining the world line, and I found a DOWN falling particle.
This is what I'm now repeating for a few weeks now.

This proves that equation (13) does not describe correctly a world line, independent of in which coordinate system it is expressed.

Where exactly this covariance goes wrong is difficult to say because it is a building - up of a lot of definitions. But fact is, that at the end of the day, when working things out from different coordinate systems, we arrive at different curves.

It would have been much easier if you would have expressed equation 13 purely as a function of the metric. Equation 13 being an equation for a world line, it must be a covariant expression, and it probably (well, surely) isn't. In other words, equation 13 does not describe a geometrical object on the manifold (unless you specify in WHICH coordinate system it must be worked out).

 

[Careful]

**
thanks for pointing out the typo. **

The typo is irrelevant.

** Indeed, to explicitly calculate a curve, I pick a coordinate system. Why do you think the construction of \tau breaks covariance? **

Look, you *pick out* a preferred tetrad and construct \tau such that it is not invariant at all under local Lorentz boosts. Now you could say, there exists a law which is locally Lorentz invariant and introduces this frame as a dynamically preferred element : then I say, give us this law (this is what I and Vanesch are asking all the time). Of course you can say, well I could give you a set of fully covariant laws in which I can dynamically distinguish this and this frame (coordinate system) : my reply would be the same (I have said somewhere else that you can even make Newtonian physics generally covariant). In both cases, you grossly demolish the spirit of the equivalence principle (and construct a dragon of a theory) and it is indeed very simple (as we both pointed out) to come up with theoretical constructs containing many different geodesics (Vanesch did that in one of his first messages). So, as both your *papers* stand now, I see no attempt to restore covariance, neither local lorentz invariance.