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

[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.

 

[George Jones]

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

Regards,
George

 

[vanesch]

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

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

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

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

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

 

[vanesch]

Hi again,

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

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

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

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

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

 

[vanesch]

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

 

[hossi]

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

Dear vanesh,

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

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

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

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

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



B.

 

[vanesch]

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

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

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

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

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

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

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

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

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

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

 

[vanesch]

Progress...

Hello Sabine and others,

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

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

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

Could you check whether this calculational scheme is correct ?

 

[vanesch]

Conclusion ...

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

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

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

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

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

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

Your Honor, I rest my case.

 

[vanesch]

for completeness...

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

 

[hossi]

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

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

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

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

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

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

Your Honor, I rest my case.

Hi vanesh,

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

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

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

B.

 

[vanesch]

Hi vanesh,

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

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

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

B.

Hi Sabine,

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

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

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

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

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

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

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

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

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

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

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

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

 

[vanesch]

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

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

 

[vanesch]

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

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

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

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

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

 

[hossi]

Hi vanesh,

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

B.

 

[hossi]

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

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

B.

 

[vanesch]

Hi vanesh,

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

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

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

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

(I use D for del)

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

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

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

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

 

[vanesch]

Hi vanesh,

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

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

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

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

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

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

Same for eta: it's a matrix.

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

So E1 is a matrix too.

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

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

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

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

And then things get a bit more involved.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[vanesch]

general solution

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

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

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

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

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

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

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

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

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

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