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

[vanesch]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[Careful]

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

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

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

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

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

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


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

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

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

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

Careful

 

[josh1]

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

 

[hossi]

Dear Careful,

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

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

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

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

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

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

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

Well, I could say the same thing.

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

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

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

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

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



B.

 

[hossi]

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

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

B.

 

[hossi]

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

Hi Rade,

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

B.

 

[josh1]

Hi Hossi,

Is there an antigravity analog of a black hole?

 

[hossi]

Dear vanesh,

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

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

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

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

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

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

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

I have no idea what you are trying to say.

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

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

B.

 

[hossi]

Hi Hossi,

Is there an antigravity analog of a black hole?

Hi Josh ,

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

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



B.

 

[Careful]

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

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

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

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


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

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

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

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

Careful

 

[josh1]

Hi again Hossi,

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

 

[hossi]

Hi Careful,

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

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

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

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

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

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

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

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

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

Good. Then take it.

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

Maybe you are not listening closely enough.

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

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

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

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

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

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

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

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

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

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

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

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

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



B.

 

[hossi]

Hi again Hossi,

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

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

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



B.

 

[josh1]

Okay hossi,

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

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

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

 

[hossi]

The footnote on page 10 of your paper reads

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

Maybe you meant that

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

Smart josh.

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

B.

 

[josh1]

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

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

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

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

 

[Careful]

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

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

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


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

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

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

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

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

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


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

Careful

 

[vanesch]

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

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

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

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

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

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

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

 

[Careful]

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

 

[vanesch]

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

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

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

 

[josh1]

Hi hossi,

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

 

[hossi]

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

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

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

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

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

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

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

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

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

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

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

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

B.

 

[Careful]

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

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

Cheers,

Careful

 

[hossi]

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

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

 

[Careful]

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

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

 

[hossi]

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

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

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



B.

 

[vanesch]

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

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

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

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

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

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

 

[Careful]

**Now we are getting somewhere **

We were there one week ago.

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

Cheers,

Careful

 

[hossi]

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

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

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

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



B.

 

[Careful]

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

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