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

[Chronos]

So, anti-gravitating particles are ruled out, if I follow the arguments correctly. Otherwise I would expect an abundance of them in saddle points between say, galactic clusters, and some very weird anti-lensing effects.

 

[vanesch]

Just to explicitly follow up:

Consider the Schwarzschild metric:

ds^2 = -c^2 (1-\frac{2GM}{c^2 r}) dt^2 + (1-\frac{2GM}{c^2 r})^{-1} dr^2 + r^2 d\Omega^2

Now, transform the coordinates:

T = t ;
<br /> X = r \sin \theta \cos \phi
Y = r \sin \theta \cos \phi
\zeta = r - R

We are interested only for things near theta = 0, for \zeta much smaller than R and for R much much bigger than the Schwarzschild radius.

This gives us then an approximate expression for the metric in these coordinates:
<br /> ds^2 = -c^2 (1 - \frac{2 G M}{c^2 R}(1-\frac{\zeta}{R})) dT^2<br /> + (1 - \frac{2 G M}{c^2 R}(1-\frac{\zeta}{R}))^{-1} d\zeta^2 + dX^2 + dY^2 <br />

Now, we can rescale \zeta slightly, so that
dZ = d\zeta (1 - \frac{2 G M}{c^2 R}(1-\frac{\zeta}{R}))^{-1/2}
by integrating from Z = 0 to Z for zeta going from 0 to zeta. This almost doesn't change the value of zeta, so we can easily replace the first order term of zeta in the dT coefficient by Z without harm (it are higher order effects).

And then we obtain, in the coordinates X,Y,Z,T:
<br /> ds^2 = -c^2 (1 - \frac{2 G M}{c^2 R}(1-\frac{Z}{R})) dT^2<br /> + dZ^2 + dX^2 + dY^2 <br />

Now, this is the same metric as the Rindler metric
<br /> ds^2 = -c^2 (1 + g/c^2 z)^2 dt^2 + dz^2 + dx^2 + dy^2<br />

after a small rescaling by a constant factor of (1+ 2GM/c^2R) of T (the time dilatation due to the gravitational well we're in), when we expand the Rindler metric for small g/c^2 z,

on the condition that we identify g with + G M/R^2
(which is exactly the inverse of the gravitational acceleration in the Newtonian limit at the Earth surface).

So the coordinate system X,Y,Z,T is the coordinate system of an observer at the Earth's surface. And its metric corresponds in a good approximation, to the metric of an upward accelerated observer with acceleration +g in the +Z direction in a flat spacetime.

Next, we can transform this into:

x = X, y = Y, z = Z - 1/2 g T^2
t = T

This is (in the Newtonian limit) the coordinate system of the observer in the falling elevator. As you see, it has a meaning, and is retraceable from the Schwarzschild coordinates.

Now, if you work out the metric in (x,y,z,t), you obtain, to a good approximation, a Lorentzian metric in a FINITE patch, near X=Y=Z=T=0, in the coordinates x,y,z,t.

 

[Careful]

Actually, after some conversation with Patrick, I am afraid that there is no difference at all between the gravitating and antigravitating particle which is easily seen since the map from the normal tangent and cotangent bundle to the underlined bundles also properly transforms the application of a covariant derivative on a vectorfield and so on. Hence, the geodesics remain unchanged.

This appears also logical from the physical point of view since one would think different physical objects to be associated to new physical fields.

Cheers,

Careful

 

[hossi]

Guys, you are really being tough on me. I will think over it, hope I have time at the weekend (sorry, have to do some butter-and-bread physics).

Vanesh, you don't even try to follow my arguments and repeatedly claim I said things I never said. E.g. I never claimed "that one of these two coordinate systems does not exist over a local patch", I tried to make clear the difference between a LOB and a free-falling frame. My answer to your equivalence principle problem is still the same as in the very beginning.

Chronos, your argument does not rule out anything, whether anti-g matter is observable or not certainly depends on the amount.

Careful, thanks for being careful, give me some days to sort out my thoughts.

Thanks to all of you - I am serious, I appreciate the feedback, even if depressing

B.

 

[vanesch]

Vanesh, you don't even try to follow my arguments and repeatedly claim I said things I never said. E.g. I never claimed "that one of these two coordinate systems does not exist over a local patch", I tried to make clear the difference between a LOB and a free-falling frame. My answer to your equivalence principle problem is still the same as in the very beginning.

Dear Sabine,

I'm in fact not so much discussing your paper, I'm discussing the simple fact that you claim that there can be particles falling UP at the surface of the earth, that this is not in contradiction with the equivalence principle, and that this is a gravitational effect.
These elements, in themselves are contradictory, and this is what I tried to discuss. Now, you bring in your paper, I looked a bit at it, but it is not really discussing this up-falling particle.
I'm by far not an expert on GR, I know some of it, but what you bring up is rather elementary if you have some physical insight in the basic elements of GR that I fail to see how you cannot try to explain this more carefully.

Because, at first sight, what you propose is the gravitational equivalence, of, say, somebody who has written a sophisticated treatment on thermodynamics and comes to the conclusion that, after all, one can violate the efficiency of a Carnot engine, but that this is not in contradiction with the second law of thermodynamics. If someone, who is not really versed in sophisticated versions of non-equilibrium thermodynamics, simply explains that he thinks that these two statements are impossible, you need to be able to explain that carefully.
I'm not an expert on sophisticated differential geometry at all, I'll grant you that without problems. But I do know the basics of GR well enough to know that you cannot have, at the same time, a particle falling up on the surface of the earth, and respect of the equivalence principle.

I'm not talking about any DYNAMICS of GR yet, I'm simply talking about the kinematical description. I'm now at the level of someone who knows his thermodynamics well enough to understand fully the derivation of the Carnot engine from the second law, I present you the reasoning that leads, IMO to the contradiction of your proposal, and I haven't seen any rebuttal to it.

Now, for the moment, I'm of course thinking you make an error somewhere. But I can conceive the fact that *I* am making an elementary mistake. In both cases, one of both will learn something. There's the saying: "If you think you argue with an idiot, chances are, the other thinks the same" So why should you bother talking with an idiot ?
The reason is that if you want to have any chance for your idea to be valued, you need to be able to explain it to idiots ! If you can't, your idea is worthless, in a way.

Let's put things on a row. You can maybe point out where you disagree, but what I propose is really quite simple and elementary in GR. I wouldn't be able to do sophisticated things in it anyways.

1) there is a 4-dim manifold, called spacetime, and I have a way of describing a finite patch around me, with 4 coordinates. I can of course consider different ways to describe these coordinates and then there is of course a smooth mapping between these two sets of coordinates (4 functions of 4 variables).

2) concerning gravity, all I need to know is given by the metric tensor expressed in my 4 coordinates, over this finite patch. All what is "in gravity" is encoded in this metric. As this is a tensor, its transformation under a smooth coordinate transformation is given by the normal rules.

1) and 2) are, in my opinion, the essence of the equivalence principle. Maybe you disagree, but this is how it is normally presented everywhere.
Point 2 is very important, btw. It says that all you need to know is encoded in a finite local patch of the metric. So your "antigravity" will have to be able to be deduced from this, and that's what I've been asking you to do. If it is not deducible from just the form of the metric in a finite local patch, then this doesn't qualify as "gravity" - I'll come to that.Mind you that 1) and 2) still doesn't say anything about any motion observed in a coordinate frame of any particle or anything. We come to this now:

3) a particle describes a world line, that is, a map from R to the manifold. This means that, if you know the world line's description in one coordinate set, that its description in another set simply follows the coordinate transformation.

This isn't strictly necessary of course. If "particles" cannot cause events (like a click in a detector) but are intermediate, abstract quantities, such as, say, zero crossings of a wave or something, this doesn't need to be the case.

You claim that your anti-grav. particle, however, follows a world line.

And, finally:

4) the world line of a normal particle is given by a geodesic of the metric tensor mentioned in 2).

Again, this doesn't need to be the case, but for normal particles, this is the case.

And now we come to your claim (for which your paper might be an inspiration, but which is, in itself, IMO, contradictory - which is what I want to discuss).

You claim, that an anti-grav particle (let's call it a Sabinon ) does the following:

1) it has a world line
2) it falls UPWARDS at the surface of the Earth (with 1 g)
3) in a flat spacetime, it follows the geodesics by the metric tensor (= Minkowski metric ==> uniform straight motion).

Well, I do not need to read sophisticated papers to know that these 3 conditions in themselves cannot be true, at least, if:

4) it is a gravitational effect, so that (see point 2) its motion can be derived purely from the local metric over a finite patch of spacetime

I have exposed zillions of times now the reasoning behind the derivation of a contradiction, which is, in summary:
at the surface of the earth, we are nearly in a flat piece of spacetime, so according to 3) the sabinon has to follow the geodesics nearly.
This can be made explicit by transforming explicitly to a coordinate frame where the Minkowski metric is essentially correct (the falling elevator), considering that it describes a world line, and transforming back to the coordinate system at the surface of the earth. It's what I've been doing in more and more explicit steps, and is a rather elementary exercise.

Your statement is simply that this is somehow not true, but that's not good enough for me. I would like to see how you derive EXPLICITLY the world line of a sabinon in a coordinate system at the surface of the earth, BY USING ONLY A SMALL PATCH OF SPACETIME WITH THE METRIC ON IT which can be approached very well by a flat spacetime, and show that it falls UP ; because I think that I proved that this is impossible.

Now, by you failing to give me this explicit derivation, I had to GUESS what you might do. I can think of different ways to have a sabinon fall up:

1) introduce an extra field. For instance, I could introduce an EM field, have all matter have a positive charge proportional to their mass, and then I'd find out that with the right choice, there's an extra repulsion on the particle. But that's not "anti - gravity". I suspect you to do this, in fact. I suspect your "tau" to be related to such an extra field. The only way for you to prove me wrong here is by DEDUCING the 'falling up' from the metric, and the metric alone, IN A SMALL PATCH around the particle (and not GLOBALLY).

2) use the "geodesics" of the inverse metric. At a certain point, I thought that that was what you tried to do. This works, but the price to pay is that a sabinon has no world line anymore. It's "geodesics" are now observer dependent. This is this famous "particle that accelerates away from you, the more you accelerate towards it".

But, again, I think I've shown enough that, if you stick to 1) and 2) (so that its equation of motion is derivable from the metric and the metric alone over a finite patch of spacetime), that you cannot give me a way which allows you to have a world line of a particle that falls up at the surface of the earth, because I proved the opposite (I think).

This is why I ask you to tell us how this derivation goes about, in the examples I gave you (where I gave you the patch of metric). It has in fact not much to do with your paper as such, but with your claim of anti-gravitating particles respecting the equivalence principle. Apparently you base your claim on what you write in your paper, but I don't see how.

 

[Careful]

Hi Vanesch,

I thought the latter was your version (and I have some comments which might be useful):

**
1) there is a 4-dim manifold, called spacetime, and I have a way of describing a finite patch around me, with 4 coordinates. I can of course consider different ways to describe these coordinates and then there is of course a smooth mapping between these two sets of coordinates (4 functions of 4 variables).

2) concerning gravity, all I need to know is given by the metric tensor expressed in my 4 coordinates, over this finite patch. All what is "in gravity" is encoded in this metric. As this is a tensor, its transformation under a smooth coordinate transformation is given by the normal rules.
**

The equivalence principle rather consists out of many more parts :
(a) inertial mass = passive gravitational mass, or : the motion of a (electrically) neutral test particle is independent upon the internal structure of the particle (although this version is open for obvious criticism)
(b) covariance principle : the laws of nature have to be written in tensorial form (criticism : no spinors )

(a) + (b) = weak equivalence principle

(c) the result of a local, non gravitational experiment is independent of the speed of the free falling reference system in which it is done (local Lorentz invariance)

(d) the result of a local and non - gravitational experiment is independent of the place and time in which it is performed (local position invariance).

All four together form the strong equivalence principle (or Einstein equivalence principle) which is broken in many ``gravitational´´ theories. For abbreviation, let's speak about WEP and SEP and let me show how violation of SEP - but not of WEP - has some bearing upon your point (2).

For example (d) is very strong and practically restricts severely the coupling of matter to gravitation (no coupling to the curvature tensor). Take the simple case of a point particle, with the standard lagrangian:

m int( dt sqrt( g( dx/dt , dx/dt ) )

one could change this to

int (dt m sqrt( 1 + a R( x(t) ) ) sqrt( g( dx/dt , dx/dt ) )

with the dimension of a = length^2. Actually, in this way you can get the one particle schroedinger equation out of Weyl geometry. The latter lagrangian violates (d) but not WEP. SEP is also violated by the inflaton for example.

Another way to violate (d) would be the addition of a torsion field to the connection (this again was one of the old geometric attempts to incoorporate QM / there were of course other attempts by adding an anti symmetric part to the metric tensor). Both examples violate your second principle : the former since it is not sufficient to know just the metric to determine the coupling to the gravitational field while no other physical field is present, the latter since we add an extra physical field.

Now, torsion does not contribute to the geodesic equation, hence I thought in the beginning she might have added some curvature form to the action of the anti gravitating particle. But of course, she does none of these : basically what happens in the paper is that a rather unconventional isomorphism (that shuffles through contractions and so on) is set up between the two bundles (hence nothing changes).

Furthermore, it needs to be said that such changes (involving the curvature) in the equation of the free particle will involve violations of causality (cfr the de Broglie mass problem). Anyway, I hope I added something here to your understanding of the different versions of the equivalence principle.

 

[vanesch]

one could change this to

int (dt m sqrt( 1 + a R( x(t) ) ) sqrt( g( dx/dt , dx/dt ) )

with the dimension of a = 1/length^2. Actually, in this way you can get the one particle schroedinger equation out of Weyl geometry. The latter lagrangian violates (d) but not WEP. SEP is also violated by the inflaton for example.

I fail to see how this violates my "2)" that the local equation of motion is not derivable from the local metric over a finite patch. After all, under the hypothesis of no torsion, I can DERIVE the Riemann tensor from the metric tensor (over a finite patch of course, not in 1 point).
So it seems that this R(x(t)) can be calculated, if the metric is given over a finite patch, no ? From g, I calculate the Riemann tensor, and hence the Ricci scalar (that's what R(x(t)) is, no ?). So R(x(t)) is just a complicated expression of the metric and its derivatives, or am I wrong here ?

 

[Careful]

I fail to see how this violates my "2)" that the local equation of motion is not derivable from the local metric over a finite patch. After all, under the hypothesis of no torsion, I can DERIVE the Riemann tensor from the metric tensor (over a finite patch of course, not in 1 point).
So it seems that this R(x(t)) can be calculated, if the metric is given over a finite patch, no ? From g, I calculate the Riemann tensor, and hence the Ricci scalar (that's what R(x(t)) is, no ?). So R(x(t)) is just a complicated expression of the metric and its derivatives, or am I wrong here ?

It violates your (2) in the sense that even if the metric is the only dynamical variable involved in gravitation, then still the Einstein equations (``that what is in gravity´´) are not fixed. If your statement was kinematical, then there is not contradiction. But if you meant the latter, then the concerns you expressed towards the a priori impossibility for a distinction between gravitating and anti gravitating matter are false. The fact that they are the same in THIS case requires more that simply your version of the (weak + (c)) equivalence principle.

 

[vanesch]

It violates your (2) in the sense that even if the metric is the only dynamical variable involved in gravitation, then still the Einstein equations (``that what is in gravity´´) are not fixed. If your statement was kinematical, then there is not contradiction.

I'm only involved here with the kinematics, indeed, with small test particles.
I didn't mean to imply that the Einstein equations are fixed!

EDIT: but even then, I don't understand your remark. After all, the lagrangian IS now expressed purely in terms of the metric and its derivatives, so how does this not fix the dynamics ?

 

[Careful]

I'm only involved here with the kinematics, indeed, with small test particles.
I didn't mean to imply that the Einstein equations are fixed!

Ok, then there is no misunderstanding I only wanted to say why I was careful in the beginning with my comments towards the existence of possible differences due to tidal effects or not.



Careful

 

[Careful]

Reply to your EDIT: how to build the right hand side of the Einstein equations ? You start out with the energy momentum tensor for the dynamical object at hand in Minkowski and apply the minimal substitution principle (this garantuees that (d) is satisfied). So, the SEP + the knowledge of the dynamics in Minkowski fixes the dynamical coupling to gravity. However, if you abandon (d), this is not the case anymore. Concretely, the free particle has a unique coupling to gravitation given SEP, but given WEP one has more exotic possibilities such as the one I gave. Both alternatives are ``free particles´´ but they respond differently to a gravitational field.

 

[Rade]

...I'm in fact not so much discussing your paper, I'm discussing the simple fact that you claim that there can be particles falling UP at the surface of the earth, that this is not in contradiction with the equivalence principle, and that this is a gravitational effect...

I had mentioned before the possibility of "up" and "down" particle interactions between gravity "down interaction" and antigravity "up interaction" when you consider the particles that interact to be clusters of nucleons (protons and neutrons). Thus a matter cluster [PNP] (helium-3) may bind with an antimatter cluster [NP] (anti deuteron) via a gravity + antigravity interaction--but of course there is neither mathematics nor experiments ever attempted to falsify such a claim--thus remains an open question for future scientific study. Who has time to develop the mathematics for such interaction of rotating nucleon clusters using gravity and antigravity ?

 

[Chronos]

Pardon me for being dense, but does not an anti-gravitating particle wreak havoc on the energy conservancy principle? It looks very unphysical to me.

 

[vanesch]

What I understand of Sabine's work is that she arrives at a potentially different equation of motion for a "free" anti-gravity particle (further on called a sabinon ), than the usual geodesic which is postulated for a normal particle in GR.
I do not mind this! The problem I have is not so much that there could be a different equation of motion than the geodesic. The problem I have with her proposal is that this equation of motion seems not to be derivable from the metric in a local patch of spacetime. Now, or she admits this - and then I have no problem with it - but it is hard to see how she can claim then not to shatter to pieces ENTIRELY the equivalence principle. For instance, you could introduce also instantaneous action in GR, using an extra field that tells you how to make this unambiguous (essentially an ether theory then).
Or she claims that this is not the case, that you CAN derive this equation of motion from the local metric over a finite patch ; and then I don't believe her and want to see this derivation. The reason why I don't believe her is that there is, to a good approximation, no difference between the metric at the surface of the earth, and the metric of a uniformly accelerated observer in flat space. So both should find the SAME equation of motion (up to eventually some SMALL correction). The small difference between both metrics resides in tidal effects which are present at the Earth surface, and which are not present for a uniformly accelerated observer, but these effects are 1) very small compared to g and 2) more importantly, can be made as small as I wish, by going to a bigger planet, keeping the 1 g and having less and less tidal effects.
It is difficult to imagine that for observers with two metrics who are nearly the same (surface of earth/big planet versus accelerated observer in flat space), one will find a particle that falls UP and the other will find a particle that falls DOWN.
So that's why I want to see the explicit derivation.
I suspect that the whole "trick" is the "fixation" of this famous "tau". Because in the paper is discussed how tau *transforms*, but not how you calibrate it (in what frame, and why ?). If there is some arbitariness in this choice, then this tau is a NEW FIELD that is not derivable from the metric, locally. This can very well be: that she fixes this tau totally differently in a Schwarzschild metric than in a Rindler metric.
But this then, comes down to what I claimed: that the equation of motion is NOT derivable from a local finite patch of metric, in which case not much remains from the equivalence principle and in which case we have in fact a kind of ether theory.

Another possibility exists, in which the actual equation of motion of a sabinon does NOT indicate that it falls up at the surface of the earth. But then I don't know why this is called "antigravity" and it goes against Sabine's claim. I didn't check it starting from the elements of her paper. The path between the metric and the equation of motion is rather long (and involves this tau).

And maybe I'm wrong, in which case I'm in for a big surprise and I'll learn something, which is also positive, and Sabine will have had an exercise in explaining her idea.

But in the mean time I remain sceptical, for the reason I set out, that it will be possible, from a given metric over a finite patch of space, to derive an upward falling particle equation of motion.

 

[hossi]

Rindler and Einstein's elevator

So why should you bother talking with an idiot ?
The reason is that if you want to have any chance for your idea to be valued, you need to be able to explain it to idiots ! If you can't, your idea is worthless, in a way.

Hi vanesh,

I take it very seriously to make my work accessible to as many people as possible. I genuinely try the best I can, this is part of the reason why I am here in this forum (another reason is that I like arguing).

Let me start again with the equivalence principle. A particle locally experiences gravitation the same way as acceleration in flat space. Place your particle in Einstein's elevator. In a free falling frame in a gravitational field that means, when you accelerate the elevator with the appropriate acceleration, the particle will not not notice anything. For an a-particle the elevator has to be accelerated in the oppiside direction. Both free-falling frames are not identical, but for both seperately, the eq. principle holds. That's why I speak of a relaxation of the equivalence principle, not of a generalization. You are of course completely right that in a gravitational field, there is no elevator that could make BOTH particles at once not notice it. That is exactly the point. Both particles will be indistinguishable if and only if spacetime is globally flat.

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

The answer to your question is of course yes, the motion of the a-particle can be derived from the local metric (over a patch of spacetime). Just that this is not exactly the easiest way to do it. The reason is that in a global coordinate system, the geodesics are related by g being the inverse of g, in a local frame, the relation between both is not that apparent (as careful has found, more about this later).

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

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

Is not in space-time coordinates (see e.g. MTW 6.18), therefore you can't just take the inverse and get the geodesics. But I guess I don't get the point here, if you have an accelerated observer in flat space, he apparently does not move on a geodesic.

Does that help?

B.

 

[hossi]

Pardon me for being dense, but does not an anti-gravitating particle wreak havoc on the energy conservancy principle? It looks very unphysical to me.

Indeed the energy conservation is an important point. You have to look at the covariant version. The conservation law for the kinetic energy is modified - essentially because the covariant derivative you use has to be appropriate for the a-particle. A usual particle that falls down in a graviational field gains kinetic energy. An a-particle gains kinetic energy in the same gravitational field, when it falls up. (This actually is identical to the equations of motions).

B.

 

[hossi]

Hi Careful,

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

The essence of the approach I have proposed is that there are two ways a field can behave under coordinate trafos. The one is mapped to the other by \tau. Whether you like that or not, it is handy. I personally don't like it, but I couldn't come up with anything better. Note that the elements of both spaces don't transform alike. How could they possibly have the same properties under parrallel transport?! They can't (unless spacetime is globally flat.)

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

You can therefore derive two curves following from them, the usual geodesic and Eq (38).

Now what you have done is you have made a coordinate transformation on the manifold from \partial to \tau \partial, and derived the geodesic motion in that system. I am not even sure what that system means. It's neither a coordinate system nor a LOB. But that is not essential. The point is that of course this transformation does not change the geodesic equation. It just makes it look more ugly (as you point out, the commutators don't vanish, neither is the metric locally constant). What you have to consider instead is taking the derivatives \tau \nabla \tau^{-1} (which have the same commutation behaviour as the usual ones).

Thanks for the brain stretch.

B.

 

[hossi]

I had mentioned before the possibility of "up" and "down" particle interactions between gravity "down interaction" and antigravity "up interaction" when you consider the particles that interact to be clusters of nucleons (protons and neutrons). Thus a matter cluster [PNP] (helium-3) may bind with an antimatter cluster [NP] (anti deuteron) via a gravity + antigravity interaction--but of course there is neither mathematics nor experiments ever attempted to falsify such a claim--thus remains an open question for future scientific study. Who has time to develop the mathematics for such interaction of rotating nucleon clusters using gravity and antigravity ?

I completely fail to see how anti-gravity would bind the system. Wouldn't it in the contrary push it apart? Also, in my model the both types of particles don't interact except for the gravitational interaction. Indeed, it turned out to be impossible - all other coupling terms (electro, strong, etc) either violate Lorentz- or gauge-invariance (or both). That has seriously depressed me for quite some while (I mean, how could we ever detect the a-particles?!). Thus, I don't see why such a study would be interesting, or what model it should be based on. I personally doubt you can cook up a reasonable setup for the investigation.



B.

 

[Rade]

I completely fail to see how anti-gravity would bind the system. Wouldn't it in the contrary push it apart?

Thank you for the time of your response. The model I study predicts that there is a neutral coexistence between gravity and anti-gravity, with gravity related to "matter" clusters, and anti-gravity related to "anti-matter" clusters. Because the two clusters must have different mass, they bind. This differs from the common understanding that matter and antimatter when they meet annihilate. From your answer I take it then that you are not aware of any research that has attempted to bind matter helium-3 with antimatter deuteron. I guess I do not understand why such an experiment would not be of interest mathematically and/or theoretically. If the model I study is correct, when these two isotopes meet, the reaction should not yield complete annihilation energy. The reason being that anti-gravitation is present within the system, coexisting with gravity at the sub-atomic level.

 

[hossi]

Thank you for the time of your response. The model I study predicts that there is a neutral coexistence between gravity and anti-gravity, with gravity related to "matter" clusters, and anti-gravity related to "anti-matter" clusters. Because the two clusters must have different mass, they bind.

Hi Rade,

I understand the matter clusters, and the anti-grav. matter clusters. But what force do they bind with? Strong? Electric?

This differs from the common understanding that matter and antimatter when they meet annihilate. From your answer I take it then that you are not aware of any research that has attempted to bind matter helium-3 with antimatter deuteron. I guess I do not understand why such an experiment would not be of interest mathematically and/or theoretically. .

Wait, wait, are we talking anti-matter (conjugation of electic charge), or are we talking anti-gravitating-matter (negative gravitational mass), or both? I know that there is effort going into making clusters out of anti-matter (like nuclei etc), which is certainly interesting to see whether they have - or don't have - the same properties as the usual nuclei. But where does the anti-g come in? And, how do you avoid it annihilates into nothing with the usual matter (that better shouldn't happen, or you could also make them both out of nothing).

Besides this, what is the model you study, and could you give me some references?



B.

 

[twin]

Hi Sabine,

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

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


Twin

 

[Careful]

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

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

I was already way past that issue.

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

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

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

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


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

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

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

Cheers,

Careful

 

[Careful]

Hi Sabine,

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

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


Twin

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

Cheers,

Careful

 

[vanesch]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[hossi]

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

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

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

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

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



B.

 

[hossi]

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

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

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

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

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

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

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



B.

 

[hossi]

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

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

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

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

If your answer is yes[...]

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

So I take it that this is impossible.

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

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

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

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

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

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

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

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

B.

 

[Careful]

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

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

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

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



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

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




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

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

 

[hossi]

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

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

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

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

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

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

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

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

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

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



B.

 

[Rade]

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

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