# Anti-gravitation discussion

1. Apr 24, 2006

### hossi

Dear vanesh,

I have decided to move the anti-gravitation discussion into a new thread. At least in this regard josh is right, it's gotten a bit off-topic in the old one.

For those who want to follow, discussion is about the paper

http://arxiv.org/abs/gr-qc/0508013" [Broken]

No, it has not. (Actually, that offends me because that is what I criticize most about the works by Frederic Henry-Couannier etal. They do have two metrics, which is imo completely unphysical. They should at least try to interpret this).

The manifold has one metric. The different set of connection coefficients corresponds to the different properties of the quantity to be transported, not to any different properties of the manifold. If it's too implausible to think about anti-gravitation, think about transporting a spinor. It has a well defined transport-behaviour. The connection you use for this however, is not that of a vector. It's different, because it transforms under another representation than the vector.

Same with the anti-gravitating particle. It has a different transformation behaviour under general diffeomorphism than a usual particle. Therefore, the covariant derivative (and the connection coefficients appearing in it) are different. They can be derived directly from the metric and the transformation law of the particle, but they are not identical to the usual Christoffelsymbols.

Best,

B.

Last edited by a moderator: May 2, 2017
2. Apr 24, 2006

Dear Hossi. Do you see any reason, from theory, why a matter nucleon cluster of a proton + neutron [NP]+ (also called deuteron) would NOT be able to bind with an antimatter cluster [PNP]- (also called antimatter helium-3) ? There is an "off standard model" of the atomic nucleus that predicts such, but at present no experimental evidence either pro or con as far as I know. That is, as far as I know, no one has ever tried to experimentally bind deuteron plus anti helium-3--have they ? The model predicts the two would bind via superposition of gravity plus anti-gravity forces at the sub-atomic level. Thank you for any comments you may have.

3. Apr 24, 2006

### hossi

thanks for the thoughts. My comment is that this has nothing to do with my model. Maybe you find something in

"[URL [Broken] existence of antigravity
Authors: Dragan Slavkov Hajdukovic[/URL]

Sorry, I can't be more helpful in this regard.
Best,

B.

Last edited by a moderator: May 2, 2017
4. Apr 25, 2006

### garrett

Hi Sabine,
I'm looking at your paper again (ahh, the odd things physicists do to relax) and I'm encouraged that you have a reasonable looking action (23)
$$S = \int d^4 x \, \sqrt{-g} \left[ G R + L + \underline{L} \right]$$
including matter and anti-gravitating matter Lagrangians. I like simple examples. The Lagrangian for a run-of-the-mill point mass on a parameterized trajectory, $q^\mu(\tau)$, is
$$L(x) = \frac{m}{2} \int d\tau \, g_{\mu \nu}(x) \frac{dq^\mu}{d\tau} \frac{dq^\nu}{d\tau} \frac{1}{\sqrt{-g}} \delta^4 \left( x - q(\tau) \right)$$
So, what's the Lagrangian, $\underline{L}$, supposed to be for an anti-gravitating point mass?

Last edited: Apr 25, 2006
5. Apr 25, 2006

### vanesch

Staff Emeritus
Ok, I looked a bit more at your paper, and here's what I think is the essence of my comment:

what else, is this g_underscore, which you introduce on the bottom of page 4, but ANOTHER METRIC ?
In other words, LOCALLY, how do I derive g_underscore from g - which must be possible if it is not *another* metric, but a quantity derivable from the metric.

And it is my guess that this is what is not possible: to find a LOCAL way to derive g_underscore from g.
Now, globally, I can understand it (more or less), but the whole idea is that all quantities of use should be expressible locally. And I have the impression that this local transformation from g into g_underscore is what you cannot do.
And, not to far from my giant black hole, g is essentially flat (diag(-1,-1,-1,1) in a certain patch with certain coordinates, called inertial coordinates). There's a SMALL correction to that, which are tidal effects: maybe you succeed in using this small effect to find your g_underscore, I don't know. But as I can make this as small as I wish (by making my black hole bigger and bigger, and to go at about larger and larger distances), it is my impression that you have NO WAY of deriving *locally* g_underscore from g.

So let's express my question differently:
Imagine I give you a local PATCH of manifold, by giving you locally a coordinate system, and in that coordinate system, g. I've given you now entirely the metrical structure over this patch, so you know about everything there is to know about gravity in this patch.
What's your g_underscore in this patch ?

Yes, but if you give me an INFINITESIMAL vector transformation, I can derive from it, the corresponding spinor transformation (for one). So this is LOCALLY defined. And second, as I said, spinors are *auxilliary* quantities from which we can MAKE tensor quantities, but aren't directly observable. Only their composed tensor quantities are eventually observable.

I don't think they can be derived LOCALLY from the metric. You need the GLOBAL metric for that, no ?

Last edited by a moderator: May 2, 2017
6. Apr 25, 2006

### Careful

Patrick,

The metric with the underscores is actually the inverse of the original one (with transposed indices). Sabine : you might want to change transformation under general diffeomorphisms by general *coordinate* transformations; it is important to keep these apart, since the latter provide the DEFINING property of the bundle at hand, while the former give an active transformation on the bundle. In this context, the notion of transpose you use is *not* intrinsic (which would require a metric), but relative to two coordinate systems: therefore G^T is not equal to G^{-1} for a Lorentz transformation (between two inertial coordinate systems), but G^{-1} = M G^T M where M = diag(-1,1,1,1) and an example is easily provided by a boost in the (t,z) direction where G = G^T /= G^{-1} (albeit the two representations are clearly equivalent). Actually, it is very easy (and much more clear) to explicitely construct the bundle TM-underline from TM. Anyway, I have to further read the paper (and redo the math), but an immediate worry of mine is that one would expect much more antigravitating matter to be present than gravitating one IF this suggestion were to provide a solution for the cosmological constant (which is actually the dominant energy source in the universe). Hence, why don't we see it ??

Cheers,

Careful

Last edited: Apr 25, 2006
7. Apr 25, 2006

### vanesch

Staff Emeritus

But then it is not *anti-gravity*, in the sense that, say, at the earth surface, the thing would not fall UPWARD with an acceleration of 1 g!

Because, at the earth surface, in a FREE FALLING FRAME, the original metric is essentially the Lorentz metric diag(-1,-1,-1,1), so its inverse would be too, and hence the geodesics would be (to a good approximation) uniform motion straight lines ; in other words, the "anti gravity" particles would FALL to earth too. The tiny deviation would be the tidal effects, which would, indeed, be different. But in a falling elevator, these tidal effects are TINY as compared to the 2 g upward acceleration these anti-gravity particles are supposed to have in your falling elevator.

Last edited: Apr 25, 2006
8. Apr 25, 2006

### vanesch

Staff Emeritus
However, I see a serious problem with this. Because, after all, the "1 g downward acceleration" which we observe at the earth's surface is the cumulative effect of tiny tidal effects from infinity (assuming asymptotic flat spacetime), and if all these accumulated tidal effects are inverse for the antigravity particle, it would fall UP and not DOWN.

So this is only in a way understandable, if these antigravity particles accelerate AWAY from you the more you accelerate TOWARDS them. When you are in uniform motion wrt to such a particle, then they remain in uniform motion. When you accelerate in direction X, they accelerate twice as hard in direction X. When you accelerate in direction -Y, they accelerate twice as hard in direction -Y.
(and that's exactly what they are supposed to do when they follow the inverse metric under general coordinate transformations).

Mmmm.... funny stuff...

I think I understand now, because this explains the "paradox" I had.

For a free falling observer, falling towards earth, the anti-gravity particle falls also along with him (is in uniform motion). For an observer at earth's surface (which accelerates upwards with respect to this inertial observer), the particle, due to its funny transformation property, accelerates 2 g upwards wrt the 1 g downwards of the inertial observer, which makes it fall UPWARD with 1 g.

And for an observer accelerating upwards with 3 g in a rocket lifting off the earth, the particle is seen to accelerate upwards with 5 g (2 x 3 - 1).

A really funny particle !

Last edited: Apr 25, 2006
9. Apr 25, 2006

### vanesch

Staff Emeritus
To follow up: the error in my thinking was that these anti-gravity particles have a worldline (map from R into the manifold M). They don't. Their "world line" is observer-dependent (in a given coordinate system they do have a "world line" but the events that correspond to them are not coordinate-independent).

As an exercise:
consider an anti-gravity particle of 1 kg at rest for an observer A at rest wrt the earth's surface, which this observer "drops" at (his) t = 0. "event of dropping" = event E1.
Consider an observer B, in free fall passing by observer A at event E1 (and his clock is then also at t' = 0).
Consider a third observer C, shooting off his rocket at event E1, with an upward acceleration wrt the earth surface of 2 g. His t"=0 for this event.
As at t = t' = t" = 0, the three observers are at E1, and as relativistic effects are minimal, we can give them (almost) identical coordinate frames x,y and z, with z = "up".

Exercise: what is the observed motion ("coordinate world line") for the anti-gravity particle for each of these observers, in other words, what's
z(t), z'(t') and z"(t") (assuming that the particle doesn't undergo any lateral displacement) ?

My guess is that z(t) = + g/2 t^2 (particle "falls up")
z'(t') = 0 (particle stays at rest)
z"(t") = + 3 g/2 t"^2 (particle falls up inside the rocket)

Note that these three parametrizations do not describe a world line in the usual sense, but 3 different world lines.

Exercise 2: assume that event E1 is at 20 m above ground level. With what event corresponds (as seen by the three observers A, B and C) the event "the particle crosses ground level" ? If it occurs.

With what event corresponds "the particle reaches 2000 m above ground level ?

Ground level: A: never
B: ground level is reached when the observer reaches it, at t' = 2 seconds
C: never

2000 m above ground level:
A: t = 20 s
B: never
C: t" = 8.2 seconds (Sqrt[200/3])

Or am I wrong ?

10. Apr 25, 2006

### hossi

Hi vanesh,

I will have to read your comments more carfully, but have to go to a seminar soon. So, just some comments. The quantity g is not a metric. It is a scalar product on the TM. It does not measure any distances in space-time. Yes, it can be derived from the metric. When the metric g of spacetime is known, so is g.

(There was an explicit example in the paper which computed the geodesic motion in a Schwarzschild-Background, but the editor complained that the paper was too long, so I had to cut it.)

Indeed, the metric is the inverse of the usual metric (because g is symmetric). The particle in the earth's gravitational field falls up, essentially because 1/(1-2M/r) ~ 1+2M/r

The anti-gravitating particle's world lines are not observer dependent. They are, of course, if the observer has a mass (as you indicate). But that is also the case in usual GR. You don't usually take into account the attraction between the particle and the observer when you compute a geodesic.

Best,

B.

11. Apr 25, 2006

### hossi

Good question. I have asked myself the same thing, and I am afraid I can't answer it. I have some ongoing work on the matter, but so far its interesting mainly in the early universe. Though the anti-g matter becomes important again in the late stages of the expansion, it does not behave like a cosmlogical constant.

There are however several points in the whole scenario that I have not really thought through. If I had 5 postdocs, I would have no problem keeping them busy

Another interesting question about the Cosm. Const. is how the anti-g would affect the vacuum-energy in a quantized version. Make that 6 postdocs.

Best,

B.

12. Apr 25, 2006

### Careful

**Good question. I have asked myself the same thing, and I am afraid I can't answer it. I have some ongoing work on the matter, but so far its interesting mainly in the early universe. Though the anti-g matter becomes important again in the late stages of the expansion, it does not behave like a cosmlogical constant.**

Well 6 postdocs is of course anyone's wet dream, but I would like you to comment upon my remark about the transpose, since it has some bearing upon the construction of the map \tau between TM and \underline{TM}. In my calculation \tau is nothing but the metric g, so in a global flat coordinate system this would be the minkowski metric, which is different from the identity map (as you claim).

Cheers,

Careful

13. Apr 25, 2006

### hossi

Well, where do you get the -1 from? \tau makes a transposition, so in a global flat coordinate system the determinante is =1. How does that agree with (-1,1,1,1)?

Indeed, in my calculation, \tau is not the metric. I admit, it does not become clear in the paper, but you get \tau as follows: convert g into a locally flat coordinate system by means of the tetrad field e (one space-time index, one internal index). Apply two of those (one for each index) to the local tau, which is just the identity (1,1,1,1). This gives you the coordinate dependent \tau. In case the metric is diagonal, \tau has the same entries as the metric, but without the minus sign for the 00-component.

Best,

B.

PS: Thanks for the comment on general diff. and general coord. trafos. I actually wasn't aware of the difference can you explain more?

Last edited: Apr 25, 2006
14. Apr 25, 2006

### hossi

Hi garrett,

another excellent question of yours. I have had the same question and, as you might have noticed, successfully avoided answering it in the paper. Indeed, I didn't need a Lagrangian for the anti-g point particle, since I derived the equation of motion from the field equations by using the stress-energy-tensor of a point particle. An approach that I find more minimalistic, since the action for the geodesic motion is (im most textbooks) postulated as an additional assumption - which is unneccessary, since geodesic motion follows from the field equations.

It should be possible, however, to derive this stress-energy-tensor from an appropriate Lagrangian, which then should also give the curve by direct variation. I kind of suspect that it's possible to answer the question of one looks into how it is usually done for the point-particle. But I was happy to have found 2 different ways to get the same curve, and wasn't really desperate for a 3rd one.

Anyway, if you come to any conclusions, I would be happy to hear them!

I can tell you though what the Lagrangian for the anti-g ultrarelativistic fluid is. Its - \rho, which amazingly goes very well with the analysis in

"[URL [Broken] Homogeneous Scalar Field and the Wet Dark Sides of the Universe
Alberto Diez-Tejedor, Alexander Feinstein
gr-qc/0604031[/URL]

(though they discard part of their results as unphysical because of negative energies...) I don't think it's completely trivial. At least for the fluid, I had to think about the relation between the usual matter and the anti-g one for some while.

Best,

B.

Last edited by a moderator: May 2, 2017
15. Apr 26, 2006

### Careful

Ah, sorry, we actually agreed, the diagonal tau's are indeed just transposition, but the horizontal ones involve only g and g^{-1} and transposition. But you don't need the tetrad at all to define all this. The difference between a diffeomorphism and a local coordinate transformation is crucial in gravitation and both determine the same amount of gauge freedom. Concretely, let Vbe a vectorfield and f a diffeomorphism, then (f_{*} V)^{\alpha} (f(p)) = ((Df)(p) V(p))^{\alpha} where Df : TM_p -> TM_{f(p)}.

Last edited: Apr 26, 2006
16. Apr 26, 2006

### vanesch

Staff Emeritus
So I take it that you and Careful agree, that in a specific coordinate system, the tensor g_underscore equals the transposed inverse matrix of g, right ?

This is in fact what I applied.

Right, this is in Schwarzschild coordinates. Now, if we place ourselves in a patch close to the surface of the earth, this metric reduces, in the coordinate frame of an observer fixed at the surface, to:

ds^2 = -(1 + gz) dt^2 + dx^2 + dy^2 + dz^2, right ?

(if we leave out tidal effects).

Which corresponds to the metric of an upward accelerating observer in flat space.

So, if we inverse the metric tensor in this system, we obtain:

(underscore) ds^2 = -(1 - gz) dt^2 + dx^2 + dy^2 + dz^2

and I understand that the "world lines" of your anti-gravity particle are the geodesics if we take THIS to be the "metric" (although I understand that you don't want to call it that way, it is just a way of calculating the world line).
And indeed, in this coordinate system, the world line of your antigrav particle has an UPWARD acceleration of g, as it should. And this is of course also what you would find in a Schwarzschild coordinate system.

This corresponds to my observer A.

But to a free-falling observer, the metric is:

ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

If we inverse THIS metric to find g_underscore, of course it doesn't change (Lorentz metric). So in THIS frame, the "anti geodesics" are straight lines, and the particle FALLS WITH THE FRAME. It "sees a flat metric".

This was my observer B. But you've never calculated this, because it are not Schwarzschild coordinates :-)

Nonono, I was not talking about any mass of the observer. It was the anti-gravity particle which had a mass of 1 kg, or just any mass you wanted.

And my observer C, in his rocket, has a metric of:

ds^2 = -(1+3gz) dt^2 +dx^2 + dy^2 + dz^2

If I inverse this metric, in order to find g_underscore, I find:

(underscore) ds^2 = - (1 - 3 gz) dt^2 + dx^2 + dy^2 + dz^2

And the "anti-geodesics" in this frame correspond to a particle falling UPWARD with 3 g (in this frame, so falling upward with 5 g wrt the guy on the ground).

I would like to know what goes wrong in this reasoning if you disagree with it.

Of course, my basic point is this:
in a GIVEN coordinate system (t,x,y,z), with a given metric ds^2, written out by the tensor g, the steps one needs to perform to find the trajectory of an antigravity particle are this:

1) take the matrix of g, and inverse it: this is now, if I understand well, the matrix representation of g_underscore in this coordinate frame.

2) if you pretend g_underscore to be a metric, find the corresponding geodesics of g_underescore

3) these geodesics are the "world lines" of the anti-grav particles in this coordinate system, let's call them "anti geodesics", or "anti-world lines".

And now, it turns out, that the hence obtained world line is dependent on the frame in which this procedure is executed, and I tried to illustrate that with observers which are in uniform acceleration wrt eachother.

A way to see this is: take a flat spacetime, and an inertial observer.
The anti-world lines are the same as the world lines.
Now, consider an accelerated observer, with metric:
- (1 + g z)^2 dt^2 + dx^2 + dy^2 + dz^2
The normal geodesics are particles, accelerating DOWNWARDS in this frame ("falling particles").
The underscore metric is -(1- gz)^2 dt^2 + dx^2 + dy^2 + dz^2
and the geodesics of THIS metric are UPWARD ACCELERATING (because of the sign change of the g, which came from the inversion and the fact that gz is small compared to 1) ; so our anti-grav particles are accelerating UPWARDS in this frame.

It should be clear that particles that are in uniform motion (or at rest) wrt an observer, and accelerate upward in the frame of an upward accelerating observer, do not describe the same world line. Nevertheless, this is what I understand of your proposal.

If not, you should explain me how a uniformly accelerating observer, with metric
-(1+gz) dt^2 + dx^2 + dy^2 + dz^2
(and this is the only stuff you need to know) calculates the trajectory, in his coordinate system, of an antigravity particle.

Last edited: Apr 26, 2006
17. Apr 26, 2006

### hossi

Hi vanesh,

I am afraid you are making the same mistake again and again. The quantity g is not a metric. I never said so. I never said you get the motion of the anti-gravitating particle by calculating the Christoffel-symbols wrt g. I have no idea what comes out when you do that, and don't see why I should argue about it.

As I pointed out earlier, the free-falling frame for the usual particle is not the same as the free-falling frame of the anti-g particle. That was why I made sure we have the same notion of 'local' - it includes an infinitesimal sourrounding.

Indeed, these are not coordinates at all. Meaning, the dt, dx etc you have above are no 1-forms and thus the coefficients not suitable to compute the connection.

He computes the metric g of the manifold in whatever coordinate frame. From this he gets \tau and g. Since the observer is smart, he has read my paper and is able to compute the connection coefficients for the anti-g particle. From this he calculates the trajectory that he can convert back and forth in any coordinate system as he wants to.

Best,

B.

18. Apr 26, 2006

### hossi

Hi Careful,

thanks for pointing this out. You are right, I don't need the tetrad. But it's very convenient and it is the easiest way to get the \tau. Also, you need it anyway for spinor fields. Best,

B.

19. Apr 26, 2006

### hossi

So you can derive the transportation of the anti-graviational particle. You do it as you do it for the usual field. In curved space the crucial point is that you have to take care of the derivation of the local basis since it is not constant - this gives you essentially the connection coefficients. For the usual particle, its the derivation of the basis in TM, for the anti-g particle its the basis in TM, for the spinor its some basis in the appropriate bundle.

I don't see how it matters for the definition of the connection whether these things are observable or not.

Best,

B.

20. Apr 26, 2006

### Careful

Hmm, the coordinate expressions are pretty easy too Anyway, I was calculating a bit today on some point which worried me from the beginning, and that is the expression of the anti-connection as the being the anti-christoffel symbol (which has very different transformation properties from the usual connection symbol). Concretely, it appears to me that the basis \underline{\partial} _{\underline{\alpha}} is NOT commuting, hence one woud not expect the connection coefficients to be symmetric in the bottom indices. Therefore, I worked out the transformation rules of the anti-connection using the *defining* properties of the anti-covariant derivative alone; the latter calculation confirmed my suspicion.

So perhaps I missed something, perhaps not.

Cheers,

Careful