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

[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"

see also garrett's smart comments on http://backreaction.blogspot.com/2006/04/anti-gravitation.html"

Exactly. That's my whole point, because LOCALLY there is no difference between this patch of manifold and a patch of manifold in deep space, concerning its metrical structure. BOTH are essentially flat, you see. So there is NO WAY in which to derive this OTHER curve, if the only thing that is given, is the metric.
The metric is THE SAME in the two cases, but the curves are DIFFERENT.


[...] a DIFFERENT set of connection coefficients corresponds to a DIFFERENT metric. So we now have TWO different metrics on our manifold. Is this what you are after ? But, it is a strange manifold who has two different metrics !

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.



B.

 

[Rade]

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.

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.

 

[hossi]

Thank you for any comments you may have.

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

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

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


B.

 

[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?

 

[vanesch]

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"

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 ?

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.

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.

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.

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

 

[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

 

[vanesch]

Patrick,

The metric with the underscores is actually the inverse of the original one (with transposed indices).

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.

 

[vanesch]

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.

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 !

 

[vanesch]

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 ?

 

[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.



B.

 

[hossi]

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 ??

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.



B.

 

[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

 

[hossi]

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).

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.



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?

 

[hossi]

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?

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 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.



B.

 

[Careful]

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.



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?

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)}.

 

[vanesch]

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.

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.

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

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 :-)

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.

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 each other.

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.

 

[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.

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".

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.

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

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.

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.

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.



B.

 

[hossi]

But you don't need the tetrad at all to define all this.

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.

B.

 

[hossi]

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.

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.

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.



B.

 

[Careful]

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.

B.

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

 

[vanesch]

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.

Well, then I don't see how you you set up the equations of motion for an antigravity particle in a given coordinate system. I thought that the equation of motion was the equation of the 'geodesic' you obtain from the g_underscore matrix (which is the inverse of g).

We agree that such a particle has, to an observer standing still at the surface of the earth, an equation of motion which corresponds to a particle falling UPWARD with 1 g, right ?
So its equation of motion here is z(t) = z0 + g/2 t^2 in non-relativistic approximation, right ?

Again, I'd like to ask you, if I give you the coordinate system
(X,Y,Z,T), and I tell you that the metric is:

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

how do I get the equation of motion of the antigrav. particle in this coordinate system ?

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.

This is what I don't understand, because a FREEFALLING frame is locally a lorentz frame, and you said that in a lorentz frame, the antigravity particle is in uniform motion.

It has the metric ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

If I give you the coordinate set (t,x,y,z) and its metric ds^2 = -dt^2 + dx^2 + dy^2 + dz^2, how do you set up the equation of motion of the antigravity particle in this coordinate set ?

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.

what are you saying ? That the coordinate set in a free falling elevator is not a coordinate set ?
And that the coordinate set of an astronaut in a rocket is not a coordinate set ?

How about Rindler coordinates which describe exactly that ?

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.

Well, because I'm not smart, could you show me how this goes for the above two coordinate sets ? (T,X,Y,Z) and (t,x,y,z) with their metric ?

But I'll give it a try:
I have (T,X,Y,Z). Now, I realize that with a transformation, I can turn this into a lorentz frame (t,x,y,z). Here, tau = 1 and g = eta, right ? So in this frame, anti-gravity particles follow a straight and uniform motion, right ?
I take your word for it that their world lines have coordinate independent meaning, so in this case, when I go back to T,X,Y,Z, I see the anti-gravity particles fall like any other, is that correct ?

But what's the difference between THIS case, and the case of our free falling elevator at Earth's surface (which has ALSO a local lorentz frame with metric eta, and in which I take it, we set tau = 1). If now, in this falling elevator, our anti-gravity particles have a uniform straight line motion, and (I take your word for it) their world line is observer-independent, well THEN THEY FALL WITH THE ELEVATOR wrt to an observer at the Earth surface, no ?
And as such, they fall DOWN, and not UP ?

Nevertheless, you must not agree with this, so can you show me, how you derive the equations of motion in the two coordinate sets ?

 

[hossi]

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.

Hi Careful,

you are confusing me . The connection coefficients are not symmetric in the bottom indices, because one of the indices belongs to the 'direction' (i.e. is a space-time index), whereas the other belongs to the basis (i.e. is internal and underlined).

B.

PS: I have computed the coefficients for the Schwarzschild and FRW metric, and they are indeed not symmetric in the bottom indices. I had to be very careful...

 

[hossi]

Well, then I don't see how you you set up the equations of motion for an antigravity particle in a given coordinate system. I thought that the equation of motion was the equation of the 'geodesic' you obtain from the g_underscore matrix (which is the inverse of g).

Never said so. I just abbreviated my calculation saying "The particle in the Earth's gravitational field falls up, essentially because 1/(1-2M/r) ~ 1+2M/r."

Again, I'd like to ask you, [...] how do I get the equation of motion of the antigrav. particle in this coordinate system ?

I have already answered this question. See above.

This is what I don't understand, because a FREEFALLING frame is locally a lorentz frame, and you said that in a lorentz frame, the antigravity particle is in uniform motion.

It has the metric ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

If I give you the coordinate set (t,x,y,z) and its metric ds^2 = -dt^2 + dx^2 + dy^2 + dz^2, how do you set up the equation of motion of the antigravity particle in this coordinate set ?

what are you saying ? That the coordinate set in a free falling elevator is not a coordinate set ?
And that the coordinate set of an astronaut in a rocket is not a coordinate set ?

How about Rindler coordinates which describe exactly that ?

Rindler coordinates are in globally flat space. What you are mixing up are two different notions of localities. You have locally a tangential space at one point. In this space belonging to the point you can always pick a local orthonormal basis (LOB). In this LOB the metric is Minkowskian. If you take all the LOBs together over various points, this does NOT give you a coordinate system on the manifold. The reason being that the basis of tangential space varies with the point. That was what I was saying. The transformation between the space-time basis and the LOB is the tetrad, see careful's post above.

The locality you mean above is in a sourrounding of the point, which includes an infinitesimal region, or transport of the basis. This infinitesimal transport depends on the properties of the basis, hence is different whether you are talking about elements of TM or TM.

Here, tau = 1 and g = eta, right ? So in this frame, anti-gravity particles follow a straight and uniform motion, right ?

tau=1 in the LOB which belongs to one point, therefore no motion, no curve not even infinitesimal. You want to consider a local sourrounding you have to take care to adequately transform \tau (which is not constant) and which indeed transforms the properties of the usual trafo into that of the anti-g particle (this is what is is defined to do).

Nevertheless, you must not agree with this, so can you show me, how you derive the equations of motion in the two coordinate sets ?

Same question as before, same answer as before. What do you want? I have a list with the connection coefficients for the Schwarzschild-metric, looks nasty, I can sent it to you if you like? Then you can compute the curves - I only did the Newtonian limit.

B.

 

[Careful]

Hi Careful,

you are confusing me . The connection coefficients are not symmetric in the bottom indices, because one of the indices belongs to the 'direction' (i.e. is a space-time index), whereas the other belongs to the basis (i.e. is internal and underlined).

B.

PS: I have computed the coefficients for the Schwarzschild and FRW metric, and they are indeed not symmetric in the bottom indices. I had to be very careful...

No, I am aiming at the connection expressed in formula 16 of your paper. All indices are underlined and the right hand side is clearly symmetric in the bottom ones. I calculated the transformation behavior of THIS connection from the general defining properties, and it doesn't match the symmetry.

Cheers,

Careful

 

[hossi]

No, I am aiming at the connection expressed in formula 16 of your paper.

Thanks. I will have to think about it. I never used (16), instead I used (17). I will check it.

B.

 

[Careful]

Thanks. I will have to think about it. I never used (16), instead I used (17). I will check it.

B.

That 17 follows from 16 is trivial of course (and that it is not symmetric in the bottom indices too). However, 16 can only be used under very restrictive conditions which I do not believe to apply here.



Careful

 

[hossi]

That 17 follows from 16 is trivial of course (and that it is not symmetric in the bottom indices too). However, 16 can only be used under very restrictive conditions which I do not believe to apply here.



Careful

I see, there should be the additional terms in (16) and (17) from the commutators of the basis. I have to think about it in more details, but you are right, at first sight I don't see why these should vanish. Luckily, since I have never used the symmetry in the lower indices (the symbols with one usual and one underlined bottom index never have them), it does not affect any other conclusions. I will let you know what comes out of it.

B.

 

[Careful]

I see, there should be the additional terms in (16) and (17) from the commutators of the basis. I have to think about it in more details, but you are right, at first sight I don't see why these should vanish. Luckily, since I have never used the symmetry in the lower indices (the symbols with one usual and one underlined bottom index never have them), it does not affect any other conclusions. I will let you know what comes out of it.

B.

Right, but it will affect the calculations of specific anti-geodesics though.

Careful

 

[hossi]

Right, but it will affect the calculations of specific anti-geodesics though.

Right. Though I suspect that for specific cases they might vanish, since the metric (and so \tau) most often has additional symmetries. Thanks again B.

 

[vanesch]

Rindler coordinates are in globally flat space. What you are mixing up are two different notions of localities. You have locally a tangential space at one point. In this space belonging to the point you can always pick a local orthonormal basis (LOB). In this LOB the metric is Minkowskian. If you take all the LOBs together over various points, this does NOT give you a coordinate system on the manifold. The reason being that the basis of tangential space varies with the point.

But this is silly ! What you are essentially claiming, is that an observer in an elevator DOES NOT HAVE A COORDINATE SYSTEM. That he doesn't have meter sticks and that he cannot have a clock. Because that is exactly what Rindler coordinates do:
Rindler coordinates DO assign 4 numbers to events in the environment of another event in a smooth way, and THAT IS ALL YOU NEED for a coordinate system. Look at p. 172 in MWT if you want to: paragraph 6.6 "The local coordinate system of an accelerated observer" where the tetrad is extended to a finite patch.

Now, I know that Rindler coordinates do not cover the ENTIRE space, but we only need a local patch (just as the coordinate system of an observer in an elevator only needs a few meters and a few seconds of extension).

The point is, that over these few meters and seconds, in a free falling elevator, this coordinate system is essentially minkowskian ;
while at the surface of the earth, over a few meters and a few seconds, this coordinate system is essentially like a Rindler coordinate system.
And yes, at the surface of the earth, spacetime is essentially flat (there are tiny tidal effects which we can neglect entirely).

I fail to see how you can claim that one of these two coordinate systems does not exist over a local patch.

And now we come of course to my essential difficulty for your claim (for which I do not even need to read your paper - and which I have since the beginning):
if, in a flat spacetime, your particles behave as normal particles, and if they are seen on an observer-independent world line, then, at the surface of the earth, they should fall down. Because in the coordinate system with minkowski coordinates, they are moving uniformly, and this coordinate system is the one of a falling elevator.
Because, locally, spacetime IS flat to a good approximation, at the Earth's surface (and only, an observer at its surface is accelerated upward, and hence uses Rindler like coordinates).

In other words, my claim, from the beginning, is:
your anti-gravity particle cannot fall upward at the surface of the Earth if 1)spacetime geometry is respected (meaning, we still have our manifold with coordinates on it and so on, and its trajectory is a world line on that manifold) and 2) if it is in uniform motion in the minkowski coordinate system in a flat spacetime and 3) if, as it should, its equation of motion is determined locally, based upon the only gravitational structure that is present: the metric (hence, that there is NOT *another* field present, like, say, an EM field).

You cannot get around that, no matter how much differential geometry you are going to use.

The simple reason being the one I'm trying to tell you from the beginning: we ARE in such a situation at the Earth's surface: spacetime IS flat (to a good approximation) in a patch around a point at the surface, in a falling elevator we ARE in the minkowski coordinate frame. So if the particle is in uniform motion (or at rest) in this frame, and it traces out a world line which is observer-independent, then this world line is, for an observer at the surface of the earth, a downward falling particle (with the elevator).

Now, you insisting more on the fact that it fell UPWARD, I concluded then that, in a given coordinate frame (= way of assigning 4 numbers to events in a local patch, such that this is a smooth way) with a given metric (the 16 numbers of the g-tensor in this coordinate frame), what you somehow did was to take the INVERSE g-tensor, and write down the equation of motion that corresponds to "geodesics" when pretending that this inverse g was a metric. Clearly, the curves hence obtained are NOT observer independent, BUT they allow you, to have, for certain observers, to see a particle fall upwards at the surface of the earth. But this is apparently NOT what you do.

Let's ask a different question, now that you claim that certain coordinate sets are not "good" coordinates:

Let's take an observer at the surface of the earth, and let's say he takes his z-axis upward, and an x and y-axis in the horizontal plane. Let's say that he has a clock in his pocket. Like in mechanics 101.
Does this, or doesn't this, qualify as a coordinate system over a local patch of spacetime ? What's the metric like ? I'd guess it pretty much looks like the metric of a Rindler coordinate system.
Next, take a second observer, place him in a falling elevator which crosses our first one at event E (x=0,y=0,z=0,t=0). Let's say that this observer, in his falling elevator, has meter sticks and sets up his X,Y,Z system, with a clock in his pocket, giving T. Does this, or doesn't this, qualify also as a coordinate system over this patch of spacetime ? What's the metric in THIS coordinate system ? I'd guess it is pretty much Minkowskian.

I do not need more than a few meters and a few seconds of spacetime to see a particle fall UP or DOWN.

Because if suddenly, coordinate systems in elevators, trains and rockets don't qualify anymore as coordinate systems, one would wonder how Einstein came originally to his theory of relativity !

tau=1 in the LOB which belongs to one point, therefore no motion, no curve not even infinitesimal. You want to consider a local sourrounding you have to take care to adequately transform \tau (which is not constant) and which indeed transforms the properties of the usual trafo into that of the anti-g particle (this is what is is defined to do).

Yes, but in the falling elevator, the coordinate system extends over meters, and the metric is Minkowskian over meters and seconds. That's not "in one point", but over a local patch and that's good enough to establish, that in this system, the particle is moving uniformly (or at rest), for a few seconds.
It has now traced out its world line over this patch, and it is falling down. The only non-Minkowskian deviations are tiny tidal effects, but no matter in which way you apply them, the bulk of the motion is with the elevator.

This is the thing I latched upon from the beginning, because *this* is the physical essence of the equivalence principle.

Same question as before, same answer as before. What do you want? I have a list with the connection coefficients for the Schwarzschild-metric, looks nasty, I can sent it to you if you like? Then you can compute the curves - I only did the Newtonian limit.

I would like you to do the Newtonian limit in the coordinate system of our falling elevator (*), for 3 seconds, with the particle initially at rest.

To me, this looks simple, because you have 1) essentially flat spacetime and 2) a minkowski metric.

Once you have (a short piece of) the world line in this system (which I guess to be : X = 0, Y=0, Z = 0 as a function of T :-), I'd like to see how this is seen from an observer, fixed at the surface of the earth.

(*) this local coordinate system consists of orthogonal meter sticks and the eigentime of the person in the elevator ; take the Z-axis up.

 

[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.