Anti-gravitation discussion

  • Thread starter hossi
  • Start date
198
0
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]

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

vanesch said:
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.

Best,

B.
 
Last edited by a moderator:
R

Rade

hossi said:
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.
 
198
0
Rade said:
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 [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:
412
45
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)
[tex]S = \int d^4 x \, \sqrt{-g} \left[ G R + L + \underline{L} \right] [/tex]
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, [itex]q^\mu(\tau)[/itex], is
[tex]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) [/tex]
So, what's the Lagrangian, [itex]\underline{L}[/itex], supposed to be for an anti-gravitating point mass?
 
Last edited:

vanesch

Staff Emeritus
Science Advisor
Gold Member
5,007
16
hossi said:
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]
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 ?
 
Last edited by a moderator:
1,667
0
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:

vanesch

Staff Emeritus
Science Advisor
Gold Member
5,007
16
Careful said:
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.
 
Last edited:

vanesch

Staff Emeritus
Science Advisor
Gold Member
5,007
16
vanesch said:
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 !
 
Last edited:

vanesch

Staff Emeritus
Science Advisor
Gold Member
5,007
16
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 ?
 
198
0
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.
 
198
0
Careful said:
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 :wink:

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.
 
1,667
0
**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
 
198
0
Careful said:
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.

Best,

B.

PS: Thanks for the comment on general diff. and general coord. trafos. I actually wasn't aware of the difference :blushing: can you explain more?
 
Last edited:
198
0
garrett said:
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)
[tex]S = \int d^4 x \, \sqrt{-g} \left[ G R + L + \underline{L} \right] [/tex]
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, [itex]q^\mu(\tau)[/itex], is
[tex]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) [/tex]
So, what's the Lagrangian, [itex]\underline{L}[/itex], 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 [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:
1,667
0
hossi said:
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 :blushing: 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)}.
 
Last edited:

vanesch

Staff Emeritus
Science Advisor
Gold Member
5,007
16
hossi said:
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 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:
198
0
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.

vanesch said:
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.

vanesch said:
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.


vanesch said:
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.

Best,

B.
 
198
0
Careful said:
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. Best,

B.
 
198
0
hossi said:
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.
vanesch said:
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.

Best,

B.
 
1,667
0
hossi said:
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.
Hmm, the coordinate expressions are pretty easy too :smile: 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

Staff Emeritus
Science Advisor
Gold Member
5,007
16
hossi said:
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.
:confused: 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 ?
 
Last edited:
198
0
Careful said:
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 :confused: . 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). Best,

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...
 
198
0
vanesch said:
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."

vanesch said:
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.

vanesch said:
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 ?

:confused: 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.



vanesch said:
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).

vanesch said:
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. Best,

B.
 
1,667
0
hossi said:
Hi Careful,

you are confusing me :confused: . 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). Best,

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
 
198
0
Careful said:
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. Best,

B.
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top