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Homogeneous gravitational field and the geodesic deviation

by sergiokapone
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sergiokapone
#1
Jan24-13, 12:41 AM
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In General Relativity (GR), we have the _geodesic deviation equation_ (GDE)

$$\tag{1}\frac{D^2\xi^{\alpha}}{d\tau^2}=R^{\alpha}_{\beta\gamma\delta} \frac{dx^{\beta}}{d\tau}\xi^{\gamma}\frac{dx^{\delta}}{d\tau}, $$

see e.g. [Wikipedia](http://en.wikipedia.org/wiki/Geodesi...ation_equation) or [MTW](http://en.wikipedia.org/wiki/Gravitation_%28book%29).

Visually, this deviation can be imagined, to observe the motion of two test particles in the presence of a spherically symmetric mass.In the case of a homogeneous gravitational field, such as a geodesic deviation should not be, because acceleration due to gravity equal at every point.

Clearly, such a field can be realized in a uniformly accelerated frame of reference. In this case, all components of the curvature tensor will be zero, and the equation (1) correctly states that the deviation will not be surviving.

But if a homogeneous field will be created by infinite homogeneous flat layer, in this case, the components of the curvature tensor are non-zero, then by (1) will be a deviation. It turns out that such fields, even though they are homogeneous, can be discerned.
I think this situation is paradoxical. Is there an explanation?
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tiny-tim
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Jan24-13, 06:40 AM
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hi sergiokapone!

yes, the geodesic deviation in a uniform field should be zero
Quote Quote by sergiokapone View Post
But if a homogeneous field will be created by infinite homogeneous flat layer, in this case, the components of the curvature tensor are non-zero
are they?
sergiokapone
#3
Jan24-13, 07:21 AM
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Quote Quote by tiny-tim View Post
hi sergiokapone!

yes, the geodesic deviation in a uniform field should be zero


are they?
Equation of motion in the metric of this type:
$$ds^2=(1-8gz)^{-1/4}dt^2-(1-8gz)^{1/2}(dx^2+dy^2)-(1-8gz)^{-5/4}dz^2$$
in the Newtonian limit becomes as equation of motion in homogeneous field. However, this metric has a non-zero curvature. The metric of this type was obtained in 1971 by ukrainian astronomer Bogorodsky, in the case of an infinite plane. The article, unfortunately, in Russian.
In English, a derivation Bogorodsky's metrics can be found in this article http://arxiv.org/pdf/gr-qc/0202058.pdf
In his article, Bogorodsky gets two solutions, one of them has no curvature, and enters the corresponding transformation of the Minkowski metric, and another - a curvature. A very interesting fact.

Bill_K
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Jan24-13, 09:16 AM
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Homogeneous gravitational field and the geodesic deviation

The metric of this type was obtained in 1971 by ukrainian astronomer Bogorodsky, in the case of an infinite plane. The article, unfortunately, in Russian.
If Bogorodsky had checked the literature (always a good idea!) he would have found that such very simple solutions depending on only one variable z had been enumerated by Kasner back in 1925.
sergiokapone
#5
Jan24-13, 09:23 AM
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Quote Quote by Bill_K View Post
If Bogorodsky had checked the literature (always a good idea!) he would have found that such very simple solutions depending on only one variable z had been enumerated by Kasner back in 1925.
But it does not matter now. I wonder why, in a uniform field of this type is the deviation of geodesics.
Bill_K
#6
Jan24-13, 10:53 AM
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I wonder why, in a uniform field of this type is the deviation of geodesics.
Isn't it just because the Riemann tensor is nonzero? Do you want us to come up with an intuitive reason why it's nonzero?

Well, by reflection symmetry there are geodesics in which a particle "falls" in the z direction, keeping the coordinate values x = const and y = const. Then from the (1−8gz)1/2(dx2+dy2) term in the metric, the distance between two such neighboring particles changes as they fall. (It would be much harder to explain if it did not!)
sergiokapone
#7
Jan24-13, 12:00 PM
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Quote Quote by Bill_K View Post
Isn't it just because the Riemann tensor is nonzero? Do you want us to come up with an intuitive reason why it's nonzero?
Ok, can we call such a field as homogeneous in the GR-sense?
bcrowell
#8
Jan24-13, 12:54 PM
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You may find this relevant: http://www.lightandmatter.com/html_b...tml#Section7.4

You say "homogeneous," but do you really want something isotropic as well as homogeneous?

The Ionescu paper's description of Bogorodskii's work is completely coordinate-based. To some extent this is inevitable in discussions of the GR equivalent of a Newtonian uniform field, since the Newtonian notion of a field is coordinate-dependent. However, one really wants to classify especially symmetric spacetimes according to their *intrinsic* symmetry, not their symmetry when expressed in some coordinates.

Quote Quote by Bill_K View Post
If Bogorodsky had checked the literature (always a good idea!) he would have found that such very simple solutions depending on only one variable z had been enumerated by Kasner back in 1925.
A general Kasner metric lacks the high degree of symmetry we'd like for a truly uniform field. By fiddling with metrics of the Kasner form, you can get the Petrov metric described in the link above. The Petrov metric is basically the vacuum spacetime with the highest intrinsic symmetry you can get without having it be Minkowski space.

Quote Quote by sergiokapone View Post
Visually, this deviation can be imagined, to observe the motion of two test particles in the presence of a spherically symmetric mass.In the case of a homogeneous gravitational field, such as a geodesic deviation should not be, because acceleration due to gravity equal at every point.

Clearly, such a field can be realized in a uniformly accelerated frame of reference. In this case, all components of the curvature tensor will be zero, and the equation (1) correctly states that the deviation will not be surviving.

But if a homogeneous field will be created by infinite homogeneous flat layer, in this case, the components of the curvature tensor are non-zero, then by (1) will be a deviation. It turns out that such fields, even though they are homogeneous, can be discerned.
I think this situation is paradoxical. Is there an explanation?
The Newtonian result is that an infinite, flat sheet of mass has a uniform field on both sides. But I don't see any reason to think that anything similar holds in GR. It's not even obvious how to state such a notion in GR. For example, what does it mean for the sheet to be "flat?" A cylinder of dust is intrinsically flat, and the Petrov metric can be interpreted as the field of a certain rotating cylinder of dust. "Uniform field" would have to be translated into some appropriately coordinate-independent langage in GR, i.e., it would have to become something like a statement about the number of Killing vectors. This is the kind of criterion on which the Petrov metric becomes the best candidate for a uniform field in GR.
pervect
#9
Jan24-13, 02:45 PM
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Quote Quote by sergiokapone View Post
Equation of motion in the metric of this type:
$$ds^2=(1-8gz)^{-1/4}dt^2-(1-8gz)^{1/2}(dx^2+dy^2)-(1-8gz)^{-5/4}dz^2$$
in the Newtonian limit becomes as equation of motion in homogeneous field.
I throw this metric into GRTensor and find all the components of R_{ab} are zero. (Except perhaps at the origin, where I suspect R is techinically undefined). R_{abcd) isn't zero, though.

[add]
So this is in fact a solution to EInstein's equation, and it does have curvature. But I have to agree that it's unclear if it represents an "infinite flat sheet". For instance, you see tidal forces in the x directions (R_txtx is nonzero). That's not something you'd expect in an infinite flat sheet.
sergiokapone
#10
Jan24-13, 03:39 PM
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Quote Quote by bcrowell View Post
The Newtonian result is that an infinite, flat sheet of mass has a uniform field on both sides. But I don't see any reason to think that anything similar holds in GR. It's not even obvious how to state such a notion in GR. For example, what does it mean for the sheet to be "flat?" A cylinder of dust is intrinsically flat, and the Petrov metric can be interpreted as the field of a certain rotating cylinder of dust. "Uniform field" would have to be translated into some appropriately coordinate-independent langage in GR, i.e., it would have to become something like a statement about the number of Killing vectors. This is the kind of criterion on which the Petrov metric becomes the best candidate for a uniform field in GR.
bcrowell, thank you! The question becomes clear to me.

Quote Quote by pervect View Post
]
So this is in fact a solution to EInstein's equation, and it does have curvature. But I have to agree that it's unclear if it represents an "infinite flat sheet". For instance, you see tidal forces in the x directions (R_txtx is nonzero). That's not something you'd expect in an infinite flat sheet.
Why R_txtx? I thought that the tidal force along the x-direction must meet the components $$R^x_{yxy}$$ and $$R^x_{zxz}$$ which, according to my calculations is non-zero.
Mentz114
#11
Jan24-13, 04:06 PM
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Quote Quote by pervect View Post
I throw this metric into GRTensor and find all the components of R_{ab} are zero. (Except perhaps at the origin, where I suspect R is techinically undefined). R_{abcd) isn't zero, though.

[add]
So this is in fact a solution to EInstein's equation, and it does have curvature. But I have to agree that it's unclear if it represents an "infinite flat sheet". For instance, you see tidal forces in the x directions (R_txtx is nonzero). That's not something you'd expect in an infinite flat sheet.
I agree with your tidal calculation. The values are ( for a hovering observer)

[tex]T_{xx}=T_{yy}=\frac{2g^2}{\sqrt{1-8\,g\,z}}[/tex]

But I also found 2 Killing vectors, one each in ∂x and ∂y directions. I guess that means the potential doesn't change in those directions ?
Mentz114
#12
Jan24-13, 04:13 PM
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Quote Quote by sergiokapone View Post
Why R_txtx? I thought that the tidal force along the x-direction must meet the components $$R^x_{yxy}$$ and $$R^x_{zxz}$$ which, according to my calculations is non-zero.
If we consider a stationary frame, then only the t-component of the 4-velocity U is non-zero and in the contraction Tab= RdacbUcUd we need to look only at R0x0y as Pervect has done ( some errors with the indexes there, but you know what I mean ...)
sergiokapone
#13
Jan24-13, 04:26 PM
P: 110
Quote Quote by Mentz114 View Post
If we consider a stationary frame
Oh, I see.
sergiokapone
#14
Jan24-13, 04:30 PM
P: 110
Quote Quote by Mentz114 View Post
But I also found 2 Killing vectors, one each in ∂x and ∂y directions. I guess that means the potential doesn't change in those directions ?
Yes, I think so. The px and py components of momentum are conserved, like in Newtonian homogeneous field.
pervect
#15
Jan24-13, 04:55 PM
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Quote Quote by Mentz114 View Post
I agree with your tidal calculation. The values are ( for a hovering observer)

[tex]T_{xx}=T_{yy}=\frac{2g^2}{\sqrt{1-8\,g\,z}}[/tex]

But I also found 2 Killing vectors, one each in ∂x and ∂y directions. I guess that means the potential doesn't change in those directions ?
I suppose that depends what you mean by potential. But you're right about it being a killing vector, thus [itex]g^{xx} dx / d\lambda[/itex] should be constant along a geodesic.

This is easily confirmed, if we let our geodesic be [itex]t(\lambda)[/itex], [itex]x(\lambda)[/itex], [itex]y(\lambda)[/itex], [itex]z(\lambda)[/itex]

then

[tex]
\frac{d}{d\lambda} \left[ \frac{\dot{x}}{\sqrt{1- 8 g z(\lambda)}} \right] = \frac{1}{\sqrt{1-8 g z(\lambda)}} \left[ \ddot{x} + \frac{4 \dot{x}\dot{z} } {\sqrt{1-8 g z(\lambda)} } \right]
[/tex]

where the rhs is zero because of the geodesic equation, thus insuring that the derivative is zero and
[tex]\frac{\dot{x}}{\sqrt{1- 8 g z(\lambda)}}[/tex]

is constant , and can be considered as the x-momentum.

(I've worked this out more for the OP than Mentz, I hope it's somewhat clear. The "dots" represent derivatives with respect to [itex]\lambda[/itex].)

Now I'm scratching my head about how there can be a tidal force, but I need to get back to the leaky faucet repair :-(.
Mentz114
#16
Jan24-13, 05:22 PM
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Pervect, that's cool. The Killing vector

[tex]K_\mu= C\sqrt{1-8\,g\,z}\ \partial_x[/tex] seems to tie in, with K.U = const ( U being the geodesic ). If we identify the constant C with the rest-mass m, we get the conserved momentum exactly.

I solved the Killing equations [itex]K_{(a;b)}=0[/itex] to get K, and C is a constant of integration.
Mentz114
#17
Jan24-13, 05:26 PM
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Quote Quote by sergiokapone View Post
Yes, I think so. The px and py components of momentum are conserved, like in Newtonian homogeneous field.
Right. Strange about the tidal forces. I must be misinterpreting something.
sergiokapone
#18
Jan24-13, 05:38 PM
P: 110
Quote Quote by pervect View Post
where the rhs is zero because of the geodesic equation, thus insuring that the derivative is zero and
[tex]\frac{\dot{x}}{\sqrt{1- 8 g z(\lambda)}}[/tex]

is constant , and can be considered as the x-momentum.

(I've worked this out more for the OP than Mentz, I hope it's somewhat clear. The "dots" represent derivatives with respect to [itex]\lambda[/itex].)
Strange result. From geodesics eqn via mometums:
$$m\frac{dp_{\beta}}{d\lambda}=\frac{1}{2}g_{\nu \alpha, \beta}p^{\nu}p^{\alpha}$$
I see that there are conserve such values: $$p_t=g_{tt} \dot{t},$$ $$p_x=g_{xx} \dot{x}=\sqrt{1-8gz}\dot{x}$$ and $$p_y=g_{yy} \dot{y}=\sqrt{1-8gz}\dot{y}$$


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