# Homogeneous gravitational field and the geodesic deviation

• sergiokapone
In summary, the conversation discusses the geodesic deviation equation in General Relativity and how it can be applied to a uniform gravitational field. It is noted that in a homogeneous field, such as a uniformly accelerated frame of reference, the curvature tensor is zero and the geodesic deviation is also zero. However, in the case of an infinite homogeneous flat layer, the components of the curvature tensor are non-zero, leading to a non-zero geodesic deviation. The conversation also touches on the concept of a "uniform field" in GR and how it is difficult to define in a coordinate-independent manner.
sergiokapone
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/Geodesic_deviation_equation) or [MTW](http://en.wikipedia.org/wiki/Gravitation_(book)).

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?

hi sergiokapone!

yes, the geodesic deviation in a uniform field should be zero
sergiokapone said:
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?

tiny-tim said:
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.

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.

Bill_K said:
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.

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!)

Bill_K said:
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?

You may find this relevant: http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html#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.

Bill_K said:
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.

sergiokapone said:
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.

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sergiokapone said:
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.

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.

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bcrowell said:
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.

pervect said:
]
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.

pervect said:
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.

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)

$$T_{xx}=T_{yy}=\frac{2g^2}{\sqrt{1-8\,g\,z}}$$

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 ?

sergiokapone said:
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 ...)

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Mentz114 said:
If we consider a stationary frame
Oh, I see.

Mentz114 said:
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.

Mentz114 said:
I agree with your tidal calculation. The values are ( for a hovering observer)

$$T_{xx}=T_{yy}=\frac{2g^2}{\sqrt{1-8\,g\,z}}$$

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 $g^{xx} dx / d\lambda$ should be constant along a geodesic.

This is easily confirmed, if we let our geodesic be $t(\lambda)$, $x(\lambda)$, $y(\lambda)$, $z(\lambda)$

then

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

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

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 $\lambda$.)

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 :-(.

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Pervect, that's cool. The Killing vector

$$K_\mu= C\sqrt{1-8\,g\,z}\ \partial_x$$ 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 $K_{(a;b)}=0$ to get K, and C is a constant of integration.

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sergiokapone said:
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.

pervect said:
where the rhs is zero because of the geodesic equation, thus insuring that the derivative is zero and
$$\frac{\dot{x}}{\sqrt{1- 8 g z(\lambda)}}$$

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 $\lambda$.)

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 $$p_y=g_{yy} \dot{y}=\sqrt{1-8gz}\dot{y}$$

sergiokapone said:
Strange result. From geodesics eqn via mometums:
$$m\frac{dp_{\nu}}{d\lambda}=\frac{1}{2}g_{\nu \alpha, \beta}p^{\alpha}p^{\beta}$$
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 $$p_y=g_{yy} \dot{y}=\sqrt{1-8gz}\dot{y}$$

Are you sure ?

The conserved quantity found from the Killing vector is just $m\dot{x}$. (I've edited my post above).

No luck with the faucet :-(. But basically what seems to be happening is that if $\dot{x}$ starts out as zero, it remains zero. Thus in this case, x remains constant along a geodesic. However, the separation between neighboring geodesics (both of which have constant x) changes with time for a free falling observer, due to the g_xx and g_yy metric coefficient dependence on z which changes in time. Hence, there really is a tidal force in the free-fall geodesic Fermi frame.

What's really needed to give some intuitive significance to the metric is to calculate some Fermi-normal coordinates. However, this will probably wind up to be a real pain-in-the-rear to do.

I edited my geodesics eqn, due to wrong indexes)

Mentz114 said:
Are you sure ?

The conserved quantity found from the Killing vector is just $m\dot{x}$. (I've edited my post above).

Maybe I'm wrong, but I do not see where (indices corrected). Take a look.

I just want to round off the tidal stuff by remarking that the components of the tidal tensor calculated in the comoving frame field are

$$T_{xx}=T_{yy}=\frac{2\,{g}^{2}}{{\left( 1-8\,g\,z\right) }^{\frac{3}{4}}},\ \ \ T_{zz}=-\frac{4\,{g}^{2}}{{\left( 1-8\,g\,z\right) }^{\frac{3}{4}}}$$

which shows that the 'ball of dust' in free-fall would become squished in the z-direction and expand in the x, y-directions but the volume is preserved, with Taa=0

Mentz114 said:
I just want to round off the tidal stuff by remarking that the components of the tidal tensor calculated in the comoving frame field are

$$T_{xx}=T_{yy}=\frac{2\,{g}^{2}}{{\left( 1-8\,g\,z\right) }^{\frac{3}{4}}},\ \ \ T_{zz}=-\frac{4\,{g}^{2}}{{\left( 1-8\,g\,z\right) }^{\frac{3}{4}}}$$

which shows that the 'ball of dust' in free-fall would become squished in the z-direction and expand in the x, y-directions but the volume is preserved, with Taa=0

My Maple14 calculation for $$T_{zz}=-\frac{4\,{g}^{2}}{{\left( 1-8\,g\,z\right) }^{\frac{9}{4}}}$$

Code:
restart;
> with( tensor ):
> coord := [t, x, y, z]:
> g_compts := array(symmetric,sparse, 1..4, 1..4):
> g_compts[1,1] := (1-8*ge*z)^(-1/4): g_compts[2,2] := -(1-8*ge*z)^(1/2):
> g_compts[3,3] := -(1-8*ge*z)^(1/2):    g_compts[4,4] := -(1-8*ge*z)^(-5/4):
> g := create( [-1,-1], eval(g_compts));

>
> tensorsGR(coord,g,contra_metric,det_met, C1, C2, Rm, Rc, R, G, C);
> display_allGR (coord,g,contra_metric, det_met, C1, C2, Rm, Rc, R, G, C);

sergiokapone said:
I edited my geodesics eqn, due to wrong indexes)
Maybe I'm wrong, but I do not see where (indices corrected). Take a look.

This could just be terminology, but your px is not a constant so it cannot be a conserved quantity as it stands. But $p_x\cdot K_x = m\dot{x}$ is a constant.

sergiokapone said:
My Maple14 calculation for $$T_{zz}=-\frac{4\,{g}^{2}}{{\left( 1-8\,g\,z\right) }^{\frac{9}{4}}}$$
I assume that is in the coordinate basis. My last Tzz is in the comoving frame basis.

My coordinate basis calculation agrees with yours if I let U=∂t. But one could argue that U=1/(√g00)∂t is more physical because it takes into account the gravitational time dilation.

Anyway, it looks like Maple is correct. Thanks for doing the calculation, it's good to have a check on my results.

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Mentz114 said:
I just want to round off the tidal stuff by remarking that the components of the tidal tensor calculated in the comoving frame field are

$$T_{xx}=T_{yy}=\frac{2\,{g}^{2}}{{\left( 1-8\,g\,z\right) }^{\frac{3}{4}}},\ \ \ T_{zz}=-\frac{4\,{g}^{2}}{{\left( 1-8\,g\,z\right) }^{\frac{3}{4}}}$$

which shows that the 'ball of dust' in free-fall would become squished in the z-direction and expand in the x, y-directions but the volume is preserved, with Taa=0

I think you were right and I was wrong...my expression has a sign error after you differentiate it. (Unless I made another sign error! I havaen't had as much time to duoble check as I'd like...)

Mentz114 said:
This could just be terminology, but your px is not a constant so it cannot be a conserved quantity as it stands. But $p_x\cdot K_x = m\dot{x}$ is a constant.

I got their conserved quantities from the equations of motion and see no reason to doubt. conserved quantity is defined as $p^x \cdot K_x = const$. Since $p^x=\dot{x}$ - contravariant component of the momentum, $K_x=\sqrt{1-8gz}$ - Killing vector, as you find, conserved quantity, to be the same as in my case, namely,$p_x=\sqrt{1-8gz}p^x$-covariant component of the momentum. If the conserved quantities in your and my case, get different, it is strongly depressed.

let's see how I got. From the eqn
$$m\frac{dp_{\beta}}{d\lambda}=\frac{1}{2}g_{\nu \alpha, \beta}p^{\nu}p^{\alpha}$$
....
$$m\frac{dp_{0}}{d\lambda}=\frac{1}{2}g_{00, 0}p^{0}p^{0}+\frac{1}{2}g_{11, 0}p^{1}p^{1}+...$$
since "gee's" does not depend on time, we find
$$m\frac{dp_{0}}{d\lambda}=0$$
hence, it turns
$$p_{0} =const$$ - covariant component is constant.
Similarly, you can find other conserved quantities. This method finds the correct conserved quantities in the case of the Schwarzschild metric, and I see no reason why he account may not work here.

sergiokapone said:
I got their conserved quantities from the equations of motion and see no reason to doubt. conserved quantity is defined as $p^x \cdot K_x = const$. Since $p^x=\dot{x}$ - contravariant component of the momentum, $K_x=\sqrt{1-8gz}$ - Killing vector, as you find, conserved quantity, to be the same as in my case, namely,$p_x=\sqrt{1-8gz}p^x$-covariant component of the momentum. If the conserved quantities in your and my case, get different, it is strongly depressed.
You're right, I used p_x instead of p^x when contracting with the Killing co-vector.

## 1. What is a homogeneous gravitational field?

A homogeneous gravitational field refers to a gravitational field that has the same strength and direction at every point in space. This means that all objects in the field experience the same acceleration due to gravity regardless of their mass or position.

## 2. How does a homogeneous gravitational field affect the motion of objects?

In a homogeneous gravitational field, objects will follow the path of a geodesic, which is a straight line in curved spacetime. This means that objects will accelerate towards the source of the gravitational field at a constant rate, regardless of their initial velocity or direction.

## 3. What is geodesic deviation in relation to a homogeneous gravitational field?

Geodesic deviation refers to the change in distance between two objects that are initially at rest with respect to each other and then released in a gravitational field. In a homogeneous gravitational field, the distance between the objects will increase at a constant rate due to the curvature of spacetime.

## 4. How is a homogeneous gravitational field different from a non-homogeneous gravitational field?

A homogeneous gravitational field has the same strength and direction at every point in space, while a non-homogeneous gravitational field varies in strength and direction at different points. This means that objects in a non-homogeneous gravitational field will experience different accelerations depending on their position.

## 5. What are some real-life examples of a homogeneous gravitational field?

One example of a homogeneous gravitational field is the gravitational field on the surface of the Earth. Another example is the gravitational field surrounding a spherical object, such as a planet or star, where the strength and direction of the field are the same at every point on the surface. In both cases, objects will experience the same acceleration due to gravity regardless of their mass or position.

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