Register to reply 
Homogeneous gravitational field and the geodesic deviationby sergiokapone
Tags: geodesic deviation 
Share this thread: 
#1
Jan2413, 12:41 AM

P: 110

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 nonzero, 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? 


#2
Jan2413, 06:40 AM

Sci Advisor
HW Helper
Thanks
P: 26,148



#3
Jan2413, 07:21 AM

P: 110

$$ds^2=(18gz)^{1/4}dt^2(18gz)^{1/2}(dx^2+dy^2)(18gz)^{5/4}dz^2$$ in the Newtonian limit becomes as equation of motion in homogeneous field. However, this metric has a nonzero 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/grqc/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. 


#4
Jan2413, 09:16 AM

Sci Advisor
Thanks
P: 4,160

Homogeneous gravitational field and the geodesic deviation



#5
Jan2413, 09:23 AM

P: 110




#6
Jan2413, 10:53 AM

Sci Advisor
Thanks
P: 4,160

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}(dx^{2}+dy^{2}) 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!) 


#7
Jan2413, 12:00 PM

P: 110




#8
Jan2413, 12:54 PM

Emeritus
Sci Advisor
PF Gold
P: 5,598

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 coordinatebased. 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 coordinatedependent. However, one really wants to classify especially symmetric spacetimes according to their *intrinsic* symmetry, not their symmetry when expressed in some coordinates. 


#9
Jan2413, 02:45 PM

Emeritus
Sci Advisor
P: 7,659

[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. 


#10
Jan2413, 03:39 PM

P: 110




#11
Jan2413, 04:06 PM

PF Gold
P: 4,087

[tex]T_{xx}=T_{yy}=\frac{2g^2}{\sqrt{18\,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 ? 


#12
Jan2413, 04:13 PM

PF Gold
P: 4,087




#13
Jan2413, 04:26 PM

P: 110




#14
Jan2413, 04:30 PM

P: 110




#15
Jan2413, 04:55 PM

Emeritus
Sci Advisor
P: 7,659

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{18 g z(\lambda)}} \left[ \ddot{x} + \frac{4 \dot{x}\dot{z} } {\sqrt{18 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 xmomentum. (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 :(. 


#16
Jan2413, 05:22 PM

PF Gold
P: 4,087

Pervect, that's cool. The Killing vector
[tex]K_\mu= C\sqrt{18\,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 restmass 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. 


#17
Jan2413, 05:26 PM

PF Gold
P: 4,087




#18
Jan2413, 05:38 PM

P: 110

$$m\frac{dp_{\beta}}{d\lambda}=\frac{1}{2}g_{\nu \alpha, \beta}p^{\nu}p^{\alpha}$$ I see that there are conserve such values: $$p_t=g_{tt} \dot{t},$$ $$p_x=g_{xx} \dot{x}=\sqrt{18gz}\dot{x}$$ and $$p_y=g_{yy} \dot{y}=\sqrt{18gz}\dot{y}$$ 


Register to reply 
Related Discussions  
Geodesic deviation  Introductory Physics Homework  0  
Geodesic deviation  Special & General Relativity  1  
Geodesic deviation  Special & General Relativity  2  
Equilibrium temperature distribution of ideal gas in homogeneous gravitational field  Classical Physics  4  
Geodesic Deviation  Differential Geometry  4 