# Homogeneous gravitational field and the geodesic deviation

by sergiokapone
Tags: geodesic deviation
PF Gold
P: 4,087
 Quote by sergiokapone 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).
 Emeritus Sci Advisor P: 7,657 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.
P: 110
I edited my geodesics eqn, due to wrong indexes)

 Quote by Mentz114 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.
 PF Gold P: 4,087 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
P: 110
 Quote by Mentz114 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 $$T_{zz}=-\frac{4\,{g}^{2}}{{\left( 1-8\,g\,z\right) }^{\frac{9}{4}}}$$
 P: 110 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);
PF Gold
P: 4,087
 Quote by sergiokapone 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.
PF Gold
P: 4,087
 Quote by sergiokapone 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.
Emeritus
P: 7,657
 Quote by Mentz114 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...)
P: 110
 Quote by Mentz114 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.
 P: 110 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.
PF Gold
P: 4,087
 Quote by sergiokapone 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.

 Related Discussions Introductory Physics Homework 0 Special & General Relativity 1 Special & General Relativity 2 Classical Physics 4 Differential Geometry 4