Homogeneous gravitational field and the geodesic deviation

sergiokapone

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

Mentz114

Are you sure ?

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

pervect

No luck with the faucet :-(. But basically what seems to be happening is that if [itex]\dot{x}[/itex] 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.

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.

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

[tex]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}}}[/tex]

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 T^a_a=0

sergiokapone

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

sergiokapone

Mentz114

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

Mentz114

I assume that is in the coordinate basis. My last T_zz 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.

pervect

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...)

sergiokapone

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

sergiokapone

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.

Mentz114

You're right, I used p_x instead of p^x when contracting with the Killing co-vector.