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Homogeneous gravitational field and the geodesic deviationby sergiokapone
Tags: geodesic deviation 
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#19
Jan2413, 05:41 PM

PF Gold
P: 4,087

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


#20
Jan2413, 05:42 PM

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P: 7,596

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 freefall geodesic Fermi frame.
What's really needed to give some intuitive significance to the metric is to calculate some Ferminormal coordinates. However, this will probably wind up to be a real painintherear to do. 


#21
Jan2413, 05:46 PM

P: 110

I edited my geodesics eqn, due to wrong indexes)



#22
Jan2413, 06:06 PM

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
[tex]T_{xx}=T_{yy}=\frac{2\,{g}^{2}}{{\left( 18\,g\,z\right) }^{\frac{3}{4}}},\ \ \ T_{zz}=\frac{4\,{g}^{2}}{{\left( 18\,g\,z\right) }^{\frac{3}{4}}}[/tex] which shows that the 'ball of dust' in freefall would become squished in the zdirection and expand in the x, ydirections but the volume is preserved, with T^{a}_{a}=0 


#23
Jan2413, 06:11 PM

P: 110




#24
Jan2413, 06:14 PM

P: 110




#25
Jan2413, 06:16 PM

PF Gold
P: 4,087




#26
Jan2413, 06:20 PM

PF Gold
P: 4,087

My coordinate basis calculation agrees with yours if I let U=∂t. But one could argue that U=1/(√g_{00})∂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. 


#27
Jan2413, 08:23 PM

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#28
Jan2513, 02:42 AM

P: 110




#29
Jan2513, 02:55 AM

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. 


#30
Jan2513, 10:22 AM

PF Gold
P: 4,087




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