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Strange geodesic in Schwartzschild metric 
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#1
Feb2210, 01:15 PM

P: 256

The following curve is geodesic in Schwardschild metric:
[tex] \tau \mapsto [(12m/r_0)^{1/2}\tau,r_0,0,0][/tex]. The tangent vector is: [tex] [(12m/r_0)^{1/2},0,0,0] [/tex], its length is 1 and its product with killing vector [tex]\partial_t [/tex] is equal: [tex] (12m/r_0)^{1/2} = \textrm{const}[/tex]. So the body lays at rest in gravitational field  why it's possible?? In newtonian limit it's impossible  the body which does not rotate around a star cannot have constant radious. 


#2
Feb2210, 01:49 PM

Sci Advisor
P: 1,891

Constant energy is necessary for a geodesic, not sufficient. Plug this tangent vector in the equation of motion, you'll get [itex]\ddot r \neq 0[/itex].
Btw., it's Schwarzschild. 


#3
Feb2210, 02:07 PM

Mentor
P: 6,229

[tex]\mathbf{u} = \left( u^t , u^r, u^\theta, u^\phi, \right) = \left( \left( 1  \frac{2m}{r_0} \right)^{\frac{1}{2}}, 0, 0, 0 \right), [/tex] then the 4acceleration is given by [tex] \begin{equation*} \begin{split} \mathbf{a} &= \nabla_{\mathbf{u}} \mathbf{u} \\ &= u^\alpha \nabla_{\partial_\alpha} \left( u^\beta \partial_\beta \right) \\ &= u^\alpha \left( \nabla_{\partial_\alpha} \left( u^\beta \right) \partial_\beta + u^\beta \nabla_{\partial_\alpha} \left( \partial_\beta \right) \right) \\ &= \left( u^t \right)^2 \Gamma^\mu {}_{tt} \partial_\mu \end{split} \end{equation*} [/tex] which is nonzero. 


#4
Feb2310, 02:11 PM

Mentor
P: 6,229

Strange geodesic in Schwartzschild metric
Using [tex]0 = \Gamma^t {}_{tt} = \Gamma^\theta {}_{tt} = \Gamma^\phi {}_{tt}[/tex] and [tex]\Gamma^r {}_{tt} = \left( 1  \frac{2m}{r_0} \right) \frac{m}{r_0^2}[/tex] gives [tex]\mathbf{a} = \left( 0, \frac{m}{r_0^2}, 0, 0 \right)[/tex] with magnitude [tex]a = \left( 1  \frac{2m}{r_0} \right)^{\frac{1}{2}} \frac{m}{r_0^2}[/tex] Taking [itex]r_0[/itex] to be much larger that the Schwarzschild radius, and restoring [itex]c[/itex] and [itex]G[/itex] gives [tex]a = \frac{Gm}{r_0^2}[/tex]. Consequently, such a hovering observer experiences normal Newtonian weight. 


#5
Feb2310, 02:15 PM

PF Gold
P: 4,087

George, that's very instructive, thank you. I'm still reading Lee's book and I've bookmarked this thread.



#6
Feb2310, 03:38 PM

P: 665

[tex]m=GM/c^2,[/tex] where [tex]M[/tex] is the mass of mass of gravitating body. But I respect George's style and accept it as another alternative. AB 


#7
Feb2310, 05:47 PM

Sci Advisor
PF Gold
P: 1,843




#8
Feb2410, 05:06 AM

P: 665

[tex]m=GM.[/tex] AB 


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