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Strange geodesic in Schwartzschild metric

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paweld
#1
Feb22-10, 01:15 PM
P: 256
The following curve is geodesic in Schwardschild metric:
[tex] \tau \mapsto [(1-2m/r_0)^{-1/2}\tau,r_0,0,0][/tex].
The tangent vector is: [tex] [(1-2m/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] (1-2m/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.
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Ich
#2
Feb22-10, 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.
George Jones
#3
Feb22-10, 02:07 PM
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P: 6,229
Quote Quote by paweld View Post
The following curve is geodesic in Schwardschild metric:
[tex] \tau \mapsto [(1-2m/r_0)^{-1/2}\tau,r_0,0,0][/tex].
The tangent vector is: [tex] [(1-2m/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] (1-2m/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.
This not a geodesic. If

[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 4-acceleration 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 non-zero.

George Jones
#4
Feb23-10, 02:11 PM
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Strange geodesic in Schwartzschild metric

Quote Quote by George Jones View Post
[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]
I was waiting for comments before finishing this off.

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.
Mentz114
#5
Feb23-10, 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.
Altabeh
#6
Feb23-10, 03:38 PM
P: 665
Quote Quote by George Jones View Post
I was waiting for comments before finishing this off.

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.
A very confusing thing here is the use of [tex]m[/tex] to denote both half of the Schwarzschild redius and the mass of gravitating body! I think in textbooks whose authers prefer using the notation [tex]2m[/tex] instead of [tex]r_s[/tex] to symbolize the Schwarzschild redius, they later use

[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
DrGreg
#7
Feb23-10, 05:47 PM
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P: 1,843
Quote Quote by Altabeh View Post
A very confusing thing here is the use of [tex]m[/tex] to denote both half of the Schwarzschild redius and the mass of gravitating body! I think in textbooks whose authers prefer using the notation [tex]2m[/tex] instead of [tex]r_s[/tex] to symbolize the Schwarzschild redius, they later use

[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
There is a convention that many authors of advanced texts use, as well as choosing units of distance and time such that c=1, they also choose units of mass such that G=1. It can cause confusion to persons unfamiliar with it.
Altabeh
#8
Feb24-10, 05:06 AM
P: 665
Quote Quote by DrGreg View Post
There is a convention that many authors of advanced texts use, as well as choosing units of distance and time such that c=1, they also choose units of mass such that G=1. It can cause confusion to persons unfamiliar with it.
But you didn't notice that George put a G in the last equation which means the convention that I probably seem to have forgotten leads to

[tex]m=GM.[/tex]

AB


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