Strange geodesic in Schwartzschild metric

[paweld]

The following curve is geodesic in Schwardschild metric:
\tau \mapsto [(1-2m/r_0)^{-1/2}\tau,r_0,0,0].
The tangent vector is: [(1-2m/r_0)^{-1/2},0,0,0], its length is 1 and its
product with killing vector \partial_t is equal: (1-2m/r_0)^{1/2} = \textrm{const}. 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.

 

[Ich]

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

 

[George Jones]

The following curve is geodesic in Schwardschild metric:
\tau \mapsto [(1-2m/r_0)^{-1/2}\tau,r_0,0,0].
The tangent vector is: [(1-2m/r_0)^{-1/2},0,0,0], its length is 1 and its
product with killing vector \partial_t is equal: (1-2m/r_0)^{1/2} = \textrm{const}. 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

\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),<br />

then the 4-acceleration is given by

<br /> \begin{equation*}<br /> \begin{split}<br /> \mathbf{a} &amp;= \nabla_{\mathbf{u}} \mathbf{u} \\<br /> &amp;= u^\alpha \nabla_{\partial_\alpha} \left( u^\beta \partial_\beta \right) \\<br /> &amp;= u^\alpha \left( \nabla_{\partial_\alpha} \left( u^\beta \right) \partial_\beta + u^\beta \nabla_{\partial_\alpha} \left( \partial_\beta \right) \right) \\<br /> &amp;= \left( u^t \right)^2 \Gamma^\mu {}_{tt} \partial_\mu<br /> \end{split}<br /> \end{equation*}<br />

which is non-zero.

 

[George Jones]

<br /> \begin{equation*}<br /> \begin{split}<br /> \mathbf{a} &amp;= \nabla_{\mathbf{u}} \mathbf{u} \\<br /> &amp;= u^\alpha \nabla_{\partial_\alpha} \left( u^\beta \partial_\beta \right) \\<br /> &amp;= u^\alpha \left( \nabla_{\partial_\alpha} \left( u^\beta \right) \partial_\beta + u^\beta \nabla_{\partial_\alpha} \left( \partial_\beta \right) \right) \\<br /> &amp;= \left( u^t \right)^2 \Gamma^\mu {}_{tt} \partial_\mu<br /> \end{split}<br /> \end{equation*}<br />

I was waiting for comments before finishing this off.

Using

0 = \Gamma^t {}_{tt} = \Gamma^\theta {}_{tt} = \Gamma^\phi {}_{tt}

and

\Gamma^r {}_{tt} = \left( 1 - \frac{2m}{r_0} \right) \frac{m}{r_0^2}

gives

\mathbf{a} = \left( 0, \frac{m}{r_0^2}, 0, 0 \right)

with magnitude

a = \left( 1 - \frac{2m}{r_0} \right)^{-\frac{1}{2}} \frac{m}{r_0^2}

Taking r_0 to be much larger that the Schwarzschild radius, and restoring c and G gives

a = \frac{Gm}{r_0^2}.

Consequently, such a hovering observer experiences normal Newtonian weight.

 

[Mentz114]

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

 

[Altabeh]

I was waiting for comments before finishing this off.

Using

0 = \Gamma^t {}_{tt} = \Gamma^\theta {}_{tt} = \Gamma^\phi {}_{tt}

and

\Gamma^r {}_{tt} = \left( 1 - \frac{2m}{r_0} \right) \frac{m}{r_0^2}

gives

\mathbf{a} = \left( 0, \frac{m}{r_0^2}, 0, 0 \right)

with magnitude

a = \left( 1 - \frac{2m}{r_0} \right)^{-\frac{1}{2}} \frac{m}{r_0^2}

Taking r_0 to be much larger that the Schwarzschild radius, and restoring c and G gives

a = \frac{Gm}{r_0^2}.

Consequently, such a hovering observer experiences normal Newtonian weight.

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

m=GM/c^2,

where M is the mass of mass of gravitating body. But I respect George's style and accept it as another alternative.

AB

 

[DrGreg]

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

m=GM/c^2,

where M 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]

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

m=GM.

AB