# Strange geodesic in Schwartzschild metric

 Share this thread:
 P: 256 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.
 Sci Advisor P: 1,910 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.
Mentor
P: 6,246
 Quote by 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.
This not a geodesic. If

$$\mathbf{u} = \left( u^t , u^r, u^\theta, u^\phi, \right) = \left( \left( 1 - \frac{2m}{r_0} \right)^{-\frac{1}{2}}, 0, 0, 0 \right),$$

then the 4-acceleration is given by

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

which is non-zero.

Mentor
P: 6,246
Strange geodesic in Schwartzschild metric

 Quote by George Jones $$\begin{equation*} \begin{split} \mathbf{a} &= \nabla_{\mathbf{u}} \mathbf{u} \\ &= u^\alpha \nabla_{\partial_\alpha} \left( u^\beta \partial_\beta \right) \\ &= u^\alpha \left( \nabla_{\partial_\alpha} \left( u^\beta \right) \partial_\beta + u^\beta \nabla_{\partial_\alpha} \left( \partial_\beta \right) \right) \\ &= \left( u^t \right)^2 \Gamma^\mu {}_{tt} \partial_\mu \end{split} \end{equation*}$$
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.
 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.
P: 665
 Quote by George Jones 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
Sci Advisor
PF Gold
P: 1,847
 Quote by Altabeh 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.
P: 665
 Quote by DrGreg 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

 Related Discussions Special & General Relativity 1 Calculus 6 Differential Geometry 1 Special & General Relativity 18