- #1
Sonderval
- 234
- 11
Consider a force-free particle moving on a geodesic with four-velocity [tex]v^\nu[/tex].
The formula for the four-acceleration in any coordinate system is
[tex]
\frac{dx^\mu}{d\tau} = - \Gamma^\mu_{\nu\lambda} v^\nu v^\lambda[/tex]
Since the four-acceleration on the left side is orthogonal to the four-velocity, this implies
[tex]\Gamma^\mu_{\nu\lambda} v^\nu v^\lambda v_\mu=0[/tex]
Is this correct? I've never seen this equation anywhere. (Perhaps because it is trivial?)
It seems to imply that for any coordinate system with 0 as a time-like coordinate
[tex]\Gamma^0_{00}=0[/tex]
because I can always find a particle which is at rest in this system (i.e. with a four-velocity of (1,0,0,0) ). Is that a valid conclusion?
The formula for the four-acceleration in any coordinate system is
[tex]
\frac{dx^\mu}{d\tau} = - \Gamma^\mu_{\nu\lambda} v^\nu v^\lambda[/tex]
Since the four-acceleration on the left side is orthogonal to the four-velocity, this implies
[tex]\Gamma^\mu_{\nu\lambda} v^\nu v^\lambda v_\mu=0[/tex]
Is this correct? I've never seen this equation anywhere. (Perhaps because it is trivial?)
It seems to imply that for any coordinate system with 0 as a time-like coordinate
[tex]\Gamma^0_{00}=0[/tex]
because I can always find a particle which is at rest in this system (i.e. with a four-velocity of (1,0,0,0) ). Is that a valid conclusion?