- #1
shichao116
- 13
- 0
Hi all, I'm now reading Chap 11 of Gravitation by Wheeler, etc.
In exercise 11.7, by introducing Jacobi curvature tensor, which contains exactly the same information content as Riemann curvature tensor, we are asked to show that we can actually measure ALL components of Jacobi curvature tensor by measuring the geodesic deviation and thus we can measure all components of Riemann tensor because Riemann tensor can be expressed in terms of Jacobi tensor.
The component of Jacobi tensor: [tex]J^\mu_{\nu\alpha\beta}=J^\mu_{\nu\beta\alpha}=1/2(R^\mu_{\alpha\nu\beta}+R^\mu_{\beta\nu\alpha})[/tex], where R is Riemann tensor
The geodesic deviation equation in terms of Jacobi tensor is:
[tex](n^\mu;_\alpha u^\alpha);_\beta u^\beta+J^\mu_{\alpha\beta\gamma}u^\alpha u^\beta n^\gamma =0[/tex]
We are allowed to choose arbitrary vectors u and n, where u is the tangent vector of geodesic and n is the separation vector between fiducial geodesic and neighboring geodesic.
In other words, by choosing u and n, we can directly measure the 'relative acceleration' of geodesics using say, test particles, which is the first term in the geodesic equation above. If we choose u and n to be some basis vectors, we surely can obtain corresponding components of Jacobi tensor directly. But these components are all of form of [tex] J^\mu_{\alpha\alpha\gamma}[/tex], i.e. having the same value on the second and the third index(this is because we do derivatives to separation vector n twice in the same direction, which is by definition the deviation of geodesic with tangent vector u ). For instance, if we choose n = (1,0,0,0) and u = (0,1,0,0), we get [tex]n^\mu;_0;_0+J^\mu_{001}=0[/tex], thus obtain [tex]J^\mu_{001=-n^\mu;_0;_0}[/tex] . If we want to calculate other components, we have to choose u to be the form like (1,1,0,0). But in this way, while [tex] J^\mu_{01\gamma}[/tex] appears in the equation, [tex] J^\mu_{10\gamma}[/tex] also shows up(this is because we do derivatives to separation vector n twice in the same direction, which is by definition the deviation of geodesic with tangent vector u ). Then we have two unknowns in one equation. How to solve this problem?
(The rest 2 terms if of the form [tex] J^\mu_{00\gamma}[/tex] and [tex] J^\mu_{11\gamma}[/tex] which we already obtain directly by setting u to be basis vectors in 0 and 1 directions as indicated above)
Hope I clearly state the problem.
Thanks very much for any help.
In exercise 11.7, by introducing Jacobi curvature tensor, which contains exactly the same information content as Riemann curvature tensor, we are asked to show that we can actually measure ALL components of Jacobi curvature tensor by measuring the geodesic deviation and thus we can measure all components of Riemann tensor because Riemann tensor can be expressed in terms of Jacobi tensor.
The component of Jacobi tensor: [tex]J^\mu_{\nu\alpha\beta}=J^\mu_{\nu\beta\alpha}=1/2(R^\mu_{\alpha\nu\beta}+R^\mu_{\beta\nu\alpha})[/tex], where R is Riemann tensor
The geodesic deviation equation in terms of Jacobi tensor is:
[tex](n^\mu;_\alpha u^\alpha);_\beta u^\beta+J^\mu_{\alpha\beta\gamma}u^\alpha u^\beta n^\gamma =0[/tex]
We are allowed to choose arbitrary vectors u and n, where u is the tangent vector of geodesic and n is the separation vector between fiducial geodesic and neighboring geodesic.
In other words, by choosing u and n, we can directly measure the 'relative acceleration' of geodesics using say, test particles, which is the first term in the geodesic equation above. If we choose u and n to be some basis vectors, we surely can obtain corresponding components of Jacobi tensor directly. But these components are all of form of [tex] J^\mu_{\alpha\alpha\gamma}[/tex], i.e. having the same value on the second and the third index(this is because we do derivatives to separation vector n twice in the same direction, which is by definition the deviation of geodesic with tangent vector u ). For instance, if we choose n = (1,0,0,0) and u = (0,1,0,0), we get [tex]n^\mu;_0;_0+J^\mu_{001}=0[/tex], thus obtain [tex]J^\mu_{001=-n^\mu;_0;_0}[/tex] . If we want to calculate other components, we have to choose u to be the form like (1,1,0,0). But in this way, while [tex] J^\mu_{01\gamma}[/tex] appears in the equation, [tex] J^\mu_{10\gamma}[/tex] also shows up(this is because we do derivatives to separation vector n twice in the same direction, which is by definition the deviation of geodesic with tangent vector u ). Then we have two unknowns in one equation. How to solve this problem?
(The rest 2 terms if of the form [tex] J^\mu_{00\gamma}[/tex] and [tex] J^\mu_{11\gamma}[/tex] which we already obtain directly by setting u to be basis vectors in 0 and 1 directions as indicated above)
Hope I clearly state the problem.
Thanks very much for any help.
Last edited: