I’m a little bit confused about how to get the four-velocity components from a given metric tensor (or line element). For instance, which are the components of the four-velocity in the Schwarzschild metric? Can anybody help me?
[Physics] How to get the four-velocity components from a given metric tensor
general-relativitykinematicsmetric-tensorvectorsvelocity
Related Solutions
According to Mathematica, and assuming I haven't made any silly errors typing in the metric, I get the non-zero components of $R^\mu{}_{\nu\alpha\beta}$ to be:
{1, 2, 1, 2} -> (2 G M)/(r^2 (-2 G M + c^2 r)),
{1, 2, 2, 1} -> -((2 G M)/(r^2 (-2 G M + c^2 r))),
{1, 3, 1, 3} -> -((G M)/(c^2 r)),
{1, 3, 3, 1} -> (G M)/(c^2 r),
{1, 4, 1, 4} -> -((G M Sin[\[Theta]]^2)/(c^2 r)),
{1, 4, 4, 1} -> (G M Sin[\[Theta]]^2)/(c^2 r),
{2, 1, 1, 2} -> (2 G M (-2 G M + c^2 r))/(c^4 r^4),
{2, 1, 2, 1} -> -((2 G M (-2 G M + c^2 r))/(c^4 r^4)),
{2, 3, 2, 3} -> -((G M)/(c^2 r)),
{2, 3, 3, 2} -> (G M)/(c^2 r),
{2, 4, 2, 4} -> -((G M Sin[\[Theta]]^2)/(c^2 r)),
{2, 4, 4, 2} -> (G M Sin[\[Theta]]^2)/(c^2 r),
{3, 1, 1, 3} -> (G M (2 G M - c^2 r))/(c^4 r^4),
{3, 1, 3, 1} -> (G M (-2 G M + c^2 r))/(c^4 r^4),
{3, 2, 2, 3} -> (G M)/(r^2 (-2 G M + c^2 r)),
{3, 2, 3, 2} -> (G M)/(r^2 (2 G M - c^2 r)),
{3, 4, 3, 4} -> (2 G M Sin[\[Theta]]^2)/(c^2 r),
{3, 4, 4, 3} -> -((2 G M Sin[\[Theta]]^2)/(c^2 r)),
{4, 1, 1, 4} -> (G M (2 G M - c^2 r))/(c^4 r^4),
{4, 1, 4, 1} -> (G M (-2 G M + c^2 r))/(c^4 r^4),
{4, 2, 2, 4} -> (G M)/(r^2 (-2 G M + c^2 r)),
{4, 2, 4, 2} -> (G M)/(r^2 (2 G M - c^2 r)),
{4, 3, 3, 4} -> -((2 G M)/(c^2 r)),
{4, 3, 4, 3} -> (2 G M)/(c^2 r),
As mentioned in the comments calculating algebraic expressions for the Ricci tensor containing the metric, its inverse and its first and second derivatives is straight forward using computer algebra.
The most arbitrary metric $$g_{\alpha\beta}=g_{\beta\alpha}$$ has 10 independent components which are functions of four coordinates $\left\{x_0,x_1,x_2,x_3\right\}$: $$g_{\alpha\beta}(x_0,x_1,x_2,x_3).$$ The metric has an inverse with 10 independent components $$g^{\alpha\beta}=g^{\beta\alpha}.$$
The metric has 40 independent first partial derivatives $$g_{\alpha\beta,\gamma}$$ and 100 independent second partial derivatives (100 and not 160 because of the the symmetry of second derivatives) $$g_{\alpha\beta,\gamma\delta}=g_{\alpha\beta,\delta\gamma}.$$
Using those ingredients ($g_{\alpha\beta}$, $g^{\alpha\beta}$, $g_{\alpha\beta,\gamma}$ and $g_{\alpha\beta,\gamma\delta}$) one can compute 21 components of the Riemann tensor $R_{\alpha\beta\gamma\delta}$. One could eliminate one of those 21 components using the first Bianchi identity.
Just to give one example in this post: $R_{0102}$ has 1510 terms: 4 second derivatives and the rest are contractions of Christoffel symbols:
The Ricci tensor can be constructed from the contraction $$R_{\alpha\beta}=R^\mu_{\,\,\alpha\mu\beta}$$ so it contains the components of the inverse metric an those 21 Riemann tensors:
Writing $R_{\alpha\beta}$ out in terms of $g$ only becomes super messy in case of $R_{01}$ we are talking about 8711 terms. I have no idea how to visualize such an expression here at SE. I have uploaded a PDF (careful it is rather large) of $R_{01}$ here.
I also uploaded .m files containing all 10 independent components of $R_{\alpha\beta}$ Rij.m and the 21 components of $R_{\alpha\beta\gamma\delta}$ Rijkl.m.
As pointed out in the comment of the original question those expressions have only very very limited use. But maybe some conclusions:
- We see that a tensor notation in sum convention is a very very elegant way to formulate those expressions.
- This elegant notation masks the general complexity of those expressions.
- The explicit expressions in terms of $g$ illustrate the utmost importance of symmetries and a good coordinate choice for a given problem.
- To work with the field equations symmetries and/or advanced methods of Numerical Relativity are necessary because in a naive form the expressions and equations are just to complicated. Such a "brute force" approach of formulating/ printing out the field equations of General Relativity is doomed to fail.
Best Answer
As kalle has said, the metric and the four-velocity are not intrinsically related, you can have an object with velocity in any given direction and with any magnitude even if you fix the metric.
The thing you can do with the metric, however, is to normalize the velocity: in the $-+++$ metric convention the four-velocity of a massive object always has a norm of $u^2 = g_{\mu\nu}u^\mu u^\nu = -1$, which constrains the value of a component of the velocity as long as you know the values of the other three. For example, you can choose the three spatial components of the velocity to be whatever you like, and find out using the metric what the time component should be.
For a massless object the reasoning is the same, only with $u^2 = 0$.