Kostik said:
Weinberg starts this discussion in the middle of p. 72 by saying "The values of the metric tensor ##g_{\mu\nu}## and the affine connection ##Γ^\lambda_{\mu\nu}## at a point X in an arbitrary coordinate system ##x^\mu## PROVIDE ENOUGH INFORMATION TO DETERMINE THE LOCALLY INERTIAL COORDINATES ##\xi^\alpha## in a neighborhood of X." *This* is what I am not yet convinced of.
Thanks again for any help.
Weinberg’s statement is a
provable statement. Let x = 0 be the point in question, and define new coordinates y^{\mu} by
x^{\mu} = A^{\mu}{}_{\alpha}y^{\alpha} - \frac{1}{4} A^{\mu}{}_{\tau}B^{\tau}{}_{\alpha \beta}y^{\alpha}y^{\beta} , \ \ \ \ \ (1) where A^{\mu}{}_{\alpha} and B^{\tau}{}_{\alpha \beta} = B^{\tau}{}_{\beta \alpha} are constants to be determined from reasonable requirements. At x = 0,
any rank-2 tensor g_{\mu\nu}(x) will therefore have the following transformation law \bar{g}_{\alpha\beta}(0) = A^{\mu}{}_{\alpha} \ g_{\mu\nu}(0) \ A^{\nu}{}_{\beta} . \ \ \ \ \ \ \ \ \ \ \ \ (2) In matrix form, (2) reads \bar{g} = A^{T} \ g \ A . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)
If g_{\mu\nu}(x) is a
metric tensor on some differentiable manifold M^{n} then:
(I) it must be
symmetric, i.e., g_{\mu\nu}(x) = g_{\nu\mu}(x). And,
more importantly:
(II) In order to model our
spacetime by an
equivalent class of pairs (M^{n} , g_{\mu\nu}), g_{\mu\nu} must be a
non-degenerate metric of
Lorentz signature. (bellow I will be using the mostly minus signature).
Now,
(I) \Rightarrow \ \exists R \in O(n) such that R^{T}gR = G is diagonal, and
(II) \Rightarrow \ G = \mbox{diag}(\lambda_{0}^{2}, - \lambda_{1}^{2}, \cdots , - \lambda_{n-1}^{2}), where \lambda_{r} \neq 0 \ \forall r.
So, we have
R^{T} \ g \ R = \mbox{diag}(\lambda_{0}^{2}, -\lambda_{1}^{2}, -\lambda_{2}^{2} ,\cdots , -\lambda_{n-1}^{2}) . \ \ \ \ \ (4) Define the matrix D = \mbox{diag}(1/\lambda_{0},1/\lambda_{1}, \cdots , 1/\lambda_{n-1}), and sandwich (4) by D to obtain
DR^{T} \ g \ RD = \mbox{diag}(1, -1, -1, \cdots , -1) . Thus (RD)^{T} \ g \ (RD) = \eta . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) However, for any \Lambda \in SO(1,n-1) , we also have (RD \Lambda )^{T} \ g \ (RD \Lambda ) = \Lambda^{T} \eta \Lambda = \eta . So, we can set A^{\mu}{}_{\alpha} = (RD)^{\mu}{}_{\alpha} \ \ \mbox{mod} \ \Lambda \in SO(1,n-1) . \ \ \ \ \ \ (6) Using the choice (6) together with (5), equation (2) or (3) gives us our first desired result, that is \bar{g}_{\alpha \beta}(0) = A^{\mu}{}_{\alpha} \ g_{\mu\nu}(0) \ A^{\nu}{}_{\beta} = \eta_{\alpha\beta} . \ \ \ \ \ \ (7)
The fact that the matrix (RD) determines A up to Lorentz transformations can also be deduced from parameter-counting: there are n^{2} parameters in A, n parameters in D and \frac{1}{2}n(n-1) parameters in the O(n) matrix R. This leaves us with the n^{2} - [\frac{1}{2}n(n-1) + n] = \frac{1}{2}n(n-1) free parameters needed for a Lorentz transformation.
To complete the proof of Weinberg’s statement, we need to choose the constants B^{\mu}{}_{\alpha\beta} so that, together with the choice (6) for A^{\mu}{}_{\alpha}, the system y, as defined in (1), becomes
locally inertial system. This requires a bit of algebra which I will describe them for you. Differentiating the transformation law \bar{g}_{\alpha\beta}(y) = \frac{\partial x^{\mu}}{\partial y^{\alpha}} \frac{\partial x^{\nu}}{\partial y^{\beta}} g_{\mu\nu}(x) , with respect to y^{\gamma}, we get at x = 0
\frac{\partial \bar{g}_{\alpha\beta}}{\partial y^{\gamma}}(0) = T_{\alpha\beta\gamma}(0) - \frac{1}{2}\left(A^{\mu}{}_{\alpha}g_{\mu\nu}(0) A^{\nu}{}_{\tau}\right) B^{\tau}{}_{\beta\gamma} - \frac{1}{2} \left(A^{\mu}{}_{\tau}g_{\mu\nu}(0)A^{\nu}{}_{\beta}\right) B^{\tau}{}_{\alpha\gamma} , \ \ (8)
where we have defined the object T_{\alpha\beta\gamma}(0) \equiv A^{\mu}{}_{\alpha}A^{\nu}{}_{\beta}A^{\rho}{}_{\gamma} \frac{\partial g_{\mu\nu}}{\partial x^{\rho}}(0) . \ \ \ \ (9)
Using our first result (7), we find
\frac{\partial \bar{g}_{\alpha\beta}}{\partial y^{\gamma}}(0) = T_{\alpha\beta\gamma}(0) - \frac{1}{2} (B_{\alpha\beta\gamma} + B_{\beta\alpha\gamma}) . \ \ \ \ (8')
So, to complete the proof we need to solve the following equation for B_{\alpha\beta\gamma}
T_{\alpha\beta\gamma} = \frac{1}{2}\left(B_{\alpha\beta\gamma} + B_{\beta\alpha\gamma}\right) . \ \ \ (10)
To do this, write another 2 copy of (10) using the permutation (\alpha\beta\gamma) \to (\gamma\alpha\beta) \to (\beta\gamma\alpha), then add the first two and subtract the third one. Then, due to B_{\alpha\beta\gamma} = B_{\alpha\gamma\beta}, you obtain the final result
B_{\alpha\beta\gamma} = T_{\alpha\beta\gamma}(0) + T_{\gamma\alpha\beta} (0) - T_{\beta\gamma\alpha}(0) , with T(0) as defined by (9).
So, from the values of g_{\mu\nu}(0) and \partial g (0) \sim \Gamma (0) we were able to determine (up to Lorentz transformation) the constants A and B that are needed to make the y-system locally inertial.
