Help with Weinberg p. 72 -- time dt for a photon to travel a distance d⃗x

[Kostik]

Thanks a lot strangerep. The decomposition appears to be $A = VDV^{-1}$, not $VDV^T$, so it doesn't quite match up with our equation $g=b^T\eta b$. Of course, $A^{-1}=A^T$ for orthogonal matrices, but is $b$ orthogonal?*

It is interesing to note that the definition of a Lorentz transformation is $Λ^T\eta Λ=\eta$ so we can see that when $g=\eta$ then $b=Λ$, which does agree with the known fact that b is only defined up to a Lorentz transformation.

I'll try to tidy this up ... I think it still isn't clear enough to me ... but it does seem that Weinberg made a little bit of a jump here.

*Edit: Yes. Since g is symmetric, in fact $g=VDV^T$ where D is the diagonal matrix of eigenvalues of g, and V is the matrix of g's [orthogonal] eigenvectors. But our equation is $g=b^T\eta b$, so $b=V^T$. And again here it's obvious that b (or V) is valid up to a Lorentz transformation because $Λ^T\eta Λ=\eta$.

Again many thanks. Your last post really helped me crack this.

 

[strangerep]

The decomposition appears to be $A = VDV^{-1}$, not $VDV^T$, so it doesn't quite match up with our equation $g=b^T\eta b$. Of course, $A^{-1}=A^T$ for orthogonal matrices, but is $b$ orthogonal?*

As you note in your edit, $A = VDV^T$ is correct if $g$ is real symmetric. Additional discussion here.

[...] it does seem that Weinberg made a little bit of a jump here.

Yes -- that's an example of what I meant when I said that Weinberg tends to be more suitable as an advanced text. He requires rather more of his readers, and sometimes this isn't obvious -- as in the present case.

 

[TSny]

It's just an application of eigendecomposition, i.e.,$$A ~=~ V D V^T ~,$$where $D$ is diagonal (with entries being the eigenvalues of $A$), and the columns of $V$ are the eigenvectors of a (real, symmetric) matrix $A$. In our case: $$g ~=~ b \eta b^T ~,$$ and the "constructive" solution for $b$ is obtained by finding eigenvectors of $g$.

Note that in general, $~g ~=~ b \eta b^T ~$ is not an eigendecomposition of $g$. The diagonal elements of $\eta$ are not, in general, the eigenvalues of $g$ and the columns of $b$ are not, in general, eigenvectors of $g$ .

As an example, suppose $g = \left( \begin{array}{cc} 7 & 4 \\ 4 & 1 \\ \end{array} \right)$.

Then you can show that $g = b \eta b^T$ where $\eta = \left( \begin{array}{cc} -1 &0 \\ 0 & 1 \\ \end{array} \right)$ and $b = \left( \begin{array}{cc} 3 & 4 \\ 0 & 1 \\ \end{array} \right)$

But the eigendecomposition of $g$ is $g = VDV^T$ where $V = \left( \begin{array}{cc} 1/\sqrt{5} &2/\sqrt{5} \\ -2/\sqrt{5} & 1/\sqrt{5} \\ \end{array} \right)$ and $D = \left( \begin{array}{cc} -1 &0 \\ 0 & 9 \\ \end{array} \right)$

 

[strangerep]

Note that in general, $~g ~=~ b \eta b^T ~$ is not an eigendecomposition of $g$. [...]

Heh, I was wondering whether someone would mention that.

In GR, the equivalence principle motivates an assumption that we have a metric field with signature (-,+,+,+) at each point of spacetime. Indeed, Weinberg actually starts from $\eta$ and transforms to $g$ -- see p71. His $b$'s are then also expressed as coord transformation coefficients -- see eqn(3.2.13).

IOW, we're not really starting from an arbitrary $g$, but one with the right eigenvalues.

 

[TSny]

Strangerep, I agree. Weinberg's discussion starts from the equivalence principle. This implies that (3.2.14) must have a solution where the $b$'s are associated with the coordinate transformation as in (3.2.13).

Still, the eigenvalues of $g(X)$ need not be the diagonal elements of $\eta$.

But, as you say, you can't just start from any symmetric matrix $g$. For example, taking the determinant of each side of (3.2.14) requires that the determinant of $g$ be negative.

 

[vanhees71]

The EP just says that you can, at any point $x_0$ in spacetime always find local coordinates in which $g_{\mu \nu}(x_0)=\eta_{\mu \nu}$. This implies that the symmetric matrix at each point must have 1 negative and 3 positive eigenvalues (Sylvester's theorem).

 

[samalkhaiat]

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.