Metric signature explanation Can anyone explain what metric signature is?
I have a basic knowledge regarding tensors, btw.
Also, how is it related to fundamental understanding of general relativity?
 A: In relativity there is an invariant called the proper time, $\tau$. It's an invariant in the sense that all observers will agree on it's value. In special relativity the proper time is defined as:
$$\mathrm d\tau^2 = \mathrm ds^2 = c^2~\mathrm dt^2 - \mathrm dx^2 - \mathrm dy^2 - \mathrm dz^2$$
or
$$\mathrm d\tau^2 = -\mathrm ds^2 = -c^2~\mathrm dt^2 + \mathrm dx^2 + \mathrm dy^2 + \mathrm dz^2$$
You see both sign conventions and I've never been sure which is more generally accepted. Anyhow, you can write the equation for $\mathrm ds^2$ as a matrix equation using:
$$\mathrm ds^2 = g_{\alpha\beta}x^\alpha x^\beta$$
where $x$ is the vector $(t, x, y, z)$ and $g$ is the matrix:
$$\left( \begin{matrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right)$$
The matrix is called the metric tensor (or a representation of it) and the signature is the number of positive and negative values on the leading diagonal. In this case it's (1,3) or you often just add together the negative and positive numbers to give, in this case, just 2.
Exactly the same equation is used in general relativity, but the matrix representing the metric tensor is more complicated and generally not diagonal, so you have to diagonalise it to calculate the signature. The Wikipedia article goes into this in more detail.
The reason we're interested in the signature is that we expect spacetime to have one timelike co-ordinate and three spacelike co-ordinates, so we expect the signature to be always (1,3).
A: Signature is defined as the sign of eigenvalues, which means you must first diagonalize the metric. Afterwards flipping the signature then just means multiplying the metric in eigenbasis by $-1$. If $+g_{\mu\nu}$ has the set of eigenvalues as $(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})$ then $-g_{\mu\nu}$ will have $(-\lambda_{1},-\lambda_{2},-\lambda_{3},-\lambda_{4})$ as eigenvalues. It may seem like this could have some unprecedented effect on the underlying physics but it doesn't. This happens because of the way Einstein Field Equation is defined.
$$R_{\mu\nu} -\frac{1}{2}g_{\mu\nu}R = 8\pi G T_{\mu\nu}$$
Both $\pm g_{\mu\nu}$ satisfy the einstein field equation ($G_{\mu\nu} = 8\pi G T_{\mu\nu}$). So, whichever choice you make has no effect on the underlying physics and the covariant equations derived from them.  This happens because with the choice of positive sign in front of metric i.e. $+g_{\mu\nu}$ we have $+R_{\mu\nu}$ and $+T_{\mu\nu}$, whereas $R$ stays the invarient. Under the transformation $+g_{\mu\nu} \rightarrow -g_{\mu\nu}$, we have $+R_{\mu\nu} \rightarrow -R_{\mu\nu}$ and $+T_{\mu\nu} \rightarrow -T_{\mu\nu}$:
$$-R_{\mu\nu} +\frac{1}{2}g_{\mu\nu}R = -8\pi G T_{\mu\nu}$$
or
$$R_{\mu\nu} -\frac{1}{2}g_{\mu\nu}R = 8\pi G T_{\mu\nu}$$
I hope this clears up your doubt.
