Four-vectors and metric tensor I think it's safe to say that if $x^\mu=(x^0,x^1,x^2,x^3)$, then $x_\mu=(x^0,-x^1,-x^2,-x^3)$. But I don't really understand why one follows from the other. Could someone explain?
Also, I've been churning my head around the difference between $\eta_{\mu\nu}$ and and $\eta^{\mu\nu}$. What's the difference anyway?
I understand that $\eta_{\mu\nu}$ stands for the generic matrix element of the metric tensor $\eta = \left(\begin{array}{cccc}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right)$. That I understand. But I don't understand the utility of the upstairs notation.
 A: The first follows from simple matrix multiplication, $x_\mu = \eta_{\mu\nu}x^\nu$.
As for the difference between $\eta^{\mu\nu}$ and $\eta_{\mu\nu}$, you should remember that one of these two objects is defined as the inverse of the other. Hence there is no reason to suspect that in general $\eta^{\mathfrak {ab}} = \eta_{\mathfrak{ab}}$. Certainly this relation is false for ordinary matrices, $A_{\mathfrak i \mathfrak j} \neq A^{-1}_{\mathfrak{ij}}$. Here indices in fraktur style refer to actual components with respect to some coordinates while upright indices just indicate the type of a tensor.
In general the components $\eta^{\mathfrak{ab}}$ are functions on spacetime! However in special relativity where the metric is flat, it is possible to find coordinate systems such that they are all constant and the metric is also diagonal with entries $\pm 1$ - these are the coordinate systems associated with intertial frames. (It is possible to pass from inertial frames to intertial coordinates if and only if the metric is flat.)  In these coordinate systems indeed $\eta^{\mathfrak {ab}} = \eta_{\mathfrak{ab}}$, but this is only true in very special coordinate systems (and false in general relativity except for the special case of, well, special relativity) and in general $\eta^{\mu\nu}$ and $\eta_{\mu\nu}$ must be distinguished -- they are different objects.
