Confusion about $\partial_\mu x^\mu = 4$ Why is it that $\partial_\mu x^\mu = 4$? I thought that $\partial_\mu x^\mu$ could be expanded as
$$\partial_\mu x^\mu = -\partial_1x^1 + \partial_2x^2 + \partial_3x^3 + \partial_4x^4  \\
=-1+1+1+1\\
=2$$
However this seems to not be the case because we get 2 instead of 4 . What am I doing wrong here? Is it wrong to interpret $\partial_\mu x^\mu$ as being contracted with the metric?
 A: If you write mixed indices in your summation, no minus sign enters:
$\partial_\mu x^\mu= \frac{\partial}{\partial x^\mu}x^\mu =\delta^\mu_\mu=4$
However, if you have just upper indices, it looks differently:
$\sum_\mu \partial^\mu x^\mu = \sum_\mu \eta^{\mu\nu}\partial_\nu x^\mu = \sum_\mu \eta^{\mu\nu}\delta^\mu_\nu=\sum_\mu \eta^{\mu\mu}$
which equals 2 for signature $(-+++)$ or -2 for signature $(+---)$. Note that in this second expression the sum over $\mu$ is not an Einstein sum thus requiring an explicit summation sign. It is an "unnatural" expression while the one with mixed indices is "natural" and gives the dimension of our spacetime, 4.
The derivative with respect to $x^\mu$ is $\partial_\mu$ (with lower index!), so in the mixed index expression no metric arises. While in the upper index expression we have a $\partial^\mu$ which first needs to be expressed in terms of $\partial_\mu$ which introduces the metric tensor.
A: Note that $ \partial_{\mu} x^{\nu} = \frac{\partial x^{\nu}}{\partial x^{\mu}} = \delta^{\nu}_{\mu}$ and hence
$$\partial_{\mu} x^{\mu} = \frac{\partial x^{\mu}}{\partial x^{\mu}}  =\frac{\partial x^0}{\partial x^0} + \frac{\partial x^1}{\partial x^1} +\frac{\partial x^2}{\partial x^2} + \frac{\partial x^3}{\partial x^3}  = 4 \quad .$$
A: The mistake is that:
$A_{\mu} B^\mu = A_0 B^0+ A_1 B^1+... \neq -A_0 B^0+ A_1 B^1$.
You might be confused, because: $A \cdot B = \eta_{\mu \nu}A^{\mu} B^\nu = -A^0 B^0+ A^1 B^1+... $.
A: The minus sign in your equation is wrong. If I substitute $\mu=1$ there is no minus sign.
A: Write is in the Minkowsky metric where there the forths components are imaginary: $\partial (\text{i}x_4)/\partial(\text{i}x_4)=1$.
