This may be a dumb question, but I am relatively new to Tensor Calculus and I was wondering what this partial derivative of the metric tensor would turn out to be:
$\frac {\partial g_{µv}}{\partial x^µ} $
where $g_{µv}$ is the metric tensor
I have not seen any form of the metric tensor like $g_µ$, so would this expression exist or is there an identity or property that defines what this expression is?
Consider $\mu$ to be an index that usually runs from $0$ to $3$. In particular, in general relativity, the four-vector $x^{\mu}$ denotes the row vector with components:
$$x^{\mu} = (x^0, x^1, x^2, x^3) = (ct, x, y, z)$$
Where $c$ is the speed of light, usually taken in Planck units to $c = 1$.
Hence this is a vector.
On the other side, the metric tensor $g_{\mu\nu}$ is a "matrix" (or better: it can be represented as a matrix) which turns out to be a $4\times 4$ matrix, since the $\nu$ index runs itself from $0$ to $3$.
In order to understand what is that derivative, let's take a simple trivial and probably non existent (in terms of physics) metric tensor of the form
$$ g_{\mu\nu} = \begin{pmatrix} 0 & 0 & x^2 & 0 \\ 0 & xy & xy & 0 \\ x^2 & xy & xy & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$
Now when we want to compute a derivative $\frac{\partial g_{\mu\nu}}{\partial x^{\mu}}$, what we really want is the derivative of the $\mu\nu$ component of the tensor with respect to the $\mu$ index.
When using Einstein convention, when an index is repeated, then a hidden sum over all the indices exists. In this way the derivative you wrote represents
$$ \frac{\partial g_{\mu\nu}}{\partial x^{\mu}} = \frac{\partial g_{0\nu}}{\partial x^{0}} + \frac{\partial g_{1\nu}}{\partial x^{1}}+ \frac{\partial g_{2\nu}}{\partial x^{2}} + \frac{\partial g_{3\nu}}{\partial x^{3}}$$
But this also means
$$ \frac{\partial g_{\mu\nu}}{\partial x^{\mu}} = \frac{\partial g_{0\nu}}{\partial t} + \frac{\partial g_{1\nu}}{\partial x}+ \frac{\partial g_{2\nu}}{\partial y} + \frac{\partial g_{3\nu}}{\partial z}$$
Now you see that we have a metric tensor whose components (16 in total) are declared by the index $\nu$. Since the metric tensor is ALWAYS symmetric, what are the nonzero components here? They are:
$$g_{03},\ g_{11},\ g_{12},\ g_{20},\ g_{21},\ g_{22},\ g_{33}$$
p.s. $\mu$ stands for the ROW and $\nu$ stands for the COLUMN, hence the element $g_{02}$ is, for example, the $x^2$ in the first row.
According to what component we are serving, we can take the derivative.
For example there is no term in the tensor which depends upon both $t$ and $z$ hence those derivatives will be identically zero.
The only derivatives that exist are the ones which involve the above written nonzero elements of the tensor, with respect to $x$ and $y$.
Let take the above representation again:
$$ \frac{\partial g_{\mu\nu}}{\partial x^{\mu}} = \frac{\partial g_{0\nu}}{\partial t} + \frac{\partial g_{1\nu}}{\partial x}+ \frac{\partial g_{2\nu}}{\partial y} + \frac{\partial g_{3\nu}}{\partial z}$$
As we noticed, $z$ and $t$ derivatives are zero, hence:
$$ \frac{\partial g_{\mu\nu}}{\partial x^{\mu}} = \frac{\partial g_{1\nu}}{\partial x}+ \frac{\partial g_{2\nu}}{\partial y} $$
Now what remains? The component $g_{1\nu}$ is nonzero only for $\nu = 1$ and $2$. The components are identical and their derivative is (with respect to $x$) just $y$ as you can see.
On the other side, when we deal with $g_{2\nu}$ the only nonzero components happen for $\nu = 0, 1, 2$ but $g_{20}$ does not depend on $y$ hence the derivative will be zero.
And so on.
I hope you more or less got the point!
$g_{\mu}$
This expression would represent a vector with $4$ components, in accordance with what I wrote above.
The derivative in this case would be just the derivative of the $\mu$ component of $g$, with respect to the $\mu$ component of the coordinates vector. I never saw such a notation, and it would be uncomfortable but in layman's terms it would be
$$\frac{\partial g_{\mu}}{\partial x^{\mu}} = \frac{\partial g_{0}}{\partial t} + \frac{\partial g_{1}}{\partial x}+ \frac{\partial g_{2}}{\partial y} + \frac{\partial g_{3}}{\partial z}$$
This would however be quite incorrect to write since it's better to use different index, like for example
$$\frac{\partial g_{\mu}}{\partial x^{\nu}}$$