I want to understand what the object $\partial_\mu x^\nu$ is, but let's be a bit more general:

Suppose $(x^0,\cdots,x^3)$ and $(y^0,\cdots,y^3)$ are 4-vectors. How should I interpret the object $x^\nu y_\mu$?

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    $\begingroup$ It is a mixed rank two tensor quantity. The adjective mixed means one index is covariant while the other is contravariant. $\endgroup$ – Lewis Miller Aug 25 '16 at 15:40
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    $\begingroup$ To add to @LewisMiller's correct comment/answer: $x^\nu y_\mu$ means the tensor product of the two vectors (tensors of rank one). It is a mixed tensor of rank two, which components are all the products of $x^\nu$ and $y_\mu$ $(x^0 y_0,x^0 y_1,...,x^3 y_3)$. The tensor has 16 components and can be displayed as a $4 \otimes4$ matrix. $\endgroup$ – N0va Aug 25 '16 at 18:25
  • $\begingroup$ I think as $x^a\otimes(\sum_{c}[g_{bc}\cdot y^c])$ (no einstein, a, b given) since the indices are different $\endgroup$ – Emil Aug 25 '16 at 18:33
  • $\begingroup$ I think this is a possible definition of the derivative in the X direction at point p: For any $c(t)$ where $c(0)=p$ and $\lim_{\delta t \to 0}\frac{c(0+\delta t \cdot h)-c(0)}{\delta t} = X_p$, $\partial_X f|_p = \lim_{\delta t \to 0}\frac{f(c(0+\delta t\cdot h))-f(c(0))}{\delta t}$ (unless I've dribbled myself away). $\endgroup$ – Emil Aug 25 '16 at 18:54
  • $\begingroup$ (it seems weird to be adding/subtracting points, one probably uses atlases so that the c(t) really maps to a vector in a euclidian space, but these details are still fuzzy to me :( ) $\endgroup$ – Emil Aug 25 '16 at 19:06

Object $x^{\nu}y_{\mu}$ can be thought as two dimensional table of numbers, indexed by $\mu$ and $\nu$. So it is a matrix. Tensor with more indices is also a matrix, but multidimensional one. I bet you can imagine three dimensional array of numbers. I think this is enough to get a feeling how it works when there are even more indices. There is one catch: you cannot blindly assign numerical values to elements of the array, because these depend on choice of basis. But this should be no shock if you are familiar with changing basis with ordinary vectors. Here when changing basis, apply transition matrix to every index separately. If you are familiar with notion of covariant and contravariant vectors then it shouldn't be a shock to find out that tensor indices can also be covariant (as in $y_{\mu}$) or contravariant (as in $x^{\nu}$). When going to a different basis just apply appropariate transition matrix to every index separately, according to the height of the index.

Object $\partial_{\mu}\phi^{\nu}(x)$ is to be interpreted as follows. Once basis is choen, every component of vector $\phi^{\nu}(x)$ is a number. Hence you can clearly differentiate it wih respect to $x^{\mu}$. Therefore if basis is fixed, given $\mu$ and $\nu$ you get a function. There are some subtleties involved if you want to use different basis at every point of space (like it is usually done with curvilinear coordinates). If you want to understand these, I advise that you look at the book "The Geometry of Physics" by Frankel.

Remark relevant if you are into maths: You might think something like "hey, since numbers change when going to different basis you cannot say that tensor is an array of numbers!" You can therefore think of it as an equivalence relation of all pairs (basis, array of numbers), where two such pairs are said to be equivalent if their array of numbers are related by appropariate transition matrix.


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