# Tensors in special relativity [duplicate]

I'm trying to understand tensors, but I've come across the following question:

1. Let $T^{\mu\nu}$ by a $(2,0)$ tensor. Give the definitions of $T_\mu^{\,\nu}$, $T_{\mu\nu}$, and $T^{\mu}_{\,\nu}$. Then give explicit formulas for the following components in terms of components of $T^{\mu\nu}$: $T_{00},\,T_{10},\,T_{12},\,T_0^{\,1},\,T_1^{\,1},\,T_0^{\,0},\,T^1_{\,1}$
$$T_\mu^{\,\nu}\equiv\eta_{\mu\sigma}T^{\sigma\nu} \\ T_{\mu\nu}\equiv\eta_{\sigma\mu}\eta_{\rho\nu}T^{\sigma\rho}\\ T^\mu_{\,\nu}\equiv\eta_{\sigma\nu}T^{\mu\sigma}$$

I'm finding it hard to understand these definitions. I really don't know what these tensors represent or what it means when both the indices are raised? Also are these standard definitions in special relativity or is it more likely that they are just made up by my professor for this assignment? I've always thought that things with two indices are matrices in general.

• possible duplicate of What is a tensor? You might also want to look at my discussion of the introduction of vectors and covectors here, which is a little more specific to relativity than Ron Maimon's general answer in the duplicate link. – ACuriousMind Jul 5 '14 at 1:03

Tensors in abstract mathematics are just functions with linear arguments. In abstract index notation, the placement of indices--up vs. down--tells you whether that particular argument should be a vector or covector. For example:

$$T^{\mu \nu} \equiv T(\text{covector}, \text{covector})$$

Since both the indices are up, it means both the first and second arguments should be covectors. Down indices mean the corresponding arguments should be vectors.

Matrices are usually thought of as representing linear maps from vectors to vectors. That is, if you have a linear map $M$ that takes a vector $v$ and produces a new vector $w$, then you have $M(v) = w$, and you can build a corresponding tensor with a covector $\omega$ like so:

$$\omega[M(v)] = \text{scalar} \equiv M(\omega, v)$$

where $M(\omega,v)$ is a tensor. So there is a direct correspondence between the vector-valued linear map and the tensor; no one would look at you too funny if you said that the matrix is indeed a tensor. Any quibbling over such a statement would merely be over tiny details about what the objects actually are, or if they're actually different.

Where does the metric come in? It allows us to build up a tensor with vector inputs instead of covectors, or covector inputs instead of vectors. Again, since we can think of a matrix as a 2-index tensor instead, we can do the reverse: think of the metric as a map that converts vectors to covectors, and the inverse metric converts covectors to vectors.

So given a map that is $T(\text{covector}, \text{covector})$, denoted $T^{\mu \nu}$, you can apply the inverse metric on one argument like so:

$$T([\text{inverse metric}](\text{covector}), \text{covector}) \equiv T(\text{vector}, \text{covector})$$

We write this more concisely as

$$T^{\sigma \nu} \eta_{\mu \sigma} = {T_\mu}^\nu$$