# Is my understanding of the second-rank four-tensor correct?

1. Jan 28, 2016

### jack476

I just started reading Landau and Lifschitz' Classical Theory of Fields today and I reached the section on the spacetime four-tensor.

The explanation given was that the four-tensor is a 4x4 matrix Aik with the property that for some transformation αik, elements of Aik are related to elements of the tensor A'ml in the transformed coordinates by the equation Aik = αimαklA'ml.

Am I correct in my understanding that this means that element aik of Aik is related to element a'ml of Aml, for some i, k, m, l, by the product αimαkl?

So if I have, for instance, a'23 and I want to find a14, then i = 1, k = 4, m = 2, and l = 3, so I multiply a'23 by the product of the elements α12 and α43 of the transformation matrix α? Or am I misunderstanding that?

2. Jan 28, 2016

### andrewkirk

Not quite. Using your notation, the basis-conversion equation is

$$a_{ik}=a'_{ml}\alpha_i^m\alpha_k^l$$

This looks like it's a product, but it's actually a sum of products, because under the Einstein Summation Convention we sum over matching subscript-superscript pairs. So it actually means:

$$a_{ik}=\sum_{m=1}^n\sum_{l=1}^n a'_{ml}\alpha_i^m\alpha_k^l$$

That's why I changed the $m$ and $l$ from subscripts to superscripts, by the way.

Also, the tensor is not a matrix. The matrix is a representation of the tensor in a given basis. This might seem like a quibble, but it will help a great deal with understanding as your reading goes on. One of the most important properties of tensors is that they are not dependent on the choice of basis of the underlying vector space.

3. Jan 29, 2016

### Brian T

As said before, the tensor is not a matrix but rather a geometric object (for ex, a vector is not the same as its components). One way to think of tensors, without defining the coordinate change rules, is as a multilinear map from elements of a vector space to the reals.