Tensor trace

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 6; Problem 6.3.

The trace of a rank-2 tensor is given by the contraction {F_{\;\; i}^{i}}. In matrix terminology, it is the sum of the diagonal elements. If we start with a contravariant tensor {F^{ij}}, then we can calculate the trace as follows:

\displaystyle   F_{\;\; j}^{i} \displaystyle  = \displaystyle  g_{jk}F^{ik}\ \ \ \ \ (1)
\displaystyle  F_{\;\; i}^{i} \displaystyle  = \displaystyle  g_{ik}F^{ik} \ \ \ \ \ (2)

That is, we first lower the second index, then contract the top and bottom indices.

Since the trace contains no free index, it should be a scalar, which means it should be invariant. We can prove this by doing the transformation.

\displaystyle   g_{ik}^{\prime}F^{\prime ik} \displaystyle  = \displaystyle  \frac{\partial x^{l}}{\partial x^{\prime i}}\frac{\partial x^{m}}{\partial x^{\prime k}}g_{lm}\frac{\partial x^{\prime i}}{\partial x^{a}}\frac{\partial x^{\prime k}}{\partial x^{b}}F^{ab}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{\partial x^{l}}{\partial x^{\prime i}}\frac{\partial x^{\prime i}}{\partial x^{a}}\right)\left(\frac{\partial x^{m}}{\partial x^{\prime k}}\frac{\partial x^{\prime k}}{\partial x^{b}}\right)g_{lm}F^{ab}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \delta_{\;\; a}^{l}\delta_{\;\; b}^{m}g_{lm}F^{ab}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  g_{ab}F^{ab} \ \ \ \ \ (6)

4 thoughts on “Tensor trace

  1. Pingback: Metric tensor: trace | Physics tutorials

  2. Pingback: Tensor product: numerical example | Physics tutorials

  3. Pingback: Tensors: symmetric and anti-symmetric | Physics tutorials

Leave a Reply

Your email address will not be published. Required fields are marked *