View Full Version : What is the trace of the second-Rank tensor?
In Lorentz group in QFT, why the trace of the symmetric second-Rank tensor S^{\mu\nu} is defined as follows?
S=g_{\mu\nu}S^{\mu\nu}.
Is it just a definition or the genuine trace of the second-Rank tensor, and why?
What's the trace of a matrix?
my opinion,The trace of S^{
\mu\nu} should be S^{
00}+S^{11}+S^{22}+S^{33}
, i.e. the sum over the diagonal elements.
Now the metric tensor g_{
\mu\nu} is taken into account, the trace becomes
S^{
00}-S^{11}-S^{22}-S^{33}.
The problem is that I don't know why
genneth
Feb23-09, 08:30 AM
This seems to be a common point of misunderstanding:
Matrices properly correspond to tensors with one index up and one index down.
The fact that books often just write the metric with all up or all down as a matrix is just confusing.
The basic definition of the trace of a matrix is simply the sum of its diagonal elements.
However, in order to make the trace invariant under a generalized rotation, the metric is included.
Thank Cristo, genneth, and clem.
You are correct, it is described in general relativity.
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