Contraction with the metric vanishes

  • #1
897
53

Main Question or Discussion Point

This question is probably silly, but suppose I have a contraction of the form ##g_{\mu \nu} C^{\mu \nu} = 0## where ##C^{\mu \nu}## is a tensor* and ##g_{\mu \nu}## is the metric tensor. Can I say that it must vanish for any ##g_{\mu \nu}##, and since in the most general case all ##g_{\mu \nu}## are non zero, then necessarely ##C^{\mu \nu} = 0##?

*##C^{\mu \nu}## is a symmetric tensor.
 
Last edited:

Answers and Replies

  • #2
Ibix
Science Advisor
Insights Author
6,422
5,072
Since ##g_{\mu\nu}## is symmetric, ##C^{\mu\nu}## being antisymmetric is enough, I think.
 
  • Like
Likes kent davidge
  • #3
897
53
Since ##g_{\mu\nu}## is symmetric, ##C^{\mu\nu}## being antisymmetric is enough, I think.
Yes. However it turns out that my ##C^{\mu \nu}## is also symmetric (unfortunately!)
 
  • #4
DrGreg
Science Advisor
Gold Member
2,281
856
This question is probably silly, but suppose I have a contraction of the form ##g_{\mu \nu} C^{\mu \nu} = 0## where ##C^{\mu \nu}## is a tensor* and ##g_{\mu \nu}## is the metric tensor. Can I say that it must vanish for any ##g_{\mu \nu}##, and since in the most general case all ##g_{\mu \nu}## are non zero, then necessarely ##C^{\mu \nu} = 0##?

*##C^{\mu \nu}## is a symmetric tensor.
No. Counterexample: ##C^{\mu \nu} = K^\mu K^\nu ## where ##\mathbf{K}## is any (non-zero) null vector.
 
  • Like
Likes Ibix and kent davidge
  • #5
haushofer
Science Advisor
Insights Author
2,302
673
This question is probably silly, but suppose I have a contraction of the form ##g_{\mu \nu} C^{\mu \nu} = 0## where ##C^{\mu \nu}## is a tensor* and ##g_{\mu \nu}## is the metric tensor. Can I say that it must vanish for any ##g_{\mu \nu}##, and since in the most general case all ##g_{\mu \nu}## are non zero, then necessarely ##C^{\mu \nu} = 0##?

*##C^{\mu \nu}## is a symmetric tensor.
From a Linear Algebra point of view, you take the trace. A traceless matrix is not necessarily the zero matrix.
Also, do the counting: ##g_{\mu \nu} C^{\mu \nu} = 0## puts one constraint on your tensor ##C^{\mu \nu}##, while in D dimensions a general tensor ##C^{\mu \nu}## has ##D^2## components. The condition that ##C^{\mu \nu} = C^{[\mu \nu]}##, i.e. ##C^{\mu \nu}## is antisymmetric, means ##C^{(\mu \nu)} = 0##, which are ##\frac{1}{2}D(D+1)## constraints. So ##g_{\mu \nu} C^{\mu \nu} = 0## cannot imply that ##C^{(\mu \nu)} = 0##. Of course, ##C^{(\mu \nu)} = 0## does imply that ##g_{\mu \nu} C^{\mu \nu} = g_{\mu \nu} C^{(\mu \nu)} = 0##; you get a linear combination of zeroes.

So, to answer your question: no, most definitely not.
 
  • Like
Likes kent davidge

Related Threads on Contraction with the metric vanishes

Replies
4
Views
3K
Replies
4
Views
8K
Replies
3
Views
2K
Replies
6
Views
638
Replies
7
Views
1K
Replies
2
Views
650
  • Last Post
Replies
7
Views
2K
Replies
13
Views
3K
Top