Vanishing of Contraction with Metric Tensor

  • Context: Undergrad 
  • Thread starter Thread starter kent davidge
  • Start date Start date
  • Tags Tags
    Contraction Metric
Click For Summary

Discussion Overview

The discussion revolves around the contraction of a symmetric tensor ##C^{\mu \nu}## with the metric tensor ##g_{\mu \nu}##, specifically examining the implications of the equation ##g_{\mu \nu} C^{\mu \nu} = 0##. Participants explore whether this condition necessitates that ##C^{\mu \nu}## must be zero, considering various mathematical perspectives and examples.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant suggests that if ##g_{\mu \nu} C^{\mu \nu} = 0## for all non-zero ##g_{\mu \nu}##, then it follows that ##C^{\mu \nu} = 0##.
  • Another participant argues that since ##g_{\mu\nu}## is symmetric, an antisymmetric ##C^{\mu\nu}## would suffice to satisfy the equation, implying that ##C^{\mu \nu}## could be non-zero.
  • A later reply points out that if ##C^{\mu \nu}## is symmetric, the condition cannot imply that ##C^{\mu \nu} = 0##, providing a counterexample with a non-zero null vector.
  • Another participant elaborates on the linear algebra perspective, noting that a traceless matrix does not necessarily have to be the zero matrix and discusses the constraints imposed by the conditions on the tensor components.

Areas of Agreement / Disagreement

Participants express disagreement regarding whether the condition ##g_{\mu \nu} C^{\mu \nu} = 0## necessitates that ##C^{\mu \nu} = 0##. Multiple competing views remain, with some arguing for the necessity of ##C^{\mu \nu}## being zero and others providing counterexamples and alternative interpretations.

Contextual Notes

Participants highlight the complexity of the relationship between the symmetry of the tensors involved and the implications of the contraction, indicating that the discussion is nuanced and dependent on the definitions and properties of the tensors.

kent davidge
Messages
931
Reaction score
56
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:
Physics news on Phys.org
Since ##g_{\mu\nu}## is symmetric, ##C^{\mu\nu}## being antisymmetric is enough, I think.
 
  • Like
Likes   Reactions: kent davidge
Ibix said:
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!)
 
kent davidge said:
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   Reactions: Ibix and kent davidge
kent davidge said:
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   Reactions: kent davidge

Similar threads

  • · Replies 9 ·
Replies
9
Views
717
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 9 ·
Replies
9
Views
1K