Is the Momentum-Energy Tensor Always Symmetric?

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Discussion Overview

The discussion centers on the symmetry properties of the momentum-energy tensor (or stress-energy tensor) in the context of scalar fields and tensor algebra. Participants explore the conditions under which tensors are symmetric or antisymmetric, and the implications of these properties in theoretical physics.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions how the momentum-energy tensor is "clearly" symmetric based on its definition involving derivatives and the metric.
  • Another participant asserts that the tensor is symmetric because the metric is symmetric and the product of derivatives is also symmetric.
  • A participant discusses the necessity of symmetrizing and antisymmetrizing tensors, indicating that it helps identify which components remain when interacting with other tensors.
  • Further elaboration is provided on how any generic tensor can be expressed as a sum of its symmetric and antisymmetric parts, and the implications of this for products of tensors.
  • It is noted that the product of a symmetric tensor with an antisymmetric tensor results in zero, which has implications for tensor operations.

Areas of Agreement / Disagreement

Participants generally agree on the definitions and properties of symmetric and antisymmetric tensors, but there is ongoing exploration regarding the implications and applications of these properties in specific contexts.

Contextual Notes

The discussion includes assumptions about the nature of tensors and their products, which may not be universally applicable without further context or definitions. The exploration of when to symmetrize or antisymmetrize tensors is not fully resolved and may depend on specific scenarios.

physlad
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I was reading about the momentum-energy tensor (or stress-energy tensor), at one point the author says,
"
\theta^{\mu\nu} = (\partial^\mu\phi)(\partial^\nu\phi) - g^{\mu\nu}L

This is clearly symmetric in \mu and \nu."

\theta^{\mu\nu}: is the stress-energy tensor
\phi is a scalar field
g^{\mu\nu} is the metric (+---)
L is the lagrangian density

My question (I'm not an expert in tensors) is how do you see that it's "clearly" symmetric? another silly question: when do we need to symmetrize and antisymmetrize tensors?

Please, tell me guys if this isn't the right place for my question.
 
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It's symmetric because:

- g is symmetric since it's the metric, thus the second term is symmetric.
- The first term is the product of two derivatives, so is symmetric (i.e. \partial^0\phi \partial^1\phi \equiv \partial^1\phi\partial^0\phi, and similarly for other values of mu and nu.)
 
physlad said:
... when do we need to symmetrize and antisymmetrize tensors?...

Sometimes one has a product of a symmetric tensor with another tensor which is not symmetric nor antisymmetric, then one can show that the antisymmetric part of the second is killed by the first, the same thing occurs for the antisymmetric case, this is why we need to antisymmetrise and symmetrise tensors: to see which part remains and which part is killed...
 
Suppose to have a generic tensor T, which isn't symmetric or antisymmetric.
You can ALWAYS write it as sum of its symmetric part with its antisymmetric part.

e.g. for a rank-2 tensor

T^{\mu\nu}=\frac{1}{2}(T^{\mu\nu}+T^{\nu\mu}) + \frac{1}{2}(T^{\mu\nu}-T^{\nu\mu})

On the right-hand side, the first term \left(\bullet+\bullet\right) is a symmetric tensor, while the second \left(\bullet-\bullet\right) is antisymmetric!


Hence, in general: T=T_S+T_A.


Now...if you have a product between tensors, and you know that one of them is explicitly symmetric S or antisymmetric A...you can write:

ST=ST_S+ST_A

AT=AT_S+AT_A

Of course the product between a symmetric and an antisymmetric tensor is zero, i.e. SA=AS=0!

And thus

ST=ST_S+ST_A=ST_S

AT=AT_S+AT_A=AT_A.
 

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