High School Tensor Contraction with Metric for Stress Energy: Explained

[dsaun777]

Can you contract any part of the stress energy tensor with the metric? Say if you had four components Tu1 and contracted that with g^u1 would that produce an invariant?

 

[Dale]

No. That is not a valid tensor operation.

 

[dsaun777]

No. That is not a valid tensor operation.

It has to be the full tensor in order to contract?

 

[Ibix]

You mean, what is $g_{a1}T^{a1}$, where summation over $a$ is implied? It's the 1,1 component of $g_{ab}T^{ac}=T_b{}^c$, and components of tensors are not invariants.

 

[dsaun777]

You mean, what is $g_{a1}T^{a1}$, where summation over $a$ is implied? It's the 1,1 component of $g_{ab}T^{ac}$, and components of tensors are not invariants.

Thank you that's what was thinking.

 

[dsaun777]

You mean, what is $g_{a1}T^{a1}$, where summation over $a$ is implied? It's the 1,1 component of $g_{ab}T^{ac}=T_b{}^c$, and components of tensors are not invariants.

Which by itself would be just the 1,1 component corresponding to pressure?

 

[Ibix]

Which by itself would be just the 1,1 component corresponding to pressure?

Not strictly. The contraction of your stress-energy tensor twice with the same basis vector is the pressure across a surface perpendicular to that basis vector.

If you are using an orthonormal basis then that contraction is algebraically equal to the 1,1 component of the tensor. But that's not a coordinate-independent statement and it's sloppy to say that "such and such a component is such and such an observable". It's acceptable-but-sloppy if you specify an orthonormal basis, and technically wrong if you don't.

 

[dsaun777]

Not strictly. The contraction of your stress-energy tensor twice with the same basis vector is the pressure across a surface perpendicular to that basis vector.

If you are using an orthonormal basis then that contraction is algebraically equal to the 1,1 component of the tensor. But that's not a coordinate-independent statement and it's sloppy to say that "such and such a component is such and such an observable". It's acceptable-but-sloppy if you specify an orthonormal basis, and technically wrong if you don't.

A coordinate independent statement would have to be the full contracted tensor ie. The trace?

 

[Ibix]

No - or rather, the trace isn't the only coordinate-free thing you can construct. Any contraction that has no free indices will do. $u^au^bT_{ab}$ is the energy density measured by an observer with 4-velocity $u$, for example.

If you pick an orthonormal basis that treats that observer as at rest then $u^t=1$ and all other components are zero. Then the contraction above is numerically equal to $T_{tt}$. But that's a bit like saying (in Newtonian physics) that if my velocity is 1 m/s then my momentum is equal to my mass. Numerically it's fine, but it's very sloppy - and isn't even numerically fine if I switch to a unit system where 1 m/s isn't 1 unit.

 

[dsaun777]

Do you know of any sources that describe the traces and invariants of the stress energy

 

[Ibix]

Note that $G^{ab}=8\pi T^{ab}$, so discussion of the Einstein tensor applies to the stress-energy tensor too.

MTW covers the contractions of the stress-energy tensor with various combinations of basis vectors/one-forms. I don't know about other contractions. $T^a{}_a$ and $g_{ab}T^{ab}$ are the only ones I can think of immediately. The trace is, of course, simply related to the Ricci scalar, but I don't know what you might use the other for.

 

[PeterDonis]

I don't know about other contractions. $T^a{}_a$ and $g_{ab}T^{ab}$ are the only ones I can think of immediately.

Those are the same thing, they're both the trace. The other obvious one is $T^{ab} T_{ab}$.

 

[Ibix]

Those are the same thing, they're both the trace. The other obvious one is $T^{ab} T_{ab}$.

D'oh! Of course they are - I'm just lowering an index ($g_{ab}T^{ac}=T_b{}^c$) then contracting the free indices on that. Do you know if your other one is useful for anything? Edit: it kind of looks like a "modulus-squared" of the tensor, if that makes sense.

 

[PeterDonis]

Do you know if your other one is useful for anything?

I have not seen it used for anything in the case of the stress-energy tensor, but in the case of the EM field tensor $F_{ab}$, the corresponding invariant $F^{ab} F_{ab}$ is used often.

 

[robphy]

Higher powers of tensors may be useful for calculating "principal invariants", associated with eigenvalues and the Cayley-Hamilton Theorem
https://en.wikipedia.org/wiki/Invariants_of_tensors
https://www.physicsforums.com/threads/em-tensor-invariants.676181/post-4295385

 

[haushofer]

Can you contract any part of the stress energy tensor with the metric? Say if you had four components Tu1 and contracted that with g^u1 would that produce an invariant?

Why don't you transform the components and check explicitly? :)

 

[dsaun777]

Why don't you transform the components and check explicitly? :)

It gets tricky when you change the type of matter you want

 

[PeterDonis]

It gets tricky when you change the type of matter you want

The type of matter doesn't matter. You can easily demonstrate using the general tensor transformation laws that, for example, $g_{a1} T^{a1}$ is not an invariant, regardless of the specific forms of $g$ or $T$.

If you are unable to do this, then I would strongly suggest that you spend some time learning tensor algebra and developing some facility with standard tensor operations. Sean Carroll's online lecture notes on GR have a good introductory treatment of this in the early chapters.

 

[dsaun777]

The type of matter doesn't matter. You can easily demonstrate using the general tensor transformation laws that, for example, $g_{a1} T^{a1}$ is not an invariant, regardless of the specific forms of $g$ or $T$.

If you are unable to do this, then I would strongly suggest that you spend some time learning tensor algebra and developing some facility with standard tensor operations. Sean Carroll's online lecture notes on GR have a good introductory treatment of this in the early chapters.

Too tricky

 

[PeterDonis]

Too tricky

You labeled this thread as "I" level. What I suggested to you is well within "I" level. If you really find that too tricky, then you should (a) label your threads as "B" level (and I will be happy to re-label this one as "B" level if you wish); (b) be prepared to be told that your questions cannot be answered, at least not with anything more than a brief "yes" or "no", at your level of understanding; and (c) be prepared to have a lot of your threads closed very quickly because there is no point in discussion if you can't make use of it.

 

[PeterDonis]

I will be happy to re-label this one as "B" level

And in fact I have just done so.

be prepared to have a lot of your threads closed very quickly because there is no point in discussion if you can't make use of it

And I have just done that to this thread as well.