The discussion centers on the contraction of the stress-energy tensor with the metric tensor, specifically addressing whether partial contractions yield invariants. It is established that contracting only part of the stress-energy tensor, such as ##g_{a1}T^{a1}##, does not produce an invariant. The full tensor must be contracted for valid tensor operations, and the contraction of the stress-energy tensor with the same basis vector represents pressure across a surface. Additionally, the trace of the stress-energy tensor, represented as ##T^a{}_a## or ##g_{ab}T^{ab}##, is highlighted as a key invariant.
PREREQUISITESPhysicists, mathematicians, and students of general relativity who seek to deepen their understanding of tensor operations and the properties of the stress-energy tensor.
It has to be the full tensor in order to contract?Dale said:No. That is not a valid tensor operation.
Thank you that's what was thinking.Ibix said: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.
Which by itself would be just the 1,1 component corresponding to pressure?Ibix said: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.
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.dsaun777 said:Which by itself would be just the 1,1 component corresponding to pressure?
A coordinate independent statement would have to be the full contracted tensor ie. The trace?Ibix said: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.
Ibix said:I don't know about other contractions. ##T^a{}_a## and ##g_{ab}T^{ab}## are the only ones I can think of immediately.
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 said:Those are the same thing, they're both the trace. The other obvious one is ##T^{ab} T_{ab}##.
Ibix said:Do you know if your other one is useful for anything?
Why don't you transform the components and check explicitly? :)dsaun777 said: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?
It gets tricky when you change the type of matter you wanthaushofer said:Why don't you transform the components and check explicitly? :)
dsaun777 said:It gets tricky when you change the type of matter you want
Too trickyPeterDonis said: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 said:Too tricky
PeterDonis said:I will be happy to re-label this one as "B" level
PeterDonis said: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