# What is the coordinate free stress-energy-momentum tensor

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• brombo
In summary, the stress-energy-momentum tensor is a linear map from a 4-vector to a 4-vector that gives the energy/momentum contained in a unit volume. It can be interpreted as the density of energy-momentum seen by an observer.f

#### brombo

Without regard to a coordinate system (I only wish to consider special relativity) the stress-energy-momentum tensor defines a linear transformation from a 4-vector to a 4-vector. Let T be the linear transformation then b = T(a), a and b are 4-vectors. What is the physical meaning of a and b or for a and b arbitrary vectors what is the physical meaning of a⋅T(b).

It depends on your choice of four-vector, I'd say. Take for example the energy-momentum tensor of a dust and contract it with the four-velocity. What do you get?

• vanhees71
I am not sure what you mean but if you look for example at the Einstein-Hilbert Wikipedia page you can get a definition for the stress-energy tensor in terms of the variation of the lagrangian matter density w.r.t your metric.

For a Lagrangian, ##\mathcal{L}##, in which the ##\phi_{i}## are vector fields the stress-energy-momentum tensor is give by the following coordinate free geometric algebra expression -
$$T(n) = \dot{\nabla}\left < \dot{\phi}_{i}\partial_{\nabla\phi_{i}}\mathcal{L}n\right >-n\mathcal{L}$$
where ##\nabla## is the 4-vector gradient, ##n## a 4-vector, ##\left < A \right >## the scalar part of the multivector ##A##, and the overdot indicating that the partial derivatives of ##\nabla## only operate on ##\phi_{i}##. The question is what is the physical meaning of ##T(n)##?

Without regard to a coordinate system (I only wish to consider special relativity) the stress-energy-momentum tensor defines a linear transformation from a 4-vector to a 4-vector. Let T be the linear transformation then b = T(a), a and b are 4-vectors. What is the physical meaning of a and b or for a and b arbitrary vectors what is the physical meaning of a⋅T(b).

If you have MTW's "Gravitation", look at page 131.

If you let u be the 4-velocity of some observer, then T(u), where T is the stress-energy tensor regarded as a linear map in the manner you suggest, can be regarded as the density of energy-momentum seen by that observer - i.e the amount of energy/momentum contained in a unit volume.

MTW suggests other useful physical interpretations for the stress energy tensor. The other definitions don't use the notion of the tensor as a linear map from a vector to a vector, ##T^a{}_b## as you ask, but rather regard it as a linear map from two vectors to a scalar ##T_{ab}##. It's a bit of a digression to give them all (as well as being more work), so I won't give these interpretations here, unless you are curious and ask.

For a Lagrangian, ##\mathcal{L}##, in which the ##\phi_{i}## are vector fields the stress-energy-momentum tensor is give by the following coordinate free geometric algebra expression -
$$T(n) = \dot{\nabla}\left < \dot{\phi}_{i}\partial_{\nabla\phi_{i}}\mathcal{L}n\right >-n\mathcal{L}$$
where ##\nabla## is the 4-vector gradient, ##n## a 4-vector, ##\left < A \right >## the scalar part of the multivector ##A##, and the overdot indicating that the partial derivatives of ##\nabla## only operate on ##\phi_{i}##. The question is what is the physical meaning of ##T(n)##?
Wow, talking about disgusting notation :D

Instead of giving fancy-pancy coordinate free definition, let's start with the stresstensor of a dust. It is given by

$$T_{ab} = \rho u_a u_b$$

where rho is the energy density and u is the 4-velocity of the dust-particles. If you contract this with the four velocities twice, you get

$$T_{ab} u^a u^b = \rho$$

That's a start of an interpretation, right?

• vanhees71
Thank you pervect. Your answer referencing MTW was exactly what I was looking for.