- #1

- 13

- 0

- A
- Thread starter brombo
- Start date

- #1

- 13

- 0

- #2

haushofer

Science Advisor

- 2,471

- 857

- #3

- 55

- 13

- #4

- 13

- 0

$$T(n) = \dot{\nabla}\left < \dot{\phi}_{i}\partial_{\nabla\phi_{i}}\mathcal{L}n\right >-n\mathcal{L}$$

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)##?

- #5

- 9,953

- 1,134

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.

- #6

haushofer

Science Advisor

- 2,471

- 857

Wow, talking about disgusting notation :D

$$T(n) = \dot{\nabla}\left < \dot{\phi}_{i}\partial_{\nabla\phi_{i}}\mathcal{L}n\right >-n\mathcal{L}$$

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)##?

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

[tex]

T_{ab} = \rho u_a u_b

[/tex]

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

[tex]

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

[/tex]

That's a start of an interpretation, right?

- #7

- 13

- 0

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

- Last Post

- Replies
- 7

- Views
- 3K

- Replies
- 6

- Views
- 5K

- Replies
- 16

- Views
- 8K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 902

- Last Post

- Replies
- 9

- Views
- 4K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 654

- Replies
- 4

- Views
- 2K