Interpretation of [itex]\partial_\nu T^{\mu \nu}[/itex]

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SUMMARY

The discussion centers on the interpretation of the equation \partial_\nu T^{\mu \nu} = Q^\mu, specifically in the context of the dust energy-momentum tensor defined as T^{\mu \nu} = \rho U^\mu U^\nu, where U is the four-velocity and \rho is the energy density. Participants explore the algebraic manipulation of this equation to derive the spatial vector Q^i and its physical significance. The conversation emphasizes the continuity equation and the relationship between the components of the energy-momentum tensor and the vector Q, highlighting the need to understand the physical meaning of T^{\mu \nu} for proper interpretation.

PREREQUISITES
  • Understanding of the energy-momentum tensor in general relativity.
  • Familiarity with the concept of four-velocity and its role in relativistic physics.
  • Knowledge of continuity equations in the context of fluid dynamics.
  • Basic algebraic manipulation of tensor equations.
NEXT STEPS
  • Study the physical interpretation of the energy-momentum tensor T^{\mu \nu} in general relativity.
  • Learn about the continuity equation and its applications in fluid dynamics.
  • Investigate the derivation and implications of the spatial components Q^i = \partial_\mu T^{\mu i}.
  • Explore the relationship between stress tensors and energy-momentum tensors in classical mechanics.
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Physicists, particularly those specializing in general relativity and fluid dynamics, as well as students seeking to deepen their understanding of energy-momentum tensors and their physical implications.

mjordan2nd
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I am working on Sean Carroll's problem 1.8.

If \partial_\nu T^{\mu \nu}=Q^\mu, what physically does the spatial vector Q^i represent? Use the dust energy momentum tensor to make your case.

The dust energy momentum tensor is

T^{\mu \nu}= \rho U^\mu U^\nu,

where U is the four-velocity and \rho the energy density in the rest frame.

Trying to manipulate \partial_\nu \rho U^\mu U^\nu directly didn't give me anything I could reasonably interpret. I then tried to look at U_\mu Q^\mu following the example in the textbook to see if I could possibly factor the final expression into giving me an answer I could interpret. What I got was

U_\mu \partial_\nu \rho U^\mu U^\nu=-U^\nu \partial_\nu \rho-\rho \partial_\nu U^\nu=U_\mu Q^\mu.

I can't factor four-velocity out of the second term since it is inside the derivative, so I'm not too sure that I can simplify this into something that I can interpret for Q.

I also tried to consider the part of this equation that was orthogonal to the four velocity. To do this, I multiplied the expression by the following projection vector:

P^\sigma_\nu=\delta^\mu_\nu+U^\sigma U_\nu.

This gives

P^\sigma_\mu Q^\mu = P^\sigma_\mu \partial_\nu (\rho U^\mu U^\nu),

P^\sigma_\mu \partial_\nu (\rho U^\mu U^\nu) =U^\nu \rho \partial_\nu U^\sigma,

P^\sigma_\mu Q^\mu =Q^\sigma + U^\sigma U_\mu Q^\mu,

U^\nu \rho \partial_\nu U^\sigma = Q^\sigma + U^\sigma U_\mu Q^\mu.

From here, if we plug in -U^\nu \partial_\nu \rho-\rho \partial_\nu U^\nu=U_\mu Q^\mu into the final expression above, we get the original expression: \partial_\nu T^{\mu \nu}=Q^\mu. This indicates that I have done all of the algebra correctly, however it doesn't help me interpret what the spatial part of Q is physically. Can anyone help me out on where to go from here?
 
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really how did you end up [in your 2nd equation] with somethin with no free-index when you began with something with 1 free index?
 
ChrisVer said:
really how did you end up [in your 2nd equation] with somethin with no free-index when you began with something with 1 free index?

Not sure what you mean when you say second equation. If it's the one I think you mean, it's because I'm projecting the equation along the four-velocity.
 
But that doesn't allow you to change the number of free indices in an equation.
 
Ahh, I forgot to write a factor up there.
George Jones said:
But that doesn't allow you to change the number of free indices in an equation.

Ahh, I forgot to write a factor up there. Is it better now?
 
Well, have you been introduced to what physically a energy-momentum tensor T^{\mu \nu} stands for?
Like what its components are...[I guess that's a better way to make use of the dust E-M tensor].
If you get what the T stands for, it's straightforward to get a physical interpretation of its derivative [which will give you something like a "continuity equation": a relation of the flow with time and space of the T quantitiy].
In particular you want to look at:
Q^i = \partial_\mu T^{\mu i}
So you'll only need to know what's T^{i0} and T^{ij}.
 
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The problem is that I don't know how to isolate Q in my equations above when trying to look at the dust energy-momentum tensor.
 
mjordan2nd said:
The problem is that I don't know how to isolate Q in my equations above when trying to look at the dust energy-momentum tensor.
what do you mean by isolate?
 
I mean you have:
Q^i = \partial_\mu T^{\mu i}
Q^i = \partial_0 T^{0 i } + \partial_j T^{ji}
which is 3 equations:
Q^1 = \partial_0 T^{0 1} + \partial_j T^{j1}
Q^2 = \partial_0 T^{0 2} + \partial_j T^{j2}
Q^3 = \partial_0 T^{0 3} + \partial_j T^{j3}
I gave you hints when I referred to a continuity equation... Qs are supposed to play the role of sources/sinks in the equation... the only thing you need to interpret then are the Energy-Momentum components [which you can obtain directly from the dust equation you gave]...as an example I tell you that the T^{01}= \rho U^1 U^0 = \gamma^2 \rho u^1 = \rho' u^1 is the flux of energy density along the 1-axis [pretty much proportional to the 1-momentum]. T[ij] can be a little more tricky at least if you haven't seen a stress tensor from your mechanics course (or in general the flux of a vector along the 3 directions which would be the gradF for the whole derivative quantity).
 
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