# Interpretation of $\partial_\nu T^{\mu \nu}$

• A
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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ChrisVer
Gold Member
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?

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.

George Jones
Staff Emeritus
Gold Member
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.
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?

ChrisVer
Gold Member
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.

ChrisVer
Gold Member
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?

ChrisVer
Gold Member
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 refered 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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