- #1
mjordan2nd
- 177
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I am working on Sean Carroll's problem 1.8.
The dust energy momentum tensor is
[tex]T^{\mu \nu}= \rho U^\mu U^\nu,[/tex]
where [itex]U[/itex] is the four-velocity and [itex]\rho[/itex] the energy density in the rest frame.
Trying to manipulate [itex]\partial_\nu \rho U^\mu U^\nu[/itex] directly didn't give me anything I could reasonably interpret. I then tried to look at [itex]U_\mu Q^\mu[/itex] 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
[tex]U_\mu \partial_\nu \rho U^\mu U^\nu=-U^\nu \partial_\nu \rho-\rho \partial_\nu U^\nu=U_\mu Q^\mu[/tex].
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:
[tex]P^\sigma_\nu=\delta^\mu_\nu+U^\sigma U_\nu.[/tex]
This gives
[tex]P^\sigma_\mu Q^\mu = P^\sigma_\mu \partial_\nu (\rho U^\mu U^\nu),[/tex]
[tex]P^\sigma_\mu \partial_\nu (\rho U^\mu U^\nu) =U^\nu \rho \partial_\nu U^\sigma, [/tex]
[tex]P^\sigma_\mu Q^\mu =Q^\sigma + U^\sigma U_\mu Q^\mu,[/tex]
[tex]U^\nu \rho \partial_\nu U^\sigma = Q^\sigma + U^\sigma U_\mu Q^\mu.[/tex]
From here, if we plug in [itex] -U^\nu \partial_\nu \rho-\rho \partial_\nu U^\nu=U_\mu Q^\mu[/itex] into the final expression above, we get the original expression: [itex]\partial_\nu T^{\mu \nu}=Q^\mu[/itex]. 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?
If [itex]\partial_\nu T^{\mu \nu}=Q^\mu[/itex], what physically does the spatial vector [itex]Q^i[/itex] represent? Use the dust energy momentum tensor to make your case.
The dust energy momentum tensor is
[tex]T^{\mu \nu}= \rho U^\mu U^\nu,[/tex]
where [itex]U[/itex] is the four-velocity and [itex]\rho[/itex] the energy density in the rest frame.
Trying to manipulate [itex]\partial_\nu \rho U^\mu U^\nu[/itex] directly didn't give me anything I could reasonably interpret. I then tried to look at [itex]U_\mu Q^\mu[/itex] 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
[tex]U_\mu \partial_\nu \rho U^\mu U^\nu=-U^\nu \partial_\nu \rho-\rho \partial_\nu U^\nu=U_\mu Q^\mu[/tex].
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:
[tex]P^\sigma_\nu=\delta^\mu_\nu+U^\sigma U_\nu.[/tex]
This gives
[tex]P^\sigma_\mu Q^\mu = P^\sigma_\mu \partial_\nu (\rho U^\mu U^\nu),[/tex]
[tex]P^\sigma_\mu \partial_\nu (\rho U^\mu U^\nu) =U^\nu \rho \partial_\nu U^\sigma, [/tex]
[tex]P^\sigma_\mu Q^\mu =Q^\sigma + U^\sigma U_\mu Q^\mu,[/tex]
[tex]U^\nu \rho \partial_\nu U^\sigma = Q^\sigma + U^\sigma U_\mu Q^\mu.[/tex]
From here, if we plug in [itex] -U^\nu \partial_\nu \rho-\rho \partial_\nu U^\nu=U_\mu Q^\mu[/itex] into the final expression above, we get the original expression: [itex]\partial_\nu T^{\mu \nu}=Q^\mu[/itex]. 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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