Timeline for Pressure and density using a general Lagrangian
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| May 24, 2016 at 19:08 | vote | accept | titanium | ||
| Feb 19, 2015 at 3:10 | history | edited | DanielSank | CC BY-SA 3.0 |
title capitalization
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| Sep 2, 2014 at 8:13 | history | edited | Qmechanic♦ |
edited tags
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| Aug 1, 2014 at 17:07 | comment | added | Zo the Relativist | Also, if your metric is Minkowski, $T_{\mu\nu}$ has to be zero. | |
| Jul 9, 2014 at 5:35 | history | tweeted | twitter.com/#!/StackPhysics/status/486745488847536128 | ||
| Jul 2, 2014 at 12:36 | comment | added | mastrok | I guess you have a sign problem since a stable field $\phi=const$ should give negative $w=p/\rho$. | |
| Jul 2, 2014 at 7:48 | answer | added | mastrok | timeline score: 2 | |
| Jul 2, 2014 at 7:37 | history | edited | titanium | CC BY-SA 3.0 |
I solved the problem. I have the solution for those who might find it useful.
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| Jul 2, 2014 at 6:03 | history | edited | titanium | CC BY-SA 3.0 |
deleted 3 characters in body
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| Jul 2, 2014 at 5:50 | comment | added | user10851 | btw, the upper/lower split indices on the metric are always $g^\mu_\nu = \delta^\mu_\nu$ by definition (no negative signs, no matter your sign convention) | |
| Jul 2, 2014 at 5:30 | comment | added | titanium | I have just edited the original question. $T_{\mu\nu}$ is isotropic because $phi$ only depends on $t$, and f is a function of $\phi$ and $\dot{\phi}$. Sorry for the confusion. | |
| Jul 2, 2014 at 5:28 | history | edited | titanium | CC BY-SA 3.0 |
added 83 characters in body
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| Jul 2, 2014 at 4:50 | comment | added | mastrok | You have assume that $T_{\mu\nu}$ is isotropic. However, this is not always true for arbitrary functional form $f$ of $\phi$ and $\partial_{\mu} \phi$. | |
| Jul 2, 2014 at 4:03 | history | edited | Danu | CC BY-SA 3.0 |
edited body; edited body; added 11 characters in body; edited tags
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| Jul 2, 2014 at 3:27 | history | asked | titanium | CC BY-SA 3.0 |