Thanks for the comments. I agree with your conclusions. I made a few cosmetic changes to:
http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids
it now derives the continuity equation with "rho", not "n*m".
I think we answered all my questions...
You are right. Also with your definition of E in your previous post. I wrote everything in detail here:
http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids
and it seems everything nicely fits together. Even all gammas play well...
I am still not comfortable with those \gamma factors -- because the kinetic energy is generated by one such gamma factor. So if we are comoving with the fluid, there is no kinetic energy (that makes sense), but I think we should do the analysis in the laboratory frame (as you said), and thus we...
Also a related question -- the nonrelativistic energy E is composed of the kinetic and an internal one.
The kinetic energy can be obtained by:
\rho c^2 \gamma={\rho c^2\over \sqrt{1-{v^2\over c^2}}} = \rho c^2 \left(1 + {v^2\over 2 c^2}+\cdots\right) = \rho c^2 + {1\over 2} \rho v^2 +...
Yes, that's my idea too --- but now we need to get rid of the two terms on the right --- the continuity equation for rho --- who do we derive that those are 0?
Excellent, I really appreciate your help. You derived the equation that I have here, eq. (3):
http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#relativistic-derivation-of-the-energy-equation
But as I write there (and also above in this thread), the problem...
I obtained the book. Weinberg essentially derives the same equation as Schutz:
u^\alpha \partial_\alpha S = 0
which says, that the flow conserves specific entropy. This equation is obtained from a thermodynamic relation, e.g. eq. 2.10.18 in Weinberg:
k T d \sigma = pd \left(1\over...
No, I appreciate the suggestion. :) I run out of all ideas myself, so any fresh insight from you and others is appreciated.
The way I understand it is that there are 3 nonrelativistic equations (continuity, momentum and energy) and they follow from the conservation law of the stress energy...
Just look at the first equation, take the d L / dx^\mu from the left hand side, put it to the right hand side, and then insert it into the brackets, where you differentiate with respect to \nu, so you also have to insert the \delta_\mu^\nu. then you get the second equation. It's a standard...
Then I get the following equation:
\partial_t (\rho u^i) + \partial_j (\rho u^i u^j + p\delta^{ij}) = 0
which is just the Euler momentum equation:
{\partial (\rho{\bf u})\over\partial t} + \nabla \cdot
(\rho {\bf u}{\bf u}^T) + \nabla p = 0
and that is fine...
Thanks for the link and the hint.
So I think the energy equation must follow from this:
\dot{\varrho} + (\varrho+p)\nabla_au^a = 0
but that's where I got stuck, because if I do the nonrelativistic approximation, I have to introduce units, so I get...
Hi,
I would like to start from the stress energy tensor for the perfect fluid:
T^{\mu\nu}=\begin{pmatrix} \rho c^2 & 0 & 0 & 0\cr 0 & p & 0 & 0\cr 0 & 0 & p & 0\cr 0 & 0 & 0 & p\cr\end{pmatrix}
where \rho is the mass density and p is the pressure, and I would like to derive the...