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I am trying to solve an exercise in Sean Carroll's GR book "Spacetime and Geometry". Basically we need to derive the stress-energy tensor of a perfect fluid (ie $T^{\mu\nu}=(\rho +p)U^{\mu}U^{\nu} + p\eta^{\mu\nu}$) from the stress-energy tensor of a discrete set of particles (ie $T^{\mu\nu}=\sum_a \frac{p^{\mu}_a p^{\nu}_a}{p^0_a}\delta^{(3)}(\mathbf x - \mathbf x^{(a)})$), under the hypothesis of isotropy.

I managed to get in for the $T^{00}$ component and the $T^{0i}$ components, by replacing $p^{\mu}$ by $p^0$ a trivial sum appears: energy density for the 00-component and momentum density the 0i-components (vanishing by isotropy). But I am still struggling with the pure spatial part, I was thinking of substituting the sum by an integral, then the non diagonal part vanishes by isotropy again. Could it be something like: $\sum_a = \int d^3x \rho(x)$? Because then I need a relation that relates the density distribution $\rho$ and $p^{\mu}$ to the pressure (more exactly the definition of the pressure from these two concepts).

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There are two points I wish to highlight. 1) Simple substitution of the sum by an integral would not work and is not justified. However, one should switch from microscopical quantities to macroscopical by doing avaraging over 4-volume, throughout which interparticle distances and times can be considered small.

Macroscopic stress-energy tensor will be then: $$ {\bf T}^{\mu\nu}=\dfrac{1}{\Delta V_4}\int_{\Delta V_4}T^{\mu\nu}d V=\dfrac{1}{\sqrt{-g} d^3x^i dx^0}\int_{\Delta V_4}T^{\mu\nu} \sqrt{-g} d^3x^i dx^0 $$ Then a)in $T^{\mu\nu}$ only delta-functions depend on x, b) metric determinant g is a macroscopic quantity, is constant over selected volume and can also be taken away from the integra. One arrives then at: $$ {\bf T}^{\mu\nu}=\dfrac{1}{d^3x^i dx^0}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}\int_{\Delta V_4} \delta^{(3)}({\bf x}-{\bf x}^{(a)}) d^3 x^i dx^0 = \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}. $$ In the last expression the sum is taken over the particles which have world lines passing through $\Delta V_4$ (we ignore the fact that some particles could have left or entered the volume through its 3-boundary, as there are much less of them then the particles inside the volume).

Now the expression ${\bf T}^{\mu\nu}= \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}$ can be more comfortably treated.

2) The symmetry considerations themselves. Consider the conponenent of the macroscopic tensor:

${\bf T}^{0 0} = \dfrac{1}{d^3x^i}\sum_a p_a^0 \equiv \rho $

${\bf T}^{i 0} = \dfrac{1}{d^3x^i}\sum_a p_a^i $. As the sum $\sum_a p_a^i$ of 3-vectors is taken over a macrospoic volume, the result should result in a macroscopic 3-vector. However, if this vector was not zero, it would violate isotropy, which states that there exists no preferable direction. Hence ${\bf T}^{i 0} = 0$

${\bf T}^{i j} = \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^i p_a^j}{p_a^0}$. As just before, the sum should produce a symmetric macroscopic 3-tensor of second order. But all symmetric 3-tensors are defined by 3 eigenvectors. If eigenvalues are non-degenerate, then there exists 3 preferred directions (3 eigenvectors), if eigenvalues are single degenerate, then there are 2 preferred directions etc. No preferred directions correspond to the case when the matrix has all eigenvalues equal, that is when the matrix is proportional to kronecker delta. The coefficient of proportionality is the pressure: ${\bf T}^{i j} \equiv P \delta^{ij}$

Expressing $\delta^{ij}$ as $\eta^{ij}+U^0 U^0$ ($U^i$ is zero by symmetry considerations, and $U^0$ is hence equal to unity), and $T^{00}$ as $\rho U^0 U^0$, one arrives at the final expression for $T$.

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Thanks a lot for the 'rigorous' derivation. –  toot Mar 1 '12 at 22:26
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