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Wikipedia reports this expression for the stress-energy tensor of a perfect fluid in general relativity

$$T^{\mu \nu} = \left(\rho + \frac{p}{c^2} \right) U^{\mu} U^\nu + p g^{\mu \nu}, $$

where $\rho$ is the rest-frame mass density, $p$ pressure, and $U$ the four velocity.

Do you know a reference where I could find how this expression is derived?

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Before answering, please see our policy on resource recommendation questions. Please write substantial answers that detail the style, content, and prerequisites of the book, paper or other resource. Explain the nature of the resource so that readers can decide which one is best suited for them rather than relying on the opinions of others. Answers containing only a reference to a book or paper will be removed!

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    $\begingroup$ Deep's answer seems nice. Is there a reason that you really need this to be a resource recommendation rather than just an ordinary question? Normally the reason for asking something as a resource recommendation is because you don't think it would be feasible to answer it within the SE format, e.g., if you want an entire book on a subject. $\endgroup$ – Ben Crowell Jun 16 '18 at 18:47
  • $\begingroup$ @Ben Crowell: Yes, given how vast the subject can be, it would be preferable for me to have the answer as a resource recommendation. $\endgroup$ – Marc Schroeder Jun 19 '18 at 11:49
  • $\begingroup$ An elegant but mathematically rigorous derivation is given in "The Large Scale Structure of Space-time" by Hawking and Ellis. $\endgroup$ – Richard Jul 9 at 7:49
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We adopt the system of units in which speed of light is 1.

Components of the stress tensor $T^{\alpha\beta}$ physically mean the following: $T^{00}$ is the energy density, $T^{0j}$ is the energy flux across the spatial-surface $x_j=$ constant ($j=1,2,3$), $T^{i0}$ is the density of $i$-th component of momentum, and $T^{ij}$ is the $i$-th component of momentum flux across the spatial-surface $x_j=$ constant ($i,j=1,2,3$). Normal momentum flux ($T^{ij}$ for $i=j$) causes normal stress on the fluid element and the others ($T^{ij}$ for $i\neq j$) cause shear stress on the fluid element.

An ideal fluid is one whose viscosity and conductivity are zero. Consider an elemental volume of ideal fluid in its MCRF (momentarily co-moving reference frame). Since conductivity is zero, there is no energy flux into or out of it, which implies $T^{0j}=0$. Since there is no viscosity, it doesn't experience shear stresses, therefore $T^{ij}=0$ when $i\neq j$. Further the statement that the fluid has no viscosity is a frame-independent statement, so $T^{ij}=0$ when $i\neq j$ in any reference frame, and so the matrix $T^{ij}$ must be diagonal in all reference frames. This is possible only if $T^{ij}=p\delta^{ij}$ in which $\delta^{ij}$ is the identity tensor and $p$ is a scalar called pressure. If we denote the energy density by $\rho$, then the stress tensor $T^{\alpha\beta}$ in the MCRF of the fluid element is: $$\begin{bmatrix} \rho & 0 & 0& 0\\ 0 & p& 0& 0\\ 0 & 0& p& 0\\ 0 & 0& 0& p \end{bmatrix}$$ This can be simplified as: $$\begin{bmatrix} \rho & 0 & 0& 0\\ 0 & p& 0& 0\\ 0 & 0& p& 0\\ 0 & 0& 0& p \end{bmatrix}= \begin{bmatrix} \rho+p & 0 & 0& 0\\ 0 & 0& 0& 0\\ 0 & 0& 0& 0\\ 0 & 0& 0& 0 \end{bmatrix}+ \begin{bmatrix} -p & 0 & 0& 0\\ 0 & p& 0& 0\\ 0 & 0& p& 0\\ 0 & 0& 0& p \end{bmatrix}\\ \Rightarrow\quad T^{\alpha\beta}=(\rho+p)(\mathbf{e}_0\mathbf{e}_0)^{\alpha\beta}+p\eta^{\alpha\beta}$$

in which $\eta^{\alpha\beta}$ is the metric tensor. The unit vector in the time direction (of the MCRF of the fluid element) $\mathbf{e}_0$ is nothing but its 4-velocity $\mathbf{U}$. Therefore the dyadic $\mathbf{e}_0\mathbf{e}_0=\mathbf{U}\mathbf{U}$, whose component is $(\mathbf{U}\mathbf{U})^{\alpha\beta}=U^\alpha U^\beta$. Thus we have: $$T^{\alpha\beta}=(\rho+p)U^\alpha U^\beta+p\eta^{\alpha\beta}$$

Reference: General Relativity by B. Schutz.

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    $\begingroup$ You shouldn't write \begin{equation}T^{\alpha\beta}=(\rho+p)\mathbf{e}^0\mathbf{e}^0+p\eta^{\alpha\beta}.\end{equation} For consistency, this should be the tensor \begin{equation}\mathbf{T} =(\rho+p)\mathbf{U} \otimes \mathbf{U} +p \, \boldsymbol{\eta}\end{equation} instead, or in other words \begin{equation}T^{\alpha\beta} \, \mathbf{e}_{\alpha} \otimes \mathbf{e}_{\beta} =(\rho+p)\mathbf{e}_0 \otimes \mathbf{e}_0+p \, \eta^{\alpha\beta} \, \mathbf{e}_{\alpha} \otimes \mathbf{e}_{\beta}.\end{equation} $\endgroup$ – Cham Jun 16 '18 at 13:24

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