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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    $\begingroup$ 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. $\endgroup$
    – user4552
    Jun 16, 2018 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. $\endgroup$ Jun 19, 2018 at 11:49
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    $\begingroup$ An elegant but mathematically rigorous derivation is given in "The Large Scale Structure of Space-time" by Hawking and Ellis. $\endgroup$
    – Richard
    Jul 9, 2019 at 7:49
  • $\begingroup$ Related physics.stackexchange.com/q/694491/226902 $\endgroup$
    – Quillo
    Feb 22, 2022 at 22:01
  • $\begingroup$ Moreover: your stress-energy tensor is one for perfect fluids but also viscous fluids that only have bulk viscosity have the same stress-energy tensor (i.e. fluids with bulk viscosity but no heat conduction and no shear viscosity, see e.g. doi.org/10.48550/arXiv.2003.04609 . Related: physics.stackexchange.com/q/647825/226902 $\endgroup$
    – Quillo
    May 4, 2022 at 13:22

1 Answer 1


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, 2018 at 13:24
  • $\begingroup$ @Cham - This is becoming a big issue for me. I tried to convert the referenced Wiki page into Mathematica code. I want to be able to solve the EFEs for either natural units or SI units. When working in natural units, $U=\{0,0,0,1\}$, but when when working in SI units, I defined $U=\{0,0,0,c\}$ This seemed to cause a mess. Are we supposed to be multiplying $(\rho + P)$ by the product of the velocity vectors, or by the cross product of the basis vectors? $\endgroup$
    – Quark Soup
    Feb 22, 2021 at 21:32
  • $\begingroup$ @GluonSoup: it's the outer product of the "comoving fluid vector" cross product is not even well-defined in four dimensions. $\endgroup$ Jul 30, 2021 at 7:33

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