There is an image in the Wikipedia about the stress-energy tensor:

enter image description here

I have a rough understanding of the stress tensor: you imagine cutting out a tiny cube from the fluid and form a matrix out of the forces on each side of it: it's not hard to see that the forces that push the faces outwards so the x direction of the force on the x side, the y direction of the force on the y side, etc. is the pressure, while other forces directions are the shear stress (these can be non-zero is some jelly-like substance).

Now the first question: Why is the lower half of the spatial part is the momentum flux, while only the upper one is the shear stress on the image? Is the image wrong? Isn't the spatial part describe the classical stress tensor?

The second question is: what's the intuition behind the temporal part?

So if I have a Minkowski spacetime and cut out a cube of it, then it's quite hard to see why "pressure" on the temporal face means density, also why "shear stresses" on the temporal face translate to momentum density, not speaking of the "temporal force components" the spatial faces...

Also stresses are measured in Pascals, mass density is measured in $kg/m^3$ while momentum density is in the units of $Ns/m^3$. These units doesn't seem to be compatible, but still they are in the same matrix, why?

EDIT: It has been pointed out that momentum flux refer to the entire blue part, it indeed have a lighter frame I didn't notice. Then yet another subquestion: what's the difference between the momentum density and momentum flux?

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    $\begingroup$ Note: the entire blue (plus green) box is the momentum flux. $\endgroup$
    – lemon
    Commented May 14, 2015 at 16:39
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    $\begingroup$ All of the elements of $T^{ij}$ (with $i,j\in\{1,2,3\}$) are part of the momentum flux, not just "the lower half." Similarly, it is all the $T^{i,j}$ for $i\neq j$ that constitute the shear stress. $\endgroup$
    – Kyle Kanos
    Commented May 14, 2015 at 16:47
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    $\begingroup$ Related: physics.stackexchange.com/q/28875 $\endgroup$
    – Kyle Kanos
    Commented May 14, 2015 at 16:49
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    $\begingroup$ $pressure=mass~density\times c^2=momentum~density/c$ Since $c=1$ they are all the same $\endgroup$
    – Jim
    Commented May 14, 2015 at 19:19
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    $\begingroup$ The confusion here is due to the fact that on universities academics try to teach students. But having knowledge of a subject is something completely different from transferring that knowledge! Most academics have no knowledge of didactics at all.Leaving out the "c'' from equations (for no obvious reason) is probably the best example of this. Good luck, students! ;-) $\endgroup$ Commented Feb 16, 2020 at 4:33

4 Answers 4


For any matter/energy distribution we can in principle assemble it from point particles. So the stress-energy tensor of the whole system can be expressed as a sum of the stress-energy tensors of the point particles. The reason this helps is that the stress-energy of a point particle is very simple. It is:

$$ T^{\alpha\beta}({\bf x},t) = \gamma m v^\alpha v^\beta $$

at the position of the particle and zero everywhere else. The variable $v$ is the velocity vector $(c, \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt})$ i.e. it is the derivative of the position with respect to coordinate time (not proper time).

Expressed this way it's obvious that all the entries in the stress-energy tensor have the same dimensions of $ML^2T^{-2}$ (divide by $L^3$ to turn it into a density). So the only remaining question is how the ensemble properties like momentum density and pressure emerge from the point particle description.

The $T^{00}$ element is easy since that's just $\gamma mc^2$, which is the energy. So add up all the point particles and you get the total energy.

The $T^{i0}$ elements look like $\gamma mv^ic$, so add up all the point particles and you get the total momentum multiplied by the velocity in the time direction. Likewise the non-diagonal $T^{ij}$ elements give the total momentum multiplied by the velocity in the $j$ direction. Both are momentum fluxes.

The diagonal elements (other than $T^{00}$) look like a kinetic energy $\gamma m(v^i)^2$. If you consider an ensemble of particles with random velocities (e.g. thermal velocities) then the kinetic energy is simply related to the pressure, and that's why the diagonal terms are effectively a pressure.

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    $\begingroup$ Isn't $\gamma m c^2$ the total energy? I knew the rest energy is only $mc^2$. $\endgroup$
    – Calmarius
    Commented May 15, 2015 at 10:25
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    $\begingroup$ @Calmarius: oops, yes. I'm too used to working in the rest frame where $\gamma$ is one :-) $\endgroup$ Commented May 15, 2015 at 10:38
  • $\begingroup$ Hm, does anyone know why then does wikipedia say that the $T^{00}& term is the mass density instead of the energy density (while keeping the interpretations of the others the same)? It seems that how they define it, the tensor doesn't have the same units for different components :S $\endgroup$
    – guillefix
    Commented Jun 7, 2015 at 20:44
  • $\begingroup$ While I'm reading this I'm thinking of the WAMP-picture of the universe. When you look at it you see a distribution of temperatures (but correct me if i'm wrong). Is this distribution related to the stress-energy-momentum tensor? And if so, how? $\endgroup$ Commented Feb 14, 2020 at 22:21
  • $\begingroup$ @TL;DR: sorry, but none of the diagonal components are orange; they are green ;-) $\endgroup$ Commented Feb 15, 2020 at 15:58

TL;DR: It may be helpful to think of the above-diagonal orange components as "energy flux" rather than "momentum density"; if you do this, the interpretations in terms of shears and pressures become more natural.

Here's another way to think of the stress-energy tensor. First, you're hopefully familiar with the notion of the energy-momentum four-vector: $p^\mu = (E/c, p_x, p_y, p_z)$. Each one of the components of this quantity is conserved.

Second, you have hopefully come across some form of the continuity equation. This is a statement about conservation of some quantity that can flow through space. If this quantity has a density $\rho$ and a flux density $\vec{J}$, then we have $$ \frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot \vec{J} = 0. $$ What this says, effectively, is that if the quantity [foo] is flowing out of a region of space (i.e., $\vec{\nabla} \cdot \vec{J} \neq 0$ at a particular point), then the density of [foo] must be changing at that point (and with the opposite sign). This is, for example, how charge conservation is enforced in classical electrodynamics; if we integrate the equation over some region and use the divergence theorem, we get $$ \frac{dQ_\text{enc}}{dt} + \oint \vec{J} \cdot d\vec{a} = 0, $$ i.e., if there's a net current flux through the surface, then the charge enclosed is changing.

The stress-energy tensor can be thought of as combining these two notions. If we want $E$ to be conserved, for example, and we allow energy to be spread out over space, then it must obey a law like $$ \frac{\partial}{\partial t}\text{(energy density)} = - \vec{\nabla} \cdot \text{(energy flux)} $$ and if we want momentum to be conserved, then each of the momentum components $x$, $y$, $z$ must also satisfy a similar law: $$ \frac{\partial}{\partial t}\text{($p_x$ density)} = - \vec{\nabla} \cdot \text{($p_x$ flux)} $$

If we look at the $T^{1 \mu}$ components of the stress-energy tensor, though, we have precisely these quantities! $T^{10}$ is the momentum density; and $T^{11}$, $T^{12}$, and $T^{13}$ are the fluxes of $x$-momentum in the $x$-, $y$-, and $z-$direction respectively. To see why this is, note that if $x$-momentum is fluxing in the $x$-direction across a surface, this means that the objects on the other side of the surface are experiencing a force in the $x$-direction (since $\vec{F} = d \vec{p}/dt$); and since momentum flux is just momentum per time per area, $T^{11}$ is just a pressure in the $x$-direction. By the same logic, if $x$-momentum is fluxing in the $y$-direction across a surface, this would correspond to a shear stress (a force is being exerted parallel to the surface rather than perpendicular to it.)

So that explains why the green components are pressures, the dark blue are shears, and the components $T^{10}$, $T^{20}$, and $T^{30}$ are momentum densities. But what about energy conservation? Well, if we try to do the same here, we can identify $T^{00}$ as the energy density; but under this interpretation $T^{01}$, $T^{02}$, and $T^{03}$ are more naturally thought of as energy flux rather than momentum density. It is apparently just a fact about the universe that these two quantities are equal to each other; at the very least, given the symmetries we know about between space, time, energy, and momentum, it should seem plausible that this is true.

Finally: all four of the above equations can be expressed pretty compactly as $$ \frac{\partial T^{\mu 0}}{\partial t} + \partial_i T^{\mu i} = 0 $$ or even more compactly as $$ \frac{\partial T^{\mu \nu}}{\partial x^\nu} = 0. $$ Thus, we have a nice tensorial relation expressing the conservation of energy and momentum in our system.

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    $\begingroup$ I like the physics use of the term "foo" from programming discussions -- don't mind if I steal this. $\endgroup$ Commented Nov 27, 2018 at 13:44
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    $\begingroup$ See The Feynman Lectures on Physics, Volume II, Section 27-6 Field Momentum[1] for a similar, though simpler, discussion, in which the stress-energy tensor is not mentioned by name! [1]: feynmanlectures.caltech.edu/II_27.html#Ch27-S6 $\endgroup$ Commented Jan 15, 2019 at 15:33

First, it is not the stress tensor $T^{μν}$ that has these interpretations; but rather (as already made clear in your picture) the stress tensor density $𝔗^ρ_ν$ that does.

Second, the natural form for the tensor density, before any indexes are raised or lowered by any metric, is not rank (2,0) or rank (0,2), but rank (1,1). The stress tensor is related to this by $T^{ρμ} = 𝔗^ρ_ν g^{νμ}/\sqrt{|g|}$, or in its more usual form as $T_{μν} = g_{μρ} 𝔗^ρ_ν/\sqrt{|g|}$, where $g_{νμ}$ is the metric, $g$ its determinant and $g^{νμ}$ its inverse.

This is the quantity that appears in the continuity equation $∂_ρ 𝔗^ρ_ν = 0$ (or, in curvilinear coordinates and/or curved space-times $∂_ρ 𝔗^ρ_ν = Γ_{ρν}^μ 𝔗^ρ_μ$), where $Γ_{ρν}^μ$ is the Levi-Civita connection associated with the metric $g_{μν}$). The 3-current $T_ν = 𝔗^ρ_ν ∂_ρ ˩ d^4 x$ is that associated with the momentum component $p_ν$. More generally, the 3-currents are associated with vector flows associated with a vector field $Δx^μ = Δ^ν ∂_ν$ ($ξ = ξ^ν ∂_ν$ is more commonly seen in the literature in place of $Δ$, but my version is more evocative), the corresponding 3-current being $T_Δ = Δ^ν 𝔗^ρ_ν ∂_ρ ˩ d^4 x$. This would be associated with the momentum "component" $p_Δ = Δ^ν p_ν$.

The stress-tensor density is an adjustment of the canonical stress tensor density, which I will denote $𝔓^ρ_ν$: $$𝔗^ρ_ν = 𝔓^ρ_ν + ∂_μ 𝔭^{ρμ}_ν,$$ by another tensor density satisfying the property $𝔭^{ρμ}_ν = -𝔭^{μρ}_ν$, which ensures that both versions of the stress tensor have the same divergence: $∂_ρ 𝔗^ρ_ν = ∂_ρ 𝔓^ρ_ν$. This adjustment is chosen so as to make the stress tensor symmetric; i.e. $T^{μν} = T^{νμ}$, or in any of the other equivalent forms: $𝔗^ρ_ν g^{νμ} = 𝔗^μ_ν g^{νρ}$, $g_{μρ} 𝔗^ρ_ν = g_{νρ} 𝔗^ρ_μ$ or $T_{μν} = T_{νμ}$. This is actually a condition on the metric, which generally aligns the principal axes of the metric and its inverse with the left and right eigenvectors of the stress tensor density. So, if you were actually given the stress tensor density first, then 6 of the 10 components of the metric would be determined, leaving only the 4 scaling units for each of the 4 principal directions.

For a field theory derived from an action integral $S = \int 𝔏(q,v) d^4 x$, involving a Lagrangian density $𝔏(q,v)$ that is a function of a field with components $q^A$ and gradients $v^A_μ = ∂_μ q^A$, the canonical stress tensor density is given by $$𝔓^ρ_ν = \frac{∂𝔏}{∂v^A_ρ} v^A_ν - δ^ρ_ν 𝔏.$$ The integral, itself, is taken with respect to the coordinates $\left(x^0,x^1,x^2,x^3\right)$, which in Cartesian form would be $(t,x,y,z)$, with $d^4 x = dt dx dy dz$.

As for the dimensions, $S$ has the dimension $[S] = H$ of action, which is $H = ML^2/T$, where $M$, $L$ and $T$ respectively denote the dimensions of mass, length and time duration.

For the coordinates $x^μ$, let $[μ] = \left[x^μ\right]$ denote the corresponding dimension. Correspondingly, we have: $[dx^μ] = [μ]$ and $[∂_μ] = 1/[μ]$. Let $Ω = [d^4 x] = [0][1][2][3]$ denote the dimension of $d^4x$. For Cartesian coordinates or warped versions thereof, we have $\left[x^0\right] = T$, $\left[x^1\right] = \left[x^2\right] = \left[x^3\right] = L$, and thus $Ω = T L^3$. For cylindrical coordinates (or those similar to it) it would be $Ω = T L^2$ while for spherical coordinates (or similar), it would be $Ω = T L$, so it actually depends on what type of coordinate grid you have.

Let $A = \left[g_{μν} dx^μ dx^ν\right]$ denote the dimension of the line element associated with the metric. Then, for the metric, its determinant and its inverse, we have $\left[g_{μν}\right] = A/([μ][ν])$, $[\sqrt{|g|}] = A^2/Ω$ and $\left[g^{μν}\right] = [μ][ν]/A$. In particular, we also have $\sqrt{|g|} d^4 x = A^2$, independent of the type of coordinate grid you have. The normal convention is to treat the metric as a proper distance metric, in which case $A = L^2$. If it were a proper time metric, we would instead have $A = T^2$.

The Lagrangian density has the dimension $[𝔏] = H/Ω$. Thus, we find: $$\left[𝔗^μ_ν\right] = \left[𝔓^ρ_ν\right] = \frac H Ω \frac {[ρ]}{[ν]}.$$

With Cartesian or Cartesian-like coordinates, this reduces to: $$\begin{align} \left[𝔗^0_0\right] &= \frac H Ω &= \frac{M}{LT^2} &= \frac{1}{L^3}\frac{ML^2}{T^2}, \\ \left[𝔗^i_0\right] &= \frac H Ω \frac L T &= \frac{M}{T^2} &= \frac{1}{L^3}\frac {ML^2}{T^2}\frac{L}{T}, \\ \left[𝔗^0_j\right] &= \frac H Ω \frac T L &= \frac{M}{L^2T} &= \frac{1}{L^3}\frac{ML}{T}, \\ \left[𝔗^i_j\right] &= \frac H Ω &= \frac{M}{LT^2} &= \frac{1}{L^3}\frac{ML}{T}\frac{L}{T}, \end{align}$$ for $i,j = 1,2,3$. These are, respectively, the dimensions for energy density, energy flux density, momentum density and momentum flux density. In the last case, we also have: $$\frac{M}{LT^2} = \frac{1}{L^2} \frac{ML}{T^2},$$ which is the dimension for pressure.

For the stress tensor in either rank (2,0) or rank (0,2) form, the respective dimensions are: $$\left[T^{μν}\right] = \frac{H}{A^2}\frac{[μ][ν]}{A}, \hspace 1em \left[T_{μν}\right] = \frac{H}{A^2}\frac{A}{[μ][ν]},$$ and the physics literature is both confused and inconsistent on this, because of the widespread failure in it to distinguish between the different forms of the stress tensor and their densities. These dimensions depend on what convention is adopted for the metric. With Cartesian-like coordinates and adopting the convention $A = L^2$, we get: $$\left[T^{μν}\right] = \frac{M[μ][ν]}{TL^4}, \hspace 1em \left[T_{μν}\right] = \frac{M}{T[μ][ν]}.$$ For $i,j = 1,2,3$, this gives rise to the following: $$\begin{align} \left[T^{00}\right] &= \frac{MT}{L^4}, &\left[T^{i0}\right] &= \frac{M}{L^3} = \left[T^{0j}\right], &\left[T^{ij}\right] &= \frac{M}{TL^2}, \\ \left[T_{00}\right] &= \frac{M}{T^3}, &\left[T_{i0}\right] &= \frac{M}{T^2L} = \left[T_{0j}\right], &\left[T_{ij}\right] &= \frac{M}{TL^2}. \end{align}$$ Their dimensions are not connected directly to any of those for the corresponding densities and flux densities, simply because these quantities are not densities at all; and it is a mistake to regard them as such. You're okay if you use light-seconds for time, and use the coordinate $x^0 = ct$, where $c$ denotes the vacuum light speed, while retaining the convention $A = L^2$. Then, all the components in all of the forms have the dimensions of momentum-density. But that won't be coherent for cylindrical-like or spherical-like coordinate grids, since the angular coordinates are normally considered to be in radians which is, effectively, dimensionless.

One of the biggest places this confusion between tensor and tensor density leads to the wrong results or equations - and one of the most widespread errors in the Physics literature (especially the on-line versions, as of the past few years) concerns the coupling coefficient for Einstein's equation. The dimensions for the Levi-Civita connection, defined by the conditions: $$∂_μ g_{νρ} = g_{νσ} Γ^σ_{μρ} + g_{σρ} Γ^σ_{μν}, \hspace 1em Γ^ρ_{μν} = Γ^ρ_{νμ},$$ are readily seen to be: $$\left[Γ^ρ_{μν}\right] = \frac{[ρ]}{[μ][ν]}.$$ The curvature coefficients, defined by: $$R^ρ_{σμν} = ∂_μ Γ^ρ_{νσ} - ∂_ν Γ^ρ_{μσ} + Γ^ρ_{μα} Γ^α_{νσ} - Γ^ρ_{να} Γ^α_{μσ}$$ have the dimensions: $$\left[R^ρ_{σμν}\right] = \frac{[ρ]}{[σ][μ][ν]}.$$ The Ricci tensor $R_{μν} = R^ρ_{μρν}$ has dimensions: $$\left[R_{μν}\right] = \frac{1}{[μ][ν]}.$$ The curvature scalar $R = g^{μν} R_{μν}$ has dimension $$[R] = \frac{1}{A} = \frac{1}{L^2},$$ the last equation applying if we adopt the convention $A = L^2$.

The Einstein-Hilbert Lagrangian density for the gravitational field has the form: $$𝔏(g) = k \sqrt{|g|} R,$$ for some coefficient $k$ that is determined by other means to be $1/(16πG)$, up to powers of vacuum light speed $c$, where $G$ is Newton's coefficient. Upon substitution, we get: $$[𝔏(g)] = \frac{H}{Ω} = [k] \frac{A^2}{Ω} \frac{1}{A} ⇒ [k] = \frac{H}{A} = \frac{M}{T},$$ again, adopting the convention that $A = L^2$. This is independent of what the dimensions of the individual coordinates, since the $Ω$'s cancel out.

The coefficient is normally written as $k = 1/(2κ)$. Therefore, $[κ] = T/M$ and $κ = 8πG$ up to powers of $c$.

Since $[G] = L^3/(MT^2)$, then $M/T = \left[c^3/G\right]$. Therefore, the coupling coefficient is $k = c^3/(16πG)$ and $κ = 8πG/c^3$, not the $k = c^4/(16πG)$ or $κ = 8πG/c^4$ that you see running around on the net - and even on the wall over at Leiden! The literature is confused and you not only see both $c^4$ and $c^3$ in the denominator for $κ$ (or numerator for $k$), but sometimes even $c^2$ (e.g. in Einstein's "The Meaning Of Relativity").

We can also check this, more directly, on the Einstein Equations: $$G_{μν} = R_{μν} - \frac 1 2 g_{μν} R = κ T_{μν},$$ where $G_{μν}$ is the Einstein tensor, defined above, we get: $$\frac{1}{[μ][ν]} = \left[G_{μν}\right] = \left[κT_{μν}\right] = [κ]\frac{M}{T[μ][ν]} ⇒ [κ] = \frac{T}{M} = \left[\frac{8πG}{c^3}\right].$$ Thus, the correct form of that equation is: $$G_{μν} = R_{μν} - \frac 1 2 g_{μν} R = \frac{8πG}{c^3} T_{μν}.$$


if c=1 it all works. Else you think in 3D. c is the speed of time in terms of space. Wouldn't we make that 1, if we want to scale 4 dimensions? Our axis can be labeled anything, but why not scale them 1,1. To do that set c=1. if you want a trickier graph , set c=2, but there's no point in that, because c is the slope of the axis, so let c=1. Better yet set c=Sqrt(-1), since the metric tensor subtracts c^2 in the measurement of squared movement. 1,1,1,-c^2 is the trace. Square root to solve for the SE tensor, then either make t the imaginary by setting c=1, or make it real by setting c=i. These equations say it's imaginary, which it is, since it is a timelike dimension. let me know when t'=i, maybe at Sqrt(2)c, then you might have changed everything, since t is now likely space like. Would you then experience the extension of the euclidean space where time is a distance? IDK. The metric result would become negative, thus imaginary hyper-distance is being accounted for. Where would you move then? The sum of the integral of hyper-distance squared, would be rewinding? If that pythagorean theorem holds for every path, would time travel work? IDK. Let c=1 and these things will be much easier.


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