# Intuitive understanding of the elements in the stress-energy tensor

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

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?

• Note: the entire blue (plus green) box is the momentum flux. May 14, 2015 at 16:39
• 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. May 14, 2015 at 16:47
• May 14, 2015 at 16:49
• $pressure=mass~density\times c^2=momentum~density/c$ Since $c=1$ they are all the same
– Jim
May 14, 2015 at 19:19
• 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! ;-) Feb 16, 2020 at 4:33

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.

• Isn't $\gamma m c^2$ the total energy? I knew the rest energy is only $mc^2$. May 15, 2015 at 10:25
• @Calmarius: oops, yes. I'm too used to working in the rest frame where $\gamma$ is one :-) May 15, 2015 at 10:38
• Hm, does anyone know why then does wikipedia say that the $T^{00}& term is the mass density instead of the energy density (while keeping the interpretations of the others the same)? It seems that how they define it, the tensor doesn't have the same units for different components :S Jun 7, 2015 at 20:44 • While I'm reading this I'm thinking of the WAMP-picture of the universe. When you look at it you see a distribution of temperatures (but correct me if i'm wrong). Is this distribution related to the stress-energy-momentum tensor? And if so, how? Feb 14, 2020 at 22:21 • @TL;DR: sorry, but none of the diagonal components are orange; they are green ;-) Feb 15, 2020 at 15:58 TL;DR: It may be helpful to think of the above-diagonal orange components as "energy flux" rather than "momentum density"; if you do this, the interpretations in terms of shears and pressures become more natural. Here's another way to think of the stress-energy tensor. First, you're hopefully familiar with the notion of the energy-momentum four-vector:$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.

• I like the physics use of the term "foo" from programming discussions -- don't mind if I steal this. Nov 27, 2018 at 13:44
• 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 Jan 15, 2019 at 15:33

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.