# What is the explanation for the energy momentum tensor for incoherent dust?

I've read about the energy momentum tensor for incoherent dust in several sources such as this page and Einstein's own book "The Meaning or Relativity". They say something like:

The simplest energy-momentum tensor that can be constructed from these two dust quantities (the momentum $\rho$ and vector $v$) is the following: $T^{ab} = \rho v^a v^b$

I don't understand the physical justification for multiplying $v$ by itself to form a two dimensional energy-momentum tensor. I've read the derivation here but it is far from simple. Can someone provide an explanation?

It might be helpful to start with non relativistic hydrodynamics and ask for the change of the momentum of a fluid in time. Let $\rho$ be the mass density of an incompressible fluid. If $\rho v$ is the momentum per volume, then we can write: $$\frac{\partial(\rho v)}{\partial t}=\rho \frac{\partial v}{\partial t}+\frac{\partial \rho}{\partial t} v.$$ The first term on the right hand side can be processed by the Euler equation: $$\frac{\partial v^i}{\partial t}=-v^k \frac{\partial v^i}{\partial x^k}-\frac{1}{\rho}\frac{\partial p}{\partial x^i}$$ while $p$ is the pressure. The second term in the first equation can be replaced by the continuity equation: $$\frac{\partial \rho}{\partial t}=-\frac{\partial (\rho v^k)}{\partial x^k}.$$ Then, after substitution of both and a few algebraic manipulations, we get: $$\frac{\partial(\rho v^i)}{\partial t}=-\frac{\partial p}{\partial x^i}-\rho\, v^k\frac{\partial v^i}{\partial x^k}-v^i\frac{\partial(\rho v^k)}{\partial x^k}.$$ Now, after applying the product rule on the right hand side we obtain the quadratic term of the velocity: $$\frac{\partial(\rho v^i)}{\partial t}=-\frac{\partial p}{\partial x^i}- \frac{\partial }{\partial x^k}(\rho\,v^iv^k).$$ The first term on the right hand side can be expressed with help of the Kronecker delta, i.e.: $$\frac{\partial(\rho v^i)}{\partial t}=-\frac{\partial }{\partial x^k}(\delta^{ik}p+\rho\,v^iv^k).$$ By applying the notation $$T^{ik}:=\delta^{ik} p +\rho\,v^i v^k$$ we obtain the representation: $$\frac{\partial(\rho v^i)}{\partial t}=-\frac{\partial }{\partial x^k}T^{ik}.$$ This result is easily generalized to the relativistic case.