I'm currently reading Blandford and Thorne's applications of classical physics here. They define the stress tensor as a geometric object that gives the force when a vector surface is inserted into once of it's slots. (Eq 1.32)

$F\left( - \right) = T\left( { - ,\vec \Sigma } \right)$

Then they go on to describe what this expression is in terms of its components in eqn 1.33. From this definition of the stress tensor its obvious how ${T_{jk}}$ is the $j$ component of force across a unit surface perpendicular to $\hat{e}_k$. What I'm confused by is the fact that ${T_{jk}}$ is also the $j$-component of momentum that crosses a unit area which is perpendicular to $\hat{e}_k$, per unit time. I don't see how these two components are representing the same thing. The $j$ component of force measures the rate of change of the $j$ component of momentum not it's flux through a surface.

As a specific example, consider a collection of particles moving with a uniform velocity in the $\hat{e}_1$ direction. Applying the definition in terms of momentum fluxes, there's a net flux of $x$-momentum across a unit surface perpendicular to $\hat{e}_1$. Therefore $T_{11}$ is some non zero value. But applying the definition in terms of force I conclude that since the particles are moving uniformly, there is no change in momentum. Therefore there is no force and all components of $T$ are zero.

How are these two definitions equivalent and in particular how do I resolve this apparent contradiction?

  • $\begingroup$ I don't have time to write a full answer, but to address your first question: if region A exerts a force on region B, then because of Newton's third law the decrease in momentum in A is equal to the increase in momentum in B, so we say there is a flow. $\endgroup$
    – Javier
    Dec 29, 2016 at 4:25
  • $\begingroup$ I'm confused on the last statement you made. How does newton's third law imply a flow? $\endgroup$ Dec 30, 2016 at 5:15


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