I am reading through Fundamentals of Plasma Physics by Bellan and came across this definition of the pressure tensor:
I am wondering how the off-diagonal terms vanish in the case of an isotropic distribution, and how this can be shown mathematically.
Moreover, why would the integral of e.g. $v_x' v_y' f_\sigma dv'$ not be equal to 0 for an anisotropic distribution?
I will first reference the following answer on how to define velocity moments of a velocity distribution function (VDF), $f\left( v_{x}, v_{y}, v_{z} \right)$: https://physics.stackexchange.com/a/218643/59023
We can also define a Maxwellian VDF following the answer https://physics.stackexchange.com/a/257592/59023 as: $$ \begin{align} f\left( v_{x}, v_{y}, v_{z} \right) & = f\left( v_{x} \right) \ f\left( v_{y} \right) \ f\left( v_{z} \right) \tag{0a} \\ & = \prod_{ k = x,y,z } \ A_{k} \ e^{^{\displaystyle - \left( \frac{ v_{k} - v_{ok} }{ V_{Tk} } \right)^{2} }} \tag{0b} \end{align} $$ where the total normalization factor is given by: $$ A_{x} \ A_{y} \ A_{z} \ = \frac{ n_{s} }{ \pi^{3/2} \ V_{Tx} \ V_{Ty} \ V_{Tz} } \tag{1} $$ where $n_{s}$ is the number density [e.g., cm-3] of species $s$, $V_{Ts}$ is the thermal speed [e.g., km s-1], here the most probable speed, of species $s$, and $v_{ok}$ is the bulk flow drift velocity (i.e., first velocity moment) in the kth direction.
As discussed in the first link above, each factor of $v_{j}$ multiplied with $f\left( v_{x}, v_{y}, v_{z} \right)$ in an integrand increases the order of the velocity moment. So if we integrated $v_{x} \ f\left( v_{x}, v_{y}, v_{z} \right)$ over all velocity space, we'd get $n_{s} \ v_{ox}$. Since the integral of $f\left( v_{x}, v_{y}, v_{z} \right)$ over all velocity space just returns $n_{s}$ (well this is how we define/construct the VDF so that the normalization parameter is such that this answer follows), then any constant times $f\left( v_{x}, v_{y}, v_{z} \right)$ integrated over all velocity space just returns $n_{s}$ times that constant.
The pressure tensor is unique in that you do not multiply by the dyad $v_{j} \ v_{k}$. Rather, you use what is called the peculiar velocity, where the kth vector component is defined as $\left( v_{k} - v_{ok} \right)$. Since the distributive property applies here, we would be able to integrate each of these parts separately. The second peculiar velocity vector component is pulled outside the kth vector component integral, so it doesn't matter.
Thus, we have the following: $$ \begin{align} \int_{-\infty}^{\infty} dv_{z} \int_{-\infty}^{\infty} dv_{y} \int_{-\infty}^{\infty} dv_{x} \ v_{x} \ f\left( v_{x}, v_{y}, v_{z} \right) & = n_{s} \ v_{ox} \tag{2a} \\ \int_{-\infty}^{\infty} dv_{z} \int_{-\infty}^{\infty} dv_{y} \int_{-\infty}^{\infty} dv_{x} \ v_{ox} \ f\left( v_{x}, v_{y}, v_{z} \right) & = n_{s} \ v_{ox} \tag{2b} \\ \int_{-\infty}^{\infty} dv_{z} \int_{-\infty}^{\infty} dv_{y} \int_{-\infty}^{\infty} dv_{x} \ \left( v_{x} - v_{ox} \right) \ f\left( v_{x}, v_{y}, v_{z} \right) & = 0 \tag{2c} \end{align} $$ Note that $v_{j}$ are abscissa of $f\left( v_{x}, v_{y}, v_{z} \right)$, not specific velocity vectors. The solution shown in Equation 2c should make sense since the integrand is done over abscissa transformed (Galilean transformation) into a frame where distribution is at rest along that vector coordinate. Suppose we multiply $\left( v_{y} - v_{oy} \right)$ to the integrand in Equation 2c, what then? Well this factor would move outside of the $\int dv_{x}$ integral, i.e., it would not affect the results of that first integral. Thus, the result is still zero.
As an aside, the off-diagonal terms in the pressure tensor, sometimes called the strain tensor or the viscosity terms, represent momentum transport across boundaries. That is, the $P_{ij}$ term in the pressure tensor, where $i \neq j$, would represent mean momentum transport rate in the ith direction across the jth planar surface in a frame of reference moving at velocity $\mathbf{v}_{o}$.