As Ross encourages us to make more quantitative posts, I’ve decided to make my second post this week about the Navier-Stokes equation. The text book doesn’t mention the equation explicitly, as it involves rather complicated calculus. But I’ll try to explain it in simple terms. The equation looks like:
Rho(dv/dt + v•(del)v)=f-(del)p+(del)T
Where rho is the density of the liquid, v is the flow velocity, p is the pressure, f is the force per unit volume acting on the liquid, and T is the stress tensor (A tensor is the general term for a vector or matrix. So a vector is a tensor in 1 dimension and a matrix is a 2 dimensional tensor). (del) is a differential operator; in three dimensions, (del)u=du/dx+du/dy+du/dz. (del) u is a vector that points in the direction of maximum slope in u.
This equation is a non-linear partial differential equation. The solutions to such equations are not numerical answers but functions which satisfy the equation. It takes significant mathematical skill to handle equations like these, which is I suppose why this equation is not covered in detail in Nelson. To make things more complicated, like the n-body problem I mentioned in a comment to an earlier post, this is another equation which has no known analytical solution.
Despite the Navier-Stokes equation being very difficult to solve, we can still understand what it means. First, I’m going to verify it passes dimensional analysis.
Rho has units kg/m^3, and v has units m/s. Therefore rho(dv/dt) has units kg/m^3 x m/s/s = kgm/s^2/m^3 = N/m^3. (del)v has units m/s/m, so Rho(v•(del)v) also has units kg/m^3 x m/s x m/m/s= kg/m^3 x m/s/s= N/m^3. Pressure has units N/m^2, so (del)p has units N/m^3/m= N/m^3. Page 172 tells us stress is force/unit area, so stress has the same units as pressure, so the units of (del)T must be N/m^3 also. Finally, force per unit volume must have units N/m^3. So this equation is dimensionally correct; each term has units N/m^3.
Consider multiplying every term by the unit volume. The right side of the equation describes the forces acting on the liquid. The density term becomes a term describing the mass of the volume of liquid, and the remaining term in the left side of the equation describes in a complicated way the rate of change of the velocity of the flow, over the whole liquid. This is effectively the acceleration of the flow. So you can see that the Navier-Stokes equations is really just Newton’s second law of motion, f=ma, but applied to a continuous field of matter.