Is drag coefficient in fluid dynamics a constant in every direction?

Question

I am referring the book 'Stochastic Processes in Cell Biology'. In this book the author says that Fokker Plank equation in isotropic diffusion: $$\frac{\partial p(x,t)}{\partial t}=-\frac{F}{\gamma}\frac{\partial p(x,t)}{\partial x}+D\frac{{\partial}^2 p(x,t)}{{\partial x}^2}$$ $\frac{F}{\gamma}=v$ where F is force, $\gamma$ is drag coefficient and $v$ is terminal velocity. For three dimensions the same book refer to Fokker Plank equation as: $$\frac{\partial p(\textbf{X},t)}{\partial t}=-\nabla\cdot [\frac{\textbf{F}}{\gamma}p(\textbf{X},t)]+D \nabla^2 p(\textbf{X},t), \label{FPE3}$$ where $\nabla^2$ is Laplacian operator. Now my question is, can I write $\frac{\textbf{F}}{\gamma}$ as equivalent terminal velocity in three dimensions which can be vectorized as:

$$\frac{\textbf{F}}{\gamma}=\textbf{v}=(u \textbf{i}+v \textbf{j}+w \textbf{k}),$$ Basically is drag coefficient a constant which on dividing force can give a three dimensional velocity?

Answer 1

Unless the environment is isotropic, you will need to use a diffusion tensor. This goes for drag as well, because the drag coefficient is intimately related to the diffusion coefficient.

If the force is the gradient of a potential energy function, $F=-\nabla V$, then the system should approach thermodynamic equilibrium, $p\propto \exp (-\beta V)$, which must be a time-independent solution to the F-P equation. This requires that ${{\gamma }^{-1}}=\beta D=D/kT$, in your notation.

With anisotropic and/or position-dependent diffusion, the equation should actually read: $$\dot{p}={{\partial }_{i}}[{{D}_{ij}}(\beta p{{\partial }_{j}}V+{{\partial }_{j}}p)]={{\partial }_{i}}[{{D}_{ij}}(-\beta p{{F}_{j}}+{{\partial }_{j}}p)]$$

Answer 2

Realize that a system model is just that, a model, not the real system itself. What you are suggesting I believe is extending the model cited to potentially better fit reality (measurement).

Since we know from experiments that drag coefficient can change with geometry of the body then one could reason indeed extending the coefficient in three dimensions may be an improvement on the model if the body is asymmetric among its principle axes. You might be able to answer your question by solving the Navier Stokes equations for your particular problem, but more than likely it would require a numerical solution, and so experiment would better provide the supporting or otherwise discouraging result.