# When to use the material derivative?

1. Apr 7, 2017

### joshmccraney

Hi PF!
When doing a force balance in fluid mechanics, $\sum F = D_t(mV)$. This equation typically results to the Navier-Stokes equations. I'm reading a the following problem:

For small particles at low velocities, the first (linear) term in Stokes’ drag law implies $F = kV$, where $k$ is a constant. Suppose a particle of mass $m$ is constrained to move horizontally.

They then solve this equation for the particle's velocity by taking $-kV = m\frac{dV}{dt}$. My question is, how do we know the acceleration term is $\frac{dV}{dt}$ instead of $\frac{\partial V}{\partial t} + V \frac{dV}{dx}$?

2. Apr 7, 2017

### Staff: Mentor

It depends on whether your velocity depends only on time or on location as well. The material derivative is a total derivative, that depends on time and space. E.g. if you have a flow in a pipe, it only depends on time and not on the location whithin the pipe, that is $\frac{dV}{dt}$.

Wikipedia explains by an example of a material derivation the following situation:
Given a temperature distribution on the surface of a lake which warms up, e.g. due to sunlight (time component $\frac{\partial V}{\partial t}$) and due to warmer inflows (local component $\frac{d V}{d x}$). The partial derivative w.r.t. time describes the change in temperature for a w.r.t. the banks standing observer in the lake, who can only observe the change in time at his fixed place. But since the water also gets warmer along certain directions, an observer in a floating boat would measure these additional changes.

3. Apr 7, 2017

### joshmccraney

But viscous forces in a pipe slow flow down as the radial location increases, so $\frac{dV}{dr}\neq0$.

Good analogy!