The time derivative of the absolute value of a gradient.

Question

I am interested in finding out the time rate of change of the absolute value of the density gradient, such that the directional change of the density gradient does not affect the final sign of the time rate of change.

For a density field $\rho=\rho(\vec{x},t)$, the governing density equation is given by

$\frac{\partial\rho}{\partial t} = g(\vec{x},t) $

what is

$\frac{\partial\lvert\nabla\rho\rvert}{\partial t}=?$

Answer 1

Lets assume things are nice enough for all derivatives to commute. Then $\partial_t ∇ \rho(x,t) = ∇ g(x,t)$, whence $$ \partial_t[|∇ \rho|^2] = \partial_t [∇ \rho · ∇ \rho ] = 2 ∇ \rho · \partial_t ∇ \rho = 2 ∇ \rho · ∇ g$$ at the same time, $$\partial_t[|∇ \rho|^2] = 2|∇ \rho | · \partial_t|∇ \rho|$$ So if we define $\xi$ as the unit vector in the direction of the (hopefully nonvanishing) gradient $∇ \rho$,

$$ \partial_t | ∇ \rho | = \xi · ∇ g$$