- #1

HJ Farnsworth

- 128

- 1

In Griffiths E&M, 3rd. Ed., on page 214, the following is part of the derivation of the continuity equation (the same derivation is shown on the Wikipedia article for the current density, under the continuity equation section: http://en.wikipedia.org/wiki/Current_density)

[itex]\int[/itex][itex]\nabla[/itex][itex]\cdot[/itex]

**J**d[itex]\tau[/itex]=-[itex]\frac{d}{dt}[/itex][itex]\int[/itex][itex]\rho[/itex]d[itex]\tau[/itex]=-[itex]\int[/itex][itex]\frac{\partial}{\partial t}[/itex][itex]\rho[/itex]d[itex]\tau[/itex],

where the integral is over the volume. Why isn't the relation

[itex]\int[/itex][itex]\nabla[/itex][itex]\cdot[/itex]

**J**d[itex]\tau[/itex]=-[itex]\frac{d}{dt}[/itex][itex]\int[/itex][itex]\rho[/itex]d[itex]\tau[/itex]=-[itex]\int[/itex][itex]\frac{d}{dt}[/itex][itex]\rho[/itex]d[itex]\tau[/itex]?

Ie., in commutating the integral and the derivative, why is it switched to a partial derivative ∂/∂t rather than keeping the total derivative d/dt?

Furthermore, how would the [itex]\frac{\partial x}{\partial t}[/itex], [itex]\frac{\partial y}{\partial t}[/itex], and [itex]\frac{\partial z}{\partial t}[/itex] terms in the total derivative [itex]\frac{d\rho}{dt}[/itex]=[itex]\frac{\partial \rho}{\partial t}[/itex]+[itex]\frac{\partial \rho}{\partial x}[/itex][itex]\frac{\partial x}{\partial t}[/itex]+[itex]\frac{\partial \rho}{\partial y}[/itex][itex]\frac{\partial y}{\partial t}[/itex]+[itex]\frac{\partial \rho}{\partial z}[/itex][itex]\frac{\partial z}{\partial t}[/itex] be interpreted?

I think the answer to my first question might be that those terms are 0, but it's hard to think about whether that's true since I am having trouble interpreting those terms in this equation.

Anyway, thanks for any help that you can give.

-HJ Farnsworth