Is this true?
${\frac{d\nabla}{dt} = 0}$
And if it is, then would this be different: ${\frac{d\nabla}{dt} . E = 0}$
What would be the derivative of the divergence with respect to time?
0
$\begingroup$
derivatives
vector-analysis
divergence
-
4It doesn't really make sense to apply a derivative to a differential operator.. Differential operators really only apply to functions. We like to use shorthand notation and apply them to other operators, but it should only ever be thought of as acting on functions. – 2017-02-20
-
0It is because I want to make the following derivative: $ \frac{d \nabla . E}{dt} = \frac{d \nabla}{dt} . E + \nabla . \frac{dE}{dt} $ – 2017-02-20
-
0And I think the first term of the right size is 0, but is it? – 2017-02-20
1 Answers
0
If your coordinate system is stationary then $[\partial_{t}, \nabla]=0$. I.e. $\partial_{t}\nabla\cdot\bf{E}=\nabla\cdot(\partial_{t}\bf{E})$. If you are in the comooving coordiantes, like you always find yourself to be in hydrodynamics, then the above is not tru anymore.