What would be the derivative of the divergence with respect to time?
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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.
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Comments
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Ivan over 1 year
Is this true?
${\frac{d\nabla}{dt} = 0}$And if it is, then would this be different: ${\frac{d\nabla}{dt} . E = 0}$
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Cameron Williams over 6 yearsIt 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.
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Ivan over 6 yearsIt is because I want to make the following derivative: $ \frac{d \nabla . E}{dt} = \frac{d \nabla}{dt} . E + \nabla . \frac{dE}{dt} $
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Ivan over 6 yearsAnd I think the first term of the right size is 0, but is it?
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