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I'm currently reading the chapter "Potentials and Fields" in Griffiths Electrodynamics, 4th edition.

I actually had a doubt which is more concerned with vector algebra. So basically the equation written in Griffiths is:

$F$ = $\frac{dp}{dt}$ = $q$($E$ + $v$ $\times$ B) = $q$ [-$\nabla$ $V$ - $\frac{\partial A}{\partial t}$ + v $\times$ ($\nabla$ $\times$ $A$)]

Now we know $\nabla(A \cdot B) = A \times (\nabla \times B) + B \times (\nabla \times A) + (A \cdot \nabla)B + (B \cdot \nabla)A $

It contains 4 terms in RHS. But while writing this equation for $v$ and $A$, Griffiths only write two terms.

$ \nabla(v \cdot A) = v \times (\nabla \times A) + (v \cdot \nabla)A $

Why are the other two terms ignored($ A \times(\nabla \times v) + (A \cdot \nabla)v$ )

I understand that if v is independent of position then the curl of v and the divergence of v is zero, but is that always the case? Because the formula is written in generalization.

Mayank Jha
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1 Answers1

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$v$ only depends on time and not on position, whereas $A$ is a field and function of position and time.

my2cts
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  • Is it always true?(that v is independent of position) we can have a situation in which we have velocity dependent on position of the particle? – Mayank Jha Mar 19 '20 at 10:16
  • $v$ is only defined at the position of the particle. It depends only on time. It does not have a gradient or rotation. – my2cts Mar 19 '20 at 11:34