[Electromagnetism] Force on a moving charge expression

[carlosbgois]

Homework Statement

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The total force on a moving charge q with velocity v is given by \mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B}) Using the scalar and vector potentials, show that \mathbf{F}=q[-\nabla\phi-\frac{d\mathbf{A}}{dt}+\nabla(\mathbf{A}\cdot\mathbf{v})]

Homework Equations

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(1) \mathbf{E}=-\frac{d\mathbf{A}}{dt}-\nabla\phi
(2) \mathbf{B}=\nabla\times\mathbf{A}
(3) \mathbf{v}\times(\nabla\times\mathbf{A})=\nabla(\mathbf{v}\cdot\mathbf{A})-\mathbf{A}(\mathbf{v}\cdot\nabla)

The Attempt at a Solution

\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})=q[-\nabla\phi-\frac{d\mathbf{A}}{dt}+\mathbf{v}\times\mathbf{B}]

Now I need to show that

\mathbf{v}\times\mathbf{B}=\nabla(\mathbf{A}\cdot\mathbf{v})

I tried applying (3) but didn't know where to go from there.

 

[TSny]

Homework Equations

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(1) \mathbf{E}=-\frac{d\mathbf{A}}{dt}-\nabla\phi

Here, the time derivative should be a partial derivative $\frac{\partial \mathbf{A}}{\partial t}$ ; i.e., it denotes the rate of change of $\mathbf{A}$ at a fixed point of space. But in the final expression that you want to get to for the force, the derivative is the "convective" time derivative $\frac{d\mathbf{A}}{dt}$ ; i.e., it denotes the rate of change of $\mathbf{A}$ as you move along with the particle. You will need to relate the two types of derivatives. See http://www.continuummechanics.org/cm/materialderivative.html

(3) \mathbf{v}\times(\nabla\times\mathbf{A})=\nabla(\mathbf{v}\cdot\mathbf{A})-\mathbf{A}(\mathbf{v}\cdot\nabla)

The last term is not quite written correctly. In your way of writing it, the Del operator has nothing to act upon.

 

[loops496]

As TSny said \mathbf{E}=-\nabla \phi - \partial_t \mathbf{A} nowusing the triple product identity:
\mathbf{v}\times(\nabla \times \mathbf{A}) = \nabla(\mathbf{A} \cdot \mathbf{v}) - \mathbf{A}(\mathbf{v} \cdot \nabla)
Which in the Lorentz equation:
\mathbf{F} = q \left[-\nabla \phi - \partial_t \mathbf{A} + \nabla(\mathbf{A} \cdot \mathbf{v}) - \mathbf{A}(\mathbf{v} \cdot \nabla) \right]
or
\mathbf{F} = q \left[-\nabla \phi - [\partial_t + (\mathbf{v} \cdot \nabla) ] \mathbf{A} + \nabla(\mathbf{A} \cdot \mathbf{v})\right]
where the term in the square bracket acting on \mathbf{A} is called the convective derivative, viz. (\partial_t + (\mathbf{v} \cdot \nabla) ) \mathbf{A} = \frac{d \mathbf{A}}{dt} as required.

 

[carlosbgois]

Got it! Thank you all.