# [Electromagnetism] Force on a moving charge expression

1. Aug 29, 2015

### carlosbgois

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Aug 29, 2015

### TSny

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

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

Last edited: Aug 29, 2015
3. Aug 31, 2015

### 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.

4. Sep 8, 2015

### carlosbgois

Got it! Thank you all.