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Homework Help: [Electromagnetism] Force on a moving charge expression

  1. Aug 29, 2015 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. The attempt at a solution


    Now I need to show that


    I tried applying (3) but didn't know where to go from there.
  2. jcsd
  3. Aug 29, 2015 #2


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    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
  4. Aug 31, 2015 #3
    As TSny said [itex]\mathbf{E}=-\nabla \phi - \partial_t \mathbf{A}[/itex] nowusing the triple product identity:
    [tex]\mathbf{v}\times(\nabla \times \mathbf{A}) = \nabla(\mathbf{A} \cdot \mathbf{v}) - \mathbf{A}(\mathbf{v} \cdot \nabla)[/tex]
    Which in the Lorentz equation:
    [tex]\mathbf{F} = q \left[-\nabla \phi - \partial_t \mathbf{A} + \nabla(\mathbf{A} \cdot \mathbf{v}) - \mathbf{A}(\mathbf{v} \cdot \nabla) \right] [/tex]
    [tex]\mathbf{F} = q \left[-\nabla \phi - [\partial_t + (\mathbf{v} \cdot \nabla) ] \mathbf{A} + \nabla(\mathbf{A} \cdot \mathbf{v})\right] [/tex]
    where the term in the square bracket acting on [itex]\mathbf{A}[/itex] is called the convective derivative, viz. [itex](\partial_t + (\mathbf{v} \cdot \nabla) ) \mathbf{A} = \frac{d \mathbf{A}}{dt}[/itex] as required.
  5. Sep 8, 2015 #4
    Got it! Thank you all.
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