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In "What the electromagnetic vector potential describes", E. J. Konopinski asserts:

Operational definitions of $\phi$, $\mathbf{A}$ should now be expected to stem from the equation of motion (2) when it is reexpressed in terms of the field description by the potentials, through substitutions from (1):

(3) $\frac{d}{dt} (M\mathbf{v} + q\mathbf{A}) = -\nabla q(\phi - \mathbf{v} · \mathbf{A})$

The referenced equations are:

(1) $\mathbf{B} = \nabla × \mathbf{A}, \mathbf{E} = -\nabla \phi - \frac{∂\mathbf{A}}{∂t}$

(2) $\frac{d }{dt} (M\mathbf{v}) = q(\mathbf{E} + \mathbf{v} × \mathbf{B})$

He doesn't show the identities used in these substitutions. What is the derivation?

Sancol.
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James Bowery
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  • Looks like they did the substitutions of (1) into (2) and used some vector calculus. – Kyle Kanos Oct 18 '15 at 23:33
  • Also I'm pretty sure the total derivative $d/dt = \partial /\partial t + \mathbf{v} \cdot \nabla$ is involved for the vector potential. – Javier Oct 19 '15 at 00:01

1 Answers1

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If you rewrite equation (2) with all the details you get

$$ \left.\frac{d(M\vec v(t))}{dt}\right|_{t=t_0}=q\left(\vec E(\vec r(t_0),t_0)+ \vec v(t_0)\times \vec B(\vec r(t_0),t_0)\right)$$

And taking the ith component (and using $\epsilon_{ijk}$ and the summation convention to write the cross product) you get:

$$M\frac{d^2r_i}{dt^2}=q\left(E_i+ \epsilon_{ijk}v_jB_k\right)$$

Next, using the potentials you get $q\left(E_i+ \epsilon_{ijk}v_jB_k\right)=$

$$q\left(-\partial_i\Phi -\partial_tA_i+ \epsilon_{ijk}v_j\epsilon_{klm}\partial_lA_m\right)=$$ $$q\left(-\partial_i\Phi -\partial_tA_i+ \epsilon_{ijk}\epsilon_{klm}v_j\partial_lA_m\right)=$$ $$q\left(-\partial_i\Phi -\partial_tA_i+ \epsilon_{kij}\epsilon_{klm}v_j\partial_lA_m\right)=$$ $$q\left(-\partial_i\Phi -\partial_tA_i+ (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})v_j\partial_lA_m\right)=$$ $$q\left(-\partial_i\Phi -\partial_tA_i+v_j\partial_iA_j-v_l\partial_lA_i\right)=$$ $$q\left(-\partial_i\Phi -\partial_tA_i+\partial_iv_jA_j-v_l\partial_lA_i\right)=$$ $$q\left(-\partial_i\Phi+\partial_iv_jA_j-\partial_tA_i-v_l\partial_lA_i\right)=$$ $$q\left(-\partial_i(\Phi-v_jA_j)-(\partial_tA_i+v_l\partial_lA_i)\right)=$$ $$q\left(-\partial_i(\Phi-v_jA_j)-d_tA_i\right)=$$ $$q\left(-\partial_i(\Phi-\vec v\cdot \vec A)-d_tA_i\right).$$

So $$Md_{tt}r_i=q\left(-\partial_i(\Phi-\vec v\cdot \vec A)-d_tA_i\right).$$

So $$Md_{t}v_i=-\partial_i(q\Phi-q\vec v\cdot \vec A)-qd_tA_i$$

$$Md_{t}v_i+qd_tA_i=-\partial_i(q\Phi-q\vec v\cdot \vec A)$$

$$d_{t}(Mv_i+qA_i)=-\partial_i(q\Phi-q\vec v\cdot \vec A).$$

So

$$\frac{d(M\vec v_i+q\vec A)}{dt}=-\vec \nabla (q\Phi-q\vec v\cdot \vec A).$$

And remember that first line where I wrote "all the details" ... they should really be in every single line. Many of the lines are straightforward, some use identities and some hinge on those details I only wrote once and in particular all those unwritten sums are just over spatial degrees of freedom and sometimes you have to consider $\vec A=\vec A(x,y,z,t)$ as a field in spacetime and sometimes it is vector function of time $\vec A=\vec A(t)=\vec A(\vec r(t),t)$ and do not let the notation confuse you.

I would never call these operational definitions, in particular the thing evolving on the left hand side is the vector function of time, not the field. And the thing on the right hand side is the spatial field dotted with a globally uniform velocity field (a velocity that doesn't depend on space, just on time).

So in particular these are not dynamical equations for the potential even though from the left hand side it might look so.

Timaeus
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  • The first convention is otherwise known as the Levi-Civita symbol cross product in Einstein notation. https://en.wikipedia.org/wiki/Levi-Civita_symbol#Cross_product_.28two_vectors.29 – James Bowery Oct 22 '15 at 00:35