# Deriving the expression of the Lorentz force using the language of a mathematician

This is a continuation of my previous “complaining” post on the language difference of mathematician vs physicists regarding the same thing.

First keeping in mind that non-conservative only matters in macroscopic level, in atom level every type of force is conservative. Maxwell Faraday’s equation implies that the electric field $\boldsymbol{E}$ is non-conservative when the magnetic field $\boldsymbol{B}$ varies in time, and is not expressible as the gradient of a scalar field, and not subject to the gradient theorem (path independence of a gradient field) since its rotational is not zero.

Here the goal is to derive the equation of motion from Lagrangian of a particle moving in an electromagnetic field with charge $q$, which is described by the Lagrangian

$L=\frac{m\boldsymbol{v}^2}2+\frac qc\boldsymbol{v}\cdot\boldsymbol{A}(\boldsymbol{r},t)-q\phi(\boldsymbol{r},t).$

The electric and magnetic fields $(\boldsymbol{E}(\boldsymbol{r},t)$ and $\boldsymbol{B}(\boldsymbol{r},t)$ correspondingly$)$ are related to “potentials” $\boldsymbol{A}(\boldsymbol{r},t)$ and $\phi(\boldsymbol{r},t)$ introduced above as follows:

$\boldsymbol{E}=-\frac1c\frac{\partial\boldsymbol{A}}{\partial t}-\nabla\phi,\qquad\boldsymbol{B}=\nabla\times\boldsymbol{A}.$

The equation of motion obtained from the Lagrangian basically is the Euler-Lagrange equation:

$\displaystyle\frac{\partial L}{\partial {\boldsymbol{r} }} - \frac{\mathrm d }{\mathrm dt}\left( \frac{\partial L}{\partial {\boldsymbol{r}'}}\right) =0, \tag{1}$

whereas the $\dfrac{\mathrm d }{\mathrm dt}$ represents taking the total derivative (time derivative, plus the convective derivative w.r.t. the spacial variable, or so to speak). Yet prime only stand for the time derivative usually.

Plugging $L$ into $(1)$ we have: (the usual postulate is that $\boldsymbol{v}$ is not explicitly a function of $\boldsymbol{r}$), and sing the identity

$\boldsymbol{v}\times(\nabla\times\boldsymbol{A}) =\nabla(\boldsymbol{v}\cdot\boldsymbol{A})-(\boldsymbol{v}\cdot\nabla)\boldsymbol{A},$

we have

\begin{aligned} \frac{\partial L}{\partial {\boldsymbol r}} &= \nabla \left( \frac{q}{c} \boldsymbol{v}\cdot \boldsymbol{A} - q\phi\right) \\ &= q\left(\frac{1}{c} \nabla(\boldsymbol{v}\cdot \boldsymbol{A}) - \nabla \phi \right) \\ &= q\left(\frac{1}{c} \boldsymbol{v}\times( \nabla\times\boldsymbol{A}) + \frac{1}{c} (\boldsymbol{v}\cdot \nabla) \boldsymbol{A} - \nabla \phi \right). \end{aligned}

And

$\frac{\partial L}{\partial {\boldsymbol{r'}}} = \frac{\partial L}{\partial {\boldsymbol{v}}} = m\boldsymbol{v} + \frac{q}{c}\boldsymbol{A}.$

Now taking the total derivative:

$\frac{\mathrm d }{\mathrm dt}\left( \frac{\partial L}{\partial {\boldsymbol r'}}\right) = \frac{\partial }{\partial t}\left( \frac{\partial L}{\partial {\boldsymbol r'}}\right) + \left(\frac{\mathrm d \boldsymbol{r} }{\mathrm dt}\cdot\nabla\right)\left( \frac{\partial L}{\partial {\boldsymbol r'}}\right).$

Notice the total derivative of $\boldsymbol{r} = (x(t),y(t),z(t))$ is $\boldsymbol{r}’$, hence:

$\frac{\mathrm d }{\mathrm dt}\left( \frac{\partial L}{\partial {\boldsymbol{r'}}}\right) = \frac{\partial }{\partial t}\left(m\boldsymbol{v} + \frac{q}{c}\boldsymbol{A}\right) +(\boldsymbol{v}\cdot \nabla ) \left(m\boldsymbol{v} + \frac{q}{c}\boldsymbol{A}\right) \\ = m\frac{\partial^2 \boldsymbol{r}}{\partial t^2} + \frac{q}{c}\frac{\partial \boldsymbol{A}}{\partial t} + \frac{q}{c}(\boldsymbol{v}\cdot \nabla )\boldsymbol{A},$

where we used that the convective derivative of $\boldsymbol{v} = \boldsymbol{v}(t)$ is zero.

Now put everything back to $(1)$, we have:

$\displaystyle m\frac{\partial^2 \boldsymbol{r}}{\partial t^2} = - \frac{q}{c}\frac{\partial \boldsymbol{A}}{\partial t} +\frac{q}{c} \boldsymbol{v}\times( \nabla\times\boldsymbol{A} )- q\nabla \phi. \tag{2}$

Equation $(2)$ is basically the motion equation of the Lagrangian.

Express the force acting on the particle in terms of electric and magnetic fields only (i.e. the equation of motion should have the form of the Newton’s second law and contain fields $\boldsymbol{E}$ and $\boldsymbol{B}$ but not the ‘potentials’ $\boldsymbol{A}$ and $\phi$).

If the potentials are replaced using the $\boldsymbol{E}$ and $\boldsymbol{B}$ respectively, $(2)$ will become: $m\frac{\partial^2 \boldsymbol{r}}{\partial t^2} = q(\boldsymbol{E} + \boldsymbol{v}\times \boldsymbol{B}).$ This is nothing but the famous Lorentz force formula, a.k.a. the Newton second law for a particle moving through an electromagnetic field.

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