# Peeter Joot's Blog.

• ## Recent Comments

 ivor on Just Energy Canada nasty busin… A final pre-exam upd… on An updated compilation of note… Anon on About peeterjoot on About Anon on About
• ## People not reading this blog: 6,973,738,433 minus:

• 132,815 hits

# Archive for February 8th, 2011

## PHY450H1S. Relativistic Electrodynamics Lecture 10 (Taught by Prof. Erich Poppitz). Lorentz force equation energy term, and four vector formulation of the Lorentz force equation.

Posted by peeterjoot on February 8, 2011

[Click here for a PDF of this post with nicer formatting]

Covering chapter 3 material from the text [1].

Covering lecture notes pp. 74-83: gauge transformations in 3-vector language (74); energy of a relativistic particle in EM field (75); variational principle and equation of motion in 4-vector form (76-77); the field strength tensor (78-80); the fourth equation of motion (81)

# What is the significance to the gauge invariance of the action?

We had argued that under a gauge transformation

\begin{aligned}A_i \rightarrow A_i + \frac{\partial {\chi}}{\partial {x^i}},\end{aligned} \hspace{\stretch{1}}(2.1)

the action for a particle changes by a boundary term

\begin{aligned}- \frac{e}{c} ( \chi(x_b) - \chi(x_a) ).\end{aligned} \hspace{\stretch{1}}(2.2)

Because $S$ changes by a boundary term only, variation problem is not affected. The extremal trajectories are then the same, hence the EOM are the same.

## A less high brow demonstration.

With our four potential split into space and time components

\begin{aligned}A^i = (\phi, \mathbf{A}),\end{aligned} \hspace{\stretch{1}}(2.3)

the lower index representation of the same vector is

\begin{aligned}A_i = (\phi, -\mathbf{A}).\end{aligned} \hspace{\stretch{1}}(2.4)

Our gauge transformation is then

\begin{aligned}A_0 &\rightarrow A_0 + \frac{\partial {\chi}}{\partial {x^0}} \\ -\mathbf{A} &\rightarrow -\mathbf{A} + \frac{\partial {\chi}}{\partial {\mathbf{x}}}\end{aligned} \hspace{\stretch{1}}(2.5)

or

\begin{aligned}\phi &\rightarrow \phi + \frac{1}{{c}}\frac{\partial {\chi}}{\partial {t}} \\ \mathbf{A} &\rightarrow \mathbf{A} - \boldsymbol{\nabla} \chi.\end{aligned} \hspace{\stretch{1}}(2.7)

Now observe how the electric and magnetic fields are transformed

\begin{aligned}\mathbf{E} &= - \boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \\ &\rightarrow - \boldsymbol{\nabla} \left( \phi + \frac{1}{{c}}\frac{\partial {\chi}}{\partial {t}} \right) - \frac{1}{{c}}\frac{\partial {}}{\partial {t}} \left( \mathbf{A} - \boldsymbol{\nabla} \chi \right) \\ \end{aligned}

Sufficient continuity of $\chi$ is assumed, allowing commutation of the space and time derivatives, and we are left with just $\mathbf{E}$

For the magnetic field we have

\begin{aligned}\mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{A} \\ &\rightarrow \boldsymbol{\nabla} \times (\mathbf{A} - \boldsymbol{\nabla} \chi) \\ \end{aligned}

Again with continuity assumptions, $\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \chi) = 0$, and we are left with just $\mathbf{B}$. The electromagnetic fields (as opposed to potentials) do not change under gauge transformations.

We conclude that the $\{A_i\}$ description is hugely redundant, but despite that, local $\mathcal{L}$ and $H$ can only be written in terms of the potentials $A_i$.

## Energy term of the Lorentz force. Three vector approach.

With the Lagrangian for the particle given by

\begin{aligned}\mathcal{L} = - mc^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} + \frac{e}{c} \mathbf{A} \cdot \mathbf{v} - e \phi,\end{aligned} \hspace{\stretch{1}}(2.9)

we define the energy as

\begin{aligned}\mathcal{E} = \mathbf{v} \cdot \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} - \mathcal{L}\end{aligned} \hspace{\stretch{1}}(2.10)

This is not necessarily a conserved quantity, but we define it as the energy anyways (we don’t really have a Hamiltonian when the fields are time dependent). Associated with this quantity is the general relationship

\begin{aligned}\frac{d{{\mathcal{E}}}}{dt} = -\frac{\partial {\mathcal{L}}}{\partial {t}},\end{aligned} \hspace{\stretch{1}}(2.11)

and when the Lagrangian is invariant with respect to time translation the energy $\mathcal{E}$ will be a conserved quantity (and also the Hamiltonian).

Our canonical momentum is

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = \gamma m \mathbf{v} + \frac{e}{c} \mathbf{A}\end{aligned} \hspace{\stretch{1}}(2.12)

So our energy is

\begin{aligned}\mathcal{E} = \gamma m \mathbf{v}^2 + \frac{e}{c} \mathbf{A} \cdot \mathbf{v} - \left( - mc^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} + \frac{e}{c} \mathbf{A} \cdot \mathbf{v} - e \phi \right).\end{aligned}

Or

\begin{aligned}\mathcal{E} = \underbrace{\frac{m c^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}}}_{({*})} + e \phi.\end{aligned} \hspace{\stretch{1}}(2.13)

The contribution of $({*})$ to the energy $\mathcal{E}$ comes from the free (kinetic) particle portion of the Lagrangian $\mathcal{L} = -m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}}$, and we identify the remainder as a potential energy

\begin{aligned}\mathcal{E} = \frac{m c^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} + \underbrace{e \phi}_{\text{"potential"}}.\end{aligned} \hspace{\stretch{1}}(2.14)

For the kinetic portion we can also show that we have

\begin{aligned}\frac{d}{dt} \mathcal{E}_{\text{kinetic}} =\frac{m c^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} = e \mathbf{E} \cdot \mathbf{v}.\end{aligned} \hspace{\stretch{1}}(2.15)

To show this observe that we have

\begin{aligned}\frac{d}{dt} \mathcal{E}_{\text{kinetic}} &= m c^2 \frac{d\gamma}{dt} \\ &= m c^2 \frac{d}{dt} \frac{1}{{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}}} \\ &= m c^2 \frac{\frac{\mathbf{v}}{c^2} \cdot \frac{d\mathbf{v}}{dt}}{\left(1 - \frac{\mathbf{v}^2}{c^2}\right)^{3/2}} \\ &= \frac{m \gamma \mathbf{v} \cdot \frac{d\mathbf{v}}{dt}}{1 - \frac{\mathbf{v}^2}{c^2}}\end{aligned}

We also have

\begin{aligned}\mathbf{v} \cdot \frac{d{\mathbf{p}}}{dt} &= \mathbf{v} \cdot \frac{d{{}}}{dt} \frac{m \mathbf{v}}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} \\ &= m\mathbf{v}^2 \frac{d{{\gamma}}}{dt} + m \gamma \mathbf{v} \cdot \frac{d{\mathbf{v}}}{dt} \\ &= m\mathbf{v}^2 \frac{d{{\gamma}}}{dt} + m c^2 \frac{d{{\gamma}}}{dt} \left( 1 - \frac{\mathbf{v}^2}{c^2} \right) \\ &= m c^2 \frac{d{{\gamma}}}{dt}.\end{aligned}

Utilizing the Lorentz force equation, we have

\begin{aligned}\mathbf{v} \cdot \frac{d{\mathbf{p}}}{dt} = e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right) \cdot \mathbf{v} = e \mathbf{E} \cdot \mathbf{v}\end{aligned} \hspace{\stretch{1}}(2.16)

and are able to assemble the above, and find that we have

\begin{aligned}\frac{d{{(m c^2 \gamma)}}}{dt} = e \mathbf{E} \cdot \mathbf{v} \end{aligned} \hspace{\stretch{1}}(2.17)

# Four vector Lorentz force

Using $ds = \sqrt{ dx^i dx_i }$ our action can be rewritten

\begin{aligned}S &= \int \left( -m c ds - \frac{e}{c} u^i A_i ds \right) \\ &= \int \left( -m c ds - \frac{e}{c} dx^i A_i \right) \\ &= \int \left( -m c \sqrt{ dx^i dx_i} - \frac{e}{c} dx^i A_i \right) \\ \end{aligned}

$x^i(\tau)$ is a worldline $x^i(0) = a^i$, $x^i(1) = b^i$,

We want $\delta S = S[ x + \delta x ] - S[ x ] = 0$ (to linear order in $\delta x$)

The variation of our proper length is

\begin{aligned}\delta ds &=\delta \sqrt{ dx^i dx_i } \\ &= \frac{1}{{ 2 \sqrt{ dx^i dx_i }}} \delta (dx^j dx_j)\end{aligned}

Observe that for the numerator we have

\begin{aligned}\delta (dx^j dx_j) &= \delta ( dx^j g_{jk} dx^k ) \\ &= \delta ( dx^j ) g_{jk} dx^k + dx^j g_{jk} \delta ( dx^k ) \\ &= \delta ( dx^j ) g_{jk} dx^k + dx^k g_{kj} \delta ( dx^j ) \\ &= 2 \delta ( dx^j ) g_{jk} dx^k \\ &= 2 \delta ( dx^j ) dx_j \end{aligned}

\paragraph{TIP:} If this goes too quick, or there is any disbelief, write these all out explicitly as $dx^j dx_j = dx^0 dx_0 + dx^1 dx_1 + dx^2 dx_2 + dx^3 dx_3$ and compute it that way.

For the four vector potential our variation is

\begin{aligned}\delta A_i = A_i(x + \delta x) - A_i = \frac{\partial {A_i}}{\partial {x^j}} \delta x^j = \partial_j A_i \delta x^j\end{aligned} \hspace{\stretch{1}}(3.18)

(i.e. By chain rule)

Completing the proper length variations above we have

\begin{aligned}\delta \sqrt{ dx^i dx_i } &= \frac{1}{{ \sqrt{ dx^i dx_i }}} \delta (dx^j) dx_j \\ &= \delta (dx^j) \frac{d{{x_j}}}{ds} \\ &= \delta (dx^j) u_j \\ &= d \delta x^j u_j\end{aligned}

We are now ready to assemble results and do the integration by parts

\begin{aligned}\delta S &= \int \left( -m c d (\delta x^j) u_j- \frac{e}{c} d (\delta x^i) A_i - \frac{e}{c} dx^i \partial_j A_i \delta x^j\right) \\ &= {\left. \left( -m c (\delta x^j) u_j - \frac{e}{c} (\delta x^i) A_i \right)\right\vert}_a^b+\int \left( m c \delta x^j d u_j+ \frac{e}{c} (\delta x^i) d A_i - \frac{e}{c} dx^i \partial_j A_i \delta x^j\right) \\ \end{aligned}

Our variation at the endpoints is zero ${\left.{{\delta x^i}}\right\vert}_{{a}} = {\left.{{\delta x^i}}\right\vert}_{{b}} = 0$, killing the non-integral terms

\begin{aligned}\delta S &= \int \delta x^j\left( m c d u_j+ \frac{e}{c} d A_j - \frac{e}{c} dx^i \partial_j A_i \right).\end{aligned}

Observe that our differential can also be expanded by chain rule

\begin{aligned}d A_j = \frac{\partial {A_j}}{\partial {x^i}} dx^i = \partial_i A_j dx^i,\end{aligned} \hspace{\stretch{1}}(3.19)

which simplifies the variation further

\begin{aligned}\delta S &= \int \delta x^j\left( m c d u_j+ \frac{e}{c} dx^i ( \partial_i A_j - \partial_j A_i )\right) \\ &= \int \delta x^j ds\left( m c \frac{d u_j}{ds}+ \frac{e}{c} u^i ( \partial_i A_j - \partial_j A_i )\right) \\ \end{aligned}

Since this is true for all variations $\delta x^j$, which is arbitrary, the interior part is zero everywhere in the trajectory. The antisymmetric portion, a rank 2 4-tensor is called the electromagnetic field strength tensor, and written

\begin{aligned}\boxed{F_{ij} = \partial_i A_j - \partial_j A_i.}\end{aligned} \hspace{\stretch{1}}(3.20)

In matrix form this is

\begin{aligned}{\left\lVert{ F_{ij} }\right\rVert} = \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0.\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.21)

In terms of the field strength tensor our Lorentz force equation takes the form

\begin{aligned}\boxed{\frac{d{{(m c u_i)}}}{ds} = \frac{e}{c} F_{ij} u^j.}\end{aligned} \hspace{\stretch{1}}(3.22)

# References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

## Energy term of the Lorentz force equation.

Posted by peeterjoot on February 8, 2011

[Click here for a PDF of this post with nicer formatting]

# Motivation.

In class this week, the Lorentz force was derived from an action (the simplest Lorentz invariant, gauge invariant, action that could be constructed)

\begin{aligned}S = - m c \int ds - \frac{e}{c} \int ds A^i u_i.\end{aligned} \hspace{\stretch{1}}(1.1)

We end up with the familiar equation, with the exception that the momentum includes the relativistically required gamma factor

\begin{aligned}\frac{d (\gamma m \mathbf{v})}{dt} = e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(1.2)

I asked what the energy term of this equation would be and was answered that we would get to it, and it could be obtained by a four vector minimization of the action which produces the Lorentz force equation of the following form

\begin{aligned}\frac{du^i}{d\tau} \propto e F^{ij} u_j.\end{aligned} \hspace{\stretch{1}}(1.3)

Let’s see if we can work this out without the four-vector approach, using the action expressed with an explicit space time split, then also work it out in the four vector form and compare as a consistency check.

# Three vector approach.

## The Lorentz force derivation.

For completeness, let’s work out the Lorentz force equation from the action 1.1. Parameterizing by time we have

\begin{aligned}S &= -m c^2 \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} \gamma \left( 1, \frac{1}{{c}} \mathbf{v}\right) \cdot (\phi, \mathbf{A}) \\ &= -m c^2 \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \int dt \left( \phi - \frac{1}{{c}} \mathbf{A} \cdot \mathbf{v} \right)\end{aligned}

Our Lagrangian is therefore

\begin{aligned}\mathcal{L}(\mathbf{x}, \mathbf{v}, t) = -m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \phi(\mathbf{x}, t) + \frac{e}{c} \mathbf{A}(\mathbf{x}, t) \cdot \mathbf{v}\end{aligned} \hspace{\stretch{1}}(2.4)

We can calculate our conjugate momentum easily enough

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = \gamma m \mathbf{v} + \frac{e}{c} \mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.5)

and for the gradient portion of the Euler-Lagrange equations we have

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{x}}} = -e \boldsymbol{\nabla} \phi + e \boldsymbol{\nabla} \left( \frac{\mathbf{v}}{c} \cdot \mathbf{A} \right).\end{aligned} \hspace{\stretch{1}}(2.6)

Utilizing the convective derivative (i.e. chain rule in fancy clothes)

\begin{aligned}\frac{d}{dt} = \mathbf{v} \cdot \boldsymbol{\nabla} + \frac{\partial {}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(2.7)

This gives us

\begin{aligned}-e \boldsymbol{\nabla} \phi + e \boldsymbol{\nabla} \left( \frac{\mathbf{v}}{c} \cdot \mathbf{A} \right) = \frac{d(\gamma m \mathbf{v})}{dt} + \frac{e}{c} (\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{A}+ \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}},\end{aligned} \hspace{\stretch{1}}(2.8)

and a final bit of rearranging gives us

\begin{aligned}\frac{d(\gamma m \mathbf{v})}{dt} =e \left( -\boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}}\right)+ \frac{e}{c} \left( \boldsymbol{\nabla} \left( \mathbf{v} \cdot \mathbf{A} \right) - (\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{A}\right).\end{aligned} \hspace{\stretch{1}}(2.9)

The first set of derivatives we identify with the electric field $\mathbf{E}$. For the second, utilizing the vector triple product identity [1]

\begin{aligned}\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b} (\mathbf{a} \cdot \mathbf{c}) - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c},\end{aligned} \hspace{\stretch{1}}(2.10)

we recognize as related to the magnetic field $\mathbf{v} \times \mathbf{B} = \mathbf{v} \times (\boldsymbol{\nabla} \times \mathbf{A})$.

## The power (energy) term.

When we start with an action explicitly constructed with Lorentz invariance as a requirement, it is somewhat odd to end up with a result that has only the spatial vector portion of what should logically be a four vector result. We have an equation for the particle momentum, but not one for the energy. In tutorial Simon provided the hint of how to approach this, and asked if we had calculated the Hamiltonian for the Lorentz force. We had only calculated the Hamiltonian for the free particle.

Considering this, we can only actually calculate a Hamiltonian for the case where $\phi(\mathbf{x}, t) = \phi(\mathbf{x})$ and $\mathbf{A}(\mathbf{x}, t) = \mathbf{A}(\mathbf{x})$, because when the potentials have any sort of time dependence we do not have a Lagrangian that is invariant under time translation. Returning to the derivation of the Hamiltonian conservation equation, we see that we must modify the argument slightly when there is a time dependence and get instead

\begin{aligned}\frac{d}{dt} \left( \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} \cdot \mathcal{L} - \mathcal{L} \right) + \frac{\partial {\mathcal{L}}}{\partial {t}} = 0.\end{aligned} \hspace{\stretch{1}}(2.11)

Only when there is no time dependence in the Lagrangian, do we have our conserved quantity, what we label as energy, or Hamiltonian.

From 2.5, we have

\begin{aligned}0 &= \frac{d}{dt} \left( \left( \gamma m \mathbf{v} + \frac{e}{c} \mathbf{A} \right) \cdot \mathbf{v} +m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} + e \phi - \frac{e}{c} \mathbf{A} \cdot \mathbf{v}\right) - e \frac{\partial {\phi}}{\partial {t}} + \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \cdot \mathbf{v} \\ \end{aligned}

Our $\mathbf{A} \cdot \mathbf{v}$ terms cancel, and we can combine the $\gamma$ and $\gamma^{-1}$ terms, then apply the convective derivative again

\begin{aligned}\frac{d}{dt} \left( \gamma m c^2 \right) &= - e \left( \mathbf{v} \cdot \boldsymbol{\nabla} + \frac{\partial {}}{\partial {t}} \right) \phi + e \frac{\partial {\phi}}{\partial {t}} - \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \cdot \mathbf{v} \\ &= - e \mathbf{v} \cdot \boldsymbol{\nabla} \phi - \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \cdot \mathbf{v} \\ &= + e \mathbf{v} \cdot \left( - \boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \right).\end{aligned}

This is just

\begin{aligned}\frac{d}{dt} \left( \gamma m c^2 \right) = e \mathbf{v} \cdot \mathbf{E},\end{aligned} \hspace{\stretch{1}}(2.12)

and we find the rate of change of energy term of our four momentum equation

\begin{aligned}\frac{d}{dt}\left( \frac{E}{c}, \mathbf{p}\right) = e \left( \frac{\mathbf{v}}{c} \cdot \mathbf{E}, \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(2.13)

Specified explicilty, this is

\begin{aligned}\frac{d}{dt}\left( \gamma m \left( c, \mathbf{v} \right) \right)= e \left( \frac{\mathbf{v}}{c} \cdot \mathbf{E}, \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(2.14)

While this was the result I was looking for, once written it now stands out as incomplete relativistically. We have an equation that specifies the time derivative of a four vector. What about the spatial derivatives? We really ought to have a rank two tensor result, and not a four vector result relating the fields and the energy and momentum of the particle. The Lorentz force equation, even when expanded to four vector form, does not seem complete relativistically.

With $u^i = dx^i/ds$, we can rewrite 2.14 as

\begin{aligned}\partial_0 (\gamma m u^i) = e \left( \frac{\mathbf{v}}{c} \cdot \mathbf{E}, \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(2.15)

If we were to vary the action with respect to a spatial coordinate instead of time, we should end up with a similar equation of the form $\partial_i (\gamma m u^i) = ?$. Having been pointed at the explicitly invariant result, I wonder if those equations are independent. Let’s defer exploring this, until at least after calculating the result using a four vector form of the action.

# Four vector approach.

## The Lorentz force derivation from invariant action.

We can rewrite our action, parameterizing with proper time. This is

\begin{aligned}S = -m c^2 \int d\tau \sqrt{ \frac{dx^i}{d\tau} \frac{dx_i}{d\tau} }- \frac{e}{c} \int d\tau A_i \frac{dx^i}{d\tau}\end{aligned} \hspace{\stretch{1}}(3.16)

Writing $\dot{x}^i = dx^i/d\tau$, our Lagrangian is then

\begin{aligned}\mathcal{L}(x^i, \dot{x^i}, \tau)= -m c^2 \sqrt{ \dot{x}^i \dot{x}_i }- \frac{e}{c} A_i \dot{x}^i\end{aligned} \hspace{\stretch{1}}(3.17)

The Euler-Lagrange equations take the form

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {x^i}} = \frac{d}{d\tau} \frac{\partial {\mathcal{L}}}{\partial {\dot{x}^i}} .\end{aligned} \hspace{\stretch{1}}(3.18)

Our gradient and conjugate momentum are

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {x^i}} &= - \frac{e}{c} \frac{\partial {A_j}}{\partial {x^i}} \dot{x}^j \\ \frac{\partial {\mathcal{L}}}{\partial {\dot{x}^i}} &= -m \dot{x}_i - \frac{e}{c} A_i.\end{aligned} \hspace{\stretch{1}}(3.19)

With our convective derivative taking the form

\begin{aligned}\frac{d}{d\tau} = \dot{x}^i \frac{\partial {}}{\partial {x^i}},\end{aligned} \hspace{\stretch{1}}(3.21)

we have

\begin{aligned}m \frac{d^2 x_i}{d\tau^2} &= \frac{e}{c} \frac{\partial {A_j}}{\partial {x^i}} \dot{x}^j- \frac{e}{c} \dot{x}^j \frac{\partial {A_i}}{\partial {x^j}} \\ &=\frac{e}{c} \dot{x}^j \left( \frac{\partial {A_j}}{\partial {x^i}} -\frac{\partial {A_i}}{\partial {x^j}} \right) \\ &=\frac{e}{c} \dot{x}^j \left( \partial_i A_j - \partial_j A_i\right) \\ &=\frac{e}{c} \dot{x}^j F_{ij}\end{aligned}

Our Prof wrote this with indexes raised and lowered respectively

\begin{aligned}m \frac{d^2 x^i}{d\tau^2} = \frac{e}{c} F^{ij} \dot{x}_j .\end{aligned} \hspace{\stretch{1}}(3.22)

Following the text [2] he also writes $u^i = dx^i/ds = (1/c) dx^i/d\tau$, and in that form we have

\begin{aligned}\frac{d (m c u^i)}{ds} = \frac{e}{c} F^{ij} u_j.\end{aligned} \hspace{\stretch{1}}(3.23)

## Expressed explicitly in terms of the three vector fields.

### The power term.

From 3.23, lets extract the $i=0$ term, relating the rate of change of energy to the field and particle velocity. With

\begin{aligned}\frac{d{{}}}{d\tau} = \frac{dt}{d\tau} \frac{d}{dt} = \gamma \frac{d{{}}}{dt},\end{aligned} \hspace{\stretch{1}}(3.24)

we have

\begin{aligned}\frac{d{{(m \gamma \frac{dx^i}{dt})}}}{dt} = \frac{e}{c} F^{ij} \frac{d{{x_j}}}{dt}.\end{aligned} \hspace{\stretch{1}}(3.25)

For $i=0$ we have

\begin{aligned}F^{0j} \frac{d{{x_j}}}{dt} = -F^{0\alpha} \frac{d{{x^\alpha}}}{dt} \end{aligned} \hspace{\stretch{1}}(3.26)

That component of the field is

\begin{aligned}F^{\alpha 0} &=\partial^\alpha A^0 - \partial^0 A^\alpha \\ &=-\frac{\partial {\phi}}{\partial {x^\alpha}} - \frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} \\ &= \left( -\boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \right)^\alpha.\end{aligned}

This verifies the result obtained with considerably more difficulty, using the Hamiltonian like conservation relation obtained for a time translation of a time dependent Lagrangian

\begin{aligned}\frac{d{{(m \gamma c^2 )}}}{dt} = e \mathbf{E} \cdot \mathbf{v}.\end{aligned} \hspace{\stretch{1}}(3.27)

### The Lorentz force terms.

Let’s also verify the signs for the $i > 0$ terms. For those we have

\begin{aligned}\frac{d{{(m \gamma \frac{dx^\alpha}{dt})}}}{dt} &= \frac{e}{c} F^{\alpha j} \frac{d{{x_j}}}{dt} \\ &= \frac{e}{c} F^{\alpha 0} \frac{d{{x_0}}}{dt}+\frac{e}{c} F^{\alpha \beta} \frac{d{{x_\beta}}}{dt} \\ &= e E^\alpha- \sum_{\alpha \beta} \frac{e}{c} \left( \partial^\alpha A^\beta - \partial^\beta A^\alpha\right)v^\beta \\ \end{aligned}

Since we have only spatial indexes left, lets be sloppy and imply summation over all repeated indexes, even if unmatched upper and lower. This leaves us with

\begin{aligned}-\left( \partial^\alpha A^\beta - \partial^\beta A^\alpha \right) v^\beta &=\left( \partial_\alpha A^\beta - \partial_\beta A^\alpha \right) v^\beta \\ &=\epsilon_{\alpha \beta \gamma} B^\gamma\end{aligned}

With the $v^\beta$ contraction we have

\begin{aligned}\epsilon_{\alpha \beta \gamma} B^\gamma v^\beta = (\mathbf{v} \times \mathbf{B})^\alpha,\end{aligned} \hspace{\stretch{1}}(3.28)

leaving our first result obtained by the time parameterization of the Lagrangian

\begin{aligned}\frac{d{{(m \gamma \mathbf{v})}}}{dt} = e \left(\mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(3.29)

This now has a nice symmetrical form. It’s slightly disappointing not to have a rank two tensor on the LHS like we have with the symmetric stress tensor with Poynting Vector and energy and other similar terms that relates field energy and momentum with $\mathbf{E} \cdot \mathbf{J}$ and the charge density equivalents of the Lorentz force equation. Is there such a symmetric relationship for particles too?

# References

[1] Wikipedia. Triple product — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 7-February-2011]. http://en.wikipedia.org/w/index.php?title=Triple_product&oldid=407455209.

[2] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.