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## 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

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.