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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

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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)


\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}


\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)


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


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