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# Posts Tagged ‘lagrangian’

## Two body harmonic oscillator in 3D, and 2D diamond lattice vibrations

Posted by peeterjoot on January 7, 2014

# Abridged harmonic oscillator notes

[This is an abbreviation of more extensive PDF notes associated with the latter part of this post.]

# Motivation and summary of harmonic oscillator background

After having had some trouble on a non-1D harmonic oscillator lattice problem on the exam, I attempted such a problem with enough time available to consider it properly. I found it helpful to consider first just two masses interacting harmonically in 3D, each displaced from an equilibrium position.

The Lagrangian that described this most naturally was found to be

\begin{aligned}\mathcal{L} = \frac{1}{2} m_1 \left( \dot{\mathbf{r}}_1 \right)^2+\frac{1}{2} m_2 \left( \dot{\mathbf{r}}_2 \right)^2- \frac{K}{2} \left( \left\lvert {\mathbf{r}_2 - \mathbf{r}_1} \right\rvert - a \right)^2.\end{aligned} \hspace{\stretch{1}}(2.1)

This was solved in absolute and displacement coordinates, and then I moved on to consider a linear expansion of the harmonic potential about the equilibrium point, a problem closer to the exam problem (albeit still considering only two masses). The equilibrium points were described with vectors $\mathbf{a}_1, \mathbf{a}_2$ as in fig. 2.1, where $\Delta \mathbf{a} = \left\lvert {\Delta \mathbf{a}} \right\rvert (\cos \theta_1, \cos\theta_2, \cos\theta_3)$.

fig 2.1: Direction cosines relative to equilibrium position difference vector

Using such coordinates, and generalizing, it was found that the complete Lagrangian, to second order about the equilibrium positions, is

\begin{aligned}\mathcal{L} = \sum_j \frac{m_i}{2} \dot{u}_{ij}^2 -\frac{K}{2} \sum_{i j} \cos\theta_i \cos\theta_j \left( u_{2 i} - u_{1 i} \right)\left( u_{2 j} - u_{1 j} \right).\end{aligned} \hspace{\stretch{1}}(2.2)

Evaluating the Euler-Lagrange equations, the equations of motion for the displacements were found to be

\begin{aligned}\begin{aligned}m_1 \ddot{\mathbf{u}}_1 &= K \widehat{\Delta \mathbf{a}} \left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u} \right) \\ m_2 \ddot{\mathbf{u}}_2 &= -K \widehat{\Delta \mathbf{a}} \left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u} \right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.3)

or

\begin{aligned}\boxed{\begin{aligned}\mu \Delta \ddot{\mathbf{u}} &= -K \widehat{\Delta \mathbf{a}} \left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u} \right) \\ m_1 \ddot{\mathbf{u}}_1 + m_2 \ddot{\mathbf{u}}_2 &= 0.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(2.4)

Observe that on the RHS above we have a projection operator, so we could also write

\begin{aligned}\mu \Delta \ddot{\mathbf{u}} = -K \text{Proj}_{\widehat{\Delta \mathbf{a}}} \Delta \mathbf{u}.\end{aligned} \hspace{\stretch{1}}(2.5)

We see that the equations of motion for the displacements of a system of harmonic oscillators has a rather pleasant expression in terms of projection operators, where we have projections onto the unit vectors between each pair of equilibrium position.

# A number of harmonically coupled masses

Now let’s consider masses at lattice points indexed by a lattice vector $\mathbf{n}$, as illustrated in fig. 2.2.

fig 2.2: Masses harmonically coupled in a lattice

With a coupling constant of $K_{\mathbf{n} \mathbf{m}}$ between lattice points indexed $\mathbf{n}$ and $\mathbf{m}$ (located at $\mathbf{a}_\mathbf{n}$ and $\mathbf{a}_\mathbf{m}$ respectively), and direction cosines for the equilibrium direction vector between those points given by

\begin{aligned}\mathbf{a}_\mathbf{n} - \mathbf{a}_\mathbf{m} = \Delta \mathbf{a}_{\mathbf{n} \mathbf{m}}= \left\lvert {\Delta \mathbf{a}_{\mathbf{n} \mathbf{m}}} \right\rvert (\cos \theta_{\mathbf{n} \mathbf{m} 1},\cos \theta_{\mathbf{n} \mathbf{m} 2},\cos \theta_{\mathbf{n} \mathbf{m} 3}),\end{aligned} \hspace{\stretch{1}}(2.6)

the Lagrangian is

\begin{aligned}\mathcal{L} = \sum_{\mathbf{n}, i} \frac{m_\mathbf{n}}{2} \dot{u}_{\mathbf{n} i}^2-\frac{1}{2} \sum_{\mathbf{n} \ne \mathbf{m}, i, j} \frac{K_{\mathbf{n} \mathbf{m}}}{2} \cos\theta_{\mathbf{n} \mathbf{m} i}\cos\theta_{\mathbf{n} \mathbf{m} j}\left( u_{\mathbf{n} i} - u_{\mathbf{m} i} \right)\left( u_{\mathbf{n} j} - u_{\mathbf{m} j} \right)\end{aligned} \hspace{\stretch{1}}(2.7)

Evaluating the Euler-Lagrange equations for the mass at index $\mathbf{n}$ we have

\begin{aligned}\frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{u}_{\mathbf{n} k}}} =m_\mathbf{n} \ddot{u}_{\mathbf{n} k},\end{aligned} \hspace{\stretch{1}}(2.8)

and

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {u_{\mathbf{n} k}}} &= -\sum_{\mathbf{m}, i, j}\frac{K_{\mathbf{n} \mathbf{m}}}{2} \cos\theta_{\mathbf{n} \mathbf{m} i}\cos\theta_{\mathbf{n} \mathbf{m} j}\left(\delta_{i k}\left( u_{\mathbf{n} j} - u_{\mathbf{m} j} \right)+\left( u_{\mathbf{n} i} - u_{\mathbf{m} i} \right)\delta_{j k}\right) \\ &= -\sum_{\mathbf{m}, i}K_{\mathbf{n} \mathbf{m}}\cos\theta_{\mathbf{n} \mathbf{m} k}\cos\theta_{\mathbf{n} \mathbf{m} i}\left( u_{\mathbf{n} i} - u_{\mathbf{m} i} \right) \\ &= -\sum_{\mathbf{m}}K_{\mathbf{n} \mathbf{m}}\cos\theta_{\mathbf{n} \mathbf{m} k}\widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}},\end{aligned} \hspace{\stretch{1}}(2.9)

where $\Delta \mathbf{u}_{\mathbf{n} \mathbf{m}} = \mathbf{u}_\mathbf{n} - \mathbf{u}_\mathbf{m}$. Equating both, we have in vector form

\begin{aligned}m_\mathbf{n} \ddot{\mathbf{u}}_\mathbf{n} = -\sum_{\mathbf{m}}K_{\mathbf{n} \mathbf{m}}\widehat{\Delta \mathbf{a}}\left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}} \right),\end{aligned} \hspace{\stretch{1}}(2.10)

or

\begin{aligned}\boxed{m_\mathbf{n} \ddot{\mathbf{u}}_\mathbf{n} = -\sum_{\mathbf{m}}K_{\mathbf{n} \mathbf{m}}\text{Proj}_{ \widehat{\Delta \mathbf{a}} } \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}},}\end{aligned} \hspace{\stretch{1}}(2.11)

This is an intuitively pleasing result. We have displacement and the direction of the lattice separations in the mix, but not the magnitude of the lattice separation itself.

# Two atom basis, 2D diamond lattice

As a concrete application of the previously calculated equilibrium harmonic oscillator result, let’s consider a two atom basis diamond lattice where the horizontal length is $a$ and vertical height is $b$.

Indexing for the primitive unit cells is illustrated in fig. 2.3.

fig 2.3: Primitive unit cells for rectangular lattice

Let’s write

\begin{aligned}\begin{aligned}\mathbf{r} &= a (\cos\theta, \sin\theta) = a \hat{\mathbf{r}} \\ \mathbf{s} &= a (-\cos\theta, \sin\theta) = a \hat{\mathbf{s}} \\ \mathbf{n} &= (n_1, n_2) \\ \mathbf{r}_\mathbf{n} &= n_1 \mathbf{r} + n_2 \mathbf{s},\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.12)

For mass $m_\alpha, \alpha \in \{1, 2\}$ assume a trial solution of the form

\begin{aligned}\mathbf{u}_{\mathbf{n},\alpha} = \frac{\boldsymbol{\epsilon}_\alpha(\mathbf{q})}{\sqrt{m_\alpha}} e^{i \mathbf{r}_n \cdot \mathbf{q} - \omega t}.\end{aligned} \hspace{\stretch{1}}(2.13)

The equations of motion for the two particles are

\begin{aligned}\begin{aligned}m_1 \ddot{\mathbf{u}}_{\mathbf{n}, 1} &= - K_1 \text{Proj}_{\hat{\mathbf{x}}} \left( \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} - (0,1), 2} \right)- K_1 \text{Proj}_{\hat{\mathbf{x}}} \left( \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} - (1,0), 2} \right) \\ & \quad- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left( \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n}, 2} \right)- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left( \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} - (1,1), 2} \right) \\ & \quad- K_3 \sum_\pm\text{Proj}_{\hat{\mathbf{r}}} \left( \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} \pm (1,0), 1} \right)- K_4 \sum_\pm\text{Proj}_{\hat{\mathbf{s}}} \left( \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} \pm (0,1), 1} \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.14.14)

\begin{aligned}\begin{aligned}m_2 \ddot{\mathbf{u}}_{\mathbf{n}, 2} &= - K_1 \text{Proj}_{\hat{\mathbf{x}}} \left( \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} + (1,0), 1} \right)- K_1 \text{Proj}_{\hat{\mathbf{x}}} \left( \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} + (0,1), 1} \right)\\ &\quad- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left( \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n}, 1} \right)- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left( \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} + (1,1), 1} \right)\\ &\quad- K_3 \sum_\pm\text{Proj}_{\hat{\mathbf{r}}} \left( \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} \pm (1,0), 2} \right)- K_4 \sum_\pm\text{Proj}_{\hat{\mathbf{s}}} \left( \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} \pm (0,1), 2} \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.14.14)

Insertion of the trial solution gives

\begin{aligned}\begin{aligned} \omega^2 \sqrt{m_1} \boldsymbol{\epsilon}_1&= K_1 \text{Proj}_{\hat{\mathbf{x}}} \left( \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} e^{ - i \mathbf{s} \cdot \mathbf{q} } \right)+ K_1 \text{Proj}_{\hat{\mathbf{x}}} \left( \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} e^{ - i \mathbf{r} \cdot \mathbf{q} } \right) \\ &\quad+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left( \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} \right)+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left( \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} e^{ - i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} } \right) \\ &\quad+ K_3 \left( \text{Proj}_{\hat{\mathbf{r}}} \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} \right)\sum_\pm\left( 1 - e^{ \pm i \mathbf{r} \cdot \mathbf{q} } \right)+ K_4 \left( \text{Proj}_{\hat{\mathbf{s}}} \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} \right)\sum_\pm\left( 1 - e^{ \pm i \mathbf{s} \cdot \mathbf{q} } \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.15.15)

\begin{aligned}\begin{aligned}\omega^2 \sqrt{m_2} \boldsymbol{\epsilon}_2&=K_1 \text{Proj}_{\hat{\mathbf{x}}} \left( \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} e^{ + i \mathbf{r} \cdot \mathbf{q} } \right)+ K_1 \text{Proj}_{\hat{\mathbf{x}}} \left( \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} e^{ + i \mathbf{s} \cdot \mathbf{q} } \right)\\ &\quad+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left( \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} \right)+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left( \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} e^{ + i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} } \right) \\ &\quad+ K_3 \left( \text{Proj}_{\hat{\mathbf{r}}} \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} \right)\sum_\pm\left( 1 - e^{ \pm i \mathbf{r} \cdot \mathbf{q} } \right)+ K_4 \left( \text{Proj}_{\hat{\mathbf{s}}} \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} \right)\sum_\pm\left( 1 - e^{ \pm i \mathbf{s} \cdot \mathbf{q} } \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.15.15)

Regrouping, and using the matrix form $\text{Proj}_{\hat{\mathbf{u}}} = \hat{\mathbf{u}} \hat{\mathbf{u}}^\text{T}$ for the projection operators, this is

\begin{aligned}\left(\omega^2 - \frac{2}{m_1} \left( K_1 \hat{\mathbf{x}} \hat{\mathbf{x}}^T + K_2 \hat{\mathbf{y}} \hat{\mathbf{y}}^T + 2 K_3 \hat{\mathbf{r}} \hat{\mathbf{r}}^T \sin^2 (\mathbf{r} \cdot \mathbf{q}/2) + 2 K_4 \hat{\mathbf{s}} \hat{\mathbf{s}}^T \sin^2 (\mathbf{s} \cdot \mathbf{q}/2) \right)\right)\boldsymbol{\epsilon}_1 = -\left( K_1 \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T} \left( e^{ - i \mathbf{s} \cdot \mathbf{q} } + e^{ - i \mathbf{r} \cdot \mathbf{q} } \right) + K_2 \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T} \left( 1 + e^{ - i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} } \right) \right)\frac{\boldsymbol{\epsilon}_2}{\sqrt{ m_1 m_2 }}\end{aligned} \hspace{\stretch{1}}(2.0.16.16)

\begin{aligned}\left( \omega^2 - \frac{2}{m_2} \left( K_1 \hat{\mathbf{x}} \hat{\mathbf{x}}^T + K_2 \hat{\mathbf{y}} \hat{\mathbf{y}}^T + 2 K_3 \hat{\mathbf{r}} \hat{\mathbf{r}}^T \sin^2 (\mathbf{r} \cdot \mathbf{q}/2)+ 2 K_4 \hat{\mathbf{s}} \hat{\mathbf{s}}^T \sin^2 (\mathbf{s} \cdot \mathbf{q}/2) \right) \right)\boldsymbol{\epsilon}_2 = -\left( K_1 \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T} \left( e^{ i \mathbf{s} \cdot \mathbf{q} } + e^{ i \mathbf{r} \cdot \mathbf{q} } \right) + K_2 \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T} \left( 1 + e^{ i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} } \right) \right)\frac{\boldsymbol{\epsilon}_1}{\sqrt{ m_1 m_2 }}\end{aligned} \hspace{\stretch{1}}(2.0.16.16)

As a single matrix equation, this is

\begin{aligned}A = K_1 \hat{\mathbf{x}} \hat{\mathbf{x}}^T + K_2 \hat{\mathbf{y}} \hat{\mathbf{y}}^T + 2 K_3 \hat{\mathbf{r}} \hat{\mathbf{r}}^T \sin^2 (\mathbf{r} \cdot \mathbf{q}/2)+ 2 K_4 \hat{\mathbf{s}} \hat{\mathbf{s}}^T \sin^2 (\mathbf{s} \cdot \mathbf{q}/2)\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

\begin{aligned}B = e^{ i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2 }\left( { K_1 \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T} \cos\left( (\mathbf{r} - \mathbf{s}) \cdot \mathbf{q}/2 \right)+ K_2 \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T} \cos\left( (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2 \right)} \right)\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

\begin{aligned}0 =\begin{bmatrix}\omega^2 - \frac{2 A}{m_1} & \frac{B^{*}}{\sqrt{m_1 m_2}} \\ \frac{B}{\sqrt{m_1 m_2}} & \omega^2 - \frac{2 A}{m_2} \end{bmatrix}\begin{bmatrix}\boldsymbol{\epsilon}_1 \\ \boldsymbol{\epsilon}_2\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

Observe that this is an eigenvalue problem $E \mathbf{e} = \omega^2 \mathbf{e}$ for matrix

\begin{aligned}E = \begin{bmatrix}\frac{2 A}{m_1} & -\frac{B^{*}}{\sqrt{m_1 m_2}} \\ -\frac{B}{\sqrt{m_1 m_2}} & \frac{2 A}{m_2} \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

and eigenvalues $\omega^2$.

To be explicit lets put the $A$ and $B$ functions in explicit matrix form. The orthogonal projectors have a simple form

\begin{aligned}\text{Proj}_{\hat{\mathbf{x}}} = \hat{\mathbf{x}} \hat{\mathbf{x}}^\text{T}= \begin{bmatrix}1 \\ 0\end{bmatrix}\begin{bmatrix}1 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.19a)

\begin{aligned}\text{Proj}_{\hat{\mathbf{y}}} = \hat{\mathbf{y}} \hat{\mathbf{y}}^\text{T}= \begin{bmatrix}0 \\ 1\end{bmatrix}\begin{bmatrix}0 & 1\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.19b)

For the $\hat{\mathbf{r}}$ and $\hat{\mathbf{s}}$ projection operators, we can use half angle formulations

\begin{aligned}\text{Proj}_{\hat{\mathbf{r}}} = \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T}= \begin{bmatrix}\cos\theta \\ \sin\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta\end{bmatrix}=\begin{bmatrix}\cos^2\theta & \cos\theta \sin\theta \\ \cos\theta \sin\theta & \sin^2 \theta\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1 + \cos \left( 2 \theta \right) & \sin \left( 2 \theta \right) \\ \sin \left( 2 \theta \right) & 1 - \cos \left( 2 \theta \right)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.20.20)

\begin{aligned}\text{Proj}_{\hat{\mathbf{s}}} = \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T}= \begin{bmatrix}-\cos\theta \\ \sin\theta\end{bmatrix}\begin{bmatrix}-\cos\theta & \sin\theta\end{bmatrix}=\begin{bmatrix}\cos^2\theta & -\cos\theta \sin\theta \\ -\cos\theta \sin\theta & \sin^2 \theta\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1 + \cos \left( 2 \theta \right) & -\sin \left( 2 \theta \right) \\ -\sin \left( 2 \theta \right) & 1 - \cos \left( 2 \theta \right)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.20.20)

After some manipulation, and the following helper functions

\begin{aligned}\begin{aligned}\alpha_\pm &= K_3 \sin^2 (\mathbf{r} \cdot \mathbf{q}/2) \pm K_4 \sin^2 (\mathbf{s} \cdot \mathbf{q}/2) \\ \beta_\pm &= K_1 \cos\left( (\mathbf{r} - \mathbf{s}) \cdot \mathbf{q}/2 \right) \pm K_2 \cos\left( (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2 \right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.20.20)

the block matrices of eq. 2.0.17.17 take the form

\begin{aligned}A = \begin{bmatrix}K_1 + \alpha_+ (1 + \cos\left( 2 \theta \right)) & \alpha_- \sin\left( 2 \theta \right) \\ \alpha_- \sin\left( 2 \theta \right) & K_2 + \alpha_+ (1 - \cos\left( 2 \theta \right))\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

\begin{aligned}B = e^{ i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2 }\begin{bmatrix} \beta_+ (1 + \cos \left( 2 \theta \right)) & \beta_- \sin \left( 2 \theta \right) \\ \beta_- \sin \left( 2 \theta \right) & \beta_+( 1 -\cos \left( 2 \theta \right))\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

A final bit of simplification for $B$ possible, noting that $\mathbf{r} + \mathbf{s} = 2 a (0, \sin\theta )$, and $\mathbf{r} - \mathbf{s} = 2 a(\cos\theta, 0)$, so

\begin{aligned}\beta_\pm = K_1 \cos\left( a \cos\theta q_x \right) \pm K_2 \cos\left( a \sin\theta q_y \right),\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

and

\begin{aligned}B = e^{ i a \sin\theta q_y }\begin{bmatrix} \beta_+ (1 + \cos \left( 2 \theta \right)) & \beta_- \sin \left( 2 \theta \right) \\ \beta_- \sin \left( 2 \theta \right) & \beta_+( 1 -\cos \left( 2 \theta \right))\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

It isn’t particularly illuminating to expand out the determinant for such a system, even though it can be done symbolically without too much programming. However, what is easy after formulating the matrix for this system, is actually solving it. This is done, and animated, in twoAtomBasisRectangularLatticeDispersionRelation.cdf

## Dipole Moment from constant electric field

Posted by peeterjoot on December 27, 2012

## Question: Dipole Moment from constant electric field

In [1] it is stated that the force per unit angle on a dipole system as illustrated in Fig.1 is

\begin{aligned}F_\theta = -p \mathcal{E} \sin\theta,\end{aligned} \hspace{\stretch{1}}(1.0.1)

where $\mathbf{p} = q \mathbf{r}$. The text was also referring to torques, and it wasn’t clear to me if the result was the torque or the force. Derive the result to resolve any doubt (in retrospect dimensional analysis would also have worked).

Fig1: Dipole moment coordinate

Let’s put the electric field in the $\hat{\mathbf{x}}$ direction ($\theta = 0$), so that the potential acting on charge $i$ is given implicitly by

\begin{aligned}\mathbf{F}_i = q_i \mathcal{E} \hat{\mathbf{x}} = -\nabla \phi_i = -\hat{\mathbf{x}} \frac{d{{\phi_i}}}{dx}\end{aligned} \hspace{\stretch{1}}(1.0.2)

or

\begin{aligned}\phi_i = -q_i (x_i - x_0).\end{aligned} \hspace{\stretch{1}}(1.0.3)

Our positions, and velocities are

\begin{aligned}\mathbf{r}_{1,2} = \pm \frac{r}{2} \hat{\mathbf{x}} e^{\hat{\mathbf{x}} \hat{\mathbf{y}} \theta}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\frac{d{{\mathbf{r}_{1,2}}}}{dt} = \pm \frac{r}{2} \dot{\theta} \hat{\mathbf{y}} e^{\hat{\mathbf{x}} \hat{\mathbf{y}} \theta}.\end{aligned} \hspace{\stretch{1}}(1.0.4b)

Our kinetic energy is

\begin{aligned}T = \frac{1}{{2}} \sum_i m_i \left( \frac{d{{\mathbf{r}_i}}}{dt} \right)^2 = \frac{1}{{2}} \sum_i m_i \left( \frac{r}{2} \right)^2 \dot{\theta}^2= \frac{1}{{2}} (m_1 + m_2) \left( \frac{r}{2} \right)^2 \dot{\theta}^2.\end{aligned} \hspace{\stretch{1}}(1.0.5)

For our potential energies we require the $x$ component of the position vectors, which are

\begin{aligned}x_i = \mathbf{r}_i \cdot \hat{\mathbf{x}}=\pm \left\langle{{ \frac{r}{2} \hat{\mathbf{x}} e^{\hat{\mathbf{x}} \hat{\mathbf{y}} \theta} \hat{\mathbf{x}}}}\right\rangle=\pm \frac{r}{2} \cos\theta\end{aligned} \hspace{\stretch{1}}(1.0.6)

Our potentials are

\begin{aligned}\phi_1 = -q_1 \mathcal{E} \frac{r}{2} \cos\theta + \phi_0\end{aligned} \hspace{\stretch{1}}(1.0.7a)

\begin{aligned}\phi_2 = q_2 \mathcal{E} \frac{r}{2} \cos\theta + \phi_0\end{aligned} \hspace{\stretch{1}}(1.0.7b)

Our system Lagrangian, after dropping the constant reference potential that doesn’t effect the dynamics is

\begin{aligned}\mathcal{L} = \frac{1}{{2}} (m_1 + m_2) \left( \frac{r}{2} \right)^2 \dot{\theta}^2+q_1 \mathcal{E} \frac{r}{2} \cos\theta-q_2 \mathcal{E} \frac{r}{2} \cos\theta\end{aligned} \hspace{\stretch{1}}(1.0.8)

For this problem we had two equal masses and equal magnitude charges $m = m_1 = m_2$ and $q = q_1 = -q_2$

\begin{aligned}\mathcal{L} = \frac{1}{{4}} m r^2 \dot{\theta}^2 + q r \mathcal{E} \cos\theta\end{aligned} \hspace{\stretch{1}}(1.0.9)

\begin{aligned}p_\theta = \frac{\partial {\mathcal{L}}}{\partial {\dot{\theta}}} = \frac{1}{{2}} m r^2 \dot{\theta}\end{aligned} \hspace{\stretch{1}}(1.0.10)

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\theta}} = -q r \mathcal{E} \sin\theta=\frac{d{{p_\theta}}}{dt}=\frac{1}{{2}} m r^2 \dot{d}{\theta}\end{aligned} \hspace{\stretch{1}}(1.0.11)

Putting these together, with $p = q r$, we have the result stated in the text

\begin{aligned}F_\theta = \frac{d{{p_\theta}}}{dt} = -p \mathcal{E} \sin\theta.\end{aligned} \hspace{\stretch{1}}(1.0.12)

# References

[1] E.A. Jackson. Equilibrium statistical mechanics. Dover Pubns, 2000.

## Another worked Landau pendulum problem

Posted by peeterjoot on July 14, 2012

## Question: Pendulum with support moving in line

This problem like the last, but with the point of suspension moving in a horizontal line $x = a \cos\gamma t$.

Our mass point has coordinates

\begin{aligned}p &= a \cos\gamma t + l i e^{-i\phi} \\ &= a \cos \gamma t + l i ( \cos \phi - i \sin \phi ) \\ &= ( a \cos \gamma t + l \sin \phi, l \cos \phi ),\end{aligned} \hspace{\stretch{1}}(1.10)

so that the velocity is

\begin{aligned}\dot{p} = ( -a \gamma \sin \gamma t + l \dot{\phi} \cos \phi, -l \dot{\phi} \sin \phi ).\end{aligned} \hspace{\stretch{1}}(1.11)

Our squared velocity is

\begin{aligned}\dot{p}^2 &= a^2 \gamma^2 \sin^2 \gamma t + l^2 \dot{\phi}^2 - 2 a \gamma l \dot{\phi} \sin\gamma t \cos \phi \\ &= \frac{1}{{2}} a^2 \gamma^2 \frac{d{{}}}{dt}\left( t - \frac{1}{{2 \gamma}} \sin 2 \gamma t \right) + l^2 \dot{\phi}^2 - a \gamma l \dot{\phi} ( \sin( \gamma t + \phi) + \sin(\gamma t - \phi)).\end{aligned} \hspace{\stretch{1}}(1.12)

In the last term, we can reduce the sum of sines, finding a total derivative term and a remainder as in the previous problem. That is

\begin{aligned}\dot{\phi} (\sin( \gamma t + \phi) + \sin(\gamma t - \phi)) &= (\dot{\phi} + \gamma)\sin(\gamma t + \phi) - \gamma \sin(\gamma t + \phi)+(\dot{\phi} - \gamma)\sin(\gamma t - \phi) + \gamma \sin(\gamma t - \phi) \\ &= \frac{d{{}}}{dt} \left( -\cos(\gamma t + \phi) + \cos(\gamma t - \phi) \right)+ \gamma ( \sin(\gamma t - \phi) - \sin(\gamma t + \phi) ) \\ &= \frac{d{{}}}{dt} \left( -\cos(\gamma t + \phi) + \cos(\gamma t - \phi) \right)- 2 \gamma \cos \gamma t \sin\phi.\end{aligned} \hspace{\stretch{1}}(1.13)

Putting all the pieces together and dropping the total derivatives we have the stated solution

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma^2 l \cos \gamma t \sin\phi \right) + m g l \cos\phi\end{aligned} \hspace{\stretch{1}}(1.14)

# References

[1] LD Landau and EM Lifshitz. Mechanics, vol. 1. 1976.

## Typo in Landau Mechanics problem? Nope.

Posted by peeterjoot on July 14, 2012

# Motivation

Attempting a mechanics problem from Landau I get a different answer. I wrote up my solution to see if I can spot either where I went wrong, or demonstrate the error, and then posted it to physicsforums. I wasn’t wrong, but the text wasn’t either. Here’s the complete result.

# Guts

## Question: Pendulum with support moving in circle

section 1 problem 3a of [1] is to calculate the Lagrangian of a
pendulum where the point of support is moving in a circle (figure and full text for problem in this google books reference)

The coordinates of the mass are

\begin{aligned}p = a e^{i \gamma t} + i l e^{i\phi},\end{aligned} \hspace{\stretch{1}}(1.1)

or in coordinates

\begin{aligned}p = (a \cos\gamma t + l \sin\phi, -a \sin\gamma t + l \cos\phi).\end{aligned} \hspace{\stretch{1}}(1.2)

The velocity is

\begin{aligned}\dot{p} = (-a \gamma \sin\gamma t + l \dot{\phi} \cos\phi, -a \gamma \cos\gamma t - l \dot{\phi} \sin\phi),\end{aligned} \hspace{\stretch{1}}(1.3)

and in the square

\begin{aligned}\dot{p}^2 = a^2 \gamma^2 + l^2 \dot{\phi}^2 - 2 a \gamma \dot{\phi} \sin\gamma t \cos\phi + 2 a \gamma l \dot{\phi} \cos \gamma t \sin\phi=a^2 \gamma^2 + l^2 \dot{\phi}^2 + 2 a \gamma l \dot{\phi} \sin (\gamma t - \phi).\end{aligned} \hspace{\stretch{1}}(1.4)

For the potential our height above the minimum is

\begin{aligned}h = 2a + l - a (1 -\cos\gamma t) - l \cos\phi = a ( 1 + \cos\gamma t) + l (1 - \cos\phi).\end{aligned} \hspace{\stretch{1}}(1.5)

In the potential the total derivative $\cos\gamma t$ can be dropped, as can all the constant terms, leaving

\begin{aligned}U = - m g l \cos\phi, \end{aligned} \hspace{\stretch{1}}(1.6)

so by the above the Lagrangian should be (after also dropping the constant term $m a^2 \gamma^2/2$

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma l \dot{\phi} \sin (\gamma t - \phi) \right) + m g l \cos\phi.\end{aligned} \hspace{\stretch{1}}(1.7)

This is almost the stated value in the text

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma^2 l \sin (\gamma t - \phi) \right) + m g l \cos\phi.\end{aligned} \hspace{\stretch{1}}(1.8)

We have what appears to be an innocent looking typo (text putting in a $\gamma$ instead of a $\dot{\phi}$), but the subsequent text also didn’t make sense. That referred to the omission of the total derivative $m l a \gamma \cos( \phi - \gamma t)$, which isn’t even a term that I have in my result.

In the physicsforum response it was cleverly pointed out by Dickfore that 1.7 can be recast into a total derivative

\begin{aligned}m a l \gamma \dot{\phi} \sin (\gamma t - \phi) =m a l \gamma ( \dot{\phi} - \gamma ) \sin (\gamma t - \phi) +m a l \gamma^2 \sin (\gamma t - \phi) =\frac{d{{}}}{dt}\left(m a l \gamma \cos (\gamma t - \phi) \right)+m a l \gamma^2 \sin (\gamma t - \phi),\end{aligned} \hspace{\stretch{1}}(1.9)

which resolves the connundrum!

# References

[1] LD Landau and EM Lifshitz. Mechanics, vol. 1. 1976.

Posted in Math and Physics Learning. | Tagged: , , | 2 Comments »

## Runge-Lenz vector conservation

Posted by peeterjoot on February 13, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

Notes from Prof. Poppitz’s phy354 classical mechanics lecture on the Runge-Lenz vector, a less well known conserved quantity for the 3D $1/r$ potentials that can be used to solve the Kepler problem.

# Motivation: The Kepler problem.

We can plug away at the Lagrangian in cylindrical coordinates and find eventually

\begin{aligned}\int_{\phi_0}^\phi d\phi = \int_{r_0}^r \frac{M}{m r^2} \frac{dr}{\sqrt{\frac{2}{M} ( E - U + \frac{M^2}{2 m r^2}) }}\end{aligned} \hspace{\stretch{1}}(2.1)

but this can be messy to solve, where we get elliptic integrals or worse, depending on the potential.

For the special case of the 3D problem where the potential has a $1/r$ form, this is what Prof. Poppitz called “super-integrable”. With $2N - 1 = 5$ conserved quantities to be found, we’ve got one more. Here the form of that last conserved quantity is given, called the Runge-Lenz vector, and we verify that it is conserved.

# Runge-Lenz vector

Given a potential

\begin{aligned}U = -\frac{\alpha}{r}\end{aligned} \hspace{\stretch{1}}(3.2)

and a Lagrangian

\begin{aligned}\mathcal{L} &= \frac{m \dot{r}^2}{2} + \frac{1}{{2}} \frac{M_z^2}{m r^2} - U \\ M_z &= m r^2 \dot{\phi}^2\end{aligned} \hspace{\stretch{1}}(3.3)

and writing the angular momentum as

\begin{aligned}\mathbf{M} = m \mathbf{r} \times \mathbf{v} \end{aligned} \hspace{\stretch{1}}(3.5)

the Runge-Lenz vector

\begin{aligned}\mathbf{A} = \mathbf{v} \times \mathbf{M} - \alpha \hat{\mathbf{r}},\end{aligned} \hspace{\stretch{1}}(3.6)

is a conserved quantity.

## Verify the conservation assumption.

Let’s show that the conservation assumption is correct

\begin{aligned}\frac{d}{dt} \left( \mathbf{v} \times \mathbf{M} \right)=\frac{d{{ \mathbf{v}}}}{dt} \times \mathbf{M} + \mathbf{v} \times \not{{\frac{d{{\mathbf{M} }}}{dt}}}\end{aligned} \hspace{\stretch{1}}(3.7)

Here, we note that angular momentum conservation is really $d\mathbf{M}/dt = 0$, so we are left with only the acceleration term, which we can rewrite in terms of the Euler-Lagrange equation

\begin{aligned}\frac{d}{dt} \left( \mathbf{v} \times \mathbf{M} \right)&=-\frac{1}{{m}} \boldsymbol{\nabla} U \times M \\ &=-\frac{1}{{m}} \frac{\partial {U}}{\partial {r}} \hat{\mathbf{r}} \times M \\ &=-\frac{1}{{m}} \frac{\partial {U}}{\partial {r}} \hat{\mathbf{r}} \times (m \mathbf{r} \times \mathbf{v}) \\ &=- \frac{\partial {U}}{\partial {r}} \hat{\mathbf{r}} \times (\mathbf{r} \times \mathbf{v}) \end{aligned}

We can compute the double cross product

\begin{aligned}(\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) )_i&=a_m b_r c_s \epsilon_{r s t} \epsilon_{m t i} \\ &=a_m b_r c_s \delta^{[rs]}_{i m} \\ &=a_m b_i c_m -a_m b_m c_i \end{aligned}

For

\begin{aligned}\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} -(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}\end{aligned} \hspace{\stretch{1}}(3.8)

Plugging this we have

\begin{aligned}\frac{d}{dt} \left( \mathbf{v} \times \mathbf{M} \right) &= \frac{\partial U}{\partial r} \\ left( (\hat{\mathbf{r}} \cdot \mathbf{r}) \mathbf{v}-(\hat{\mathbf{r}} \cdot \mathbf{v}) \mathbf{r} \right) \\ &= \left( \frac{\alpha}{r^2} \right)\left( r \mathbf{v}-\frac{1}{r}(\mathbf{r} \cdot \mathbf{v}) \mathbf{r} \right) \\ &= \alpha\left( \frac{\mathbf{v}}{r}-\frac{(\mathbf{r} \cdot \mathbf{v}) \mathbf{r} }{r^3}\right) \\ \end{aligned}

Now let’s look at the other term. We’ll need the derivative of $\hat{\mathbf{r}}$

\begin{aligned}\frac{d{{\hat{\mathbf{r}}}}}{dt} &=\frac{d}{dt} \frac{\mathbf{r}}{r} \\ &=\frac{\mathbf{v}}{r} + \mathbf{r} \frac{d{{\frac{1}{{r}}}}}{dt} \\ &=\frac{\mathbf{v}}{r} - \frac{\mathbf{r}}{r^2} \frac{d{{r}}}{dt} \\ &=\frac{\mathbf{v}}{r} - \frac{\mathbf{r}}{r^2} \frac{d{{ \sqrt{\mathbf{r} \cdot \mathbf{r}}}}}{dt} \\ &=\frac{\mathbf{v}}{r} - \frac{\mathbf{r}}{r^2} \frac{\mathbf{v} \cdot \mathbf{r}}{\sqrt{\mathbf{r}^2}} \\ &=\frac{\mathbf{v}}{r} - \frac{\mathbf{r}}{r^3} \mathbf{v} \cdot \mathbf{r}\end{aligned}

Putting all the bits together we’ve now verified the conservation statement

\begin{aligned}\frac{d}{dt} \left(\mathbf{v} \times \mathbf{M} - \alpha \hat{\mathbf{r}}\right)=\alpha\left( \frac{\mathbf{v}}{r}-\frac{(\mathbf{r} \cdot \mathbf{v}) \mathbf{r} }{r^3}\right) -\alpha \left( \frac{\mathbf{v}}{r} - \frac{\mathbf{r}}{r^3} \mathbf{v} \cdot \mathbf{r} \right)= 0.\end{aligned} \hspace{\stretch{1}}(3.9)

With

\begin{aligned}\frac{d}{dt} \left( \mathbf{v} \times \mathbf{M} - \alpha \hat{\mathbf{r}} \right) = 0,\end{aligned} \hspace{\stretch{1}}(3.10)

our vector must be some constant vector. Let’s write this

\begin{aligned}\mathbf{v} \times \mathbf{M} - \alpha \hat{\mathbf{r}} = \alpha \mathbf{e},\end{aligned} \hspace{\stretch{1}}(3.11)

so that

\begin{aligned}\boxed{\mathbf{v} \times \mathbf{M} = \alpha \left(\mathbf{e} + \hat{\mathbf{r}} \right).}\end{aligned} \hspace{\stretch{1}}(3.12)

Dotting 3.12 with $\mathbf{M}$ we find

\begin{aligned}\alpha \mathbf{M} \cdot \left(\mathbf{e} + \hat{\mathbf{r}} \right)&=\mathbf{M} \cdot (\mathbf{v} \times \mathbf{M}) \\ &= 0\end{aligned}

With $\hat{\mathbf{r}}$ lying in the plane of the trajectory (perpendicular to $\mathbf{M}$), we must also have $\mathbf{e}$ lying in the plane of the trajectory.

Now we can dot 3.12 with $\mathbf{r}$ to find

\begin{aligned}\mathbf{r} \cdot (\mathbf{v} \times \mathbf{M}) &= \alpha \mathbf{r} \cdot \left(\mathbf{e} + \hat{\mathbf{r}} \right) \\ &= \alpha \cdot \left( r e \cos(\phi - \phi_0) + r \right) \\ \mathbf{M} \cdot (\mathbf{r} \times \mathbf{v}) &= \\ \mathbf{M} \cdot \frac{\mathbf{M}}{m} &= \\ \frac{\mathbf{M}^2}{m} &=\end{aligned}

This is

\begin{aligned}\frac{\mathbf{M}^2}{m} = \alpha r \left( 1 + e \cos(\phi - \phi_0) \right).\end{aligned} \hspace{\stretch{1}}(3.13)

This is a kind of curious implicit relationship, since $\phi$ is also a function of $r$. Recall that the kinetic portion of our Lagrangian was

\begin{aligned}\frac{1}{{2}} m (\dot{r}^2 + r^2 \dot{\phi}^2 )\end{aligned} \hspace{\stretch{1}}(3.14)

so that our angular momentum was

\begin{aligned}M_\phi = \frac{\partial }{\partial {\dot{\phi}}} \left( \frac{1}{{2}} m r^2 \dot{\phi}^2 \right) = m r^2 \dot{\phi},\end{aligned} \hspace{\stretch{1}}(3.15)

with no $\phi$ dependence in the Lagrangian we have

\begin{aligned}\frac{d}{dt} (m r^2 \dot{\phi}) = 0,\end{aligned} \hspace{\stretch{1}}(3.16)

or

\begin{aligned}\mathbf{M} = m r^2 \dot{\phi} \hat{\mathbf{z}} = \text{constant}\end{aligned} \hspace{\stretch{1}}(3.17)

Our dynamics are now fully specified, even if this not completely explicit

\begin{aligned}\boxed{\begin{aligned}r &= \frac{M^2}{m \alpha} \frac{1}{{1 + e \cos(\phi - \phi_0)}} \\ \frac{d\phi}{dt} &= \frac{M}{ m r^2}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(3.18)

What we can do is rearrange and separate variables

\begin{aligned}\frac{1}{{r^2}} = \frac{m^2 \alpha^2}{M^4} (1 + e \cos(\phi - \phi_0))^2 = \frac{m}{M} \frac{d\phi}{dt},\end{aligned} \hspace{\stretch{1}}(3.19)

to find

\begin{aligned}t - t_0 = \frac{M^3}{m \alpha^3} \int_{\phi_0}^\phi d\phi \frac{1}{{(1 + e \cos(\phi - \phi_0))^2}}=\frac{M^3}{m \alpha^3} \int_0^{\phi - \phi_0} du\frac{1}{{(1 + e \cos u)^2}}\end{aligned} \hspace{\stretch{1}}(3.20)

Now, at least $\phi = \phi(t)$ is specified implicitly.

We can also use the first of these to determine the magnitude of the radial velocity

\begin{aligned}\frac{dr}{dt} &=-\frac{M^2}{m \alpha} \frac{1}{{(1 + e \cos(\phi - \phi_0))^2}} (-e \sin(\phi - \phi_0)) \frac{d\phi}{dt} \\ &=\frac{e M^2}{m \alpha} \frac{1}{{(1 + e \cos(\phi - \phi_0))^2}} \sin(\phi - \phi_0) \frac{M}{m r^2} \\ &=\frac{e M^3}{m^2 \alpha r^2} \frac{1}{{(1 + e \cos(\phi - \phi_0))^2}} \sin(\phi - \phi_0) \\ &=\frac{e M^3}{m^2 \alpha r^2} \left( \frac{ m r \alpha }{M^2} \right)^2 \sin(\phi - \phi_0) \\ &=\frac{e }{M } \sin(\phi - \phi_0),\end{aligned}

with this, we can also find the energy

\begin{aligned}E &= \dot{r}( m \dot{r}) + \dot{\phi} ( m r^2 \dot{\phi}) - \left( \frac{1}{{2}} m \dot{r}^2 + \frac{1}{{2}} m r^2 \dot{\phi}^2 - U \right) \\ &= \frac{1}{{2}} m \dot{r}^2 + \frac{1}{{2}} m r^2 \dot{\phi}^2 + U \\ &= \frac{1}{{2}} m \dot{r}^2 + \frac{1}{{2}} m r^2 \dot{\phi}^2 - \frac{\alpha}{r} \\ &= \frac{1}{{2}} m \frac{e^2}{M^2} \sin^2(\phi - \phi_0) + \frac{1}{{2 m r^2 }} M^2 - \frac{\alpha}{r}.\end{aligned}

Or

\begin{aligned}E= \frac{m}{2 M^2} (\mathbf{e} \times \hat{\mathbf{r}})^2 + \frac{1}{{2 m r^2 }} M^2 - \frac{\alpha}{r}.\end{aligned} \hspace{\stretch{1}}(3.21)

Is this what was used in class to state the relation

\begin{aligned}e = \sqrt{1 + \frac{2 E M^2}{m \alpha^2}}.\end{aligned} \hspace{\stretch{1}}(3.22)

It’s not obvious exactly how that is obtained, but we can go back to 3.18 to eliminate the $e^2 \sin^2 \Delta \phi$ term

\begin{aligned}E = \frac{1}{{2}} m \frac{1}{M^2} \left( e^2 - \left( \frac{M^2}{r m \alpha} - 1\right)^2 \right) + \frac{1}{{2 m r^2 }} M^2 - \frac{\alpha}{r}.\end{aligned} \hspace{\stretch{1}}(3.23)

Presumably this simplifies to the desired result (or there’s other errors made in that prevent that).

## Gauge transformation of the Dirac equation.

Posted by peeterjoot on August 21, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

In [1] the gauge transformation of the Dirac equation is covered, producing the non-relativistic equation with the correct spin interaction. There are unfortunately some sign errors, some of which self correct, and some of which don’t impact the end result, but are slightly confusing. There are also some omitted details. I’ll attempt to work through the same calculation with all the signs in the right places and also fill in some of the details I found myself wanting.

# A step back. On the gauge transformation.

The gauge transformations utilized are given as

\begin{aligned}\mathcal{E} &\rightarrow \mathcal{E} - e \phi \\ \mathbf{p} &\rightarrow \mathbf{p} - e \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(2.1)

Let’s start off by reminding ourself where these come from. As outlined in section 12.9 in [2] (with some details pondered in [3]), our relativistic Lagrangian is

\begin{aligned}\mathcal{L} = -m c^2 \sqrt{ 1 - \frac{\mathbf{u}}{c^2}} + \frac{e}{c} \mathbf{u} \cdot \mathbf{A} - e \phi.\end{aligned} \hspace{\stretch{1}}(2.3)

The conjugate momentum is

\begin{aligned}\mathbf{P} = \mathbf{e}^i \frac{\partial {\mathcal{L}}}{\partial {u^i}} = \frac{m \mathbf{u}}{\sqrt{1 - \mathbf{u}^2/c^2}} + \frac{e}{c} \mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.4)

or

\begin{aligned}\mathbf{P} = \mathbf{p} + \frac{e}{c} \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(2.5)

The Hamiltonian, which must be expressed in terms of this conjugate momentum $\mathbf{P}$, is found to be

\begin{aligned}\mathcal{E} = \sqrt{ (c \mathbf{P} - e \mathbf{A})^2 + m^2 c^4 } + e \phi.\end{aligned} \hspace{\stretch{1}}(2.6)

With the free particle Lagrangian

\begin{aligned}\mathcal{L} = -m c^2 \sqrt{ 1 - \frac{\mathbf{u}}{c^2}} ,\end{aligned} \hspace{\stretch{1}}(2.7)

our conjugate momentum is

\begin{aligned}\mathbf{P} = \frac{m \mathbf{u}}{\sqrt{ 1 - \mathbf{u}^2/c^2} }.\end{aligned} \hspace{\stretch{1}}(2.8)

For this we find that our Hamiltonian $\mathcal{E} = \mathbf{P} \cdot \mathbf{u} - \mathcal{L}$ is

\begin{aligned}\mathcal{E} = \frac{m c^2}{\sqrt{1 - \mathbf{u}^2/c^2}},\end{aligned} \hspace{\stretch{1}}(2.9)

but this has to be expressed in terms of $\mathbf{P}$. Having found the form of the Hamiltonian for the interaction case, it is easily verified that 2.6 contains the required form once the interaction fields $(\phi, \mathbf{A})$ are zeroed

\begin{aligned}\mathcal{E} = \sqrt{ (c \mathbf{P})^2 + m^2 c^4 }.\end{aligned} \hspace{\stretch{1}}(2.10)

Considering the interaction case, Jackson points out that the energy and momentum terms can be combined as a four momentum

\begin{aligned}p^a = \left( \frac{1}{{c}}(\mathcal{E} - e \phi), \mathbf{P} - \frac{e}{c}\mathbf{A} \right),\end{aligned} \hspace{\stretch{1}}(2.11)

so that the re-arranged and squared Hamiltonian takes the form

\begin{aligned}p^a p_a = (m c)^2.\end{aligned} \hspace{\stretch{1}}(2.12)

From this we see that for the Lorentz force, the interaction can be found, starting with the free particle Hamiltonian 2.6, making the transformation

\begin{aligned}\mathcal{E} &\rightarrow \mathcal{E} - e\phi \\ \mathbf{P} &\rightarrow \mathbf{P} - \frac{e}{c}\mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.13)

or in covariant form

\begin{aligned}p^\mu \rightarrow p^\mu - \frac{e}{c}A^\mu.\end{aligned} \hspace{\stretch{1}}(2.15)

# On the gauge transformation of the Dirac equation.

The task at hand now is to make the transformations of 2.13, applied to the Dirac equation

\begin{aligned}{p} = \gamma_\mu p^\mu = m c.\end{aligned} \hspace{\stretch{1}}(3.16)

The first observation to make is that we appear to have different units in the Desai text. Let’s continue using the units from Jackson, and translate them later if inclined.

Right multiplication of 3.16 by $\gamma_0$ gives us

\begin{aligned}0 &= \gamma_0 ({p} - m c) \\ &= \gamma_0 \gamma_\mu \left( p^\mu - \frac{e}{c} A^\mu \right)- \gamma_0 m c\\ &=\gamma_0 \gamma_0 \left(\frac{\mathcal{E}}{c} - \frac{e}{c} \phi \right)+\gamma_0 \gamma_a \left(p^a - \frac{e}{c} A^a \right)- \gamma_0 m c \\ &=\frac{1}{{c}} \left( \mathcal{E}- e \phi \right)-\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)- \gamma_0 m c \\ \end{aligned}

With the minor notational freedom of using $\gamma_0$ instead of $\gamma_4$, this is our starting point in the Desai text, and we can now left multiply by

\begin{aligned}({p} + m c) \gamma_0 =\frac{1}{{c}} \left( \mathcal{E} - e \phi \right)+\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)+ \gamma_0 m c.\end{aligned} \hspace{\stretch{1}}(3.17)

The motivation for this appears to be that this product of conjugate like quantities

\begin{aligned}\begin{aligned}0 &= ({p} + m c) \gamma_0 \gamma_0 ({p} - m c) \\ &=({p} + m c) ({p} - m c) \\ &= \frac{1}{{c^2}} \left( \mathcal{E} - e \phi \right)^2 -\left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right)^2 - (m c)^2 + \cdots,\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.18)

produces the the Klein-Gordon equation, plus some cross terms to be determined. Those cross terms are the important bits since they contain the spin interaction, even in the non-relativistic limit.

Let’s do the expansion.

\begin{aligned}0&= ({p} + m c) \gamma_0 \gamma_0 ({p} - m c) u \\ &=\left(\frac{1}{{c}} \left( \mathcal{E} - e \phi \right)+\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)+ \gamma_0 m c\right)\left(\frac{1}{{c}} \left( \mathcal{E}- e \phi \right)-\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)- \gamma_0 m c \right) u \\ &=\frac{1}{{c}} \left( \mathcal{E} - e \phi \right)\left(\frac{1}{{c}} \left( \mathcal{E}- e \phi \right)-\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)- \gamma_0 m c \right) u \\ &\qquad +\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)\left(\frac{1}{{c}} \left( \mathcal{E}- e \phi \right)-\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)- \gamma_0 m c \right) u \\ &\qquad + \gamma_0 m c\left(\frac{1}{{c}} \left( \mathcal{E}- e \phi \right)-\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)- \gamma_0 m c \right) u \\ &=\left(\frac{1}{{c^2}} \left( \mathcal{E} - e \phi \right)^2- \left( \boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right) \right)^2- (mc)^2\right) u\\ &\qquad + \frac{1}{{c}} \left[{\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)},{\mathcal{E} - e \phi}\right] u- m c\left\{{\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)},{ \gamma_0}\right\} u \\ &\qquad + {\gamma_0 m\left(\mathcal{E} - e \phi\right) u}- {\gamma_0 m\left(\mathcal{E} - e \phi\right) u}\\ \end{aligned}

Since $\gamma_0$ anticommutes with any $\boldsymbol{\alpha} \cdot \mathbf{x}$, even when $\mathbf{x}$ contains operators, the anticommutator term is killed.

While done in the text, lets also do the $\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)$ square for completeness. Because this is an operator, we need to treat this as

\begin{aligned}\left( \boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right) \right)^2 u&=\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)\boldsymbol{\alpha} \cdot \left(\mathbf{P} u - \frac{e}{c} \mathbf{A} u \right),\end{aligned}

so want to treat the two vectors as independent, say $(\boldsymbol{\alpha} \cdot \mathbf{a})(\boldsymbol{\alpha} \cdot \mathbf{b})$. That is

\begin{aligned}(\boldsymbol{\alpha} \cdot \mathbf{a})(\boldsymbol{\alpha} \cdot \mathbf{b})&=\begin{bmatrix}0 & \boldsymbol{\sigma} \cdot \mathbf{a} \\ \boldsymbol{\sigma} \cdot \mathbf{a} & 0\end{bmatrix}\begin{bmatrix}0 & \boldsymbol{\sigma} \cdot \mathbf{b} \\ \boldsymbol{\sigma} \cdot \mathbf{b} & 0\end{bmatrix} \\ &=\begin{bmatrix}(\boldsymbol{\sigma} \cdot \mathbf{a}) (\boldsymbol{\sigma} \cdot \mathbf{b}) & 0 \\ 0 & (\boldsymbol{\sigma} \cdot \mathbf{a}) (\boldsymbol{\sigma} \cdot \mathbf{b}) & 0 \\ \end{bmatrix} \\ \end{aligned}

The diagonal elements can be expanded by coordinates

\begin{aligned}(\boldsymbol{\sigma} \cdot \mathbf{a}) (\boldsymbol{\sigma} \cdot \mathbf{b})&=\sum_{m,n} \sigma^m a^m \sigma^n b^n \\ &=\sum_m a^m b^m+\sum_{m\ne n} \sigma^m \sigma^n a^m b^m \\ &=\mathbf{a} \cdot \mathbf{b}+i \sum_{m\ne n} \sigma^o \epsilon^{m n o} a^m b^m \\ &=\mathbf{a} \cdot \mathbf{b}+i \boldsymbol{\sigma} \cdot (\mathbf{a} \times \mathbf{b}),\end{aligned}

for

\begin{aligned}(\boldsymbol{\alpha} \cdot \mathbf{a})(\boldsymbol{\alpha} \cdot \mathbf{b})=\begin{bmatrix}\mathbf{a} \cdot \mathbf{b} + i \boldsymbol{\sigma} \cdot (\mathbf{a} \times \mathbf{b}) & 0 \\ 0 & \mathbf{a} \cdot \mathbf{b} + i \boldsymbol{\sigma} \cdot (\mathbf{a} \times \mathbf{b})\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.19)

Plugging this back in, we now have an extra term in the expansion

\begin{aligned}0&=\left(\frac{1}{{c^2}} \left( \mathcal{E} - e \phi \right)^2- \left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right)^2- (mc)^2\right) u\\ &\qquad + \frac{1}{{c}} \left[{\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)},{\mathcal{E} - e \phi}\right] u\\ &\qquad- i \boldsymbol{\sigma}' \cdot\left(\left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right) \times \left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right)\right) u\end{aligned}

Here $\boldsymbol{\sigma}'$ was defined as the direct product of the two by two identity with the abstract matrix $\boldsymbol{\sigma}$ as follows

\begin{aligned}\boldsymbol{\sigma}' =\begin{bmatrix}\boldsymbol{\sigma} & 0 \\ 0 & \boldsymbol{\sigma}\end{bmatrix}= I \otimes \boldsymbol{\sigma}\end{aligned} \hspace{\stretch{1}}(3.20)

Like the $\mathbf{L} \times \mathbf{L}$ angular momentum operator cross products this one wasn’t zero. Expanding it yields

\begin{aligned}\left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right) \times \left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right) u&=\mathbf{P} \times \mathbf{P} u+ \frac{e^2}{c^2} \mathbf{A} \times \mathbf{A} u- \frac{e}{c} \left( \mathbf{A} \times \mathbf{P} + \mathbf{P} \times \mathbf{A} \right) u \\ &=- \frac{e}{c} \left( \mathbf{A} \times (\mathbf{P} u) + (\mathbf{P} u) \times \mathbf{A} + u (\mathbf{P} \times \mathbf{A}) \right) \\ &=- \frac{e}{c} (-i \hbar \boldsymbol{\nabla} \times \mathbf{A}) u \\ &=\frac{i e \hbar}{c} \mathbf{H} u\end{aligned}

Plugging in again we are getting closer, and now have the magnetic field cross term

\begin{aligned}0&=\left(\frac{1}{{c^2}} \left( \mathcal{E} - e \phi \right)^2- \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)^2- (mc)^2\right) u\\ &\qquad + \frac{1}{{c}}\left[{\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)},{\mathcal{E} - e \phi}\right] u\\ &\qquad+ \frac{e \hbar}{c} \boldsymbol{\sigma}' \cdot \mathbf{H} u.\end{aligned}

All that remains is evaluation of the commutator term, which should yield the electric field interaction. That commutator is

\begin{aligned}\left[{\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)},{\mathcal{E} - e \phi}\right] u&={\boldsymbol{\alpha} \cdot \mathbf{P} \mathcal{E} u}- e \boldsymbol{\alpha} \cdot \mathbf{P} \phi u- \frac{e}{c} \boldsymbol{\alpha} \cdot \mathbf{A} \mathcal{E} u+ {\frac{e^2}{c} \boldsymbol{\alpha} \cdot \mathbf{A} \phi u} \\ &- {\mathcal{E} \boldsymbol{\alpha} \cdot \mathbf{P} u}+ e \phi \boldsymbol{\alpha} \cdot \mathbf{P} u+ \frac{e}{c} \mathcal{E} \boldsymbol{\alpha} \cdot \mathbf{A} u- {\frac{e^2}{c} \phi \boldsymbol{\alpha} \cdot \mathbf{A} u} \\ &=\boldsymbol{\alpha} \cdot \left( - e \mathbf{P} \phi+ \frac{e}{c} \mathcal{E} \right) u \\ &=e i \hbar \boldsymbol{\alpha} \cdot \left( \boldsymbol{\nabla} \phi+ \frac{1}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \right) u \\ &=- e i \hbar \boldsymbol{\alpha} \cdot \mathbf{E} u\end{aligned}

That was the last bit required to fully expand the space time split of our squared momentum equations. We have

\begin{aligned}0=({p} + mc)({p} - mc) u=\left(\frac{1}{{c^2}} \left( \mathcal{E} - e \phi \right)^2- \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)^2- (mc)^2- \frac{i e \hbar}{c} \boldsymbol{\alpha} \cdot \mathbf{E}+ \frac{e \hbar}{c} \boldsymbol{\sigma}' \cdot \mathbf{H}\right) u\end{aligned} \hspace{\stretch{1}}(3.21)

This is the end result of the reduction of the spacetime split gauge transformed Dirac equation. The next step is to obtain the non-relativistic Hamiltonian operator equation (linear in the time derivative operator and quadratic in spacial partials) that has both the electric field and magnetic field terms that we desire to accurately describe spin (actually we need only the magnetic interaction term for non-relativistic spin, but we’ll see that soon).

To obtain the first order time derivatives we can consider an approximation to the $(\mathcal{E} - e \phi)^2$ terms. We can get that by considering the difference of squares factorization

\begin{aligned}\frac{1}{{c^2}} ( \mathcal{E} - e \phi - m c^2) ( \mathcal{E} - e \phi + m c^2) u&=\frac{1}{{c^2}} \left(( \mathcal{E} - e \phi )^2 u - (m c^2)^2 u- {m c^2 \mathcal{E} u}+ {\mathcal{E} m c^2 u} \right) \\ &=\frac{1}{{c^2}} ( \mathcal{E} - e \phi )^2 u - (m c)^2 u\end{aligned}

In the text, this is factored, instead of the factorization verified. I wanted to be careful to ensure that the operators did not have any effect. They don’t, which is clear in retrospect since the $\mathcal{E}$ operator and the scalar $mc$ necessarily commute. With this factorization, some relativistic approximations are possible. Considering the free particle energy, we can separate out the rest energy from the kinetic (which is perversely designated with subscript $T$ for some reason in the text (and others))

\begin{aligned}\mathcal{E}&= \gamma m c^2 \\ &= m c^2 \left( 1 + \frac{1}{{2}} \left(\frac{\mathbf{v}}{c}\right)^2 + \cdots \right) \\ &= m c^2 + \frac{1}{{2}} m \mathbf{v}^2 + \cdots \\ &\equiv m c^2 + \mathcal{E}_{T}\end{aligned}

With this definition, the energy minus mass term in terms of kinetic energy (that we also had in the Klein-Gordon equation) takes the form

\begin{aligned}\frac{1}{{c^2}} ( \mathcal{E} - e \phi )^2 u - (m c)^2 u=\frac{1}{{c^2}} ( \mathcal{E}_{T} - e \phi ) ( \mathcal{E} - e \phi + m c^2) u\end{aligned} \hspace{\stretch{1}}(3.22)

In the second factor, to get a non-relativistic approximation of $\mathcal{E} - e \phi$, the text states without motivation that $e \phi$ will be considered small compared to $m c^2$. We can make some sense of this by considering the classical Hamiltonian for a particle in a field

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

We find that, in the non-relativistic limit, we have

\begin{aligned}\mathcal{E} - e \phi = m c^2 + \frac{1}{{2}} m \mathbf{v}^2 + \cdots \approx m c^2,\end{aligned} \hspace{\stretch{1}}(3.23)

and obtain the first order approximation of our time derivative operator

\begin{aligned}\frac{1}{{c^2}} ( \mathcal{E} - e \phi )^2 u - (m c)^2 u\approx\frac{1}{{c^2}} ( \mathcal{E}_{T} - e \phi ) 2 m c^2 u,\end{aligned} \hspace{\stretch{1}}(3.24)

or

\begin{aligned}\frac{1}{{c^2}} ( \mathcal{E} - e \phi )^2 u - (m c)^2 u\approx2 m ( \mathcal{E}_{T} - e \phi ).\end{aligned} \hspace{\stretch{1}}(3.25)

It seems slightly underhanded to use the free particle Hamiltonian in one part of the approximation, and the Hamiltonian for a particle in a field for the other part. This is probably why the text just mandates that $e\phi$ be small compared to $m c^2$.

To summarize once more before the final reduction (where we eliminate the electric field component of the operator equation), we have

\begin{aligned}0=({p} + mc)({p} - mc) u\approx\left(2 m ( \mathcal{E}_{T} - e \phi )- \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)^2- \frac{i e \hbar}{c} \boldsymbol{\alpha} \cdot \mathbf{E}+ \frac{e \hbar}{c} \boldsymbol{\sigma}' \cdot \mathbf{H}\right) u.\end{aligned} \hspace{\stretch{1}}(3.26)

Except for the electric field term, this is the result that is derived in the text. It was argued that this term is not significant compared to $e \phi$ when the particle velocity is restricted to the non-relativistic domain. This is done by computing the expectation of this term relative to $e \phi$. Consider

\begin{aligned}{\left\lvert{ \left\langle{{ \frac{e \hbar}{ 2 m c} \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{e \phi } }}\right\rangle }\right\rvert}\end{aligned} \hspace{\stretch{1}}(3.27)

With the velocities low enough so that the time variation of the vector potential does not contribute to the electric field (i.e. the electrostatic case), we have

\begin{aligned}\mathbf{E} = - \boldsymbol{\nabla} \phi = - \hat{\mathbf{r}} \frac{\partial {\phi}}{\partial {r}}.\end{aligned} \hspace{\stretch{1}}(3.28)

The variation in length $a$ that is considered is labeled the characteristic length

\begin{aligned}p a \sim \hbar,\end{aligned} \hspace{\stretch{1}}(3.29)

so that with $p = m v$ we have

\begin{aligned}a \sim \frac{\hbar}{m v}.\end{aligned} \hspace{\stretch{1}}(3.30)

This characteristic length is not elaborated on, but one can observe the similarity to the Compton wavelength

\begin{aligned}L_{\text{Compton}} = \frac{\hbar}{m c},\end{aligned} \hspace{\stretch{1}}(3.31)

the length scale for which Quantum field theory must be considered. This length scale is considerably larger for velocities smaller than the speed of light. For example, the drift velocity of electrons in copper is $\sim 10^{6} \frac{\text{m}}{\text{s}}$, which fixes our length scale to $100$ times the Compton length ($\sim 10^{-12} \text{m})$. This is still a very small length, but is in the QM domain instead of QED. With such a length scale consider the magnitude of a differential contribution to the electric field

\begin{aligned}{\left\lvert{\phi}\right\rvert} = {\left\lvert{\mathbf{E}}\right\rvert} \Delta x = {\left\lvert{\mathbf{E}}\right\rvert} a,\end{aligned} \hspace{\stretch{1}}(3.32)

so that

\begin{aligned}\left\langle{{ \frac{e \hbar}{ 2 m c} \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{e {\left\lvert{\phi}\right\rvert} } }}\right\rangle&=\left\langle{{ \frac{e \hbar}{ 2 m c} \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{e a {\left\lvert{\mathbf{E}}\right\rvert} } }}\right\rangle \\ &=\left\langle{{ \frac{e \hbar}{m} \frac{1}{ 2 c} \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{e \frac{\hbar }{ m v } {\left\lvert{\mathbf{E}}\right\rvert} } }}\right\rangle \\ &=\frac{1}{{2}} \frac{v}{c} \left\langle{{ \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{ {\left\lvert{\mathbf{E}}\right\rvert} } }}\right\rangle.\end{aligned}

Thus the magnitude of this (vector) expectation is dominated by the expectation of just the $\boldsymbol{\alpha}$. That has been calculated earlier when Dirac currents were considered, where it was found that

\begin{aligned}\left\langle{{\alpha_i}}\right\rangle = \psi^\dagger \alpha_i \psi = (\mathbf{j})_i.\end{aligned} \hspace{\stretch{1}}(3.33)

Also recall that (33.73) that this current was related to momentum with

\begin{aligned}\mathbf{j} = \frac{\mathbf{p}}{m c} = \frac{\mathbf{v}}{c}\end{aligned} \hspace{\stretch{1}}(3.34)

which allows for a final approximation of the magnitude of the electric field term’s expectation value relative to the $e\phi$ term of the Hamiltonian operator. Namely

\begin{aligned}{\left\lvert{ \left\langle{{ \frac{e \hbar}{ 2 m c} \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{e \phi } }}\right\rangle }\right\rvert}\sim\frac{\mathbf{v}^2}{c^2}.\end{aligned} \hspace{\stretch{1}}(3.35)

With that last approximation made, the gauge transformed Dirac equation, after non-relativistic approximation of the energy and electric field terms, is left as

\begin{aligned}i \hbar \frac{\partial {}}{\partial {t}}=\frac{1}{{2m}} \left(i \hbar \boldsymbol{\nabla} + \frac{e}{c} \mathbf{A} \right)^2- \frac{e \hbar}{2 m c} \boldsymbol{\sigma}' \cdot \mathbf{H}+ e \phi.\end{aligned} \hspace{\stretch{1}}(3.36)

This is still a four dimensional equation, and it is stated in the text that only the large component is relevant (reducing the degrees of spin freedom to two). That argument makes a bit more sense with the matrix form of the gauge reduction which follows in the next section, so understanding that well seems worthwhile, and is the next thing to digest.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] JD Jackson. Classical Electrodynamics Wiley. John Wiley and Sons, 2nd edition, 1975.

[3] Peeter Joot. Misc Physics and Math Play, chapter Hamiltonian notes. http://sites.google.com/site/peeterjoot/math2009/miscphysics.pdf.

## PHY450H1S. Relativistic Electrodynamics Lecture 24 (Taught by Prof. Erich Poppitz). Non-relativistic electrostatic Lagrangian.

Posted by peeterjoot on March 30, 2011

Covering chapter 5 \S 37, and chapter 8 \S 65 material from the text [1].

Covering pp. 181-195: the Lagrangian for a system of non relativistic charged particles to zeroth order in $(v/c)$: electrostatic energy of a system of charges and .mass renormalization.

# A closed system of charged particles.

Consider a closed system of charged particles $(m_a, q_a)$ and imagine there is a frame where they are non-relativistic $v_a/c \ll 1$. In this case we can describe the dynamics using a Lagrangian only for particles. i.e.

\begin{aligned}\mathcal{L} = \mathcal{L}( \mathbf{x}_1, \cdots, \mathbf{x}_N, \mathbf{v}_1, \cdots, \mathbf{v}_N)\end{aligned} \hspace{\stretch{1}}(2.1)

If we work t order $(v/c)^2$.

If we try to go to $O((v/c)^3$, it’s difficult to only use $\mathcal{L}$ for particles.

This can be inferred from

\begin{aligned}P = \frac{2}{3} \frac{e^2}{c^3} {\left\lvert{\dot{d}{\mathbf{d}}}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(2.2)

because at this order, due to radiation effects, we need to include EM field as dynamical.

# Start simple

\begin{aligned}S \end{aligned}

So in the non-relativistic limit, after dropping the constant term that doesn’t effect the dynamics, our Lagrangian is

\begin{aligned}\mathcal{L}(\mathbf{x}_a, \mathbf{v}_a) = \frac{1}{{2}} \sum_a m_a \mathbf{v}_a^2 - \frac{1}{{8}} \frac{m_a \mathbf{v}_a^4}{c^2}\end{aligned} \hspace{\stretch{1}}(3.3)

The first term is $O((v/c)^0)$ where the second is $O((v/c)^2)$.

Next include the fact that particles are charged.

\begin{aligned}\mathcal{L}_{\text{interaction}} = \sum_a \left( \cancel{q_a \frac{\mathbf{v}_a}{c} \cdot \mathbf{A}(\mathbf{x}_a, t)} - q_a \phi(\mathbf{x}_a, t) \right)\end{aligned} \hspace{\stretch{1}}(3.4)

Here, working to $O((v/c)^0)$, where we consider the particles moving so slowly that we have only a Coulomb potential $\phi$, not $\mathbf{A}$.

HERE: these are NOT ‘EXTERNAL’ potentials. They are caused by all the charged particles.

\begin{aligned}\partial_i F^{i l} = \frac{4 \pi}{c} j^l = 4 \pi \rho\end{aligned} \hspace{\stretch{1}}(3.5)

For $l = \alpha$ we have have $4 \pi \rho \mathbf{v}/c$, but we won’t do this today (tomorrow).

To leading order in $v/c$, particles only created Coulomb fields and they only “feel” Coulomb fields. Hence to $O((v/c)^0)$, we have

\begin{aligned}\mathcal{L} = \sum_a \frac{m_a \mathbf{v}_a^2}{2} - q_a \phi(\mathbf{x}_a, t)\end{aligned} \hspace{\stretch{1}}(3.6)

What’s the $\phi(\mathbf{x}_a, t)$, the Coulomb field created by all the particles.

\paragraph{How to find?}

\begin{aligned}\partial_i F^{i 0} = \frac{4 \pi}{c} = 4 \pi \rho\end{aligned} \hspace{\stretch{1}}(3.7)

or

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 4 \pi \rho = - \boldsymbol{\nabla}^2 \phi \end{aligned} \hspace{\stretch{1}}(3.8)

where

\begin{aligned}\rho(\mathbf{x}, t) = \sum_a q_a \delta^3 (\mathbf{x} - \mathbf{x}_a(t))\end{aligned} \hspace{\stretch{1}}(3.9)

This is a Poisson equation

\begin{aligned}\Delta \phi(\mathbf{x}) = \sum_a q_a 4 \pi \delta^3(\mathbf{x} - \mathbf{x}_a)\end{aligned} \hspace{\stretch{1}}(3.10)

(where the time dependence has been suppressed). This has solution

\begin{aligned}\phi(\mathbf{x}, t) = \sum_b \frac{q_b}{{\left\lvert{\mathbf{x} - \mathbf{x}_b(t)}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.11)

This is the sum of instantaneous Coulomb potentials of all particles at the point of interest. Hence, it appears that $\phi(\mathbf{x}_a, t)$ should be evaluated in 3.11 at $\mathbf{x}_a$?

However 3.11 becomes infinite due to contributions of the a-th particle itself. Solution to this is to drop the term, but let’s discuss this first.

Let’s talk about the electrostatic energy of our system of particles.

\begin{aligned}\mathcal{E} &= \frac{1}{{8 \pi}} \int d^3 \mathbf{x} \left(\mathbf{E}^2 + \cancel{\mathbf{B}^2} \right) \\ &= \frac{1}{{8 \pi}} \int d^3 \mathbf{x} \mathbf{E} \cdot (-\boldsymbol{\nabla} \phi) \\ &= \frac{1}{{8 \pi}} \int d^3 \mathbf{x} \left( \boldsymbol{\nabla} \cdot (\mathbf{E} \phi) - \phi \boldsymbol{\nabla} \cdot \mathbf{E} \right) \\ &= -\frac{1}{{8 \pi}} \oint d^2 \boldsymbol{\sigma} \cdot \mathbf{E} \phi + \frac{1}{{8 \pi}} \int d^3 \mathbf{x} \phi \boldsymbol{\nabla} \cdot \mathbf{E} \\ \end{aligned}

The first term is zero since $\mathbf{E} \phi$ for a localized system of charges $\sim 1/r^3$ or higher as $V \rightarrow \infty$.

In the second term

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 4 \pi \sum_a q_a \delta^3(\mathbf{x} - \mathbf{x}_a(t))\end{aligned} \hspace{\stretch{1}}(3.12)

So we have

\begin{aligned}\sum_a \frac{1}{{2}} \int d^3 \mathbf{x} q_a \delta^3(\mathbf{x} - \mathbf{x}_a) \phi(\mathbf{x})\end{aligned} \hspace{\stretch{1}}(3.13)

for

\begin{aligned}\mathcal{E} = \frac{1}{{2}} \sum_a q_a \phi(\mathbf{x}_a)\end{aligned} \hspace{\stretch{1}}(3.14)

Now substitute 3.11 into 3.14 for

\begin{aligned}\mathcal{E} = \frac{1}{{2}} \sum_a \frac{q_a^2}{{\left\lvert{\mathbf{x} - \mathbf{x}_a}\right\rvert}} + \frac{1}{{2}} \sum_{a \ne b} \frac{q_a q_b}{{\left\lvert{\mathbf{x}_a - \mathbf{x}_b}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.15)

or

\begin{aligned}\mathcal{E} = \frac{1}{{2}} \sum_a \frac{q_a^2}{{\left\lvert{\mathbf{x} - \mathbf{x}_a}\right\rvert}} + \sum_{a < b} \frac{q_a q_b}{{\left\lvert{\mathbf{x}_a - \mathbf{x}_b}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.16)

The first term is the sum of the electrostatic self energies of all particles. The source of this infinite self energy is in assuming a \underline{point like nature} of the particle. i.e. We modeled the charge using a delta function instead of using a continuous charge distribution.

Recall that if you have a charged sphere of radius $r$

PICTURE: total charge $q$, radius $r$, our electrostatic energy is

\begin{aligned}\mathcal{E} \sim \frac{q^2}{r}\end{aligned} \hspace{\stretch{1}}(3.17)

Stipulate that rest energy $m_e c^2$ is all of electrostatic origin $\sim e^2/r_e$ we get that

\begin{aligned}r_e \sim \frac{e^2}{m_e c^2}\end{aligned} \hspace{\stretch{1}}(3.18)

This is called the classical radius of the electron, and is of a very small scale $10^{-13} \text{cm}$.

As a matter of fact the applicability of classical electrodynamics breaks down much sooner than this scale since quantum effects start kicking in.

Our Lagrangian is now

\begin{aligned}\mathcal{L}_a = \frac{1}{{2}} m_a \mathbf{v}_a^2 - q_a \phi(\mathbf{x}_a, t)\end{aligned} \hspace{\stretch{1}}(3.19)

where $\phi$ is the electrostatic potential due to all \underline{other} particles, so we have

\begin{aligned}\mathcal{L}_a = \frac{1}{{2}} m_a \mathbf{v}_a^2 - \sum_{a \ne b} \frac{q_a q_b }{{\left\lvert{\mathbf{x}_a - \mathbf{x}_b}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.20)

and for the system

\begin{aligned}\mathcal{L} = \frac{1}{{2}} \sum_a m_a \mathbf{v}_a^2 - \sum_{a < b} \frac{q_a q_b }{{\left\lvert{\mathbf{x}_a - \mathbf{x}_b}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.21)

This is THE Lagrangian for electrodynamics in the non-relativistic case, starting with the relativistic action.

# What’s next?

We continue to the next order of $v/c$ tomorrow.

# References

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

## Energy term of the Lorentz force equation.

Posted by peeterjoot on February 8, 2011

# 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.

## PHY450H1S. Relativistic Electrodynamics Lecture 7 (Taught by Prof. Erich Poppitz). Action and relativistic dynamics

Posted by peeterjoot on January 26, 2011

Will now be covering chapter 2 material from the text [1].

Covering Professor Poppitz’s lecture notes: equation of motion, symmetries, and conserved quantities (energy-momentum 4 vector) from relativistic particle action [Wednesday, Jan. 26, Tuesday, Feb. 1]

These notes are also augmented by some additional notes completing an argument on page 53.

# The relativity principle

The relativity principle implies that the EOM should be expressed in 4-vector form, just like Newton’s EOM are expressed in 3-vector form

\begin{aligned}m \ddot{\mathbf{r}} = \mathbf{f}\end{aligned} \hspace{\stretch{1}}(2.1)

Observe that in coordinate form this is

\begin{aligned}m \ddot{r}^i = f^i, \qquad i = 1,2,3\end{aligned} \hspace{\stretch{1}}(2.2)

or for a rotated frame $O'$

\begin{aligned}m \ddot{r'}^i = {f'}^i, \qquad i = 1,2,3\end{aligned} \hspace{\stretch{1}}(2.3)

Need to generalize to 4 vectors, so we need 4-velocity and 4-acceleration.

Later we will study action and Lagrangian, and then relativity will require that the action be a Lorentz scalar. The analogy for a Newtonian point particle is a scalar under rotations.

## Four vector velocity

\paragraph{Definition:} Velocity s the rate of change of position in $(ct, \mathbf{x})$-space. Position means specifying both $ct$ and $\mathbf{x}$ for a point in spacetime.

PICTURE: $x^0 = ct$ axis up, and $x^1, x^2, x^3$ axis over, with worldline $x = x(\tau)$. Here $\tau$ is a parameter for the worldline, and provides a mapping for the curve in spacetime.

PICTURE: 3D vectors, $\mathbf{r}(t)$, $\mathbf{r}(t + \Delta t)$, and the difference vector $\mathbf{r}(t + \Delta t) - \mathbf{r}(t)$.

We write

\begin{aligned}\mathbf{v}(t) \equiv \lim_{\Delta t \rightarrow zero} \frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{ \Delta t}\end{aligned} \hspace{\stretch{1}}(2.4)

For four vectors we will parameterize the worldline by its “length”, with $O$ taken from some arbitrary point on it. We can also take $\tau$ to be the proper time, and the only difference will be the factor of $c$ (which becomes especially easy with the choice $c=1$ that is avoided in this class).

\begin{aligned}\frac{x^i(\tau + \Delta \tau) - x^i(\tau)}{\Delta \tau}\end{aligned} \hspace{\stretch{1}}(2.5)

We’ll take

\begin{aligned}\Delta \tau = ds = \frac{dx^i}{ds}\end{aligned} \hspace{\stretch{1}}(2.6)

This is a nice quantity, we are dividing a vector by a scalar, and thus get a four vector as a result (i.e. the result transforms as a four vector).

PICTURE: small fragment of a worldline with constant slope over the infinitesimal interval. $dx^0$ up and $dx^1$ over.

\begin{aligned}u^i \equiv \frac{dx^i}{ds}\end{aligned} \hspace{\stretch{1}}(2.7)

\begin{aligned}ds^2 &= (dx^0)^2 - (dx^1)^2 \\ &= c^2 \left( (dt)^2 - \frac{1}{{c^2}} (dx^1)^2 \right) \\ &= c^2 (dt)^2 \left( 1 - \frac{1}{{c^2}} \frac{dx^1}{dt^2} \right) \end{aligned}

Or

\begin{aligned}ds = c dt \sqrt{1 - \frac{1}{{c^2}} \frac{dx^1}{dt^2} }\end{aligned} \hspace{\stretch{1}}(2.8)

NOTE: Prof admits pulling a fast one, since he has aligned the worldline along the $x^1$ axis, however this is always possible by rotating the coordinate system.

\begin{aligned}u^0 &= \frac{dx^0}{ds} \\ &= \frac{c dt}{ c dt \sqrt{ 1 - \mathbf{v}^2/c^2} } \\ &= \frac{1}{ \sqrt{ 1 - \mathbf{v}^2/c^2} } \\ &= \gamma\end{aligned}

\begin{aligned}u^1 &= \frac{dx^1}{ds} \\ &= \frac{dx^1 }{ c dt \sqrt{ 1 - \mathbf{v}^2/c^2} } \\ &= \frac{v^1/c}{ \sqrt{ 1 - \mathbf{v}^2/c^2} } \\ &= \gamma \frac{v^1}{c}\end{aligned}

Similarily

\begin{aligned}u^2 &= \gamma \frac{v^2}{c} \\ u^3 &= \gamma \frac{v^2}{c}\end{aligned}

We’ve now unpacked the four velocity, and have

\begin{aligned}u^i = \left( \gamma, \frac{\mathbf{v}}{c} \gamma \right)\end{aligned} \hspace{\stretch{1}}(2.9)

## Length of the four velocity vector

Recall that this length is

\begin{aligned}u^i g_{ij} u^j &= u^i u_i \\ &= u_i u^i \\ &= (u^0)^2 - (u_i)^2 \\ &= \gamma^2 - \gamma^2 \frac{\mathbf{v}}{c} \cdot \frac{\mathbf{v}}{c} \\ &= \gamma^2 \left(1 - \frac{\mathbf{v}^2}{c^2} \right)\end{aligned}

The four velocity in physics is

\begin{aligned}u^i = \left( \gamma, \frac{\mathbf{v}}{c} \gamma \right)\end{aligned} \hspace{\stretch{1}}(2.10)

but in mathematics the meaning of $u^i u_i = 1$ means that this quantity is the unit tangent vector to the worldline.

## Four acceleration

In Newtonian physics we have

\begin{aligned}\mathbf{a} = \frac{\mathbf{v}}{dt}\end{aligned} \hspace{\stretch{1}}(2.11)

Our relativistic mapping of this, with $v \rightarrow u^i$ and $t \rightarrow s$, gives

\begin{aligned}w^i = \frac{d u^i}{ds}\end{aligned} \hspace{\stretch{1}}(2.12)

Geometrically $w^i$ is the normal to the worldline. This follows from $u^i g_{ij} u^j = 1$, so

\begin{aligned}\frac{d}{ds} \left( u^i g_{ij} u^j \right) &=\frac{d u^i}{ds} g_{ij} u^j +u^i g_{ij} \frac{d u^j}{ds} \\ &=\frac{d u^i}{ds} g_{ij} u^j +u^j \underbrace{g_{ji}}_{= g_{ij}} \frac{d u^i}{ds} \\ &=\frac{d u^i}{ds} g_{ij} u^j +u^j g_{ji} \frac{d u^i}{ds} \\ &=2 \frac{d u^i}{ds} g_{ij} u^j \end{aligned}

Note that we’ve utilized the fact above that the dummy summation indexes can be swapped (or changed to anything else we feel inclined to use).

The conclusion is that the dot product of the acceleration and the velocity is zero

\begin{aligned}w_i u^i = 0.\end{aligned} \hspace{\stretch{1}}(2.13)

# Relativistic action.

\begin{aligned}S_{ab} = ?\end{aligned} \hspace{\stretch{1}}(3.14)

What is the action for a worldline from $a \rightarrow b$.

We want something that has velocity dependence ($u^i$ not $\mathbf{v}$), but that is Lorentz invariant and has only first derivatives.

The relativisitic length is the simplest so we could form

\begin{aligned}\int ds u^i u_i\end{aligned} \hspace{\stretch{1}}(3.15)

but that’s not interesting since $u^i u_i = 1$. We could form

\begin{aligned}\int ds u^i \frac{u_i}{ds} = \int ds w^i u_i\end{aligned} \hspace{\stretch{1}}(3.16)

but then this is just zero.

We could form something like

\begin{aligned}\int ds \frac{w^i}{ds} u_i\end{aligned} \hspace{\stretch{1}}(3.17)

This is non zero and non-constant, but evaluating the EOM for such an action would produce a result that has higher than second order derivatives.

We are left with

\begin{aligned}S_{ab} = \text{constant} \int_a^b ds \end{aligned} \hspace{\stretch{1}}(3.18)

To fix this constant we note that if we want to minimize the action over the infinitesimal interval, then we need a minus sign. Since the Lagrangian has dimensions of energy, and the dimensions of energy times time are momentum, our action must then have dimensions of momentum. So one possible constant that fixes up our dimensions is $mc$. Construct an action with the following form

\begin{aligned}S_{ab} = - m c\int_a^b ds,\end{aligned} \hspace{\stretch{1}}(3.19)

does the job we want. Here “m” is a characteristic of the particle, which \underline{is a Lorentz scalar}. It also happens to have dimensions of mass. With $ds = c dt \sqrt{1 - \mathbf{v}^2/c^2}$, we have

\begin{aligned}S_{ab} = - m c^2 \int_{t_a}^{t_b} dt \sqrt{ 1 - \frac{1}{{c^2}} \left( \frac{d \mathbf{x}(t) }{dt} \right)^2 }\end{aligned} \hspace{\stretch{1}}(3.20)

Now everything looks like it was in classical mechanics.

\begin{aligned}S_{ab} = \int_{t_a}^{t_b} \mathcal{L}(\dot{\mathbf{x}}(t)) dt\end{aligned} \hspace{\stretch{1}}(3.21)

\begin{aligned}\mathcal{L}(\dot{\mathbf{x}}(t)) = -m c^2 \end{aligned} \hspace{\stretch{1}}(3.22)

Now find the extremum of $S$. That problem is really to compute the variation in the action that results from varying the coordinates around the stationary point, and equate that variation to zero to find the extremum

\begin{aligned}\delta S = S[\mathbf{x}(t) + \delta \mathbf{x}(t)] - S[ \mathbf{x}(t) ] = 0\end{aligned} \hspace{\stretch{1}}(3.23)

The usual condition is imposed where we have zero variation of the coordinates at the boundies of the action integral

\begin{aligned}0 = \delta \mathbf{x}(t_a) = \delta \mathbf{x}(t_b) \end{aligned} \hspace{\stretch{1}}(3.24)

Returning to our action we have

\begin{aligned}\frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{\mathbf{x}}}} = \frac{\partial {\mathcal{L}}}{\partial {\mathbf{x}}} = 0\end{aligned} \hspace{\stretch{1}}(3.25)

This last is zero because it’s a free particle with no position dependence.

\begin{aligned} 0 &= -m c^2 \frac{d}{dt} \frac{\partial}{\partial \dot{\mathbf{x}}} \sqrt{ 1 - \dot{\mathbf{x}}^2 } \\ &= -m c^2 \frac{d}{dt} \frac{- \dot{\mathbf{x}}}{\sqrt{ 1 - \dot{\mathbf{x}}^2 } } \\ &= m c^2 \frac{d}{dt} \gamma \dot\mathbf{x} \end{aligned}

So we have

\begin{aligned}\frac{d}{dt} (\gamma \dot{\mathbf{x}}) = 0\end{aligned} \hspace{\stretch{1}}(3.26)

By evaluating this, we can eventually show that we can construct a four vector equation. Doing this we have

\begin{aligned}\frac{d}{dt} (\gamma \mathbf{v}) &=\frac{d}{dt} \left( \left(1 - \mathbf{v}^2/c^2\right)^{-1/2} \mathbf{v} \right) \\ &=-2 (-1/2) \mathbf{v} (\mathbf{v} \cdot \dot{\mathbf{v}})/c^2 \left(1 - \mathbf{v}^2/c^2\right)^{-3/2} + \left(1 - \mathbf{v}^2/c^2\right)^{-1/2} \dot{\mathbf{v}} \\ &=\gamma \left( \frac{\mathbf{v} (\mathbf{v} \cdot \dot{\mathbf{v}}) }{ c^2 - \mathbf{v}^2 } + \dot{\mathbf{v}} \right)\end{aligned}

Or

\begin{aligned}\frac{\mathbf{v} (\mathbf{v} \cdot \dot{\mathbf{v}}) }{ c^2 - \mathbf{v}^2 } + \dot{\mathbf{v}} = 0\end{aligned} \hspace{\stretch{1}}(3.27)

Clearly $\dot{\mathbf{v}} = 0$ is a solution, but is it the only solution?

By dotting this with $\mathbf{v}$ we have

\begin{aligned}0 &= \frac{\mathbf{v}^2 (\mathbf{v} \cdot \dot{\mathbf{v}}) }{ c^2 - \mathbf{v}^2 } + \dot{\mathbf{v}} \cdot \mathbf{v} \\ &= (\mathbf{v} \cdot \dot{\mathbf{v}}) \left( 1 + \frac{\mathbf{v}^2}{c^2 - \mathbf{v}^2} \right) \\ &= (\mathbf{v} \cdot \dot{\mathbf{v}}) \frac{c^2}{c^2 - \mathbf{v}^2} \end{aligned}

This implies that $\dot{\mathbf{v}} = 0$ (a contraction) or that $\mathbf{v} \cdot \dot{\mathbf{v}} = 0$. To examine the perpendicularity question, let’s take cross products. This gives

\begin{aligned}0 =\frac{(\mathbf{v} \times \mathbf{v}) (\mathbf{v} \cdot \dot{\mathbf{v}}) }{ c^2 - \mathbf{v}^2 } + \dot{\mathbf{v}} \times \mathbf{v}\end{aligned} \hspace{\stretch{1}}(3.28)

We have found that $\mathbf{v} \cdot \dot{\mathbf{v}} = 0$ and $\mathbf{v} \times \dot{\mathbf{v}} = 0$. This can only mean that $\dot{\mathbf{v}} = 0$, contradicting the assumption that is non-zero. We conclude that $\dot{\mathbf{v}} = 0$ is the only solution to 3.27.

# Next time

We want to finish up and show how this results in a four velocity equation. We have

\begin{aligned}\frac{d}{dt} ( \gamma \mathbf{v}) = 0\end{aligned} \hspace{\stretch{1}}(4.29)

which is

\begin{aligned}\frac{d}{dt} ( u^\alpha ) = 0, \qquad \text{for} u^\alpha = u^1, u^2, u^3\end{aligned} \hspace{\stretch{1}}(4.30)

eventually, we will show that we also have

\begin{aligned}\frac{d}{dt} ( u^i ) = 0\end{aligned} \hspace{\stretch{1}}(4.31)

# References

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

## Classical Electrodynamic gauge interaction.

Posted by peeterjoot on October 22, 2010

# Motivation.

In [1] chapter 6, we have a statement that in classical mechanics the electromagnetic interaction is due to a transformation of the following form

\begin{aligned}\mathbf{p} &\rightarrow \mathbf{p} - \frac{e}{c} \mathbf{A} \\ E &\rightarrow E - e \phi\end{aligned} \hspace{\stretch{1}}(1.1)

Let’s verify that this does produce the classical interaction law. Putting a more familiar label on this, we should see that we obtain the Lorentz force law from a transformation of the Hamiltonian.

# Hamiltonian equations.

Recall that the Hamiltonian was defined in terms of conjugate momentum components $p_k$ as

\begin{aligned}H(x_k, p_k) = \dot{x}_k p_k - \mathcal{L}(x_k, \dot{x}_k),\end{aligned} \hspace{\stretch{1}}(2.3)

we can take $x_k$ partials to obtain the first of the Hamiltonian system of equations for the motion

\begin{aligned}\frac{\partial {H}}{\partial {x_k}} &= - \frac{\partial {\mathcal{L}}}{\partial {x_k}} \\ &= - \frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{x}_k}} \end{aligned}

With $p_k \equiv {\partial {\mathcal{L}}}/{\partial {\dot{x}_k}}$, and taking $p_k$ partials too, we have the system of equations

\begin{subequations}

\begin{aligned} \frac{\partial {H}}{\partial {x_k}} &= - \frac{d p_k}{dt}\end{aligned} \hspace{\stretch{1}}(2.4a)

\begin{aligned} \frac{\partial {H}}{\partial {p_k}} &= \dot{x}_k\end{aligned} \hspace{\stretch{1}}(2.4b)

\end{subequations}

# Classical interaction

Starting with the free particle Hamiltonian

\begin{aligned}H = \frac{\mathbf{p}}{2m},\end{aligned} \hspace{\stretch{1}}(3.5)

we make the transformation required to both the energy and momentum terms

\begin{aligned}H - e\phi = \frac{\left(\mathbf{p} - \frac{e}{c} \mathbf{A}\right)^2 }{2m} = \frac{1}{{2m}} \mathbf{p}^2 - \frac{e}{m c} \mathbf{p} \cdot \mathbf{A} + \frac{1}{{2m}} \left(\frac{e}{c}\right)^2 \mathbf{A}^2 \end{aligned} \hspace{\stretch{1}}(3.6)

From 2.4b we find

\begin{aligned}\frac{d x_k}{dt} = \frac{\partial {H}}{\partial {p_k}} = \frac{1}{{m}} \left( p_k - \frac{e}{c} A_k \right),\end{aligned} \hspace{\stretch{1}}(3.7)

or

\begin{aligned}p_k = m \frac{d x_k}{dt} + \frac{e}{c} A_k.\end{aligned} \hspace{\stretch{1}}(3.8)

Taking derivatives and employing 2.4a we have

\begin{aligned}\frac{d p_k}{dt} &= m \frac{d^2 x_k}{dt^2} + \frac{e}{c} \frac{d A_k}{dt} \\ &= -\frac{\partial {H}}{\partial {x_k}} \\ &=\frac{1}{{m}} \frac{e}{c} p_n \frac{\partial {A_n}}{\partial {x_k}} - e \frac{\partial {\phi}}{\partial {x_k}}- \frac{1}{{m}} \left(\frac{e}{c}\right)^2 A_k \frac{\partial {A_k}}{\partial {x_k}} \\ &=\frac{1}{{m}} \frac{e}{c} \left(m \frac{d x_n}{dt} + \frac{e}{c} A_n\right)\frac{\partial {A_n}}{\partial {x_k}} - e \frac{\partial {\phi}}{\partial {x_k}}- \frac{1}{{m}} \left(\frac{e}{c}\right)^2 A_k \frac{\partial {A_k}}{\partial {x_k}} \\ &=\frac{e}{c} \frac{d x_n}{dt}\frac{\partial {A_n}}{\partial {x_k}} - e \frac{\partial {\phi}}{\partial {x_k}}\end{aligned}

Rearranging and utilizing the convective derivative expansion $d/dt = (d x_a/dt) {\partial {}}/{\partial {x_a}}$ (ie: chain rule), we have

\begin{aligned}m \frac{d^2 x_k}{dt^2} &=\frac{e}{c} \frac{d x_n}{dt}\left( \frac{\partial {A_n}}{\partial {x_k}}- \frac{\partial {A_k}}{\partial {x_n}} \right) - e \frac{\partial {\phi}}{\partial {x_k}}\end{aligned} \hspace{\stretch{1}}(3.9)

We guess and expect that the first term of 3.9 is $e (\mathbf{v}/c \times \mathbf{B})_k$. Let’s verify this

\begin{aligned}(\mathbf{v} \times \mathbf{B})_k&= \dot{x}_m B_d \epsilon_{k m d} \\ &= \dot{x}_m ( \epsilon_{d a b} \partial_a A_b ) \epsilon_{k m d} \\ &= \dot{x}_m \partial_a A_b \epsilon_{d a b} \epsilon_{d k m}\end{aligned}

Since $\epsilon_{d a b} \epsilon_{d k m} = \delta_{a k} \delta_{b m} - \delta_{a m} \delta_{b k}$ we have

\begin{aligned}(\mathbf{v} \times \mathbf{B})_k&= \dot{x}_m \partial_a A_b \epsilon_{d a b} \epsilon_{d k m} \\ &=\dot{x}_m \partial_a A_b \delta_{a k} \delta_{b m} -\dot{x}_m \partial_a A_b \delta_{a m} \delta_{b k} \\ &= \dot{x}_m ( \partial_k A_m - \partial_m A_k )\end{aligned}

Except for a difference in dummy summation variables, this matches what we had in 3.9. Thus we are able to put that into the traditional Lorentz force vector form

\begin{aligned}m \frac{d^2 \mathbf{x}}{dt^2} &= e \frac{\mathbf{v}}{c} \times \mathbf{B} + e \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(3.10)

It’s good to see that we get the classical interaction from this transformation before moving on to the trickier seeming QM interaction.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.