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Posted by peeterjoot on February 3, 2012

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# Disclaimer.

Ungraded solutions to posted problem set 1 (I’m auditing half the lectures for this course and won’t be submitting any solutions for grading).

# Problem 1. Lorentz force Lagrangian.

## Evaluate the Euler-Lagrange equations.

This problem has two parts. The first is to derive the Lorentz force equation

\begin{aligned}\mathbf{F} &= q (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \\ \mathbf{E} &= -\boldsymbol{\nabla} \phi - \frac{\partial {\mathbf{A}}}{\partial {t}} \\ \mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{A}\end{aligned} \hspace{\stretch{1}}(2.1)

using the Euler-Lagrange equations using the Lagrangian

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \mathbf{v}^2 + q \mathbf{v} \cdot \mathbf{A} - q \phi.\end{aligned} \hspace{\stretch{1}}(2.4)

In coordinates, employing summation convention, this Lagrangian is

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \dot{x}_j \dot{x}_j + q \dot{x}_j A_j - q \phi.\end{aligned} \hspace{\stretch{1}}(2.5)

Taking derivatives

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\dot{x}_i}} = m \dot{x}_i + q A_i,\end{aligned} \hspace{\stretch{1}}(2.6)

\begin{aligned}\frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{x}_i}} &= m \dot{d}{x}_i + q \frac{\partial {A_i}}{\partial {t}}+ q \frac{\partial {A_i}}{\partial {x_j}} \frac{dx_j}{dt} \\ &=m \dot{d}{x}_i + q \frac{\partial {A_i}}{\partial {t}}+ q \frac{\partial {A_i}}{\partial {x_j}} \dot{x}_j\end{aligned}

This must equal

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {x_i}} = q \dot{x}_j \frac{\partial {A_j}}{\partial {x_i}} - q \frac{\partial {\phi}}{\partial {x_i}},\end{aligned} \hspace{\stretch{1}}(2.7)

So we have

\begin{aligned}m \dot{d}{x}_i &= -q \frac{\partial {A_i}}{\partial {t}}- q \frac{\partial {A_i}}{\partial {x_j}} \dot{x}_j+q \dot{x}_j \frac{\partial {A_j}}{\partial {x_i}} - q \frac{\partial {\phi}}{\partial {x_i}} \\ &=-q \left( \frac{\partial {A_i}}{\partial {t}} - \frac{\partial {\phi}}{\partial {x_i}} \right)+q v_j \left( \frac{\partial {A_j}}{\partial {x_i}} - \frac{\partial {A_i}}{\partial {x_j}} \right)\end{aligned}

The first term is just $E_i$. If we expand out $(\mathbf{v} \times \mathbf{B})_i$ we see that matches

\begin{aligned}(\mathbf{v} \times \mathbf{B})_i&=v_a B_b \epsilon_{abi} \\ &=v_a \partial_r A_s \epsilon_{rsb} \epsilon_{abi} \\ &=v_a \partial_r A_s \delta_{rs}^{[ia]} \\ &=v_a (\partial_i A_a - \partial_a A_i).\end{aligned}

A $a \rightarrow j$ substition, and comparision of this with the Euler-Lagrange result above completes the exersize.

## Show that the Lagrangian is gauge invariant.

With a gauge transformation of the form

\begin{aligned}\phi &\rightarrow \phi + \frac{\partial {\chi}}{\partial {t}} \\ \mathbf{A} &\rightarrow \mathbf{A} - \boldsymbol{\nabla} \chi,\end{aligned} \hspace{\stretch{1}}(2.8)

show that the Lagrangian is invariant.

We really only have to show that

\begin{aligned}\mathbf{v} \cdot \mathbf{A} - \phi\end{aligned} \hspace{\stretch{1}}(2.10)

is invariant. Making the transformation we have

\begin{aligned}\mathbf{v} \cdot \mathbf{A} - \phi&\rightarrow v_j \left(A_j - \partial_j \chi \right) - \left(\phi + \frac{\partial {\chi}}{\partial {t}} \right) \\ &=v_j A_j - \phi - v_j \partial_j \chi - \frac{\partial {\chi}}{\partial {t}} \\ &=\mathbf{v} \cdot \mathbf{A} - \phi- \left( \frac{d x_j}{dt} \frac{\partial {\chi}}{\partial {x_j}} + \frac{\partial {\chi}}{\partial {t}} \right) \\ &=\mathbf{v} \cdot \mathbf{A} - \phi- \frac{d \chi(\mathbf{x}, t)}{dt}.\end{aligned}

We see then that the Lagrangian transforms as

\begin{aligned}\mathcal{L} \rightarrow \mathcal{L} + \frac{d}{dt}\left( -q \chi \right),\end{aligned} \hspace{\stretch{1}}(2.11)

and differs only by a total derivative. With the lemma from the lecture, we see that this gauge transformation does not have any effect on the end result of applying the Euler-Lagrange equations.

# Problem 2. Action minimization problem for surface gravity.

Here we are told to guess at a solution

\begin{aligned}y = a_2 t^2 + a_1 t + a_0,\end{aligned} \hspace{\stretch{1}}(3.12)

for the height of a particle thrown up into the air. With initial condition $y(0) = 0$ we have

\begin{aligned}a_0 = 0,\end{aligned} \hspace{\stretch{1}}(3.13)

and with a final condition of $y(T) = 0$ we also have

\begin{aligned}0 &= a_2 T^2 + a_1 T \\ &= T( a_2 T + a_1 ),\end{aligned}

so have

\begin{aligned}y(t) &= a_2 t^2 - a_2 T t = a_2 (t^2 - T t) \\ \dot{y}(t) &= a_2 (2 t - T )\end{aligned} \hspace{\stretch{1}}(3.14)

So our Lagrangian is

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m a_2^2 (2 t - T )^2 - m g a_2 (t^2 - T t)\end{aligned} \hspace{\stretch{1}}(3.16)

and our action is

\begin{aligned}S = \int_0^T dt \left( \frac{1}{{2}} m a_2^2 (2 t - T )^2 - m g a_2 (t^2 - T t)\right).\end{aligned} \hspace{\stretch{1}}(3.17)

To minimize this action with respect to $a_2$ we take the derivative

\begin{aligned}\frac{\partial {S}}{\partial {a_2}} = \int_0^T\left( m a_2 (2 t - T )^2 - m g (t^2 - T t)\right).\end{aligned} \hspace{\stretch{1}}(3.18)

Integrating we have

\begin{aligned}0 &= \frac{\partial {S}}{\partial {a_2}} \\ &={\left.\left(\frac{1}{{6}} m a_2 (2 t - T )^3 - m g \left(\frac{1}{{3}}t^3 - \frac{1}{{2}}T t^2 \right)\right)\right\vert}_0^T \\ &=\frac{1}{{6}} m a_2 T^3 - m g \left(\frac{1}{{3}}T^3 - \frac{1}{{2}}T^3 \right)-\frac{1}{{6}} m a_2 (- T )^3 \\ &=m T^3 \left( \frac{1}{{3}} a_2 - g \left( \frac{1}{{3}} - \frac{1}{{2}} \right) \right) \\ &=\frac{1}{{3}} m T^3 \left( a_2 - g \left( 1 - \frac{3}{2} \right) \right) \\ \end{aligned}

or

\begin{aligned}a_2 + g/2 = 0,\end{aligned} \hspace{\stretch{1}}(3.19)

which is the result we are required to show.

# Problem 3. Change of variables in a Lagrangian.

Here we want to show that after a change of variables, provided such a transformation is non-singular, the Euler-Lagrange equations are still valid.

Let’s write

\begin{aligned}r_i = r_i(q_1, q_2, \cdots q_N).\end{aligned} \hspace{\stretch{1}}(4.20)

Our “velocity” variables in terms of the original parameterization $q_i$ are

\begin{aligned}\dot{r}_j = \frac{dr_j}{dt} = \frac{\partial {r_j}}{\partial {q_i}} \frac{\partial {q_i}}{\partial {t}} = \dot{q}_i \frac{\partial {r_j}}{\partial {q_i}},\end{aligned} \hspace{\stretch{1}}(4.21)

so we have

\begin{aligned}\frac{\partial {\dot{r}_j}}{\partial {\dot{q}_i}} = \frac{\partial {r_j}}{\partial {q_i}}.\end{aligned} \hspace{\stretch{1}}(4.22)

Computing the LHS of the Euler Lagrange equation we find

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {q_i}} = \frac{\partial {\mathcal{L}}}{\partial {r_j}} \frac{\partial {r_j}}{\partial {q_i}}+\frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \frac{\partial {\dot{r}_j}}{\partial {q_i}}.\end{aligned} \hspace{\stretch{1}}(4.23)

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\dot{q}_i}} = \frac{\partial {\mathcal{L}}}{\partial {r_j}} \frac{\partial {r_j}}{\partial {\dot{q}_i}}+\frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \frac{\partial {\dot{r}_j}}{\partial {\dot{q}_i}}= \frac{\partial {\mathcal{L}}}{\partial {r_j}} \frac{\partial {r_j}}{\partial {\dot{q}_i}}+\frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \frac{\partial {r_j}}{\partial {q_i}},\end{aligned} \hspace{\stretch{1}}(4.24)

but ${\partial {r_j}}/{\partial {\dot{q}_i}} = 0$, so this is just

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\dot{q}_i}} = \frac{\partial {\mathcal{L}}}{\partial {r_j}} \frac{\partial {r_j}}{\partial {\dot{q}_i}}+\frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \frac{\partial {\dot{r}_j}}{\partial {\dot{q}_i}}= \frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \frac{\partial {r_j}}{\partial {q_i}}.\end{aligned} \hspace{\stretch{1}}(4.25)

The Euler-Lagrange equations become

\begin{aligned}0 &=\frac{\partial {\mathcal{L}}}{\partial {r_j}} \frac{\partial {r_j}}{\partial {q_i}}+\frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \frac{\partial {\dot{r}_j}}{\partial {q_i}}- \frac{d{{}}}{dt} \left(\frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \frac{\partial {r_j}}{\partial {q_i}}\right) \\ &= \frac{\partial {\mathcal{L}}}{\partial {r_j}} \frac{\partial {r_j}}{\partial {q_i}}+ \not{{\frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \frac{\partial {\dot{r}_j}}{\partial {q_i}}}}- \left( \frac{d{{}}}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \right) \frac{\partial {r_j}}{\partial {q_i}}- \not{{\frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \frac{d{{}}}{dt} \frac{\partial {r_j}}{\partial {q_i}} }}\\ &=\left( \frac{\partial {\mathcal{L}}}{\partial {r_j}} -\frac{d{{}}}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}} \right) \frac{\partial {r_j}}{\partial {q_i}}\end{aligned}

Since we have an assumption that the transformation is non-singular, we have for all $j$

\begin{aligned}\frac{\partial {r_j}}{\partial {q_i}} \ne 0,\end{aligned} \hspace{\stretch{1}}(4.26)

so we have the Euler-Lagrange equations for the new abstract coordinates as well

\begin{aligned}0 = \frac{\partial {\mathcal{L}}}{\partial {r_j}} -\frac{d{{}}}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{r}_j}}.\end{aligned} \hspace{\stretch{1}}(4.27)