Web bonus problem--Week 10-11

[Victor Ivrii]

Consider Lagrangian
$$L(\mathbf{q}, \dot{\mathbf{q}})=
 \frac{m}{2}{\dot{\mathbf{q}}}^2 + e\mathbf{A}(\mathbf{q})\cdot \dot{\mathbf{q}} -V(\mathbf{q})$$ ($\mathbf{q}=(q_1,\ldots,q_n)$ and find

a. Euler-Lagrange equations

b. Generalized momenta $p_i= \frac{\partial L }{\partial \dot{q}_i}$ and generalized forces $\frac{\partial L }{\partial q_i}$

c. Hamiltonian $H(\mathbf{q},\mathbf{p})$ (look how it is defined http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter10/S10.3.html)

d. What effect has $\mathbf{A}$ ?

[Adam Gao]

$L(\mathbf{q}, \dot{\mathbf{q}})=  \frac{m}{2}{\dot{\mathbf{q}}}^2 + e\mathbf{A}(\mathbf{q})\cdot \dot{\mathbf{q}} -V(\mathbf{q})$

a.

Euler-lagrange equations are of the form

$\frac{\partial L}{\partial q} = \frac{d}{dt}\frac{\partial L}{\partial \dot q}$

Compute both sides:

$\frac{\partial L}{\partial q} = eA'(q)\cdot q - V'(q)$

$\frac{\partial L}{\partial\dot q} = m\dot q + eA(q)$

$\frac{d}{dt}\frac{\partial L}{\partial \dot q} = m\ddot{q} + eA'(q)\cdot\dot q + \frac{\partial A}{\partial t}$

so

$\frac{\partial L}{\partial q} = eA'(q)\cdot q - V'(q) = \frac{d}{dt}\frac{\partial L}{\partial \dot q} = m\ddot{q} + eA'(q)\cdot\dot q + \frac{\partial A}{\partial t}$

b.

Generalized momenta in index form can be taken from the Euler Lagrange equation quantities:

$p_i = \frac{\partial L}{\partial \dot q_i} =  m\dot q_{i} + eA(q)_{i}$

Generalized forces take the following form:

$\frac{\partial L}{\partial q_{i}} = e\frac{\partial A}{\partial q_i} - \frac{\partial V}{\partial q_i}$

c.

The Hamiltonian $H(q,p)$ is defined as $H := \sum_{k=1}^n
\frac{\partial L}{\partial \dot{q}_k} \dot{q}_k -L$

Plugging in the generalized momenta and Lagrangian we get:

$H =  \sum((m\dot q_{i} + eA(q)_{i})\dot{q}_i )- L = m\dot q^2 + eA(q)\cdot\dot{q} - \frac{m}{2}{\dot{q}}^2 - eA(q)\cdot \dot{q} + V(q) = \frac{m}{2}{\dot{q}}^2 + V(q)$

d.

$A$ has no effect on the Hamiltonian (or in physics, energy). However, A does affect generalized momenta; and its positional derivative (derivative with respect to $q$) affects generalized forces.

[Victor Ivrii]

What you wrote is an expression for an energy (correct) in Lagrangian dynamics  but it is not a Hamiltonian: in the Hamiltonian you must purge $\dot{\mathbf{q}}$ and replace it by its expression through $\mathbf{p}$. The real momentum would be indeed $m\mathbf{p}$ but we need a generalized momentum. And then $\mathbf{A}$ affects Hamiltonian.

Also, in Euler-Lagrange equations, find $\ddot{\mathbf{q}}$ and find an extra term, due to $\mathbf{A}$ ($n=3$).

Why one needs a Hamiltonian?
* To use Hamiltonian formalism
* Want to go to Quantum Mechanics? Then you quantize the Hamiltonian!

[Adam Gao]

c. and d. Correction to Hamiltonian:

First express $\dot{q}$ in terms of generalized momenta $p$.

We have $p_i = \frac{\partial L}{\partial \dot q_i} =  m\dot q_{i} + eA(q)_{i}$

so

$\dot{q}_i = \frac{p_i}{m} - \frac{eA(q)_i}{m}$

For the Hamiltonian to really be a Hamiltonian, we must replace the $\dot{q}$ term:

$H = \frac{m}{2}{\dot{q}}^2 + V(q)$

Now replace $\dot{q}_i$:

$H = \frac{m}{2}\sum_{i=1}^{n}( \frac{p_i}{m} - \frac{eA(q)_i}{m})^2 + V(q)$

I suppose $A$ affects the Hamiltonian because it affects generalized momenta.

e.

Find $\ddot{q}$ due to $A$.

The Euler-Lagrange equation is as follows:

$eA'(q) - V'(q) = m\ddot{q} + eA'(q)\dot q + \frac{\partial A}{\partial t}$

Solving for $\ddot{q}$:

$\ddot{q} = \frac{1}{m}(eA'(q) - V'(q) - A'(q)\dot{q} + \frac{\partial A}{\partial t})$

[Victor Ivrii]

Traditionally people write $$H=-\frac{1}{2m}(\mathbf{p}-e\mathbf{A})^2+V(q).$$

However $\ddot{\mathbf{q}}$ is not found correctly. Please calculate in coordinates $\ddot{q}_j$. There will be a standard term $-\frac{1}mV_{q_j}$ and another term, depending on $\mathbf{A}$.

[Tristan Fraser]

Given a Lagrangian of $$ L(q,\dot{q})  = \frac{m\dot{\vec{q}}^2}{2} + eA( \vec {q})\dot{\vec{q}} - V(q) $$ and a Hamiltonian of $$ H (q,p) =  \frac{m}{2} \sum_{i =1}^{n} (\frac{p_i - eA(q)_i)}{m})^2  + V(q) $$

To find $ \ddot{q}$:

Use the Euler Lagrange equations: $ \frac{\partial L}{\partial \vec{q}} -  \frac{d}{dt} \frac{\partial L}{\partial \dot{\vec{q}}} = 0 $, we get:

$m\dot{q_j} + e\frac{\partial A}{\partial q_j} - m\ddot{q_j} - e (\frac{\partial A_j}{\partial q_i} \dot{q_j} - e \frac{\partial A_j}{\partial t})  = 0$ then $$ m\ddot{q_j} = m\dot{q_j} + e(\frac{\partial A_j}{\partial q_j}  - \frac{\partial A_j}{\partial q_i} \dot{q_j} - \frac{\partial A_j}{\partial t} ) $$

$$  \ddot{q_j}  =  \frac{1}{m} (m\dot{q_j} + e(\frac{\partial A_j}{\partial q_j} - \frac{\partial A_j}{\partial q_i} \dot{q_j} - \frac{\partial A_j}{\partial t} )) $$

Alternatively, one can use Hamilton's equations of motion to find $\dot{q}, \ddot{q}$ ,

$$ \frac{\partial H}{\partial p}  = \dot{q}  =  - \frac{2p}{2m} + \frac{eA}{2m} $$, where taking the time derivative gives us:

$$ \ddot{q}  = - \frac{\dot{p}}{m} + \frac{e}{2m} \frac{\partial A }{\partial q} \dot{q} $$

[Victor Ivrii]

No. you need really look coordinates:
\begin{gather*}
\frac{d\ }{dt} \frac{\partial L}{\partial \dot{q}_j}=\frac{\partial L}{\partial q}_j\implies \frac{d\ }{dt} (m\dot{q}_j+A_j(q))=\sum_k A_{k ,q_j} \dot{q}_k -V_{q_j}\\
\implies m\ddot{q}_j + \sum_k A_{j,q_k}\dot{q}_k +\sum_k A_{k ,q_j} \dot{q}_k -V_{q_j}\implies
m\ddot{q}_j =\sum_k F_{jk}(q) \dot{q}_k -V_{q_j}
\end{gather*}
with anti-symmetric matrix $\mathbf{F}$:
$$
F_{jk}=A_{k ,q_j}-A_{j,q_k}.
\tag{*}
$$
If $n=3$ such matrix is associated with a vector $\mathbf{B}$, so that $\mathbf{F}\mathbf{v}= \mathbf{v}\times \mathbf{B}$, and one can see easily, that for (*) $\mathbf{B}=\nabla \times \mathbf{A}$.

Comparing with  Physics class we conclude that $\dot{\mathbf{q}}\times \mathbf{B}$ is a Lorentz force, $\mathbf{B}$ is a vector intensity of magnetic field, and $\mathbf{A}$ is a vector potential.

For $n>3$ we cannot reduce tensor intensity $\mathbf{F}$ to $\mathbf{B}$.