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Web bonus problem--Week 10-11

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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 $\mathbf{A}$.

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