# An electron is subjected to an electromagnetic field using the canonical equations solve

So I was given the following vector field:

$\vec{A}(t)=\{A_{0x}cos(\omega t + \phi_x), A_{0y}cos(\omega t + \phi_y), A_{0z}cos(\omega t + \phi_z)\}$

Where the amplitudes $A_{0i}$ and phase shifts \phi_i are constants. An electron is moving through this field with the constant initial velocity:

$\vec{v}_0=\{v_{0x},v_{0y},0\}$

The vector field pulls the electron out of a metal that occupies half-space $z<0$ causing photoelectron emission. I need to find a condition on phase \phi_z under which the electron will continue to fly away and never return back and hit the metal as well as the average drift velocity of the electron.

I started by finding the Lagrangian:

Where I shortened the initial velocity to:

$\vec{v_0}=\{\dot{x},\dot{y},0\}$

Therefore:

$L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)+\frac{e}{c}\dot{\vec{r}}\vec{A}-U(r)$

This simplifies to:

$L=\frac{1}{2}m(\dot{x}^2+\frac{e}{c}[A_x \dot{x}cos(\omega t + \phi_x)+A_y \dot{y} cos(\omega t +\phi_y)]-U(r)$

Next I found the momentum using the canonical relationship:

$p=\frac{\delta L}{\delta \dot{q}}$

This gives:

$p_x=m\dot{x} +\frac{eA_x}{c}cos(\omega t + \phi_x)$ $p_y=m\dot{y} + \frac{eA_y}{c}cos(\omega t + \phi_y)$

This lets me find the energy via the hamiltonian:

$E=H=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}$

When you plug in for $p_x$ and $p_y$ you get:

$H=\frac{1}{2m}[m\dot{x}+\frac{eA_x}{c}cos(\omega t + \phi_x)]^2+\frac{1}{2m}[m\dot{y}+cos(\omega t + \phi_y]^2$

where:

$\alpha=\frac{eA_x}{c}cos(\omega t +\phi_x)$

$\beta=\frac{eA_y}{c}cos(\omega t + \phi_y)$

We can also say:

$p_x=m\dot{x}$

$p_y=m\dot{y}$

Therefore:

$H=\frac{1}{2m}[p_x+\alpha]^2+\frac{1}{2m}[p_y+\beta]^2$

Now we can use the relation for the unknown equation:

$\frac{\delta W}{\delta q_i}=p_i$

Plugging this into the energy equation we get:

$E=\frac{1}{2m}[\frac{\delta W_x}{\delta x}+\alpha]^2+\frac{1}{2m}[\frac{\delta W_y}{\delta y}+\beta]^2$

pulling the y terms out we get:

$0=\frac{1}{2m}(\frac{\delta W_y}{\delta y})^2 + \frac{\beta}{2m}(\frac{\delta W_y}{\delta y})$

This simplifies to:

$W_y=-\beta y$

Which we plug back into the equation.

With algebra we can now simplify the energy equation to:

$2mE=(\frac{\delta W_x}{\delta x})^2+\alpha\frac{\delta W_x}{\delta x}+ \Omega$

Where $\Omega=\alpha^2+\beta^2$.

My first question is, how do I solve the above equation for $W_x$. My second question is this even the right direction I should be heading? I feel like I have some egregious error that I am not seeing.

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Why is there no ${\dot z}^{2}$ term in your lagrangian? – Jerry Schirmer Apr 23 '12 at 4:00
I guess I assumed there would be no $\dot{z}^2$ term in my lagrangian because the initial velocity was in the xy-plane. The problem still stands that in the $\frac{e}{c}\dot{\vec{r}}\vec{A}$ term there won't be any terms in the z-direction this getting rid of the only term with the phase $\phi_z$. – mnky9800n Apr 23 '12 at 4:36
You have three components to $\vec A$. You will therefore have a $\hat z$ component to ${\vec \nabla} \times {\vec A}$, and even if your initial z velocity is zero, you will still pick up a $z$ velocity after your initial timestep. – Jerry Schirmer Apr 23 '12 at 5:14
And even if you have no spatial dependence in the $A_{i}$, you will still have $\dot {vec A}\neq 0$, which will give you an nonzero $E_{z}$ – Jerry Schirmer Apr 23 '12 at 6:03
You said "..the initial velocity was in the xy-plane...". Thats the point. What you have been given is the initial condition which you need to use AFTER you have got the equation of motion for the particle. The full equations of motion will have $\dot{z}$ as well. Once you get them, you integrate them. Then you will have constants of integration. And that is when you should use these initial conditions on the velocity to fix those constants.... not in the Lagrangian itself. – Vijay Murthy Apr 23 '12 at 14:18