Landau Quantization: Classical Particle Motion in a Uniform Magnetic Field

[Dickfore]

Yes, this would be $x(t)$ that you need to substitute in the above expression and integrate again.

 

[Sonny Liston]

Integrating my Lagrangian$L = (m/2)[(v_{0})^2 +1] - (q/c)B[(x_{0} - \frac{v_{0}}{\omega} sin(t\omega)][-(v_{0})sin(t\omega)]$

Gives

$S(t) = [(m(v_{0})^2)/2]\Delta t - (q/c)B[\frac{(v_{0})x_{0}}{\omega}](cos(\omega t_{1}) - cos(\omega t_{0})) - (q/c)B[\frac{(v_{0})^2}{\omega}]((t_{1})/2 - \frac{sin(2\omega t_{1})}{4\omega} - (t_{0})/2) + \frac{sin(2\omega t_{0})}{4\omega}$

This looks more complicated than it should be..

 

[Dickfore]

Ok, do this first:

Take the lagrangian:

$$L = \frac{m v^{2}_{0}}{2} \, \left(v^{2}_{x} + v^{2}_{y} \right) - \frac{q B}{c} \, x \, v_{y}$$

substitute:

$$\begin{array}{rcl} v_{x} & = & v_{0} \, \cos{(\omega \, t)} \\ v_{y} & = & -v_{0} \, \sin{(\omega \, t)} \\ x & = & \frac{v_{0}}{\omega} \, \sin{(\omega t)}$$

and perform integration from 0 to t, by keeping in mind that:

$$\omega = \frac{q \, B}{m \, c}$$

What is the lagrangian as a function of t?

What is the integral of this from 0 to t?

 

[Sonny Liston]

I get that $L=(v_{0})^2[\frac{m}{2} + (sin(\omega t)^2)$

Have I done anything stupid so far?

 

[Dickfore]

First of all, your equation is dimensionally inconsistent. In the large parentheses, the first term has a dimension of mass, and the second term is dimensionless.

 

[Sonny Liston]

All I did was cancel the omega in the x term and the omega in the $v_{y}$ term, which I got from plugging in the provided equations.

Sigh. How did I manage to mess THAT up?

 

[Dickfore]

The vector potential term from the Lagrangian has the product:

$$x v_{x} = \frac{v_{0}}{\omega} \, \sin{(\omega t)} \, v_{0} \, \cos{(\omega t)} = \frac{v^{2}_{0}}{\omega} \sin{(\omega t)} \, \cos{(\omega t)}$$

using the double angle formula:

$$\sin{(2 \alpha)} = 2 \, \sin{(\alpha)} \, \cos{(\alpha)}$$

this product can be further simplified as:

$$\frac{v^{2}_{0}}{2 \omega} \, \sin{(2 \omega t)}$$

However, this product is also multiplied by $q \, B/c$ in the Lagrangian. Using the definition of the cyclotron frequency $\omega = q \, B/(m \, c)$, what is:

$$\frac{q B}{\omega c} = ?$$

The kinetic energy term has the sum of squares:

$$v^{2}_{x} + v^{2}_{y} = ?$$

What is this sum equal to? What is it multiplied by in the Lagrangian?

 

[Sonny Liston]

$$\frac{q B}{\omega c} = ?$$[itex]\frac{qB}{\omega c} = \frac{qBmc}{Bc} = qm[/tex]

The kinetic energy term has the sum of squares:

$$v^{2}_{x} + v^{2}_{y} = ?$$

What is this sum equal to? What is it multiplied by in the Lagrangian?

$$v^{2}_{x} + v^{2}_{y} = ((v_{0})^2)((cos(\omega t))^2 + (sin(\omega t))^2) = ((v_{0})^2)$$

This is then multiplied by $$\frac{m}{2}$$

 

[Sonny Liston]

Whaddya mean?

 

[Dickfore]

$$\frac{q B}{\omega c} = ?$$


[itex]\frac{qB}{\omega c} = \frac{qBmc}{Bc} = qm[/tex]

This is incorrect.

The kinetic energy term has the sum of squares:

$$v^{2}_{x} + v^{2}_{y} = ?$$

What is this sum equal to? What is it multiplied by in the Lagrangian?

$$v^{2}_{x} + v^{2}_{y} = ((v_{0})^2)((cos(\omega t))^2 + (sin(\omega t))^2) = ((v_{0})^2)$$

This is then multiplied by $$\frac{m}{2}$$

This is correct.

 

[Sonny Liston]

I didn't see the charge term in the original cyclotron frequency. In that case, we don't get qm, we get m.

 

[Dickfore]

I didn't see the charge term in the original cyclotron frequency. In that case, we don't get qm, we get m.

Yes. So, what is the final expression for the Lagrangian L?

 

[Sonny Liston]

$$L = \frac{m(v_{0})^2}{2} - \frac{m(v_{0})^2}{2\omega}sin(2t\omega)$$

 

[Dickfore]

$$L = \frac{m(v_{0})^2}{2} - \frac{m(v_{0})^2}{2\omega}sin(2t\omega)$$

Where did the extra $\omega$ come from in the second term? C'mon man. Concentrate on your algebra. This is taking too long.

 

[Sonny Liston]

$$S(t) = [\frac{m((v_{0})^2)}{2}]t + [\frac{m((v_{0})^2)}{2}]\frac{(cos(2t\omega) + 1}{2\omega}$$

Extra omega eliminated here. Sorry about that.

 

[Dickfore]

(Again, you kept an extra $\omega$ in the second term.) I see you corrected that.

Also:

$$\int_{0}^{t}{\sin{(2 \omega t)} \, dt} = -\left.\frac{1}{2 \omega} \, \cos{(\omega t)}\right|^{t}_{0} = \frac{1}{2\omega} \, \left[1 - \cos{(2 \omega t)}\right]$$

Finally, what will be the change in S during one period (take $t = 2\pi/\omega$ in the above formula)?

 

[Sonny Liston]

$$1 - cos (2 \omega \frac{(2\pi}{\omega}) = 0$$, so it is appropriately periodic.

 

[Dickfore]

No, the first term is proportional to $t$ and is not periodic.

 

[Sonny Liston]

Assuming that's satisfactory, may I ask you one last quick question (you are a saint, by the way. I have been unbelievably useless throughout this.) How would I go about evaluating the magnetic flux through the classical trajectory as a function of the energy?

In case you're wondering about my remarkable ignorance, I'm a math student who never took classical mechanics and is in a quantum course, so all the classical mechanics stuff is new to me. Thanks again.

 

[Dickfore]

What does the trajectory look like? What are its dimensions in terms of the quantities we have already defined? What is the magnetic flux from a homogeneous magnetic field perpendicular through a contour?

 

[Sonny Liston]

In my notes, I have that it is given by $$\frac{1}{2\pi}\oint A dx$$, and the baby E&M book I am looking at says that it can be given by the Area of the surface dotted with the B field times cos(theta), for theta the angle between the B field and the surface.

But since the B field is homogeneous in the z-direction, $$\theta = \frac{\pi}{2}$$ and cos=0, so something must be really flawed in my reasoning.

 

[Dickfore]

In my notes, I have that it is given by $$\frac{1}{2\pi}\oint A dx$$, and the baby E&M book I am looking at says that it can be given by the Area of the surface dotted with the B field times cos(theta), for theta the angle between the B field and the surface.

But since the B field is homogeneous in the z-direction, $$\theta = \frac{\pi}{2}$$ and cos=0, so something must be really flawed in my reasoning.

Yes, they are both equal due to Stokes' Theorem:

$$\phi_{m} = \int_{S}{\left(\mathbf{B} \cdot \hat{\mathbf{n}}\right) \, da} = \oint{\mathbf{A} \cdot d\mathbf{l}}$$

So, there is no factor of $1/(2\pi)$ if you use the circulation of the vector potential.

 

[Sonny Liston]

So there is nothing wrong in my reasoning and the flux actually is zero? It seems odd to ask me to evaluate it "as a function of the energy" then.

 

[Dickfore]

No, it is not zero. I didn't see the part where you say it's zero. Since $\mathbf{A}$ has only a zero component, you need to evaluate:

$$\oint{\mathbf{A} \cdot d\mathbf{l}} = \oint{A_{y} \, dy} = B \int_{0}^{T}{x(t) \, v_{y}(t)}$$

where $T = 2\pi/\omega$ is the period of the trajectory and i used $A_{y} = B \, x$.

 

[Sonny Liston]

$$B\int x(t) v_{y} (t) = B \frac{-(v_{0})^2}{\omega} \int (sin(\omega t)^2)$$

Using $$(sin(\omega t)^2) = \frac{1 - cos (2\omega t)}{2}$$ and doing the integral, I get

$$\Phi = B\pi(- \frac{v_{0}}{\omega})^2$$

 

[Dickfore]

yes. the minus sign is redundant.

 

[Sonny Liston]

I can't thank you enough for all your help.

 

[Dickfore]

no problem. nice username ;)

 

[Sonny Liston]

Btw, I don't think I understand why the magnetic flux is a function of the energy here.