Landau Quantization: Classical Particle Motion in a Uniform Magnetic Field

[Dickfore]

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

Also:

<br /> \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]<br />

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:

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

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:

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

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.