- #1
The Head
- 144
- 2
Homework Statement
Consider the energy surface
E(k)=h2((kx2 +ky2 )/ml+kz2/mt
where m_l is the transverse mass parameter and m_l is the longitudinal mass parameter. Use the equation of motion:
h(dk/dt)= -e(vXB) with v=∇k(E)/h to show that ωc=eB/(ml*mt)1/2 when the static magnetic field B lies in the x-y plane
Homework Equations
a=(1/h)(d2E(k)/dk2)(dk/dt)
ω=a/v
The Attempt at a Solution
I've tried quite a number of things and spent several hours unsuccessfully, but one route I tried was this:
v= 1/h(∇k(E))=1/h(h2)(kx(i-hat)/ml + ky(j-hat)/ml + kz(k-hat)/mt)
B is in x-y plane, so B= Bx (i-hat) + By (j-hat)
so the cross product of v & B = kz*By/(ml) (i-hat) - kz*Bx/(ml) (j-hat) +(ky*Bx - Kx*By)/(mt) (k-hat)
Now, this is where I am not sure what to do, but I figured I would try the equation:
a=(1/h)(d2E(k)/dk2)(dk/dt) with the idea that this would be centripetal acceleration = ωv, and this way I could solve for ω_c
d2E(k)/dk2= h2(((i-hat)+(j-hat))/ml +(k-hat)/mt
dk/dt=-e/h(vXB)=-e/h*kz*By/(ml) (i-hat) - kz*Bx/(ml) (j-hat) +(ky*Bx - Kx*By)/(mt) (k-hat)
I doted dk/dt with vXB to get rid of the vectors, and divided by h to get:
a=e/mtml(Bx*(kz-ky)+By(kx-kz))=ωv
Although this seems to have somewhat the form I am going for, I have two problems: I don't know how to get rid of all the different k components, and if I divide by the vector v, then I have an answer for ω with components.
Can someone provide some guidance? Thank you!