Recent content by Wavefunction

I'll have to retry it at a later date, my when I tried to do it the last time it got quite messy and it didn't work out so nice. I see your point that the lagrangian is a more general method and you don't get stuck with the assumption that ##\mathbf{N}=\mathbf{M}\times\mathbf{B}##, but I'm going...

So as for the expression on the torque, I think we were meant to just use ##\mathbf{N}=\mathbf{M}\times\mathbf{B}##, just for the sake of making the problem tractable: I'm starting to see the argument I made in my final answer a little bit more clearly. ##\mathbf{N}=\mathbf{M}\times\mathbf{B}##...

So the logic goes as follows (Assuming a constant magnetic field oriented along the z-axis) ##\frac{dV}{dt}=\gamma \mathbf{B}\cdot \frac{d\mathbf{L}}{dt}## Now the torque on a magnetic moment ##\mathbf{M}## is given by ##\mathbf{N}=\frac{d\mathbf{L}}{dt}=\mathbf{M}\times\mathbf{B}## Thus...

Hi Tsny and thank you for the reply. Eventually, I was able to get that ##\frac{dE}{dt}=\frac{dT}{dt}## since the change in angular momentum is perpendicular to the ##B## field; thus, ##\frac{dV}{dt}=0##. Once I got there, then I was able use ##\frac{dE}{dt}=\frac{dT}{dt}## to show that...

Update: I have figured out the solution. Basically, you need to think about the torque on the magnetic moment of the top due to the magnetic field; then relate this torque to the change in potential energy. Once you figure out this relationship you can show that the angle of nutation is constant...

Setup: Let ##\hat{\mathbf{e}}_1,\hat{\mathbf{e}}_2,\hat{\mathbf{e}}_3## be the basis of the fixed frame and ##\hat{\mathbf{e}}'_1,\hat{\mathbf{e}}'_2,\hat{\mathbf{e}}'_3## be the basis of the body frame. Furthermore, let ##\phi## be the angle of rotation about the ##\hat{\mathbf{e}}_3## axis...

Nevermind everyone i figured it out: a) Yes, work must be done to "ramp it up". The work will depend on the value of ##T##: Diabatic case: If the perturbation is applied suddenly, then the system has no time to adjust and establish itself. The state would have to remain the same since the...

Homework Statement A quantum particle of mass ##m## is bound in the ground state of the one-dimensional parabolic potential well ##\frac{K_0x^2}{2}## until time ##t=0##. Between time moments of ##t=0## and ##t=T## the stiffness of the spring is ramped-up as ##K(t) = K_0...

It's okay I have actually solved the question: it was suggested to me by my quantum mechanics TA to view this as a capacitor whose capacitance changes when alpha changes. Then you can use the energy stored in the capacitor and differentiate with respect to alpha in order get the torque :) ...

Homework Statement parallel-plate capacitor consists of two fixed metal semicircles of radius R and a dielectric plate (susceptibility \varepsilon). The plate is able to rotate without friction around the axis centered at the point O (axis O is perpendicular to the picture). Plate's thickness...

Awesome thanks for help everyone! :) Cheers

In region 1: \psi'_{I}(a-\varepsilon)=C_3\frac{e^{-\kappa a}}{e^{\kappa a}-e^{-\kappa a}}\kappa(e^{\kappa(a-\varepsilon)}+e^{-\kappa(a-\varepsilon)}) In region 2: \psi'_{I}(a+\varepsilon)=-C_3\kappa e^{-\kappa a} So...

\lim_{\varepsilon\rightarrow0}\int_{x=a-\varepsilon}^{a+\varepsilon}\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2}-C\delta(x-a)\psi(x)dx=\lim_{\varepsilon\rightarrow0}\int_{x=a-\varepsilon}^{a+\varepsilon}E\psi(x)dx...

Thank you I'll reformulate my problem and post my solution later :)

I'm not sure how to deal directly with the delta function potential. The way we dealt with such a potential in class was to construct such a finite square well. Though I'm curious to hear more about your approach. Thank you for the reply :)