Electric Dipole in a Magnetic Field: Conservation of energy

sparkle123
Messages
172
Reaction score
0
Please help me with the attached question.

I don't understand how you get from the first line to the second line in (2). (I'm not too familiar with working with cross products and vectors).

Also, I don't know how you get (5) from (1) and (2).

Any help would be much appreciated. Thanks! :)
 

Attachments

  • 1.PNG
    1.PNG
    11.8 KB · Views: 481
  • 2.PNG
    2.PNG
    7.1 KB · Views: 465
Physics news on Phys.org
(2) the equality Comes from \vec{v}_{cm} = \dfrac{1}{2}(\vec{v}_1+\vec{v}_2)
(5) Try to wite down the full expression E = \dfrac{1}{2}mv_{cm}^2+\dfrac{1}{2}I\omega^2
Using (1) and (2)
And take the time derivitative(here dE/dt is indicated with a small dot above). Then you would see that the energy is conserved.
 
Thanks dikmikkel! :)
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top