Recent content by Davidllerenav

I thought so, but reading the chapter again I think not because the transmited angle isn't in facr zero, but complex.

Well since both incident angles are greater than the critical angle ##\theta_c=24.4^°##, then ##\theta_t=0##.

So i do now that it is a case of total internal reflection, but i didn't get R=1 for ##\theta_i=40.5^°##. I used the Fresnel equations for both s and p-polarized light and for s I got ##r_s=\frac{n_i\cos\theta_i-n_t\cos\theta_t}{n_i\cos\theta_i+ n_t\cos\theta_t}=0.296## using ##n_i=2.42## and...

What I did was first noting that ##\hat{\vec{S}}_1\cdot\hat{\vec{S}}_2=\frac{1}{2}(\hat{\vec{S}}^2-\hat{\vec{S}}_1^2-\hat{\vec{S}}_2^2)##, but these operators don't commute with ##\hat{S}_{1_z}## and ##\hat{S}_{2_z}##, this non the decoupled basis ##\ket{s_1,s_2;m_1,m_2}## nor the coupled one...

Ok. Thanks! I'll try to do the integral then.

I guess not? If I recall correctly, the energy is constant.

It would be ##T=2\sqrt{\frac{m}{2}}\int_{a}^{2a}\frac{dr}{\sqrt{\frac{al^2}{mr^3}-\frac{l^2}{2mr^2}}}##. I'll try it and see if I can integrate it the. I have a question, isn't it problematic that we used the energy at only ##\theta=0##?

Ok, so since ##E=\frac{1}{2}m\dot{r}^2-\frac{al^2}{m}\frac{1}{r^3}+\frac{l^2}{2m}\frac{1}{r^2}##, then at ##\theta=0##, ##E=-\frac{al^2}{m}\frac{1}{8a^3}+\frac{l^2}{2m}\frac{1}{4a^2}=-\frac{l^2}{8ma^2}+\frac{l^2}{8ma^2}=0##?

Oh sorry, my bad, I thought it was a sine insted of cosine in ##r##, the correct value would be ##r=2a##.

##\theta=0## the, right? That would make ##\dot r=0## and ##r=a##, that is constant so it makes sense.

Well, since ##\dot{r}=-a\dot{\theta}\sin\theta##, then I guess that at ##\theta=\frac{\pi}{2}## would be a good choice.

So, given that ##t=\sqrt{\frac{m}{2}}\int_{r_0}^{r}\frac{dr'}{\sqrt{E-V_{eff}}}## and ##E=\frac{1}{2}m\dot{r}^2+V_{\text{eff}}##, replacing in the integral, I would have ##t=\sqrt{\frac{m}{2}}\int_{r_0}^{r}\frac{dr'}{\sqrt{\frac{1}{2}m\dot{r}^2}}=\int_{r_0}^{r}\frac{dr'}{\dot{r}}## Then...

Ok. So since the angular part of the kinetic energy is given by ##\frac{1}{2}mr^2\dot{\theta}## and ##\theta## is cyclic, then ##mr^2\dot{\theta}=l=\text{constant}##, then ##\dot{\theta}=\frac{l}{mr^2}\Rightarrow \dot{\theta}^2=\frac{l^2}{m^2r^4}##, then it implies that...

Ok, that can be done through the Lagrangian, I think. Well, that's strange. I guess since it is classical mechanics, we aren't considering the limit of speed as the speed of light?

Well, I don't have any function of E or something. Can I just write it as the sum of the kinetic energy in one dimension plus the effective potential? The problem is still solvable even if there's a singularity at the origin? Responding to your edit: Yes, I forgot that when ##\theta=\pi##, ##r=0##.