Recent content by CharlieCW

You're right, I had taken the wrong limits of integration. Sorry for the late reply by the way, I had an exam this morning. Now continuing with the equations, let's start from the beginning. Assuming ##\chi_1=\omega^2_p/\omega##, then we can substitute in the Kramers-Kronig relation as...

Thanks for your reply, Jason. Just from the form of the equations, if I had ##\chi_1=-\omega_0/\omega##, then ##\chi_2=\pi\omega_0/2##, as when taking the limits only the evaluation on zero will lead to two non-zero identical terms. Here I have no problem evaluating the limits of the integral...

I'm kind of confused on how to evaluate the principal value as it's a topic I've never seen in complex analysis and all the literature I've read so far only deals with the formal definition, not providing an example on how to calculate it properly. Therefore, I think just understanding at least...

Well I managed to solve it and I got that both the average energy and length follow a Fermi-Dirac like distribution. I think I'll post the solution during the weekend in case anyone finds it useful.

Don't worry, with your explanation I better understood the meaning of the terms in the exponential and I think I see more clearly how to deal with these kind of systems. So then my idea about considering the tension ##\tau## for the linear case was correct since, as you mentioned, it is part of...

Indeed, the term ##pV_S## is for the pressure and volume, but since the general formula was derived for a 3D recipient I was thinking about converting it to the one-dimensional case ##pV_S\rightarrow \tau L##. However, it also makes more sense that you mention to obtain the tension as ##dZ/dl##...

Indeed, in my answer I'm basically considering the top of the well as a flat disk to find an expression for the angle ##\theta##.

Since the view is 3D, you should indeed solid angles to calculate the angle of vision. First consider the case were the astronomer is outside the well. In this case, he sees the 100% of the sky (assuming you call 100% seeing the whole half hemisphere on where they're standing). So the solid...

Homework Statement Consider a polymer formed by connecting N disc-shaped molecules into a onedimensional chain. Each molecule can align either its long axis (of length ##l_1## and energy ##E_1##) or short axis (of length ##l_2## and energy ##E_2##). Suppose that the chain is subject to tension...

Well I checked similar procedure and I managed to advance the following: While I don't know if it's really useful, if we apply mechanical equilibrium before adding the charge, it's straightforward to find that ##\rho_s=\rho_l##, where ##\rho_s## and ##\rho_l## are the volumetric densities of...

Homework Statement A long straight cylinder with radius ##a## and length ##L## has an uniform magnetization ##M## along its axis. (a) Show that when its flat extreme is placed on a flat surface with infinite permeability (i.e. a ferromagnet), it adheres with a force equal to: $$F=8\pi a^2 L...

Homework Statement A conductor sphere of radius R without charge is floating half-submerged in a liquid with dielectric constant ##\epsilon_{liquid}=\epsilon## and density ##\rho_l##. The upper air can be considered to have a dielectric constant ##\epsilon_{air}=1##. Now an infinitesimal...

Thanks, I'll check if we got it in the library after the morning lectures and I'll update you if I find the solution. Edit: I checked the book, they have the proofs I needed. As for the last one, I found it here in page 4: http://mutuslab.cs.uwindsor.ca/schurko/introphyschem/handouts/mathsht.pdf

Homework Statement Let x, y and z satisfy the state function ##f(x, y, z) = 0## and let ##w## be a function of only two of these variables. Show the following identities: $$\left(\frac{\partial x}{\partial y}\right )_w \left(\frac{\partial y}{\partial z}\right )_w =\left(\frac{\partial...

Thank you for your time, I really appreciate it. Indeed I also checked from Jackson and Greiner and I read that I was free to choose ##F(r,r')## so that ##G(r,r')## is zero on the surface. After a couple of exchanges it turns out we were right: the Green function only depends on the geometry of...