Recent content by Jillds

But it can all be made simpler considering that ##\int \frac{dx}{\sqrt{x^2+a^2}}=arcosh(\frac{x}{a})+C## Then ##x=\frac{c_2}{n_0\alpha} arcosh \frac{n_0(1+\alpha y)}{c_2}+C## which is the inverse of a cosh function which is the most similar to a parabolic graph. So that would seem to work.

I managed to figure out the ##tan(arcsec(t))## ##\Leftrightarrow tan(arccos(\frac{1}{t})## We know the ##cos \theta=\frac{1}{t}## This would mean that the adjacent of a right triangle has value 1 and the hypothenuse is t. Using Pythagoras we can find the opposite of the angle. ##O^2=H^2-A^2 =...

Homework Statement On very hot days there sometimes can be a mirage seen hovering as you drive. Very close to the ground there is a temperature gradient which makes the refraction index rises with the height. Can we explain the mirage with it? Which unit do you need to extremalise? Writer the...

Thanks very much for the explanation. Yes, I understand.

Homework Statement Solve the integral ## \int_0^{3\pi} \delta (sin \theta) d\theta## Homework EquationsThe Attempt at a Solution I can rewrite ## delta (sin \theta) ## as ##\sum_{n=-\infty}^{\infty} \frac{\delta(\theta - n\pi)}{|cos (n\pi)|}=\sum_{n=-\infty}^{\infty} \delta(\theta-n\pi)## So...

Thank you very much. Yes, with the polar coordinates it's not that hard. :)

There is mention of it in later chapters, with regards to Dirac and Fourier.

You are correct. I forgot to square the constant before the exponent. My eye fell on the fact that it's a Gaussian wave packet, therefore it's a Gaussian integral, which is basically what you're epxlaining (I think). ##\int_{-\infty}^{\infty} e^{-x^2}## would be ## \sqrt{\pi}## and the ##a^2##...

Homework Statement Following gaussian wave packet: ## \psi (x)= \frac{1}{\sqrt{\sqrt{\pi a^2}}} e^{-\frac{x^2}{2a^2}}## Prove that this function is normalized. Homework Equations ## \int_{- \infty}^{\infty} |\psi (x)|^2 dx = 1## The Attempt at a Solution Is ## \frac{1}{\sqrt{\sqrt{\pi a^2}}}...

Found the solution: De Broglie: ##p = \hbar k## Classical mechanics: ## p=mv## Hence, ## j(x,t) = \frac{\hbar k}{mL} = \frac{p}{mL} = \frac{v}{L} = v \rho(x,t)## So, I started out correct by replacing the ## \frac{1}{L}## with the ## \rho##, but failed to recognize De Broglie relation and to...

I know I could rewrite the wave in its classic notation as ## \psi (x,t) = \frac{1}{\sqrt{L}} sin (kx - wt) ## I know the phase velocity is ## v_p = \lambda f = \frac{\omega}{k}## But I don't yet see how this relates to the probability density and j(x,t)... Help very much appreciated.

phase velocity : ##v_p = \frac{\lambda}{T} = \lambda f = \frac{\omega}{k} ## group velocity : ##v_g = \frac{\partial \omega}{\partial k} ##

I know there a continuity equation: ## \frac{\partial}{\partial t} \rho(x,t) + \frac{\partial}{\partial x} j(x,t) = 0 ##. Is that the one you mean? ETA: correction on the continuity equation per mfb's following comment

Homework Statement consider a particle at an interval ##[-L/2, L/2]##, described by the wave function ## \psi (x,t)= \frac{1}{\sqrt{L}}e^{i(kx-wt)}## a) Calculate the probability density ##\rho (x,t) ## and the current density ## j(x,t)## of the particle b) How can you express ## j(x,t)## as a...

Ok, I worked it out for the exponent as follows ## e^{-[\frac{(x-v_g t)^2}{4(\alpha + i\beta t)}]} \cdot e^{-[\frac{(x-v_g t)^2}{4(\alpha - i\beta t)}]}## By then adding the powers of the exponents and working out the numenators of the fractions I get the desired result. Thanks!