12:58 AM
argh, latex got me into the habit of saying nabla instead of del

1 hour later…
1:59 AM
This is the first time I have seen the "Close Votes" queue empty
Or maybe something happened to make it look empty to me?

4 hours later…
6:02 AM
@AaronStevens hi
I have a question on pressure wave and longitudinal wave

6:54 AM
I couldn't, t find why there is phase difference of pi/2 between them

1 hour later…
8:18 AM
morning

3 hours later…
10:51 AM
A non-relativistic radial wave function takes the form $R_l \approx \frac{\sin(kr - \frac{1}{2} l \pi + \delta_l)}{r}$ as $r \to \infty$ where $\delta_l$ is determined by the boundary conditions, what is a simple example where $\delta_l$ is determined
Deriving $e^{ikz} \approx \sum_l i^l (2l+1) P_l(\cos \theta) \frac{\sin(kr-\frac{1}{2}l\pi)}{kr}$ is no joke

11:12 AM
Heh I think I had to do something like that during my master thesis?
I tried to compute the path integral of a particle on a sphere
IIRC that was basically the trick

2 hours later…
1:01 PM
It's the (radial term of the) resolution of a plane wave in spherical harmonics as $r \to \infty$, would be crazy to see this stuff in a path integral

2 hours later…
3:16 PM
Wow, people actually publish papers on whether LaTeX or Word's equation editor is more efficient
I'd be interested in one with a higher sample size than 40 though

4:01 PM
@bolbteppa it's not toooooo bad IIRC
If you allow for physics handwaving a bit
There's a nice trick you can use to get rid of higher order terms

If you think about it, you have to derive/remember spherical harmonics, you have to solve the spherical bessel equation, normalizing them properly, and you have to know their asymptotic approximations, all to express the wave equation for a free particle $e^{ikz}$ in terms of free solutions in spherical coordinates

physics would have been children's play if the world is such a way humans came up with spherical coordinate systems before the cartesian one

Getting spherical harmonics from the ladder method takes ages and full of remembering huge expressions, the spherical bessel equation is absolutely not easy to solve without a trick whose origin made no sense for a long time
Living on a 'sphere' you'd think...

sometimes I am wondering, is there any situation that summing harmomic oscillators and spherical stuff gives a bad approximation. The world is surprisingly not only mathematical, but described really well in terms of oscillators

getting spherical bessel functions from the ladder method basically amounts to the Rodrigues' formula, doesn't it?
which is (obviously) rather tedious to compute but the formula itself is rather pretty

4:14 PM
For spherical bessel with $l = 0$ you solve directly with the solution being $A\sin(kr)/r + B \cos (kr)/r$ (throwing away $B$ in the free particle case), for $l > 0$ you pull off a $r^l$ factor (can be motivated nicely) then the usual thing people tell you to do is to just differentiate the resulting equation and then magically on setting $\chi_{k,l}' = r \chi_{k,l+1}$ and expanding you will get back the $\chi_{k,l+1}$ equation with $\chi_{k,l+1}$ as the solution.

to be clear, what I have in mind is this : $j_\ell(\rho)=\rho^\ell \left(-\frac{1}{\rho}\partial_\rho\right)^\ell j_0(\rho)$
which I do think is rather pretty
to make that more useful, though, I should probably express that as $j_\ell(\rho)=Lj_{\ell-1}(\rho)$ for some appropriate operator $L$

Yeah once you get the $\chi_{k,l}' = r \chi_{k,l+1}$ thing we immediately see $\chi_{k,l+1} = \frac{1}{r} \chi_{k,l}' = (\frac{1}{r} \frac{d}{dr}) \chi_{k,l} = ... = (\frac{1}{r} \frac{d}{dr})^l \chi_{k,0}$ (the minus is sometimes left out, should check this point now hmm)

yeah
the issue, I guess, is that while that formula is pretty
using it to actually compute $j_\ell$ is pretty painful

Yeah, but it's easy to use when you're trying to get the $e^{ikz}$ thing at least

point

4:23 PM
Basically if you consider the $l=-1$ case you get the same solution as in the $l=0$ case except with no $1/r$ term and an overall minus sign so you immediately see this $(\frac{1}{r} \frac{d}{dr}) \chi_{k,l} = \chi_{k,l+1}$ assumption is very natural

looks like one has $j_{\ell+1}(\rho)=(\ell/\rho-\partial_\rho)j_\ell(\rho)$
which is cute, assuming I didn't f*** up the algebra
looks like I may have. drat

In $\frac{1}{r} \frac{d^2}{dr^2} (r R_{kl} ) - \frac{l(l+1)}{r^2} R_{kl} + k^2R_{kl} = 0$ the $l = 0$ and $l = - 1$ cases knock out the $1/r^2$ term, yeah it should be $\chi_{k,l+1} = (-\frac{1}{r}\frac{d}{dr}) \chi_{k,l}$.

nice

Was looking at the associated Legendre equation hoping there was some trick to figure out the solution from say $l = 0$ and $l = - 1$ but alas not yet

hmm, according to the DLMF I actually got the recurrence relation right. neat

4:32 PM
One interesting thing that came from looking at the $l = 0,m=0$ associated Legendre case $\frac{d}{dx}[(1-x^2)\frac{d}{dx}y] = 0$ was that $y = \frac{C_1}{2} \ln|\frac{1 \pm x}{1 \mp x}| + C_2$ (forget the signs, $C_1$ and $C_2$ being constants of integration) is the explicit solution, but it's not finite at $x = \pm 1$, so the $y$ should be a constant so it's derivative is automatically knocked out in $\frac{d}{dx}[(1-x^2)\frac{d}{dx}y]$

4:48 PM
bam =)
rep-cap =)
2.4% closer to the goal!

@EmilioPisanty What's the goal? I was thinking legendary badge, but I don't see how a 2.4% would work into that

@JMac 1/(150-109)

@EmilioPisanty That seems like a strange way to track the % to me... 1% progress now would be different than 1% progress when you started.

@JMac each step gets you a larger fraction of the road closer to the end the further along you are ;-)

5:20 PM
Anyone can please, share his views on my question.

@AvnishKabaj Damn, 6 favs

I have a question on pressure wave and longitudinal wave, I couldn't, t find why there is phase difference of pi/2 between them

@yuvrajsingh Under what circumstances are a pressure wave and a longitudinal wave out of phase by a quarter-rotation? That's not familiar to me.

1 hour later…
6:54 PM
What's going on in here?

1 hour later…
7:59 PM
0

I recently asked and answered a question on the main site: What are the Energy eigenstates for a modified quantum harmonic oscillator? First of all, this question ended up being re-tagged as homework, which I understand. While the question is not the sort of question addressed in even relatively...

3 hours later…
11:27 PM
so I have a natural magnet attached to a vertical ferromagnetic surface. How can I compute the horizontal and vertical magnitude of the magnetic force? Clearly they are easily determined experimentally, but what material property of the magnet determines the horizontal and vertical components?