Recent content by MaxwellsCat

I mean they're equivalent except for the definition of distance that you use. Technically the one with script r is more correct I think, but for distances far from the dipole you approximate with r. The equation is an approximation anyway, but there are orders of correctness I guess.

I think he's using script r to denote distance from the dipole to the point in question because he's no longer dealing with distances far from the dipole where r and script r are basically equivalent.

Maybe try and look up how to evaluate integrals with absolute values inside - aside from that you're there. Also keep in mind what a Fourier transform does - it takes a function from position/time-space to frequency-space. That should tell you a little bit about what it should look like -...

So your substitution is technically correct, but I would try writing it as an exponential function - generally MUCH easier to integrate. I'm not sure that your integral is correct either, though I don't have your working. Are you familiar with how to write sin as an exponential?

So right away there's something wrong - the units should be J not J##\cdot##s, unless you meant Joules :P Does that answer make sense? In the context of the problem, does that seem reasonable?

Sorry, I should have been explicit - I meant a solid cylinder.

You're on the right track, just realize that a disc is a cylinder with vanishing height. Otherwise yeah, try it out and report back on what you end up with. It'll be important to do a sanity check on the answer that you get - that'll tell you whether you got the right answer or not most of the...

Well first notice that the absolute value gives you slightly different versions of the exponential functions that make up ##|\!\sin{ωt}|##, right? Are you supposed to derive the expression you have listed or are you supposed to just find the Fourier transform of ##|\!\sin{ωt}|##? You say both...

Due to a wonderful human who already took this class, it was suggested that I simplify ##V(δ)## by one more step: $$V(δ) = \frac{1}{2} mω^2 \bigg{(}δ + \frac{2g}{ω^2} \bigg{)}^2 -\frac{mg^2}{ω^2}$$ Because then it's just a linear addition to ##\hat{H}## and you get energy eigenvalues that are...

Duh, thanks... $$V(δ) = \frac{1}{2}mω^2(δ+\frac{g}{ω^2})^2 + mg(δ + \frac{g}{ω^2}) = \frac{1}{2}mω^2(δ^2 + \frac{2δg}{ω^2} + \frac{g^2}{ω^4}) + mg(δ + \frac{g}{ω^2})$$ $$ = \frac{1}{2}mω^2δ^2 + mgδ + \frac{1}{2}m\frac{g^2}{ω^2} + mgδ + \frac{mg^2}{ω^2}$$ and so $$V(δ) = \frac{1}{2}mω^2δ^2 +...

So first try thinking about exactly what out of phase and in phase mean. Then, you should be able to calculate how much of a wavelength is required for each condition and use triangles find what ##d## has to be. You know how far it has to go, written in terms of ##d##, just solve for ##d##...

Homework Statement 1. Harmonic Oscillator on Earth Gravity : In class, we solve the Harmonic Oscillator Problem, with a potential $$ V(x) = \frac{m ω^2 x^2}{2} \quad (1)$$ with ##ω = \frac{k}{m}## being the classical frequency. Now, assume that x is a vertical direction and that we place...

I'm Aman, I'm a second year Physics major at UChicago. I'm also currently working with microwave cavities in a lab here at the university. I've been discovering that sometimes the TA office hours are not often enough, and that I'll need a little more help with some things that give me trouble...