Recent content by TheShrike

Ah, well that clears that up. Thanks.

That's correct. So we know the identity element ##1## belongs to ##\beta(H*)##. We also know that ##N## is a subgroup, so ##1\in N##. Therefore ##\beta(H*)## contains at least one element of ##N##.

Hello, I'm following the proof for this theorem in my textbook, and there is one part of it that I can't understand. Hopefully you can help me. Here is the part of the theorem and proof up to where I'm stuck: Let ##N## be a normal subgroup of a group ##G##. Then every subgroup of the...

Yeah, I think you did. Take a look at the first line again. You clearly know ##\sqrt{A}=A^{\frac{1}{2}}##, so perhaps make that conversion for ##T## before taking the derivative. Also, you will need the chain rule for this problem.

It is! Particles that share all their properties (such as mass, charge etc) are indistinguishable. In particular, the statistical behaviour of bosons and fermions at low temperatures (e.g. Bose-Einstein condensation) is due precisely to their indistinguishable nature.

Just making sure, but does that also mean you don't know whether the standing wave is occurring at the fundamental frequency or at one of the overtones?

I'm not quite sure of the experimental set-up you have, but think about the number of nodes of the standing wave. It can be linked to the wavelength.

Challenge V: Sylvester-Gallai Theorem, solved by Mandelbroth I didn't get anywhere with it, but I'm pleased to find I was going in this very direction. I just didn't know how to show it. Also, micromass, your second link is giving me a 404 Not Found error.

He defines ##k## as a constant. So ##\frac{\partial k}{\partial R}## is the partial derivative of ##k## with respect to the variable ##R## which is zero, as he gets.

Oh, I see what you were doing now. A conflict of conventions, as vela stated. My mistake. Looks like you had the answer all along. :wink:

I suspect the problem is the choice of Lagrangian (this is the name for the function defined by kinetic energy minus potential energy), \mathcal{L}=T-V. The action is the integral you mentioned: S=\int^{t_1}_{t_0}\mathcal{L}dt Minimising this gives the true path, and as Feynman says, this is...

You're correct that your initial eigenfunctions are wrong, but your suggested corrections are not quite right either (but really close! :) ). N labels the energy states, so that the ground state is N=1 and the first excited state is N=2. Using that in your new eigenfunctions should result in...

I think there is a difference between the average velocity of the constituent particles, and the velocity of the gas as a whole. The particles are moving about randomly in all directions, even in the wind, but in the wind there is a bias towards the direction of the wind. This bias could be of...

Alright, I've given a few of these a try. Taking tiny-tim's suggestion first: I'm going to have the differential equation: v\frac{dv}{dx}= \frac{q^2}{mx^2}\Big(1+\frac{1}{\sqrt{2}}\Big) Let those constant terms be collapsed into the constant A. Then, by separation of variables, the...

Oh, I was considering the L as a constant, so that if I integrate I'd get v(t)=\Big(\frac{q}{l}\Big)^2\Big(1+\frac{1}{(\sqrt{2})^3}\Big)t+A with A an arbitrary constant. Although putting in the initial condition would lead to A=L. Using the interpretation you've given I'd have no idea...