Recent content by pondzo

Thank you very much for your reply Sam, i understand this now. I was wondering if I could contact you personally and ask you a question about an answer you posted in this thread: https://www.physicsforums.com/threads/why-is-lorentz-group-in-3d-sl-2-r.764072/

Context The following is from the book "Ideas and methods in supersymmetry and supergravity" by I.L. Buchbinder and S.M Kuzenko, pg 56-60. It is about realizing the irreducible massive representations of the Poincare group as spin tensor fields which transform under certain representations of...

Thank you guys, I knew it must have been something silly like that. Hi Ray, I really don't like the notation I use to solve these types of questions because as you say it can become confusing at times. What would your solution to a question like this look like? with particular emphasis on the...

I have come to the conclusion that any unpaired nucleons will contribute to the spin/parity, not just the ones from the hole created or the new level filled. And if it so happens that an excited state yields three or more unpaired nucleons, then the shell model is no longer reliable in...

Homework Statement Find a solution of $$\frac{1}{x^2}\frac{\partial u(x,y)}{\partial x}+\frac{1}{y^3}\frac{\partial u(x,y)}{\partial y}=0$$ Which satisfies the condition ##\frac{\partial u(x,y)}{\partial x}\big |_{y=0}=x^3## for all ##x##. The Attempt at a Solution I get the following...

Homework Statement Consider the following example from a previous exam. We are to predict the spin and parity for F(A=17,Z=9), Florine, in the ground state and the first two excited states using the shell model. Ground state: Neutrons: (1s 1/2)^2 (1p 3/2)^4 (1p 1/2)^2 Protons: (1s 1/2)^2...

Homework Statement Stopped pions provide a useful mono-energetic source of neutrinos. For a pion at rest, calculate the energy of the neutrino in the decay $$\pi^+\rightarrow \mu^++\nu_{\mu}$$ You do not need to consider the subsequent decay of the ##\mu^+## and you can assume that the...

Upon further consideration I agree with you. I think my conclusions had something to do with how I represented the generators of the ##(C_3\times C_3)## part of the semidirect product. Would it be correct to denote ##X_1=((x,1),1)## and ##X_2=((1,x),1)## and then do the crucial calculations...

Homework Statement Write ##C_3\langle x|x^3=1\rangle## and ##C_2=\langle y|y^2=1\rangle## Let ##h_1,h_2:C_2\rightarrow \text{ Aut}(C_3\times C_3)## be the following homomorphisms: $$h_1(y)(x^a,x^b)=(x^{-a},x^{-b})~;~~~~~~h_2(y)(x^a,x^b)=(x^b,x^a)$$ Put ##G(1)=(C_3\times C_3)\rtimes_{h_1}C_2...

Haruspex could you please advise me on how you might go about the problem? Which approach would you take in order to find ##V_A## etc? So I had a look at my notes, and by using method two I concluded that: ##\text{ For } r\leq a...

How would I Find the values of ##V_A## etc? I thought I might be able to via ##C=\frac{Q}{V}## but it seems like a circular problem. I thought that ##V_A,V_B=V_C## could be set to anything in a physical setup, and post #3 would be true, but judging by the replies it seems not. This is what I...

After some extra thought I agree that method three does not give a different answer. I have followed method one all the way through to answer the question and my final answer would be this: ##\phi(r)= \begin{cases} V_A, & 0\leq r\leq a\\...

Homework Statement I am not sure whether to put this in the introductory level or advanced. It seems to be relatively introductory in an electromagnetism course. A spherical conductor of radius ##a## carries a charge ##q##. It is situated inside a concentric spherical conducting shell of...

Homework Statement A particle with charge ##e## and position vector ##\vec{r}## (relative to some frame, S) moves with constant velocity ##\vec{v}##. A second charge ##e'## is moving with the same velocity ##\vec{v}## through the field generated by ##e##. If ##\vec{d}## is a vector from ##e##...

Would I do it by defining the operators as follows? ##\hat{S}^2|s,m_s\rangle=s(s+1)\hbar^2|s,m_s\rangle## ##\hat{S}_x|s,m_s\rangle=m_s\hbar|s,m_s\rangle## ##\hat{S}_+=\hat{S}_y+i\hat{S}_z## ##\hat{S}_-=\hat{S}_y-i\hat{S}_z## Which both imply that: ##\hat{S}_y=\frac{1}{2}(\hat{S}_++\hat{S}_-)##...