Recent content by mark.laidlaw19

Once in the university library I came across an incredibly fascinating physics textbook different from pretty much every other I've encountered. It wasn't for general readers, but (in my opinion) tailored for undergraduate level students. The philosophy of the book was to develop not only an...

Homework Statement Consider two particles, each can be in one of three quantum states, 0, e and 3e, and are at temperature T. Find the partition function if they obey Fermi Dirac and Bose Einstein statistics. Homework EquationsThe Attempt at a Solution I have obtained solutions to both : FD...

Homework Statement Hi all, I have a net spin operator that the problem has asked me to find:S=\frac{\hbar}{2} \left(\begin{array}{cc}\cos\alpha&\sin\alpha\\\sin\alpha&-\cos\alpha\end{array}\right) and I need to write out the matrix representation with respect to the S_z spinor basis. Homework...

questions asked explicitly for parametrization. Otherwise it would be much easier i agree

Thank you very much. I see how this will get the correct denominator, but when you twice differentiate \int_0^R{\frac{1}{a^3+x^3}}wouldn't you end up with \int_0^R \frac{12a^4-6ax^3}{(a^3+x^3)^3}\,dx which will still have an x in the numerator even after setting a=1?

Haha yes I got the first integral fine. And yes, I can see now how to get the final integral and that's a very clever technique! For the second integral, I completed the square for the denominator, which gave two factors, so I used partial fractions and integrated both. So this integral gave...

Well I try to, I've worked through it several times but end up with this (messy) answer which I'm pretty sure does not give the correct answer. 1/3a^2log(1+R/a)+√3/6a^2*(log(\frac{2R-(1+√3)a}{2R-(1-√3)a})-log|-2-√3|)-\frac{1}{6a^2}\log(\frac{R^2-aR+a^2}{R}) I am really interested in how...

Thank you for your help everyone, just letting you know I am in class right now but will reply as soon as I can. I really appreciate all your assistance

Thank you very much for your response. In our course, however, we have not yet covered the gamma, beta and zeta functions, and this leads me to believe that the integral can be done using parametrisation.

Did you mean that a^2A=aC=1? I assumed that you did, and If I solved these I get: A=1/3a^2 and B=-1/3a^2 and C=2/(3a)

Well integrating the first fraction would give me Alog|R|-Alog(0) But I'm having a bit of trouble with the second. It doesn't seem to require a trig substitution, and I don't seem to be able to get it to work by substitution. For example, if I take u=x^2-ax+a^2, then du=(2x-a)dx, but I can't...

Well, if we assume a > 0, then to factorise the quadratic factor we would need to introduce complex numbers. But this should be okay, because the integrand can be complex, even if the variable can only be real. But if I were to keep the integrand real to go along with what you're saying...

Well, i would factorise it into three linear factors and then use partial fractions, integrating each fraction separately.

We need to calculate this: [tex]\lim_{R\rightarrow +\infinity} {\int_0^R 1/(x^3+1)^3\,dx}[\tex] (sorry i am still learning how to type these properly). I have used complex analysis and contour integration to get to this stage, but I think this can be treated a a real integral. I know the answer...

I'm not sure how to get it to look proper, but I end up with one integral that has proved, for me, to be extremely difficult to find ∫dx/(x^3+1), where the terminals go from 0 to R, and we are integrating along the real axis.