Recent content by dslowik

I think your OP was asking about purely QM effects, rather than thermal effects due to collisions with surrounding particles, i.e., assuming the chair was in a vacuum at 0K. But I think the reasoning is qualitatively the same: the momentum uncertainties of each of the individual constituent...

Nugatory, Isn't there also something special about the topology of the 'extra' dimensions? i.e., if the are 'curled up' so that , e.g., if you move 1 cm along the extra dimension you return to your starting point (in both the extra dimension and in 3d space)?

Right. 1) lhs of ⇒ is correct, but r(pi/4) seems wrong. 2) Form is correct, but seems the numbers are wrong.

What you are describing (that you are attempting to do) is the CPV of the integral and, unfortunately, Cauchy beat you to it. If you look closely at the definition of Riemann integrals in terms of limits of Riemann sums over ever finer partitions, you should see that the Riemann integral does...

I think that "relevant equation", \nabla f(r)=\frac{df}{dr}\nabla r=\frac{df}{dr}\frac{\vec{r}}{r}, is not that relevant here: e^{i\vec{k}\cdot \vec{r}}\neq f(r) since r = (x^2 + y^2 + z^2)^{1/2}. Rather, \nabla f(g(\vec r))=\frac{df}{dg}\nabla g(\vec r) with f(g) = e^g and g(\vec r) = i...

The thing to do for partial fractions would be to solve: 1/(z^2+2z/x) = 1 / z(z+2/x) = A/z+B/(z+2/x) for A and B, and then it can be integrated to logs. Or, you can complete the square and you're left with an integral that yields a hyperbolic trig solution which can be expressed in terms of...

Yes, I just got out of the shower where I also realized that result must be wrong, mainly because coth(2*pi) does not equal 1! (2*pi was close enuf to infinity for me and my calculator to conclude that odd result.) thus my struggle to make sense of the T=0 relation! And, yes, I now see the...

I still don't completely understand the energy / angular momentum relation along T=0, but, When L=0, you do get ur expected period, since coth(2*pi) = 1 there. How is the rotating problem simpler in Cartesian coords?

I guess it's kindof a poorly worded question, ambiguous, and I would assume it's trying to ask: "If you put a cup, of negligible thickness(say plastic), of water in a frig, and a cup of air in same frig (and somehow kept the air in the cup, e.g., sealed the cup with cellophane), why does it...

I think your equation defining the unit normal is wrong -it should have primes(derivatives) rather than hats on the rhs. Then: 1) Close but, The tangent line should have a single parameter; is it a or t? Try to picture: you start at r(π/4) and add an arbitrary amount (a) along the tangent...

I agree with these two points. Interesting that to specify the structure/constraints of the subgroup in first point takes infinitely many generators, while the containing group has only two generators.

Here are two ideas that will lead to proofs, but I can't work through the details now. 1) Any rational, expressed in lowest terms (by dividing common factors top and bottom), num and den can be uniquely factored into primes. So any finite set thereof, has only a finite number of distinct primes...

sorry, i was futzing with some Latex preview bug in trying to edit my reply.. I think your conjecture is correct; and those elements of S which generate H are S\cap H. And once you've proven the result for \mathbb{Q'}, you could use that conjecture to show it was true for \mathbb{R'}. But...

I think your conjecture is correct; and those elements of S which generate H are S\cap H. And once you've proven the result for But then your left with finding a subgroup which can't be finitely generated, so it would have to be infinite (or H itself could finitely generate it). And even some...

... S_{tot}(N_A,V_A,N_B,V_B) being exactly additive, S(N,V)being approximately so, and the relatively small difference S_{tot}(N_A,V_A,N_B,V_B) - S(N,V) = k\ln[P(N_A,N_B)], being responsible for the result that (the additive) entropy is maximized at equilibrium.