Recent content by Timbuqtu

Suppose a solution y(x) is invertible locally, then we can define v(y)=y'(x(y)), where x(y) is the inverse of y(x). Substituting this in your d.e., you get \frac{1}{2} v \dot{v} + 2 y^{1/2} v + y^2 = 1, where \dot{v} denotes dv/dy. This is a first order d.e. which is easier to solve. Try...

Try http://www.blender.org"

Wow! 65 dollar? Here in europe it costs at least 120 euro (=176 dollar). I have immediately ordered it from amazon. Shipping to europe only costs about 3 dollars. Ok, it will take about a month for it to arrive, but a 70 euro discount is worth waiting for.

Hi Francesca. I'll try to shortly summarize some (quantum) gravity in 2+1 dimensions: The phase space of (classical) general relativity in 2+1 dimensions is much smaller than in 3+1 dimensions, because there are no local degrees of freedom (in absence of matter). The reason is that the Riemann...

I will be going to Zakopane too! They have done a great job getting this group of lecturers together. Apart from the people addressed by others, I am pleased to see Jean-Marc Schlencker is going to tell something about 2+1 quantum gravity and related math, since I'm working in that area at the...

That's almost correct, but the period is 1/2: if x increases from 0 to 1/2, then 4 \pi x increases from 0 to 2 \pi.

I haven't read that many topology books, but I liked James Munkres' "Topology" (which includes lot's of exercises).

Actually I do object against your arguments. For square matrices we have adj(A).A = det(A).I. So indeed adj(xI-A).(xI-A)=det(xI-A).I, but how do you conclude det(xI-A)=f(x) for x a matrix, instead of an element of k? I am pretty sure that's not true in general. My shorter proof...

Nevermind, adj[XI-a] is an nxn matrix. Now I understand your proof and I think it's correct, yes.

I do not really understand what you're doing. How can you multiply the (n-1)x(n-1) adj[XI-A] matrix with the nxn (XI-A) matrix?

He was Dutch :biggrin:

The gamma matrices don't change, you just defined what they are. But what happens with \partial_\mu under a Lorentz transformation?

I suppose you meant -1½, because 1 = 3 - ½ - 1½. You could 've seen this by looking at the commutation relation: \left[a_{\mathbf{p}}^r,a_{\mathbf{p'}}^{r\dagger}\right]=\delta^3(\mathbf{p}-\mathbf{p'}). Notice that the Dirac delta has units -3. Plugging this relation into the vacuum we get...

The killing equation \nabla_{(\mu}k_{\nu)}=0 is equivalent to the vanishing of the Lie-derivative of the metric along k, \mathcal{L}_k g_{\mu\nu}=0. This last equation actually just tells you that the metric does not change if you pull it along the vector field k, so k represents a symmetry of...

Another way of understanding this number is considering the symmetries belonging to the Killing fields. The simplest example of a maximally symmetric (n dim.) manifold is just Euclidian \mathbb{R}^n (or Minkowski space in case of Lorentzian metric). Independent symmetries are given by the...