Recent content by Eredir

Check out this answer from MathOverflow: http://mathoverflow.net/questions/121620/why-does-gln-have-no-spinor-representations

A commutative C* algebra is isomorphic to the algebra of continuous functions on some space (here I'm skipping the technical details), so it provides only topological information. One of the ideas of Connes' approach to non-commutative geometry is that, by adding an operator D satisfying some...

A valid reason could be that we need to impose anti-commutation relations for the Hamiltonian to be bounded below, which is a reasonable physical requirement. From the normal mode expansion \psi(x) = \sum_{p,r}\sqrt{\frac{m}{E(p)V}}[a_{r}(p)u_{r}(p)e^{-ipx} + b_{r}^{\dagger}(p)v_{r}(p)e^{ipx}]...

Thanks guys, you have been very helpful!

Thanks for your helpful reply Jaunty. Yes, I wouldn't say that they are the same metric, given that those two spaces are not homeomorphic but only locally homeorphic. But given two spaces that are homeomorphic, or better diffeomorphic, I would be inclined to say that the solutions should be...

In Newtonian gravity the spacetime is fixed, so changing the coordinate system merely changes the parametrizations for the wordlines. In this case events take place in a unique and defined manifold which is indipendent of the dynamic, while in general relativity determining the manifold...

Of course this is true, but still not having a unique solution is quite peculiar, because in the weak field limit general relativity reduces to Newtonian theory where the solution is unique. Anyway, from the original question stems this new one: given that we can find several differential...

But being a Lorentzian manifold is just a condition on the metric, that is having signature (1,3) or (3,1). The metric is an additional structure you put on a differentiable manifold to get a (pseudo-)Riemannian manifold, it does not identify a particular differentiable manifold except for...

Hi everyone, this is my first post in this nice forum. :smile: I have some confusion regarding solutions of Einstein's field equations. I have read in several places that an exact solution to the field equations is a Lorentzian manifold. Now given a stress-energy tensor T_{\mu\nu} the...