New answers tagged

0

I'd say none. If you are talking about a Riemannian cylinder, its metric (say, induced from its embedding in $\Bbb{R}^{3}$) is $$ g=dz^{2}+d\phi^{2} $$ If the two were conformally equivalent, a conformal transformation $(z,\phi)=(z(t,x),\phi(t,x))$ would pull back the metric as $$ g'=\bigg(\frac{\partial z}{\partial t}dt+\frac{\partial z}{\partial ...


3

QED is non-perturbative because it is not defined by the sum over Feynman diagrams. Standard QFT does not rely on the perturbative expansion to define scattering amplitudes, it only uses it to compute them. String theory through CFT, on the other hand, defines the string scattering amplitude through the sum over worldsheets. This is not a perturbative ...



Top 50 recent answers are included