Tag Info

New answers tagged

1

Equation (2.4.6): $T(z)X^\mu(0)\sim \frac{1}{z}\partial X^\mu(0)$ means that the RHS is the most singular term of the LHS. $T(z) = -\frac{1}{\alpha'} :\partial X^{\mu} \partial X_{\mu}:\tag{2.4.4}$ So \begin{align*} T(z)X^{\mu}(0) & =-\frac{1}{\alpha'}:\partial X^{\nu}(z)\partial X_{\nu}(z):X^{\mu}(0)\\ & =-\frac{2:\partial ...


-4

I feel like this is an over-simplification. I agree. IMHO it's a popscience analogy that's presented as a genuine explanation, but it isn't. What is lost in the actual AdS/CFT correspondence in this analogy? A dimension. You can't just discard this. If you do, you lay yourself wide open to dropping another dimension, and saying the flat 2D image is ...



Top 50 recent answers are included