Tag Info

Hot answers tagged

4

The two notions are indeed related. Take for example the Weyl anomaly of bosonic string theory: the classical (Polyakov) action $S$ is invariant under Weyl rescalings of the worldsheet metric $\gamma_{ab}$, i.e. $$S[\gamma_{ab}(\tau,\sigma)]=S[\exp(2\omega(\tau,\sigma))\gamma_{ab}(\tau,\sigma)]=S[\gamma'_{ab}(\tau,\sigma)].$$ Since this is a conformal ...


2

In a nutshell, the $1+1=2$ extra dimensions of spacetime can (in an appropriate gauge) be viewed as a longitudinal spatial direction and a temporal direction. This makes sense because of Minkowski signature of spacetime. The point is that the physical modes of the string can be identified with the 24 (8) transversal directions of the critical bosonic ...


2

Both $h$ and $\tilde{h}$ are usually called weights. Their sum, $\Delta=h+\tilde{h}$ is the (scaling) dimension of the operator, while the difference, $s=h-\tilde{h}$ is called the spin. This is due to their association with scale transformations (dilatations) and rotations, respectively. To see this, note that the dilatation operator is given by ...


1

D-branes also arise in perturbative string theory on flat backgrounds as hypersurfaces in space on which strings can end due to Dirichlet boundary conditions. The equivalence of D-branes and p-branes is completely independent from the AdS/CFT correspondence and was discovered before the correspondence by Polchinski in 1995 (see here). D-branes and p-branes ...


1

Check out Richard Feynman's QED lectures. It's not quantum gravity yet but it might be what you're after.


1

In general you cannot, but in your special case it works out. You should be aware of what the objects in your expression actually are, and how they relate to each other. $X$ is a bosonic field and as such does not feel the presence of gamma matrices at all. Your first line could be rewritten as $$ S = \int \mathrm d^2 \sigma \, \bar \epsilon \gamma^\alpha ...


1

It's the chain rule: $$ \partial_x f(y(x)) = \partial_y f(y(x)) \cdot \partial_x y(x)$$ Your vector field $A^\mu$ depends on the worldsheet coordinates only through the worldsheet coordinates $x^\mu$. Thus, when how $A^\mu$ behaves under an infinitesimal shift on the world-sheet you need to take into account how $A^\mu$ depends on $z$ - that's $\partial_z ...


1

One comes to this conclusion due to the fact that the contraction of a symmetric tensor with an antisymmetric one vanishes. Writing down the loop diagrams involves a contraction of both vertices. If you get expressions proportional to $\epsilon_{\mu\nu}\eta^{\mu\nu}$, this will be zero due to the fact that the metric is symmetric and the epsilon tensor is ...



Only top voted, non community-wiki answers of a minimum length are eligible