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1

Going to the conformal gauge is nothing but using coordinates in which the metric is diagonal (in euclidean space this is called isothermal coordinates ). Therefore in order to show that the Polyakov action is Weyl-invariant without using the conformal gauge, it is sufficient to show that the action does not dependent on the coordinate choice at all. ...


1

I want to start by clarifying what M-theory is, in relation to string theory, just so the context of my answer will be understood. These days, "string theory" encompasses five ten-dimensional superstring theories, the 11-dimensional M-theory of membranes, 26-dimensional bosonic string theory, "supercritical strings" in more than 10 dimensions, and other ...


1

Here is an outline of the reduction from the Nambu-Goto (NG) action to the light-cone (LC) formulation from a Hamiltonian perspective: The starting point is the Hamiltonian formulation of the NG string, cf. e.g. this Phys.SE post. The Hamiltonian density is of the form "Lagrange multipliers times constraints"$^1$ $$ {\cal H}~=~\lambda^{\alpha} ...


3

I) In this alternative answer we resolve the singular Hessian $H_{\mu\nu}$ of the Nambu-Goto string action by introducing two auxiliary variables from the onset, thereby indirectly showing that the Hessian $H_{\mu\nu}$ must have co-rank 2. The target space metric has $(-,+,\ldots,+)$ sign convention, and $c=1=\hbar$. Consider the extended Nambu-Goto ...


0

The renormalization $[...]_r$ of the exponentials is defined in equation $(3.6.5)$ $$ [\mathcal{F}]_r = \exp \left( \frac{1}{2} \int d^2\sigma d^2\sigma' \Delta(\sigma,\sigma')\frac{\delta}{\delta X^\mu(\sigma)}\frac{\delta}{\delta X_\mu(\sigma')}\right) \mathcal{F} $$ Here $$ \Delta(\sigma,\sigma') =\frac{\alpha'}{2} \ln d^2(\sigma,\sigma') $$ where ...


4

I) In this answer we will consider the standard Nambu-Goto (NG) string and show that the Hessian has co-rank 2. The target space metric has $(-,+,\ldots,+)$ sign convention, and $c=1=\hbar$. The NG Lagrangian density is $${\cal L}_{NG}~:=~-T_0\sqrt{{\cal L}_{(1)}}, $$ $$ {\cal L}_{(1)}~:=~-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} ...


2

Comments to the question (v2): To be specific, let us assume that the underlying 2D manifold is the Riemann sphere $S^2\cong \mathbb{C}\cup\{\infty\}$. The group of globally defined conformal transformations is the 6-dimensional group $PSL(2,\mathbb{C})$ of Moebius transformations. Mathematically speaking, one should consider the groupoid of locally ...


2

This extra factor arises from the analogy of the conformal factor $\alpha'\omega$ term in (6.2.16). The required $\omega$ is $$\omega = \ln \left ( \frac{2\pi}{\partial_\nu\vartheta_1}\right) $$ and substituting it to the exponential we get $$ \exp\left( -\frac{\alpha'}{2}\sum_ik_i^2 \cdot \ln \frac{2\pi}{\partial_\nu\vartheta_1} \right) = ...


2

Confinement cannot be rigorously shown in QCD with current techniques, because all analytic results in QCD are perturbative and the perturbative expansion breaks down at low energies where the coupling becomes strong. QCD has a negative $\beta$-function, i.e. the Yang-Mills coupling grows at lower energies and becomes weaker at high energies. But the ...


0

Below is a summary of my very limited understanding of what the elliptic genus is. I'll first give you the mathematical definition, followed by an explanation of how it appears naturally in physics. As a first exposure to the subject, it's perhaps best to not consider the most general definitions. In what follows, I will only explain what the elliptic genus ...


0

because it gives hints about the compactification string gravity is flawed because it requires you to select a compactification to calculate with it and the compactification is famously undetermined and there's no way to reverse engineer the standard model to find the compactification that correspond to it https://youtu.be/PBOwargPdJ4?t=1733 holography ...


0

I) Recall first the $\phi\phi$-Operator Product Expansion (OPE): $$\tag{A} {\cal R}\left\{\phi(z,\bar{z})\phi(w,\bar{w})\right\} ~-~: \phi(z,\bar{z})\phi(w,\bar{w}): ~=~C(z,\bar{z};w,\bar{w}) ~{\bf 1}, $$ where the contraction is assumed to be a $c$-number: $$\tag{B} C(z,\bar{z};w,\bar{w})~=~ \langle 0 | {\cal ...


0

Here $\begin{equation} \langle ik\phi(x)ik\phi(x)\rangle = \frac{\alpha'k^2}{\pi} \text{ln}(a/2R), \end{equation}$ where $a$ is an UV cutoff. Now we can write (as all the $\phi$'s are located at $x$ i.e. Radial ordered $\{\phi^n(x)\}=\phi^{n}(x)~$) $\begin{align} \{ik\phi\}^{n}(x) ~&=~ :\{ik\phi\}^n(x): +\sum_{\text{all contractions}} \\ ...


3

There are obviously differeing genus types according to which partition function in d-dimensional QFT. At the very outset of $0$-d QFT the index is the push forward in ordinary de Rahm Cohomology; in other words, the integration of differential forms. The genus as you put it, is non-existant here. In $1$-d QFT the index of the Dirac Operator is ...


0

Try here for background into the derivation of the AdS/CFT correspondence and also this thesis for an excellent introduction to the correspondance with regards to symmetry breaking. Further to this, try these slides outlining the motivations and also providing a good reading list.


0

I can recommend these lecture notes, which discuss many prerequisites for understanding the correspondence in a concise yet accessible way.


0

$\partial_a T^{ab}=0$ implies $\partial_1 T^{11}+\partial_2 T^{21}=0$. You don't sum over the uncontracted index $b$.


2

The variation $\delta F$ for any field (or degree of freedom) $F$, given an infinitesimal transformation, is always calculated as the commutator $$ \delta F = [ \bar\epsilon Q, F ] $$ where $\bar \epsilon$ is a parameter ("angle" or "shift" or some generalization) of the transformation and $Q$ is the generator. (Those may be replaced by other letters.) ...



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