In physics, the Polyakov action is an action of the two-dimensional conformal field theory describing the worldsheet of a string in string theory. It was introduced by Stanley Deser and Bruno Zumino and independently by L. Brink, P. Di Vecchia and P. S. Howe in 1976,[1] [2] and has become associated with Alexander Polyakov after he made use of it in quantizing the string in 1981.[3] The action reads:
l{S}=
T | |
2 |
\intd2\sigma\sqrt{-h}habg\mu\nu(X)\partialaX\mu(\sigma)\partialbX\nu(\sigma),
where
T
g\mu\nu
hab
hab
h
hab
\sigma
\tau
The Polyakov action must be supplemented by the Liouville action to describe string fluctuations.
N.B.: Here, a symmetry is said to be local or global from the two dimensional theory (on the worldsheet) point of view. For example, Lorentz transformations, that are local symmetries of the space-time, are global symmetries of the theory on the worldsheet.
The action is invariant under spacetime translations and infinitesimal Lorentz transformationswhere
\omega\mu=-\omega\nu
b\alpha
The invariance under (i) follows since the action
l{S}
X\alpha
\begin{align} l{S}' &={T\over2}\intd2\sigma\sqrt{-h}habg\mu\partiala\left(X\mu+
\mu | |
\omega | |
\delta |
X\delta\right)\partialb\left(X\nu+
\nu | |
\omega | |
\delta |
X\delta\right)\\ &=l{S}+{T\over2}\intd2\sigma\sqrt{-h}hab\left(\omega\mu\partialaX\mu\partialbX\delta+\omega\nu\partialaX\delta\partialbX\nu\right)+\operatorname{O}\left(\omega2\right)\\ &=l{S}+{T\over2}\intd2\sigma\sqrt{-h}hab\left(\omega\mu+\omega\delta\right)\partialaX\mu\partialbX\delta+\operatorname{O}\left(\omega2\right)\\ &=l{S}+\operatorname{O}\left(\omega2\right). \end{align}
The action is invariant under worldsheet diffeomorphisms (or coordinates transformations) and Weyl transformations.
Assume the following transformation:
\sigma\alpha → \tilde{\sigma}\alpha\left(\sigma,\tau\right).
hab(\sigma) → \tilde{h}ab=hcd(\tilde{\sigma})
\partial{\sigma | |
a}{\partial |
\tilde{\sigma}c}
\partial{\sigma | |
b}{\partial |
\tilde{\sigma}d}.
\tilde{h}ab
\partial | |
\partial{\sigma |
a}X\mu(\tilde{\sigma})
\partial | |
\partial\sigmab |
X\nu(\tilde{\sigma})= hcd\left(\tilde{\sigma}\right)
\partial\sigmaa | |
\partial\tilde{\sigma |
c}
\partial\sigmab | |
\partial\tilde{\sigma |
d}
\partial | |
\partial\sigmaa |
| ||||
X |
b}X\nu(\tilde{\sigma})= hab\left(\tilde{\sigma}\right)
\partial | |
\partial\tilde{\sigma |
a}X\mu(\tilde{\sigma})
\partial | |
\partial\tilde{\sigma |
b}X\nu(\tilde{\sigma}).
One knows that the Jacobian of this transformation is given by
J=\operatorname{det}\left(
\partial\tilde{\sigma | |
\alpha}{\partial |
\sigma\beta}\right),
\begin{align} d2\tilde{\sigma}&=Jd2\sigma\\ h&=\operatorname{det}\left(hab\right)\\ ⇒ \tilde{h}&=J2h, \end{align}
\sqrt{-\tilde{h}}d2{\sigma}=\sqrt{-h\left(\tilde{\sigma}\right)}d2\tilde{\sigma}.
\tilde{\sigma}=\sigma
Assume the Weyl transformation:
hab\to\tilde{h}ab=Λ(\sigma)hab,
\begin{align} \tilde{h}ab&=Λ-1(\sigma)hab,\\ \operatorname{det}\left(\tilde{h}ab\right)&=Λ2(\sigma)\operatorname{det}(hab). \end{align}
l{S}', | ={T\over2}\intd2\sigma\sqrt{-\tilde{h}}\tilde{h}abg\mu\nu(X)\partialaX\mu(\sigma)\partialbX\nu(\sigma), | |
={T\over2}\intd2\sigma\sqrt{-h}\left(ΛΛ-1\right)habg\mu(X)\partialaX\mu(\sigma)\partialbX\nu(\sigma)=l{S}. |
One can define the stress–energy tensor:
Tab=
-2 | |
\sqrt{-h |
\hat{h}ab=\exp\left(\phi(\sigma)\right)hab.
\phi
\deltaS | |
\delta\phi |
=
\deltaS | |
\delta\hat{h |
ab
hab
\deltaS | |
\deltahab |
=Tab=0.
\delta\sqrt{-h}=-
12 | |
\sqrt{-h} |
hab\deltahab.
\deltaS | |
\deltahab |
=
T | |
2 |
\sqrt{-h}\left(Gab-
12 | |
h |
abhcdGcd\right),
Gab=g\mu\partialaX\mu\partialbX\nu
\begin{align} Tab&=T\left(Gab-
12 | |
h |
abhcdGcd\right)=0,\\ Gab&=
12 | |
h |
abhcdGcd,\\ G&=\operatorname{det}\left(Gab\right)=
14 | |
h |
\left(hcdGcd\right)2. \end{align}
\sqrt{-h}
\sqrt{-h}=
2\sqrt{-G | |
S={T\over2}\intd2\sigma\sqrt{-h}habGab={T\over2}\intd2\sigma
2\sqrt{-G | |
However, the Polyakov action is more easily quantized because it is linear.
Using diffeomorphisms and Weyl transformation, with a Minkowskian target space, one can make the physically insignificant transformation
\sqrt{-h}hab → ηab
l{S}={T\over2}\intd2\sigma\sqrt{-η}ηabg\mu(X)\partialaX\mu(\sigma)\partialbX\nu(\sigma)={T\over2}\intd2\sigma\left(
X |
2-X'2\right),
ηab=\left(\begin{array}{cc}1&0\ 0&-1\end{array}\right)
Keeping in mind that
Tab=0
\begin{align} T01&=T10=
X |
X'=0,\\ T00&=T11=
12 | |
\left( |
X |
2+X'2\right)=0. \end{align}
Substituting
X\mu\toX\mu+\deltaX\mu
\begin{align} \deltal{S} &=T\intd2\sigmaηab\partialaX\mu\partialb\deltaX\mu\\ &=-T\intd2\sigmaηab\partiala\partialbX\mu\deltaX\mu+\left(T\intd\tauX'\deltaX\right)\sigma=\pi-\left(T\intd\tauX'\deltaX\right)\sigma=0\\ &=0. \end{align}
And consequently
\squareX\mu=ηab\partiala\partialbX\mu=0.
The boundary conditions to satisfy the second part of the variation of the action are as follows.
X\mu(\tau,\sigma+\pi)=X\mu(\tau,\sigma).
\xi\pm=\tau\pm\sigma
\begin{align} \partial+\partial-X\mu&=0,\\ (\partial+X)2=(\partial-X)2&=0. \end{align}
Thus, the solution can be written as
X\mu=
\mu | |
X | |
+ |
(\xi+)+
\mu | |
X | |
- |
(\xi-)