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  Form of Dilaton term in polarization tensor

+ 4 like - 0 dislike

The vertex operator associated with massless state is

$$V(k,\epsilon) = -\frac{2}{\alpha}\epsilon_{\mu\nu}(k)\bar{\partial}X^\mu(\bar{z})\partial X^\nu(z)e^{ik\cdot X(z,\bar{z})}$$

The polarization tensor can be decomposed into symmetric (Graviton), antisymmetric and Trace bit (Dilaton) Lust,theisen 16.9

$$\epsilon^{(h)}_{\mu\nu} = \epsilon^{(h)}_{\nu\mu},\qquad\epsilon^{(h)}_{\mu\nu} \eta^{\mu\nu} = k^{\mu}\epsilon^{(h)}_{\mu\nu} = 0, $$

$$\epsilon^{(B)}_{\mu\nu} = -\epsilon^{(B)}_{\nu\mu},\qquad k^{\mu}\epsilon^{(B)}_{\mu\nu} = 0, $$

$$\epsilon^{(D)}_{\mu\nu} = \frac{1}{\sqrt{d-2}}(\eta_{\mu\nu}-k_{\mu}\bar{k_{\nu}}-k_{\nu}\bar{k_{\mu}})$$

$\bar{k}$ is an arbitrary light like vector orthogonal to $k$. Can you please tell me why we took that particular form for the Dilaton?

This post imported from StackExchange Physics at 2014-09-07 18:45 (UCT), posted by SE-user sol0invictus
asked Sep 7, 2014 in Theoretical Physics by sol0invictus (45 points) [ no revision ]

2 Answers

+ 4 like - 0 dislike

The polarisation tensor is not unique and subject to the equivalence $\epsilon_{\mu\nu}(k) \sim \epsilon_{\mu\nu}(k) + k_\mu\xi_\nu+k_\nu\xi_\mu$ which is nothing but  diffeomorphisms in target space.   this gauge symmetry shows up in the form the existence of a null state  which can be constructed from the state given in Eq (16.11). One fixes this gauge symmetry by choosing $k^\mu \epsilon_{\mu\nu}=0$ and   $k^\nu \epsilon_{\mu\nu}=0$ (this is the analog of $\partial_\mu A^\mu=0$ in $U(1)$ gauge theories).  This can be viewed as a transversality condition on the polarisation tensor. On decomposing into a symmetric traceless, antisymmetric and trace part, this condition remains to be implemented. For the dilation, this is done by introducing an reference vector $\bar{k}$. $\bar{k}$ is chosen such that $\bar{k}^2=0$ and $\bar{k}\cdot k=1$ (not $0$ as the OP has stated). Then, $(\eta_{\mu\nu}-k_\mu\bar{k}_\nu-k_\nu\bar{k}_\mu)$ is the projector on to the space transverse to $k$. 

answered Sep 8, 2014 by suresh (1,545 points) [ revision history ]
+ 4 like - 0 dislike

The graviton represents only the traceless symmetric part of the polarization tensor. The dilaton part belongs to the symmetric part too (the trace part). The symmetric dilaton   \(\epsilon^{(D)}_{\mu\nu} (k)\)represents only \(1\)degree of freedom, so it is necessarily expressed as a function of \(k\), and it must verify the transversality condition :  \(k^\mu\epsilon^{(D)}_{\mu\nu} (k)==0\)

The general expression is then \(\epsilon^{(D)}_{\mu\nu} (k) \sim \eta_{\mu\nu} -(k_\mu \bar k_\nu + k_\nu \bar k_\mu) \)

The transversality condition implies \(k_\nu - (k.\bar k) k_\nu =0\), and, so finally \((k.\bar k)=1\)

One may add a supplementary  transversality condition : \(\bar k^\mu\epsilon^{(D)}_{\mu\nu} (k)=0\), this clearly implies \(\bar k^2=0\)

This supplementary condition is compatible with the precedent conditions, in particular the dilaton and graviton degrees of freedom are still orthogonal: \(\epsilon^{(D)}_{\mu\nu} (k) (\epsilon^{(G)}) ^{\mu\nu} (k)=0\)

answered Sep 8, 2014 by Trimok (955 points) [ no revision ]

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