Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,862 answers , 20,637 comments
1,470 users with positive rep
502 active unimported users
More ...

Form of Dilaton term in polarization tensor

+ 4 like - 0 dislike
161 views

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 (1,535 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 (950 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsO$\varnothing$erflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...