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

PO is now at the Physics Department of Bielefeld University!

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

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Symmetry of the Polyakov action?

+ 6 like - 0 dislike
1059 views

Let us look at the Polyakov action for a string moving in a spacetime with metric $g_{\mu \nu}(X)$:$$S_P = -{1\over{4\pi \alpha'}} \int d^2 \sigma \sqrt{-\gamma} \gamma^{ab} \partial_a X^\mu \partial_b X^\nu g_{\mu\nu}(X) \tag{1}$$ and suppose there exists a Killing vector $k_\mu$ in spacetime satisfying Killing's equation $$\nabla_\mu k_\nu + \nabla_\nu k_\mu = 0.\tag{2}$$ Does this lead to a symmetry of the Polyakov action?

This post imported from StackExchange Physics at 2016-01-17 15:57 (UTC), posted by SE-user Ham
asked Jan 7, 2016 in Theoretical Physics by Ham (30 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

The answer is Yes. Hints:

  1. Perform an infinitesimal variation $$ \delta X^{\mu}~=~\varepsilon K^{\mu}(X) \tag{A} $$ of the Polyakov action, where $\varepsilon$ is an infinitesimal parameter. Then $$ \partial_a \delta X^{\lambda} ~=~ \varepsilon\partial_a X^{\mu}~\partial_{\mu}K^{\lambda}, \qquad \delta G_{\mu\nu}~=~\varepsilon K^{\lambda}~\partial_{\lambda}G_{\mu\nu}.\tag{B}$$

  2. It is easier to use the definition of a Killing vector field $$0~=~({\cal L}_K G)_{\mu\nu}~=~ K^{\lambda}~\partial_{\lambda}G_{\mu\nu} + \partial_{\mu}K^{\lambda}~G_{\lambda\nu}+G_{\mu\lambda}~\partial_{\nu}K^{\lambda} \tag{C} $$ rather than the equivalent eq. (2).

This post imported from StackExchange Physics at 2016-01-17 15:57 (UTC), posted by SE-user Qmechanic
answered Jan 7, 2016 by Qmechanic (3,120 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$\hbar$y$\varnothing$icsOverflow
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).
Please complete the anti-spam verification




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

Your rights
...