• 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.


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


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,345 answers , 22,719 comments
1,470 users with positive rep
818 active unimported users
More ...

  Non-trivial components of the stress-energy tensor of the bosonic string ghost action

+ 3 like - 0 dislike

The stress-energy tensor derived from the ghost action of a bosonic string is:

$$ T_{\alpha \beta} = \frac{i}{4 \pi} \left ( b_{\alpha \gamma} \nabla_{\beta} c^{\gamma} + b_{\beta \gamma} \nabla_{\alpha} c^{\gamma} - c^\gamma \nabla_\gamma b_{\alpha \beta} - g_{\alpha \beta} b_{\gamma \delta} \nabla_\gamma c^{\delta} \right) $$

My book says, but doesn't give the proof, that in light-cone coordinates, the only two non-trivial components are the diagonal ones, given by:

$$ T_{++} = \frac{i}{4 \pi} \left ( 2b_{++} \partial_+ c^+ - c^+ \partial b_{++} \right) $$

and the same thing for $T_{--}$ but replacing all plus signs in the indices with minus signs.

But I can't derive this. I know that in light-cone coordinates, $g_{++} = g_{--} = 0$ and $g_{+-} = g_{-+} = -\tfrac{1}{2}$ and the covariant derivative is just the partial derivative.

Now for example, consider $T_{+-}$ and focus on the $c^\gamma \nabla_\gamma b_{\alpha \beta}$ term. In light-cone coordinates, it's:

$$ c^+ \partial_+ b_{+-} + c^- \partial_- b_{+-} $$

Since these are the only terms involving partial derivatives of $b$, they must kill each other for $T_{+-}$ to be zero. I know that $b$ is symmetric and traceless, and that $b$ and $c$ anti-commute, but that's not enough to produce

$$ c^+ \partial_+ b_{+-} + c^- \partial_- b_{+-} = 0 $$

I don't see how it's possible.

This post imported from StackExchange Physics at 2014-05-21 08:30 (UCT), posted by SE-user user46242
asked May 20, 2014 in Theoretical Physics by user46242 (35 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

The point is that a traceless symmetric tensor has its (+-) and (-+) components equal to zero. This follows directly from the expression of the light-cone components of a tensor in terms of its $(x^0,x^1)$-components. This implies that $T_{+-}=T_{-+}=0$ and $b_{+-}=b_{-+}=0$. Using that, it is easy to compute $T_{++}$ and $T_{--}$.

answered May 24, 2014 by 40227 (5,140 points) [ revision history ]

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:
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