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
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Why do derivatives act on vector fields on a worldsheet?

+ 1 like - 0 dislike
841 views

The covariant derivative of a vector $A^{\mu}$ at a point $x$ is defined as

$$D_z A^{\mu}=\partial_zA^{\mu}+\Gamma^{\mu}_{\rho\sigma}(x)\partial_{z}x^{\rho}A^{\sigma}$$

where Greek symbols are spacetime (dimension $d$) and $z$ is the index of a worldsheet coordinate $\sigma$.

Why doesn't the vector $A^{\mu}$ act as a scalar with respect to the $z$ derivative? How can I prove this?

Assume $A^{\mu}$ depends on $x^\mu(\sigma^b)$, i.e. $A^{\mu}(x^\mu(\sigma^b))$.

This post imported from StackExchange Physics at 2014-11-26 10:50 (UTC), posted by SE-user sol0invictus
asked Nov 26, 2014 in Theoretical Physics by sol0invictus (45 points) [ no revision ]
I edited your question to clarify it, but I'm not 100% sure I got it right. If I changed what you meant to ask, please edit it again to fix it. By the way, we'd really rather you not delete your question and post a new one. This time it's fine, but in the future you should edit the old question instead.

This post imported from StackExchange Physics at 2014-11-26 10:50 (UTC), posted by SE-user David Z
Possible duplicate by OP: physics.stackexchange.com/q/148384/2451

This post imported from StackExchange Physics at 2014-11-26 10:50 (UTC), posted by SE-user Qmechanic
Thanks for the edit. I apologize for the mess with the previous post.

This post imported from StackExchange Physics at 2014-11-26 10:50 (UTC), posted by SE-user sol0invictus

1 Answer

+ 1 like - 0 dislike

It's the chain rule: $$ \partial_x f(y(x)) = \partial_y f(y(x)) \cdot \partial_x y(x)$$ Your vector field $A^\mu$ depends on the worldsheet coordinates only through the worldsheet coordinates $x^\mu$. Thus, when how $A^\mu$ behaves under an infinitesimal shift on the world-sheet you need to take into account how $A^\mu$ depends on $z$ - that's $\partial_z A^\mu$. But you also need to take into account how the worldsheet coordinates change - which is the connection piece.

$$D_z A^\mu = \underbrace{\partial_z A^\mu}_{\text{change in } A^\mu} + \underbrace{\Gamma^\mu_{\rho \sigma}(x) \partial_z x^\rho A^\sigma}_{\text{change of what "} \mu \text{" means.}}$$

This post imported from StackExchange Physics at 2014-11-26 10:50 (UTC), posted by SE-user Neuneck
answered Nov 26, 2014 by Neuneck (110 points) [ no revision ]
So if I had a quantity $D_zX^{\mu}(\sigma)$ would it be just partial derivative with no connection. This X is the same x where my previous covariant derivative was taken

This post imported from StackExchange Physics at 2014-11-26 10:50 (UTC), posted by SE-user sol0invictus

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$\hbar$ys$\varnothing$csOverflow
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
...