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
5,352 answers , 22,785 comments
1,470 users with positive rep
820 active unimported users
More ...

  Lagrangian depends on second derivative of field

+ 3 like - 0 dislike
1692 views

In case of the gauge-fixed Faddeev-Popov Lagrangian: $$ \mathcal{L}=-\frac{1}{4}F_{\mu\nu}\,^{a}F^{\mu\nu a}+\bar{\psi}\left(i\gamma^{\mu}D_{\mu}-m\right)\psi-\frac{\xi}{2}B^{a}B^{a}+B^{a}\partial^{\mu}A_{\mu}\,^{a}+\bar{c}^{a}\left(-\partial^{\mu}D_{\mu}\,^{ac}\right)c^{c} $$

(for example in Peskin and Schrder equation 16.44)

If you expand the last term (for the ghost fields) you get: $$ \bar{c}^{a}\left(-\partial^{\mu}D_{\mu}\,^{ac}\right)c^{c} = -\bar{c}^{a}\partial^{2}c^{a}-gf^{abc}\bar{c}^{a}\left(\partial^{\mu}A_{\mu}\,^{b}\right)c^{c}-gf^{abc}\bar{c}^{a}A_{\mu}\,^{b}\partial^{\mu}c^{c} $$

And so, the Lagrangian has a term proportional to the second derivative of $c^a$.

In this case, how does one find the classical equations of motion for the various ghost fields and their adjoints?

I found the following equations of motion so far: $$ D_{\beta}\,^{dc}F^{\beta\sigma}\,^{c}=-g\bar{\psi}\gamma^{\sigma}t^{d}\psi+\partial^{\sigma}B^{d}+gf^{dac}\left(\partial^{\sigma}\bar{c}^{a}\right)c^{c} = 0 $$ $$ \partial_{\sigma}\bar{\psi}_{\alpha,\, i}i\gamma^{\sigma}-\sum_{\beta}\sum_{j}\bar{\psi}_{\beta,\, j}\left(gA_{\mu}\,^{a}\gamma^{\mu}\,_{ji}t^{a}\,_{\beta\alpha}-m\delta_{ji}\delta_{\beta\alpha}\right)=0 $$ $$ \left(i\gamma^{\mu}D_{\mu}-m\right)\psi=0 $$ $$ B^{b}=\frac{1}{\xi}\partial^{\mu}A_{\mu}\,^{b} $$ $$ \partial^{\mu}\left(D_{\mu}\,^{dc}c^{c}\right)=0 $$ $$ f^{abd}\left(\partial_{\sigma}\bar{c}^{a}\right)A^{\sigma}\,^{b}=0 $$

But it is the last equation that I suspect is false (I saw the equation $ D_\mu\,^{ad} \partial^\mu \bar{c}^d = 0 $ in some exercise sheet and I also saw the equation $D^\mu\,^{ad}\partial_\mu B^d = igf^{dbc}(\partial^\mu\bar{c}^b)D_\mu\,^{dc} c^c$ which I don't understand how they were derived.)

Any help about what are the proper equations of motions and how to get them would be much appreciated.

This post imported from StackExchange Physics at 2014-07-03 18:16 (UCT), posted by SE-user PPR
asked Jul 3, 2014 in Theoretical Physics by PPR (135 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

I) Since total divergence terms do not contribute to Euler-Lagrange (EL) equations, cf. e.g. this Phys.SE post, one could just integrate the Faddeev-Popov $\bar{c}c$ term by part so that there are no more than first derivatives present and the standard form of the EL equations applies.

II) Alternatively, in the presence of higher derivatives, the EL equations get modified with additional terms, see e.g. Wikipedia. Note that special care should be taken in the ordering of Grassmann-odd derivatives.

III) Often in field theory, because of external indices and Grassmann-odd fields, it is quite tedious to use the EL equations directly. It is often simpler to infinitesimally vary the given action $S[\phi]$,

$$\delta S~=~\int d^4x~\text{(EL-eq)}~ \delta\phi(x) +\text{(boundary terms)} . $$ and identify the EL equations as the relevant coefficient functions on the spot, so to speak.

This post imported from StackExchange Physics at 2014-07-03 18:16 (UCT), posted by SE-user Qmechanic
answered Jul 3, 2014 by Qmechanic (3,120 points) [ no revision ]
+ 1 like - 0 dislike

Replace the field $\phi(x)$ in the action by $\phi(x)+\delta\phi(x)$, expand while dropping products of the variational field, and use integration by parts to remove all derivatives from the variational field $\delta\phi(x)$. The result is an expression of the form

[original action] plus integral over [field equation] times $\delta\phi(x)$.

This procedure works no matter how many derivatives you have, and also for anticommuting fields.

answered Jul 12, 2014 by Arnold Neumaier (15,787 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:
p$\hbar$ysicsOverflo$\varnothing$
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
...