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Lagrangian depends on second derivative of field

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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 (130 points) [ no revision ]

2 Answers

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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 (2,790 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 (12,425 points) [ revision history ]

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