| Topic: |
Science > Physics |
| User: |
"Lionel Brits" |
| Date: |
25 Apr 2006 08:50:27 PM |
| Object: |
Gauge fixing question |
I have a question on gauge fixing and path integral evaluation. I hope
it can be answered here. Starting with say, the Maxwell action, we can
write the generating functional as
Z ~ int DA delta[ G[A] ] det M exp{ -i S }
where G[A] = 0 is our gauge condition, and det M is the Faddeev-Popov
determinant. The delta functional can be massaged into the exponential,
giving a term -1/2alpha G[A]^2 in the action, which is the "gauge fixing
term".
My question is that initially we enforce G[A] = 0 using the delta
function, but once this gets turned into a "gauge fixing term", it seems
that A no longer needs to satisfy this, and classically the EOMs are not
the maxwell equations. I realize that any physical quantities don't
depend on this, but I don't see where the magic happened. Why doesn't
the gauge fixing term "fix" the gauge?
Thanks to anyone who can clarify.
.
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| User: "=?iso-8859-1?B?RGHrbA==?=" |
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| Title: Re: Gauge fixing question |
26 Apr 2006 02:14:15 AM |
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I didn't understand your question very well, but I'll try to answer
just for the fun of it (I also bet you know more QFT than I do).
When one "massages the delta functional" into the exponent, what one
does is to rewrite
Z ~ \int DA \delta[F[A]] det M exp(-iS)
(1)
as
Z ~ \int DA \delta[F[A] - C(x)] det M exp(-iS)
(2)
where C(x) is an arbitrary function, this can be done since it is
equivalent to choosing a new gauge F=C instead of F = 0, and Z is gauge
invariant. We can thus write
Z ~ \int DA \int DC \delta[F[A] - C(x)] det M exp(-iS) G[C] (3)
since the result will differ from (2) by a global normalization
constant, then
Z ~ \int DA det M exp(-iS) G[F]
(4)
where a common choice is
G[C] = exp(-(i/2 \alpha) \int d^4 x [C(x)]^2
(5)
giving the gauge fixing term
L_{gf} = -(1/2\alpha) (F[A])^2
(6)
in the Lagrangian. Thus, since we used the delta function \delta[F[A] -
C(x)] in equation (2), what is enforced by this delta is really F[A] =
C(x).
Also, the classical EOM's are now the Maxwell equations with the gauge
fixed, so A indeed still satisfies the gauge-fixing condition since the
EOM's are the equations which describe A... the gauge is indeed fixed
in the Lagrangian by the gauge fixing term. This can be seen by
noticing that the EOM's derived from the total Lagrangian (including 6)
are no longer gauge invariant (this means that they are valid only in a
certain gauge, that which has been fixed.)
(all this was taken from Pokorski's book "Gauge Field Theories")
I hope all this was of some help... but then again, maybe it wasn't,
I'm still a newbie in QFT.
Lionel Brits wrote:
I have a question on gauge fixing and path integral evaluation. I hope
it can be answered here. Starting with say, the Maxwell action, we can
write the generating functional as
Z ~ int DA delta[ G[A] ] det M exp{ -i S }
where G[A] = 0 is our gauge condition, and det M is the Faddeev-Popov
determinant. The delta functional can be massaged into the exponential,
giving a term -1/2alpha G[A]^2 in the action, which is the "gauge fixing
term".
My question is that initially we enforce G[A] = 0 using the delta
function, but once this gets turned into a "gauge fixing term", it seems
that A no longer needs to satisfy this, and classically the EOMs are not
the maxwell equations. I realize that any physical quantities don't
depend on this, but I don't see where the magic happened. Why doesn't
the gauge fixing term "fix" the gauge?
Thanks to anyone who can clarify.
.
|
|
|
| User: "Lionel Brits" |
|
| Title: Re: Gauge fixing question |
26 Apr 2006 08:43:36 AM |
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Hi,
Thanks. The massaging I understand, but thanks for reiterating. It's the
rest I don't quite follow. Do you mean to say that a solution to the
EOMs satisfy the condition F[A] = 0? I realize we can't just put F[A] to
0 in the action, because then we are extremizing a different system,
BUT, say F[A] = \partial^mu A_mu then the EOMs are \partial^2 A +
(...)F[A] = 0. In that case, we are allowed to put F[A] = 0 in the EOM?
What I mean is, does this gauge fixing procedure produce *both* a new
effective lagrangian *and* the prescription that solutions satisfy F[A] = 0?
It would then seem, that for some cases, maxwell's equations survive
unscathed after setting F[A] = 0.
Regards,
- DSA
Daël wrote:
....
L_{gf} = -(1/2\alpha) (F[A])^2
(6)
in the Lagrangian. Thus, since we used the delta function \delta[F[A] -
C(x)] in equation (2), what is enforced by this delta is really F[A] =
C(x).
Also, the classical EOM's are now the Maxwell equations with the gauge
fixed, so A indeed still satisfies the gauge-fixing condition since the
EOM's are the equations which describe A... the gauge is indeed fixed
in the Lagrangian by the gauge fixing term. This can be seen by
noticing that the EOM's derived from the total Lagrangian (including 6)
are no longer gauge invariant (this means that they are valid only in a
certain gauge, that which has been fixed.)
(all this was taken from Pokorski's book "Gauge Field Theories")
I hope all this was of some help... but then again, maybe it wasn't,
I'm still a newbie in QFT.
.
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| User: "=?iso-8859-1?B?RGHrbA==?=" |
|
| Title: Re: Gauge fixing question |
26 Apr 2006 12:23:05 PM |
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I understand your question now, and my answer is: I don't know. What I
said is that since the EOM's are no longer gauge-invariant they can
only be valid in the gauge which was fixed by the gauge fixing term of
the Lagrangian, there are also other functions and quantities which are
not gauge-invariant and depend on the gauge that was fixed, such as
Green's functions. It is in that sense that the gauge is "fixed". Your
question is quite interesting, I'll have to ask around, if you arive at
an answer independently please write it here so I can read it.
.
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