Why are there several gauge fixing choice for gauge symmetry fields?

519
0
Please teach me this:
For gauge symmetry fields,only one of any elementary subconfiguration of the whole configuration covers the all physics of the field.So we need to cut off the redundant configuration.It seem to me,in a loose sense,there is only one way to cut off the redundancy(the gauge fixing condition).But in fact,they use many gauge conditions(e.g for electromagnetic field),from which we could derive different laws.Do they exhaustively cut off all the redundant configuration,and so why they lead to different laws.
Thank you very much in advanced.
 
519
0
It seem to me any gauge fixing condition does not affect the physics of the field despite the ''remaining configuration'' after the ''cutting off'' is not ''smaller'' than ''elementary subconfiguration.So,what happend if the ''remaining'' is smaller than elementary subconfiguration?
 

tom.stoer

Science Advisor
5,759
159
Look at the following example:

[tex]A = \int_0^\infty dr\,r\,a(r)[/tex]

Now we re-write this as

[tex]A = \frac{1}{2\pi}\int_0^{2\pi} d\phi \int dr\,r\,a(r) = \frac{1}{2\pi}\int d^2r\, a(r)[/tex]

We have managed to rewrite the integral such that we have a gauge symmetry

[tex]\phi \to \phi^\prime[/tex]

which corresponds to SO(2) rotations in the xy-plane. Gauge-fixing this integral means that we introduce a delta-function. The simplest way is

[tex]A \to = \frac{1}{2\pi}\int d^2r\, 2\pi\,\delta(\phi-\phi_0)\,a(r)[/tex]

But there are more general possibilities like

[tex]A \to \frac{1}{2\pi}\int d^2r\, f(\phi)\,\delta[G(\phi)]\,a(r)[/tex]

where G is a gauge fixing function and f corresponds to the Fadeev-Popov determinant. The delta-function "counts" only zeros of G, so a "good" gauge fixing function is one that has exactly on zero. f can be calculated uniquely from G using the properties of the delta function (are you familiar with them?) and requiring that

[tex]1 = \frac{1}{2\pi}\int_0^{2\pi} d\phi\, f(\phi)\,\delta[G(\phi)][/tex]

In quantum field theory we are interested in calculating gauge invariant observables A; the choice of G reflects the freedom to work in different gauges specified by G. There are technical problems like finding "good" gauge fixing functions G, but the essential idea should be clear.
 

tom.stoer

Science Advisor
5,759
159
There is also a related explanation using the Hamiltonian formulation like in classical mechanics.
 

Related Threads for: Why are there several gauge fixing choice for gauge symmetry fields?

Replies
3
Views
984
Replies
3
Views
1K
Replies
46
Views
4K
Replies
9
Views
786
  • Last Post
Replies
14
Views
3K
Replies
5
Views
935
  • Last Post
Replies
3
Views
630
Replies
29
Views
4K

Hot Threads

Top