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  Basic Question about Gauge Transformation

+ 1 like - 0 dislike

Suppose we have an action $S=S(a,b,c)$ which is a functional of the fields $a,\, b,\,$ and $c$. We denote the variation of $S$ wrt to a given field, say $a$, i.e. $\frac{\delta S}{\delta a}$, by $E_a$.

Then $S$ is gauge invariant when

$$\delta S = \delta a E_a + \delta b E_b +\delta c E_c = 0 \tag{1}$$

This gives

$$\delta c = - (\delta a E_a + \delta b E_b)/E_c   \tag{2}$$

From the above equation $\delta c$ can be obtained for arbitrary $\delta a$ and $\delta b$. It is not necessary to have a relation b/w $\,\delta a$ and $\delta b$. Doesn't this imply that there is an infinite number of gauge-transformations here? If yes, isn't that absurd?

asked Apr 2 in Theoretical Physics by anonymous [ no revision ]

1 Answer

+ 0 like - 0 dislike

The requirement $\delta S =0$ is a stationary action condition. It should be valid even for one field, say, for $a$. It has nothing to do with a gauge field definition, IMHO. Thus, one obtains the equations of motion for the field $a$: $E_a=0$.

In the variation principle the fields $a$, $b$, and $c$ are supposed to be independent and their variations too. Otherwise you have a system with constraints.

Originally, the gauge fields were called "compensating fields", find out why.

Briefly, some variables changes preserve the form (but not the solutions) of the equations: $a\to a',\;\; E_{a'}=0$. Invariance is only required for observable results of calculations, as a matter of fact.

(Worse, any reasonable variable change, even not preserving the formal form of equations, is good if it helps solve the equations.)

answered Apr 2 by Vladimir Kalitvianski (102 points) [ revision history ]
edited Apr 3 by Vladimir Kalitvianski

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