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My Question is how to explicitly move into the "Coulomb gauge" in Yang-Mills theory.

Using the answer provided by QMechanic, one can move into the "temporal gauge" for Yang-Mills fields: Gauge fixing choice for the gauge field $A_0$ .

Once in this gauge, to move into the "Coulomb gauge," does one take an explicitly time independent $g$, and demand that it satisfy: $$ i\partial^\mu (g^{-1}(\partial_\mu - i A_{\mu})g) = 0 $$

Thus giving $\partial ^{\mu} A'_{\mu} = 0$? Are we guaranteed that such a $g$ exists? Thanks.

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The Coulomb gauge is given by $\partial^i A_i$, where $i=1,2,3.$ – Frederic Brünner Jan 18 '13 at 9:16
    
@FredericBrünner sure, my last line would read $\partial^{\mu}A'_{\mu}=\partial^{i}A'_{i}=0$. I'm wondering if my condition above is the condition required to fix the gauge. – kηives Jan 18 '13 at 17:14

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