Friday, 29 December 2017

Gauge fixing choice for the gauge field $A_0$


In many situations, I have seen that the the author makes a gauge choice $A_0=0$, e.g. Manton in his paper on the force between the 't Hooft Polyakov monopole.


Please can you provide me a mathematical justification of this? How can I always make a gauge transformation such that $A_0=0$?


Under a gauge transformation $A_i$ transforms as



$$A_i \to g A_i g^{-1} - \partial_i g g^{-1},$$


where $g$ is in the gauge group.



Answer



The "gauge fixing" condition $A_0=0$ called the temporal gauge or the Weyl-gauge please see the following Wikipedia page). This condition is only a partial gauge fixing condition because, the Yang-Mills Lagrangian remains gauge invariant under time independent gauge transformations:


$A_i \to g A_i g^{-1} - \partial_i g g^{-1}, i=1,2,3$


with $g$ time independent.


However, this is not the whole story: The time derivative of $A_0$ doesn't appear in the Yang-Mills Lagrangian.Thus it is not a dynamical variable. It is just Lagrange multiplier. It's equation of motion is just the Gauss law:


$\nabla.E=0$.


One cannot obtain this equation after setting $A_0 = 0$. So it must be added as a constraint and it must be required to vanish in canonical quantization on the physical states. (This is the reason that it is called the Gauss law constraint).


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