Friday, 29 December 2017

Gauge fixing choice for the gauge field A0


In many situations, I have seen that the the author makes a gauge choice A0=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 A0=0?


Under a gauge transformation Ai transforms as



AigAig1igg1,


where g is in the gauge group.



Answer



The "gauge fixing" condition A0=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:


AigAig1igg1,i=1,2,3


with g time independent.


However, this is not the whole story: The time derivative of A0 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:


.E=0.


One cannot obtain this equation after setting A0=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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