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.


3 Answers 3


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:


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).

  • $\begingroup$ I didnt get your last bit on the Gauss law. Isnt the gauss law obtained by setting the free index as zero in the second equation of motion(obtained by varying $A_\mu$? WHy do you say : "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". I thought it is applicable to all $A_0$'s and has nothing to do with it being 0. $\endgroup$
    – user7757
    Jul 31, 2012 at 9:12
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    $\begingroup$ But if one subtitutes $A_0 = 0$ in the Lagrangian before the computation of the equations of motion, then equation of motion of $A_0$, namely, the Gauss law will be lost, because one will not have an $A_0$ in the Lagrangian to take a variation with respect to it. In this case one has to impose the Gauss law "by hand". $\endgroup$ Jul 31, 2012 at 9:49
  • $\begingroup$ (cont.), please see the following review by: ANTTI SALMELA ethesis.helsinki.fi/julkaisut/mat/fysik/vk/salmela/gausssla.pdf $\endgroup$ Jul 31, 2012 at 9:52
  • $\begingroup$ It seems the EOM of $𝐴_0$ is $\partial_j A^j =\nabla \cdot \vec{A} =0$ --- could you clarify on this? physics.stackexchange.com/q/718090/310987 Thank you so much! $\endgroup$ Jul 13, 2022 at 1:16

I) Let us choose the following convention

$$ \tag{1} A^{\prime}_{\mu}~=~ g^{-1} (A_{\mu}+id_{\mu}) g $$

for a non-abelian gauge transformation. To be concrete let us assume that the gauge group $G$ is either $U(N)$ or $SU(N)$.

II) Then OP's question becomes

Does there exists a globally defined gauge transformation $g\in G$ so that $A^{\prime}_0=0$?

Or equivalently,

Does there exists a globally defined temporal gauge?

Answer: Yes, choose e.g. the following gauge transformation as a time-ordered Wilson-line

$$ g(\vec{r},t) ~=~ \mathcal{T}\exp\left[i\int_0^t \! dt^{\prime}~A_0(\vec{r},t^{\prime})\right], \qquad t\geq 0. $$

(There is similar formula for $t\leq 0$.) This is possible if the right-hand side is well-defined, i.e. if the former temporal gauge potential $A_0$ is integrable in time.

  • $\begingroup$ +1, this was just what I was looking for. Some questions though; Why the ``time-ordered'' part? Can't we run the time integral from $-\infty$ to $t$ ? Thanks. $\endgroup$
    – kηives
    Dec 26, 2012 at 0:18
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    $\begingroup$ 1. The time-order is needed when the gauge group $G$ is non-abelian. The formula is not correct without time-ordering. 2. Yes, one can put the initial condition at $t_0=-\infty$. $\endgroup$
    – Qmechanic
    Feb 2, 2015 at 13:53

In addition to the previous answers, this thing is completely obvious in Lattice gauge theory. This shows that axial and temporal gauges are well defined, and also gives a full gauge fixing prescription removing residual gauge ambiguities. It is the lattice version of Qmechanic and Bar-Moshe's answer.

On a lattice, there is a group element for each lattice link. There is a freedom to multiply by a group element at any point. So to completely fix a gauge, all you have to do is choose a unique path to this point from some starting position, and this gives you a unique group element at each point to multiply by.

Make this position the origin, and define the path as follows:

follow the x-axis to the x-coordinate of the point, then follow parallel to the y-axis to the x,y coordinates of the point (still at z=0 t=0), then follow parallel to the z-axis to the proper z, then parallel to the t-axis. Multiplying all the group elements in the order you encounter them. Then rotate by this matrix at this point.

This sets the gauge group element in the t-direction to the identity, this is equivalent to setting the continuum gauge field to zero. It also sets the gauge field in the z-direction to zero on the t=0 surface, it sets the gauge field in the y and z direction to zero on the x-y plane, and the entire gauge field to zero along the x-axis.

This is a complete and nonperturbative lattice gauge fixing, it leaves only the global gauge group unbroken. It corresponds to axial gauge, because it is in imaginary time, but to do the temporal gauge fixing heuristically without regulator is identical as explained by Qmechanic, and there are no problems of principle (although, as Bar Moshe said, you have to be careful not to throw away the constraint equations of motion that arise from varing the gauge field the directions that violate the gauge condition).


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