# A Question about Yang-Mills Equation

The non-homogeneous part of the Yang-Mills equations is given by

$$D\star F=\star J,$$

where $$D=d+A$$ is the covariant derivative, $$\star$$ is the Hodge star and $$J$$ is the source current.

Under a gauge transformation $$U(x)$$, the above equation should stay invariant. Under a gauge transformation $$U(x)$$, $$D\star F$$ is transformed to

$$D^{\prime}\star F^{\prime},$$

where $$F^{\prime}=UFU^{-1}$$. One finds

$$D^{\prime}\star F^{\prime}=U(D\star F)U^{-1}.$$

Thus, if the equation of motion is invariant under the gauge transformation, one expects that the current $$J$$ should transform as

$$J^{\prime}=UJU^{-1}.$$

However, take the spinor-QCD as an example, the gauge current is given by

$$J^{a}_{\mu}(x)=\bar{\Psi}\gamma_{\mu}T^{a}\Psi,$$

where $$T_{a}$$ is the generator of the gauge group. Under a gauge transformation, the spinor transforms as

$$\Psi^{\prime}(x)=U(x)\Psi(x).$$

It seems to me that this spinor current doesn't transform as expected.

What is wrong?

• "if the equation of motion is invariant under the gauge transformation". Under the gauge transformation, the equation of motion is covariant, not invariant, unless it's Abelian. – MadMax Jun 25 '19 at 20:17
• @Madmax Thanks for the correction. – Libertarian Monarchist Bot Jun 25 '19 at 20:55

To resolve this, we can go back to the derivation of the (classical) equation of motion from the lagrangian. The lagrangian has the form $$L\sim c_F\text{trace}(F_{\mu\nu}F^{\mu\nu}) + c_\Psi\bar\Psi\gamma_\mu A^\mu\Psi + \cdots$$ where $$c_F$$ and $$c_\Psi$$ are coefficients whose values aren't important here and where terms that don't involve the gauge field $$A$$ are represented by "$$\cdots$$". For this question, the important details are:

• $$F=dA+A\wedge A$$ with $$A=\sum_a A_a T^a$$, where $$A_a$$ is a one-form and $$T^a$$ is a generator of the gauge group, which we can think of as a square matrix.

• The trace is over the matrix incides of the gauge group generators $$T^a$$.

• $$\Psi$$ is a column matrix in the gauge-group domain. (The spinor index of $$\Psi$$ is not important here.)

To derive the equation of motion for the gauge field, we take the variation of $$L$$ with respect to the components $$A_{\mu,a}$$ of the gauge field. To answer the question, the important detail is that this variation of $$A$$ with respect to $$A_{\mu,a}$$ is $$dx^\mu T^a$$, so the variation leaves the trace intact, with a generator $$T^a$$ left behind in place of the guage field $$A$$ in each term where an $$A_{\mu,a}$$ was removed by the variation. The resulting equation of motion has the form $$\text{trace}(T^a D\star F)\propto \star \bar\Psi\gamma T^a\Psi,$$ where spacetime indices are buried in the differential-form notation.

The quantity $$D\star F$$ transforms as stated in the OP, but that quantity is inside the trace, so there is no conflict with the transformation of $$J^a$$ that follows from $$\Psi\to U(x)\Psi$$. Neither side of the equation has any free matrix indices, just one free index $$a$$ specifying a generator of the gauge group.

If desired, we could "break the trace open" by multiplying both sides of the equation by $$T^a$$ and summing over $$a$$. That gives $$D\star F$$ on the left-hand side, as in the OP, but it also "breaks open" the matrix product on the right-hand side, giving $$\bar\Psi_j\gamma\Psi_k$$ where $$j,k$$ are matrix indices as in $$T^a_{jk}$$. Now both sides of the equation transform as $$X\to UXU^{-1}$$.

• Thank you very much for your answer！ – Libertarian Monarchist Bot Jun 23 '19 at 20:20
• To look at the equation of motion in each of its component as $\text{trace}(T^a D\star F)\propto \star \bar\Psi\gamma T^a\Psi$ is not the convenient way, since the equation of motion is covariant, rather than invariant. Instead, one should look at the covariant current as $J_{\mu}(x)=\gamma_{\mu}\Psi\bar{\Psi}$, then the covariance of the EOM is pretty obvious. – MadMax Jun 25 '19 at 20:24