How to Implement Current Conservation for SU(N) in the Adjoint Representation?

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The discussion presents a solution to implementing current conservation for SU(N) in the adjoint representation, starting with the equation D^{\mu}F_{\mu\nu} = - j_{\nu}. By differentiating this covariantly and anti-symmetrizing, it leads to the relation D^{\nu}j_{\nu} = \frac{1}{2}[D^{\mu}, D^{\nu}]F_{\mu\nu}. The covariant derivative in the adjoint representation is defined as D^{\mu}M = \partial^{\mu}M + [A^{\mu}, M], which allows for the derivation of [D^{\mu}, D^{\nu}]M = [F^{\mu\nu}, M]. Substituting M = F_{\mu\nu} into the previous equation results in D^{\nu}j_{\nu} = 0, confirming current conservation in this context. The implementation is effectively demonstrated in the adjoint representation.
Tian
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Homework Statement
In the "An introduction to Quantum Field Thoery" of Peskin and Schroeder, the equation(15.51) of the chapter 15.3 gives the classical equation of motion, so from this equation to derive the current conservation.
Relevant Equations
the classical equation of motion for SU(N), please see my picture
Here is my solution
2C869620FEBDBAA1F955AC83ADAF6638.png
 
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D^{\mu}F_{\mu\nu} = - j_{\nu}, Differentiate this covariantly and anti-symmetrized to obtain \frac{1}{2}[D^{\mu}, D^{\nu}]F_{\mu\nu} = D^{\nu}j_{\nu}. \ \ \ \ (1) Now, from the definition of the covariant derivative in the adjoint representation (acting on any matrix-valued field) D^{\mu}M \equiv \partial^{\mu}M + [A^{\mu} , M], you can show that [D^{\mu} , D^{\nu}]M = [F^{\mu\nu} ,M] Thus, for M = F_{\mu\nu}, eq(1) becomes D^{\nu}j_{\nu} = \frac{1}{2}[F^{\mu\nu} , F_{\mu\nu}] = 0.
 
Thank you veery much . It should be done in the adjoint repesentation.
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...

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