Classical chromodynamics force density vector

Question

In classical electrodynamics there is a current and potential four-vector. The covariant force density in a charge continuum according to wikipedia is

$f_\alpha=F_{\alpha\beta}J^\beta$

But I'm not sure how to generalize it to classical chromodynamics since there are eight current and potential four-vectors. Would there be eight force vectors? Or can you define it as a single vector, the derivative of the stress energy tensor which wikipedia gives as

$f^\alpha=-\partial_\beta T^{\alpha\beta}$

I am trying to find some information on the internet but everything I find seems too technical. I would like a novice explanation for someone new to field theory on how force works in classical chromodynamics.

link to wikipedia article: https://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

Answer 1

Electrodynamics as well as chromodynamics are two Yang-Mills theories. In the framework of Yang-Mills theories you can build up all the equations of classical electrodynamics and in the same manner the one for classical chromodynamics. I'm afraid that a "novice explanation for someone new to field theory" won't be easy. Yang-Mills theories are quite technical, but i'll try to just to give you a flavour.

You can construct the field strength tensor in the most general way as $$[D_\mu, D_\nu] = -igT^AF_{\mu\nu}^A$$ where $T^A$ are the generators of the underlying symmetry group and $[D_\mu, D_\nu]$ is the commutator of the covariant derivatives. By expanding the commutator you find $$F_{\mu\nu}^A = \partial_\mu A_\nu^A-\partial_\nu A_\mu^A+gf^{ABC}A_\mu^BA_\nu^C$$ You can easily see that for classical electrodynamics, the symmetry group is $U(1)$ and so the structure constants are all zero $f^{ABC} = 0$ being an abelian group, and the group is one dimensional, so that you have only one value for the internal index $A$ and so it can be dropped.

From this we can construct the lagrangian for the theory as $$\mathcal{L} = -\frac{1}{4} F_{\mu\nu}^A F^{\mu\nu}_A = -\frac{1}{4}\text{Tr}\left(F^{\mu\nu}F_{\mu\nu}\right)$$ where the trace is over the internal indices.

Using this lagrangian one can easily build the equation of motion by variating the action associated with that lagrangian. What one finds is the following equation of motion $$m\ddot{x}^\mu = \text{Tr}(QF_{\mu\nu})\dot{x}^\nu \tag{1}$$ where, agin, the trace is over internal indices and $Q^A$ are the conserved charges of the theory which will be $\dim{G}$, the dimension of the symmetry group.

Equation $(1)$ helps us define the force you're searching, which is called Wong force $$f^\mu = Q^AF^{\mu\nu}_A \dot{x}_\nu$$ You can kind of see the same pattern as in classical electrodynamics since we can think as $Q^A\dot{x}_\nu = J_\nu^A$ the current density. In this manner $$f^\mu = F^{\mu\nu}_AJ^A_\nu$$

You can find something on Wong's equation in this paper or in a chapter in the book

-Boris Kosyakov, Introduction to the Classical Theory of Particles and Fields