Talk:Yang–Mills theory

Canonic versus dynamic Yang-Mills field and continuity equations
I have just finished drafting a new section for this article, below. I will leave this in talk for a few days before posting, for review and comment.

In Yang-Mills gauge theory, the field equations which generalize Maxwell’s equations for electrodynamic may be cast in one of two interrelated forms: canonic, and dynamic.

Canonic field equation
In canonic form, one starts with the two spacetime-covariant equations $$c{{\mu }_{0}}{{j}^{\nu }}={{\partial }_{\mu }}{{F}^{\mu \nu }}$$ and $$c{{\mu }_{0}}{{p}^{\sigma \mu \nu }}={{\partial }^{\sigma }}{{F}^{\mu \nu }}+{{\partial }^{\mu }}{{F}^{\nu \sigma }}+{{\partial }^{\nu }}{{F}^{\sigma \mu }}=0$$ reviewed in the last section, uses the non-abelian field strength $${{F}_{\text{YM}}}^{\mu \nu }={{D}^{\mu }}{{G}^{\nu }}-{{D}^{\nu }}{{G}^{\mu }}={{\partial }^{\mu }}{{G}^{\nu }}-{{\partial }^{\nu }}{{G}^{\mu }}-ig\left[ {{G}^{\mu }},{{G}^{\nu }} \right]$$ with non-commuting $$\left[ {{G}^{\mu }},{{G}^{\nu }} \right]\ne 0$$ rather than $$\left[ {{A}^{\mu }},{{A}^{\nu }} \right]=0$$ as also just reviewed, and in addition, advances all of the remaining ordinary derivatives to gauge-covariant derivatives, $${{\partial }^{\mu }}\to {{D}^{\mu }}={{\partial }^{\mu }}-ig{{G}^{\mu }}/\hbar c$$. It is also helpful to use the uppercase notation $${{j}^{\nu }}\to {{J}^{\nu }}$$ and $${{p}^{\sigma \mu \nu }}\to {{P}^{\sigma \mu \nu }}$$ to denote for the electric and magnetic charge densities in the canonic equations, retaining the lowercase notation for the dynamic form to be reviewed below. With the foregoing, the Yang-Mills extension of Maxwell’s “electric” and “magnetic” equations, in canonic form are as follows:


 * $$c{{\mu }_{0}}{{J}^{\nu }}={{D}_{\mu }}{{F}_{\text{YM}}}^{\mu \nu }=\left( {{g}^{\mu \nu }}{{D}_{\sigma }}{{D}^{\sigma }}-{{D}^{\mu }}{{D}^{\nu }} \right){{G}_{\mu }}$$


 * $$c{{\mu }_{0}}{{P}^{\sigma \mu \nu }}={{D}^{\sigma }}{{F}_{\text{YM}}}^{\mu \nu }+{{D}^{\mu }}{{F}_{\text{YM}}}^{\nu \sigma }+{{D}^{\nu }}{{F}_{\text{YM}}}^{\sigma \mu }=0$$

The Yang-Mills canonic magnetic charge density, although generalized above, remains equal to zero just like the magnetic charge density in Maxwell’s electrodynamics. This is no longer because of the commutator $$\left[ {{\partial }_{\mu }},{{\partial }_{\nu }} \right]=0$$, but rather because of the Jacobi identity $$\left[ {{D}_{\sigma }},\left[ {{D}_{\mu }},{{D}_{\nu }} \right] \right]+\left[ {{D}_{\mu }},\left[ {{D}_{\nu }},{{D}_{\sigma }} \right] \right]+\left[ {{D}_{\nu }},\left[ {{D}_{\sigma }},{{D}_{\mu }} \right] \right]=0$$ combined with the further identity $$\left[ {{D}_{\sigma }},{{F}_{\text{YM}}}_{\mu \nu } \right]={{D}_{\sigma }}{{F}_{\text{YM}}}_{\mu \nu }$$, both reviewed in the Mathematical overview.

Canonic continuity equation
Applying $${{D}_{\nu }}$$ to the above charge density, the identity $${{D}_{\nu }}{{D}_{\mu }}{{F}_{\text{YM}}}^{\mu \nu }=0$$ reviewed in the Mathematical overview enables us to the calculate Yang-Mills canonic continuity relation:


 * $$\begin{align}

& c{{\mu }_{0}}{{D}_{\nu }}{{J}^{\nu }}={{D}_{\nu }}{{D}_{\mu }}{{F}_{\text{YM}}}^{\mu \nu }=\left( {{g}^{\mu \nu }}{{D}_{\nu }}{{D}_{\sigma }}{{D}^{\sigma }}-{{D}_{\nu }}{{D}^{\mu }}{{D}^{\nu }} \right){{G}_{\mu }} \\ & ={{\partial }_{\nu }}{{\partial }_{\mu }}{{F}_{\text{YM}}}^{\mu \nu }-\left( ig\left( {{G}_{\nu }}{{\partial }_{\mu }}+{{\partial }_{\nu }}{{G}_{\mu }} \right)+{{g}^{2}}{{G}_{\nu }}{{G}_{\mu }} \right){{F}_{\text{YM}}}^{\mu \nu }=\left( {{\partial }_{\nu }}{{\partial }_{\mu }}-{{V}_{\nu \mu }} \right){{F}_{\text{YM}}}^{\mu \nu }=0 \\ \end{align}$$ ,

which includes a perturbation tensor defined by:


 * $${{V}_{\mu \nu }}\equiv ig\left( {{G}_{\mu }}{{\partial }_{\nu }}+{{\partial }_{\mu }}{{G}_{\nu }} \right)+{{g}^{2}}{{G}_{\mu }}{{G}_{\nu }}$$.

The trace $$V={{V}^{\sigma }}_{\sigma }=ig\left( {{G}^{\sigma }}{{\partial }_{\sigma }}+{{\partial }^{\sigma }}{{G}_{\sigma }} \right)+{{g}^{2}}{{G}^{\sigma }}{{G}_{\sigma }}$$ of the above is the standard expression for the perturbation in the Klein–Gordon (relativistic Schrödinger) equation.

Dynamic field equation
In dynamic form, one still begins with Maxwell’s electrodynamic equations $$c{{\mu }_{0}}{{j}^{\nu }}={{\partial }_{\mu }}{{F}^{\mu \nu }}$$ and $$c{{\mu }_{0}}{{p}^{\sigma \mu \nu }}={{\partial }^{\sigma }}{{F}^{\mu \nu }}+{{\partial }^{\mu }}{{F}^{\nu \sigma }}+{{\partial }^{\nu }}{{F}^{\sigma \mu }}=0$$, and still uses the non-commuting field strength $${{F}_{\text{YM}}}^{\mu \nu }={{D}^{\mu }}{{G}^{\nu }}-{{D}^{\nu }}{{G}^{\mu }}={{\partial }^{\mu }}{{G}^{\nu }}-{{\partial }^{\nu }}{{G}^{\mu }}-ig\left[ {{G}^{\mu }},{{G}^{\nu }} \right]$$ with $$\left[ {{G}^{\mu }},{{G}^{\nu }} \right]\ne 0$$, but does nothing further. That is, one keeps the remaining derivatives ordinary and keeps the source density notation in lowercase. Consequently, in dynamic form, the Yang-Mills generalization of Maxwell's electric charge equation (Gauss and Ampere) is:

while the dynamic Yang-Mills generalization of Maxwell's magnetic charge equation (Gauss and Faraday) is:

These are simply Maxwell equations without change, aside from the promotion of $${{F}^{\mu \nu }}={{\partial }^{\mu }}{{A}^{\nu }}-{{\partial }^{\nu }}{{A}^{\mu }}$$ with $$\left[ {{A}^{\mu }},{{A}^{\nu }} \right]=0$$, to $${{F}_{\text{YM}}}^{\mu \nu }={{D}^{\mu }}{{G}^{\nu }}-{{D}^{\nu }}{{G}^{\mu }}={{\partial }^{\mu }}{{G}^{\nu }}-{{\partial }^{\nu }}{{G}^{\mu }}-ig\left[ {{G}^{\mu }},{{G}^{\nu }} \right]$$ with $$\left[ {{G}^{\mu }},{{G}^{\nu }} \right]\ne 0$$. That is, these are Maxwell’s equations for non-commuting gauge fields, with nothing else changed. However, the identity which causes the uppercase-denoted magnetic source density to vanish in the canonic equation, $${{P}^{\sigma \mu \nu }}=0$$, do not operate to vanish the lowercase-denoted magnetic source density to vanish from the dynamic equation, $${{p}^{\sigma \mu \nu }}\ne 0$$. Instead, using the “zero” from the canonic magnetic charge equation, we are able to calculate in the above that the dynamic $${{p}^{\sigma \mu \nu }}\ne 0$$ differs from zero by the index-cyclic derivatives $${{\partial }^{\sigma }}\left[ {{G}^{\mu }},{{G}^{\nu }} \right]+{{\partial }^{\mu }}\left[ {{G}^{\nu }},{{G}^{\sigma }} \right]+{{\partial }^{\nu }}\left[ {{G}^{\sigma }},{{G}^{\mu }} \right]$$ of the non-zero Yang-Mills gauge field commutator. PatentPhysicist (talk) 20:28, 28 February 2021 (UTC)

Dynamic continuity equation
To obtain the dynamic Yang-Mills continuity equation, we apply the ordinary derivative to the dynamic Yang-Mills electric charge equation above. Using the zero from the above canonic continuity equation, it is straightforward to find that:

This dynamic continuity equation is not equal to zero. Rather, this differs from zero by the double-contracted product $${{V}_{\nu \mu }}{{F}_{\text{YM}}}^{\mu \nu }$$ of the perturbation tensor with the Yang-Mills field strength tensor. Conversely, this continuity relation is equal to zero, only when the perturbation tensor $${{V}_{\nu \mu }}=0$$.

I have just added the new section above, on Canonic versus dynamic Yang-Mills field and continuity equations, to the main article. PatentPhysicist (talk) 20:10, 2 March 2021 (UTC)

I have just added a new section to this article. Feedback welcome. PatentPhysicist (talk) 16:00, 6 March 2021 (UTC)

Section deletion
Most seriously, confusion between gauge choice and renormalizability: "Of course, adding a mass by hand destroys renormalizability, so it is necessary to find a way that this can be restored. The Higgs mechanism used for the Electroweak interaction is best-known example of how to obtain a non-zero vector boson mass without sacrificing renormalizability." Adding a mass by hand to a theory renders the theory no longer gauge invariant, which is not an issue of renormalizability.

Yang-Mills theory is about massless vector bosons, so the entire section regarding Proca mass is irrelevant. The equation $$\partial_\sigma A^\sigma=0$$ is also given by the Proca equation only when there are no external currents, as seen in the fourth equation from this page (Proca action), which can be rearranged into $$\partial^\mu\partial^\nu A_\nu=(\partial_\nu\partial^\nu+m^2)A^\mu-j^\mu\neq0$$ in general, after setting several physical constants to 1.

The mistakes along with the lack of relevant citations is why I have deleted the section.--Vampyricon (talk) 04:28, 7 March 2021 (UTC)

Follow up to section deletion
I appreciate the clarification regarding the gauge choice versus renormalizability. I found something interesting on this point at https://physics.stackexchange.com/questions/394740/massive-vector-boson-propagators-and-renormalizability.

As to the Proca mass: Start with the fourth equation from this page, which you reference: Proca action. (As it happens, I am the one who added this external source equation several days ago.)
 * $$c{{\mu }_{0}}{{j}^{\nu }}=\left( {{g}^{\mu \nu }}\left( {{\partial }_{\sigma }}{{\partial }^{\sigma }}+{{m}^{2}}{{c}^{2}}/{{\hbar }^{2}} \right)-{{\partial }^{\nu }}{{\partial }^{\mu }} \right){{B}_{\mu }}.\quad (1)$$

Also, contrast the same equation without a Proca mass:


 * $$c{{\mu }_{0}}{{j}^{\nu }}=\left( {{g}^{\mu \nu }}{{\partial }_{\sigma }}{{\partial }^{\sigma }}-{{\partial }^{\nu }}{{\partial }^{\mu }} \right){{B}_{\mu }}.\quad (2)$$

If you operate with $${{\partial }_{\nu }}$$ on each side of (2), and assume flat spacetime where $$\left[ {{\partial }_{\mu }},{{\partial }_{\nu }} \right]=0$$, then you obtain:


 * $$c{{\mu }_{0}}{{\partial }_{\nu }}{{j}^{\nu }}=\left( {{\partial }^{\mu }}{{\partial }_{\sigma }}{{\partial }^{\sigma }}-{{\partial }_{\sigma }}{{\partial }^{\sigma }}{{\partial }^{\mu }} \right){{B}_{\mu }}=0,\quad (3)$$

and accordingly, the current is conserved, $${{\partial }_{\nu }}{{j}^{\nu }}=0$$, by mathematical identity.

Now go back to (1) with $$m>0$$ and operate with $${{\partial }_{\nu }}$$. Here, we obtain:


 * $$c{{\mu }_{0}}{{\partial }_{\nu }}{{j}^{\nu }}={{\partial }^{\mu }}{{\partial }_{\sigma }}{{\partial }^{\sigma }}{{B}_{\mu }}-{{\partial }^{\mu }}{{\partial }^{\nu }}{{\partial }^{\mu }}{{B}_{\mu }}+{{m}^{2}}{{c}^{2}}/{{\hbar }^{2}}{{\partial }^{\mu }}{{B}_{\mu }}=0+{{m}^{2}}{{c}^{2}}/{{\hbar }^{2}}{{\partial }^{\mu }}{{B}_{\mu }}.\quad (4)$$

That is, $${{\partial }_{\nu }}{{j}^{\nu }}={{m}^{2}}{{\partial }^{\mu }}{{B}_{\mu }}$$ with the physical constants set to 1. Beyond the identity that produces the 0 in (3), if $${{\partial }_{\nu }}{{j}^{\nu }}=0$$ is still independently required as a conservation condition when there are external currents – as I believe it is, but please correct me if you disagree – then $${{\partial }^{\mu }}{{B}_{\mu }}=0$$ still holds for $$m>0$$ and an external current. But, if I am wrong about this being required, is there anything that would prevent us from imposing $${{\partial }^{\mu }}{{B}_{\mu }}=0$$ anyway, as a gauge condition, to remove one of the four degrees of freedom from $${{B}_{\mu }}$$ to give it the required one longitudinal plus two transverse degrees of freedom of a massive vector boson?

Also, I do not agree that “Yang-Mills theory is about massless vector bosons, so the entire section regarding Proca mass is irrelevant.” Actually, Yang-Mills gauge theory starts with massless vector bosons, but a major unsolved question is why – for example in $$SU(N)=SU(3)$$ QCD with $${{N}^{2}}-1={{3}^{2}}-1=8$$  massless gluons in the adjoint representation – one also comes upon massive vector bosons / mesons which mediate those interactions, physically. Those massive mesons may be described using propagators with a Proca mass, just like any other massive vector boson. But https://www.claymath.org/sites/default/files/yangmills.pdf is about understanding how this comes about from a theory with massless gluons which should supposedly have infinite range and be free particles, but are in fact confined: “for QCD to describe the strong force successfully...[i]t must have a ‘mass gap,’ [which] is necessary to explain why the nuclear force is strong but shortranged.” And "[i]t must have 'quark [and gluon] confinement'...to explain why we never see individual quarks" and gluons. So Proca mass is certainly not irrelevant. I am happy to discuss this also. PatentPhysicist (talk) 18:18, 7 March 2021 (UTC)

Respectfully suggested improvements for Yang–Mills_theory
The Mathematical overview for Yang–Mills_theory is not wrong in any way I can see, but I respectfully suggest the following improvements:

1) This whole subsection does not contain a single external source citation.  It would be helpful to have one or more citations to support the various formulas in this subsection.  One good source which has been my bible since 1984 is, although not all of the Mathematical overview formulas are in there.  If others are aware of some good references to establish the mathematics in this subsection, they should be added.

Now as to some specifics regarding formula discussions which I believe need to be improved:

2) Although true and correct, the relation $$F_{\mu \nu}^a = \partial_\mu A_\nu^a-\partial_\nu A_\mu^a+gf^{abc}A_\mu^bA_\nu^c $$ and the commutator $$[D_\mu, D_\nu] = -igT^aF_{\mu\nu}^a$$ are merely stated, with no more.  These should be better developed, and their derivation explicitly shown.  Indented below is in the nature of what I respectfully suggest:
 * The commutator
 * $$\hbar c\left[ {{D}^{\mu }},{{D}^{\nu }} \right]=-ig{{F}^{\mu \nu }}=-ig{{T}^{a}}{{F}^{a}}^{\mu \nu }$$
 * can be derived as follows: With $$\hbar =c=1$$ and using $$\left[ {{\partial }^{\mu }},{{\partial }^{\nu }} \right]=0,$$ apply the commutator $$\left[ {{D}^{\mu }},{{D}^{\nu }} \right]$$ to operate on any field $$\phi \left( t,\mathbf{x} \right).$$  Attentive to the product rule, one may obtain:
 * $$\begin{align}

& \left[ {{D}^{\mu }},{{D}^{\nu }} \right]\phi =\left[ \left( {{\partial }^{\mu }}-ig{{A}^{\mu }} \right),\left( {{\partial }^{\nu }}-ig{{A}^{\nu }} \right) \right]\phi \\ & \quad \quad \quad \quad \ \ =-ig{{A}^{[\mu }}{{\partial }^{\nu ]}}\phi -ig{{\partial }^{[\mu }}\left( {{A}^{\nu ]}}\phi \right)-{{g}^{2}}\left[ {{A}^{\mu }},{{A}^{\nu }} \right]\phi  \\ & \quad \quad \quad \quad \ \ =-ig{{\partial }^{[\mu }}{{A}^{\nu ]}}\phi -{{g}^{2}}\left[ {{A}^{\mu }},{{A}^{\nu }} \right]\phi =-ig{{F}^{\mu \nu }}\phi \\ \end{align}.$$
 * Given that $$\left[ {{A}^{\mu }},{{A}^{\nu }} \right]=\left[ {{T }_{i}},{{T }_{j}} \right]{{A}_{i}}^{\mu }{{A}_{j}}^{\nu }=i{{f}_{ijk}}{{T }_{k}}{{A}_{i}}^{\mu }{{A}_{j}}^{\nu }\ne 0,$$ this includes the relation
 * $${{F}^{\mu \nu }}={{T }_{c}}{{F}_{c}}^{\mu \nu }={{\partial }^{[\mu }}{{A}^{\nu ]}}-ig\left[ {{A}^{\mu }},{{A}^{\nu }} \right]={{T }_{c}}\left\{ {{\partial }^{\mu }}A_{c}^{\nu }-{{\partial }^{\nu }}A_{c}^{\mu }+g{{f}_{abc}}{{A}_{a}}^{\mu }{{A}_{b}}^{\nu } \right\}$$
 * which is further seen to contain:
 * $${{F}_{a}}^{\mu \nu }={{\partial }^{\mu }}A_{a}^{\nu }-{{\partial }^{\nu }}A_{a}^{\mu }+g{{f}_{abc}}{{A}_{b}}^{\mu }{{A}_{c}}^{\nu }.$$

3) The following existing material is important, but needs improvement:

A Bianchi identity holds


 * $$(D_\mu F_{\nu \kappa})^a+(D_\kappa F_{\mu \nu})^a+(D_\nu F_{\kappa \mu})^a=0$$

which is equivalent to the Jacobi identity


 * $$[D_{\mu}, [D_{\nu},D_{\kappa}]]+[D_{\kappa},[D_{\mu},D_{\nu}]]+[D_{\nu},[D_{\kappa},D_{\mu}]]=0$$

since $$[D_{\mu},F^a_{\nu\kappa}]=D_{\mu}F^a_{\nu\kappa}$$.

Specifically, derivation of the identity $$[D_{\mu},F^a_{\nu\kappa}]=D_{\mu}F^a_{\nu\kappa}$$ which underlies other important identities (including the above magnetic monopole / Bianchi identity) should be explicitly shown. It would also be helpful to connect together $$g{{D}^{\sigma }}{{F}^{\mu \nu }}=g\left[ {{D}^{\sigma }},{{F}^{\mu \nu }} \right]=i\hbar c\left[ {{D}^{\sigma }},\left[ {{D}^{\mu }},{{D}^{\nu }} \right] \right].$$

4) It should also be pointed out that, as it is for the Electromagnetic tensor in Maxwell’s electrodynamics, the field strength trace is zero:
 * $$F={{F}^{\mu }}_{\mu }={{\partial }^{\mu }}{{A}_{\mu }}-{{\partial }_{\mu }}{{A}^{\mu }}-ig\left[ {{A}^{\mu }},{{A}_{\mu }} \right]=i\hbar c\left[ {{D}^{\mu }},{{D}_{\mu }} \right]/g=0.$$

5) Finally, continuity equations underlie charge conservation, and so are very important to include.  Specifically, just as $${{\partial }_{\nu }}{{\partial }_{\mu }}{{F}^{\mu \nu }}=0$$ is an identity central to charge conservation in Maxwell's equations, so too, $${{D}_{\nu }}{{D}_{\mu }}{{F}^{\mu \nu }}=0$$ is an identity central to charge conservation in any Yang-Mills gauge theory.  Accordingly, I respectfully suggest adding material to show the identity $${{D}_{\nu }}{{D}_{\mu }}{{F}^{\mu \nu }}=0,$$ along the lines of what is indented below:


 * We may obtain a Yang-Mills continuity identity as follows: Start with the relations $$i\hbar c\left[ {{D}^{\mu }},{{D}^{\nu }} \right]/g={{F}^{\mu \nu }}$$ and $${{D}^{\sigma }}{{F}^{\mu \nu }}=\left[ {{D}^{\sigma }},{{F}^{\mu \nu }} \right] $$, then write:


 * $$\begin{align}

& {{D}_{\nu }}{{D}_{\mu }}{{F}^{\mu \nu }}={{D}_{\nu }}\left[ {{D}_{\mu }},{{F}^{\mu \nu }} \right] \\ & =i\hbar c{{D}_{\nu }}\left[ {{D}_{\mu }},\left[ {{D}^{\mu }},{{D}^{\nu }} \right] \right]/g=i\hbar c\left[ {{D}_{\nu }}{{D}_{\mu }}\left[ {{D}^{\mu }},{{D}^{\nu }} \right]-{{D}_{\nu }}\left[ {{D}^{\mu }},{{D}^{\nu }} \right]{{D}_{\mu }} \right]/g \\ & ={{D}_{\nu }}{{D}_{\mu }}{{F}^{\mu \nu }}-{{D}_{\nu }}{{F}^{\mu \nu }}{{D}_{\mu }} \\ \end{align},$$


 * from which we deduce that $${{D}_{\nu }}{{F}^{\mu \nu }}{{D}_{\mu }}=0$$ and thus $${{D}_{\mu }}{{F}^{\mu \nu }}{{D}_{\nu }}=0.$$ The Jacobian identity $$\left[ {{D}_{\nu }},\left[ {{D}_{\mu }},\left[ {{D}^{\mu }},{{D}^{\nu }} \right] \right] \right]=0$$ combined with the foregoing relations further implies that $$\left[ {{D}_{\nu }},\left[ {{D}_{\mu }},{{F}^{\mu \nu }} \right] \right]=\left[ {{D}_{\nu }},{{D}_{\mu }}{{F}^{\mu \nu }} \right]={{D}_{\nu }}{{D}_{\mu }}{{F}^{\mu \nu }}-{{D}_{\mu }}{{F}^{\mu \nu }}{{D}_{\nu }}=0$$.  Further combining with $${{D}_{\mu }}{{F}^{\mu \nu }}{{D}_{\nu }}=0$$, we finally deduce that:


 * $${{D}_{\nu }}{{D}_{\mu }}{{F}^{\mu \nu }}=0.$$


 * This is the Yang-Mills generalization of the continuity identity $${{\partial }_{\nu }}{{\partial }_{\mu }}{{F}^{\mu \nu }}=0$$ from electrodynamics.  PatentPhysicist (talk) 17:57, 13 March 2021 (UTC)


 * I posted the above edit suggestions on March 13. There has been no talk in response one way or the other.  I will keep this here on the talk page through the weekend.  If there are no objections and nobody finds any errors in the above, I will make changes to the main page in accordance with what I have laid out above, early next week.  PatentPhysicist (talk) 14:54, 18 March 2021 (UTC)


 * This part was removed and for good reasons. Please, avoid to put such a stuff in the article. Thanks.--Pra1998 (talk) 17:47, 18 March 2021 (UTC)

Unsolved problem in physics
The box with the "Unsolved problem in physics" has no references. Nothing in List of unsolved problems in physics matches the box content.

The one item in the List article about Yang Mills had completely different content from the box. Johnjbarton (talk) 00:32, 9 April 2024 (UTC)