0
$\begingroup$

According to the semiclassical Wong equations (see page 14 of this paper), the motion of a point particle inside a gauge field theory is described by:

$$ m \frac{\text{d}p^\mu}{\text{d}\tau} = gQ^a F^{\mu\nu}_a p_\nu, \tag{1}$$

where $m$ is the rest mass, $p^\mu$ is the four-momentum, $\tau$ proper time, $g$ coupling constant, $Q^a$ componentes of the gauge charge, and $F^{\mu\nu}_a$ components of the gauge field. The gauge charges evolve according to:

$$ m \frac{\text{d}Q^a}{\text{d}\tau} = -gf^{abc} p_\mu A^\mu_a Q_c, \tag{2}$$

with $f^{abc}$ the structure constants of the gauge group, $A^\mu_a$ the components of the gauge four-potential. It is clear that for electromagnetic field equation $(1)$ is reduced to Lorentz force in tensor form:

$$ m \frac{\text{d}u^\mu}{\text{d}\tau} = q F^{\mu\nu} u_\nu, \tag{3}$$

being $q$ electric charge, and according to equation $(2)$ the electric charge "evolves" as:

$$\frac{\text{d}q}{\text{d}\tau} = 0$$

because the gauge group for electromagnetic field is the Abelian group $\text{U}(1)$ and it has $f^{abc} = 0$.

My two questions are:

  1. According to Wong equations, it seems that we need $a$ gauge charges, with $1\le a \le \dim\text{SU}(N) = N^2-1$, and it seems we need $N^2-1$ "components of gauge charge".
  2. It seems that electroweak theory based on $\text{SU}(2)$ uses two charges (electric charge, and weak hipercharge) and quantum chromodynamics based on $\text{SU}(3)$ uses three "color" charges, but what are the relations between the number of gauge charges $Q^a$ and the other charges. Is it correct to say that electroweak theory requires three gauge charges, and chromodynamics requires eight gauge charges? What are the exact definition for the two types of charges?
$\endgroup$

2 Answers 2

2
$\begingroup$

For $SU(N)$, the non-abeliean current density $j_\mu^a$ has $N^2-1$ components (ignoring the spacetime index), the same as the number of generators of the gauge group (which in turn are associated with gauge fields $A_\mu^a$. The Lagrangrian has a coupling $\sim j^\mu_a A^a_\mu$, which generalizes the $U(1)$ (QED) coupling $\sim j^\mu A_\mu$. When computing the current for a point particle, the current $j_\mu^a$ effectively gets replaced with the gauge charges in your expressions, $Q^a$.

The 3 colors associated with $SU(3)$ is the dimension of the fundamental representation $SU(3)$. This is the simplest non-trivial representation; while Nature could use more complicated representations, it has chosen not to, at least for the quarks in the Standard Model. A quark in the fundamental representation is usually represented as a 3 component vector, and the basis vectors of this 3 dimensional vector space are associated with the three colors.

Similarly, in $SU(2)$ gauge theory, the fundamental representation is a doublet, or two component vector. This is why the Higgs forms a two component doublet, or why left-handed fermions come in pairs (the left handed electron and neutrino form a doublet, and the left handed up and down quarks form a doublet).

The electric charge and weak hypercharge are scalar quantities associated with a $U(1)$ symmetry.

Finally, the gauge coupling constant $g$ is a constant in non-Abelian gauge theories that describes the strength of the interaction of the gauge and matter fields.

As you can see, when generalizing from QED to Yang-Mills, the concept of "charge" becomes more complicated. It is lifted from a single number to a vector space on which the group action is acted, and different properties of that vector space have physical importance in different contexts. The math behind this you need to learn is representation theory, particularly of Lie groups. The physical interpretation should be covered in a quantum field theory book in the part about Yang-Mills theory.

$\endgroup$
1
$\begingroup$

The gauge charge $e$ is Lie co-vector valued. For $U(1) Γ— SU(2) Γ— SU(3)$, that's 1 + 3 + 8 = 12 components in all.

The gauge charge precesses, under Wong's Equation, so that $e$, itself, is not conserved. Instead, what is conserved is $k^{-1}(e,e)$, where $k^{-1}$ is the dual of the gauge group metric $k$, which is almost never mentioned or described explicitly in the Physics literature (Weinberg being a marginal exception). So, when adapted to a Lie-algebra basis $\left(Y_a: a ∈ 𝔄 = \{1,β‹―,12\}\right)$, and the dual basis $\left(Y^a: a ∈ 𝔄\right)$, they take the component forms $e_a$, and $k^{ab}$, with the gauge group metric $k$, itself, having the component form $k_{ab}$.

The metric is assumed to be positive definite (or negative definite, different conventions for different authors) and adjoint-invariant, $k([w,u],v]) + k(u,[w,v]) = 0$, which means $f^d_{ca} k_{db} + k_{ad} f^d_{cb} = 0$ (using the Einstein summation convention here, and everywhere else below), when expressed in terms of the structure coefficients $f^c_{ab}$ that are, themselves, given by $[Y_a, Y_b] = f^c_{ab} Y_c$. If the metric is used to raise and lower indices, then defining $f_{abc} = k_{cd} f^d_{ab}$ this entails $f_{cab} + f_{cba} = 0$ which, taken in conjunction with the anti-symmetry property $f^c_{ab} = -f^c_{ba}$, already present, makes the coefficients totally anti-symmetric.

That's only possible for groups, like $U(1) Γ— SU(2) Γ— SU(3)$, if the metric is diagonal over $SU(2)$ and $SU(3)$ with these two subgroups being orthogonal with respect to the metric. The subgroup $U(1)$ can then be made orthogonal to $SU(2)$ and $SU(3)$ by adding in appropriate combinations of $SU(2)$ and $SU(3)$ - which is effectively already done by virtue of the fact that $U(1)$ isn't merely the electromagnetism $U(1)$ but also has part of $SU(2)$ mixed in with it (resulting in the $U(1)$ for weak hypercharge). The electromagnetism $U(1)$ is not orthogonal to $SU(2)$, though it is orthogonal to $SU(3)$.

The final result is that the metric reduces to the block-diagonal matrix form $k = I_1 {k_1}^2 + I_2 {k_2}^2 + I_3 {k_3}^2$, where $I_1$, $I_2$ and $I_3$ are respectively the $1Γ—1$, $3Γ—3$ and $8Γ—8$ identity matrices for $U(1)$, $SU(2)$ and $SU(3)$. The dual metric is the same, except the coefficients are each in the denominator.

So, when written in component form with index $1$ going with $U(1)$, indices $2-4$ with $SU(2)$ and $5-12$ with $SU(3)$, the square $k^{-1}(e,e)$ of the gauge charge $e$ is $$k^{ab} e_a e_b = \frac{{e_1}^2}{{k_1}^2} + \frac{{e_2}^2 + {e_3}^2 + {e_4}^2}{{k_2}^2} + \frac{{e_5}^2 + {e_6}^2 + {e_7}^2 + {e_8}^2 + {e_9}^2 + {e_{10}}^2 + {e_{11}}^2 + {e_{12}}^2}{{k_3}^2}.$$ The coefficients $k_1$, $k_2$ and $k_3$ would be in the same ratio, respectively, as $g'$, $g$ and $g_s$. (Edit: Correction: the squared coefficients $k_1^2$, $k_2^2$, $k_3^2$ would be in the same ratio as $g'$, $g$ and $g_s$, not the coefficents themselves.)

In the Standard Model spectrum, no such quadratic invariant exists that applies uniformly across the entire spectrum of quarks and leptons. This is a dead giveaway that we're missing a gauge force!

The range of possibilities is severely limited in Quantum Field Theory by the Triangle Anomaly. Only certain gauge groups and spectra may cohabit with $SU(2)$ and $SU(3)$. As it turns out, if you exclude right-handed neutrinos (and left-handed neutrinos) - or equivalently set their gauge charges to 0 - then only the weak hypercharge $U(1)$ may occur. If, on the other hand, you allow right neutrinos and left anti-neutrinos then another $U(1)$ may be included - the $U(1)$ for a force that has Baryon number minus Lepton number ($B - L$) is its quantum number.

Large matter sources, like galaxies would comprise huge unbalanced positive $B - L$ charges, that would have to be balanced and shielded out by otherwise-invisible massive clouds of negative $B - L$ charges: which is what the right neutrinos would be and is quite likely how they would interact with galaxies.

If $B - L$ is included, then there (in fact) turns out to be not just one, but two quadratic invariants you can form from the actual spectrum exhibited by the fundamental quarks and leptons. One includes a combination of squares taken from the $SU(3)$ and the $U(1)_{B-L}$ parts, and the other includes a combination of squares taken from the $SU(2)$ part and a different $U(1)$ part whose charges are hypercharge minus $B - L$, up to normalization. That $U(1)$ looks just like part of the opposite-handed analogue of $SU(2)$. Similarly, the $U(1)_{B-L}$ looks a lot like a $U(1)$ subgroup that goes into a $SU(4)$ with the $SU(3)$ part.

Bear in mind $SU(4)$ and $SO(6)$ have the same structure, as do $SU(2) Γ— SU(2)$ and $SO(4)$. Each generation of the fermion spectrum, including right neutrinos, left anti-neutrinos, counting left and right helicity separately and counting quark colors separately do, indeed, factor into the $2Β³ Γ— 2Β²$ grids characteristic of spectra, respectively, for $SO(6)$ and $SO(4)$. Correspondingly, you will find a large number of hits on $SO(6) Γ— SO(4)$ if you do any kind of search on the topic.

The argument posed here is analogous to that which Maxwell used to show that one needed a displacement current $βˆ‚πƒ/βˆ‚t$ in the Maxwell equation $βˆ‡Γ—π‡ = βˆ‚πƒ/βˆ‚t + 𝐉$ expressing the equation between the magnetic field $𝐇$ and current density $𝐉$. He argued that conservation would be broken unless this corner case were also added in. Here, too, the extra $U(1)$ for a $B-L$ force is a corner case that has to be added in, to get a charge-squared invariant that applies uniformly across the entire quark and lepton spectrum and for it to be conserved under the precession governed by Wong's equation.

To address the question of precisely what the metric $k$ is, and what the square of $e$ is, treat everything as a generalization of electromagnetism and then draw the appropriate comparisons.

The gauge group metric is what you use to contract indices $a$, $b$, etc. in the Yang-Mills Lagrangian density $$𝔏 = -ΒΌ \sqrt{|g|} g^{μρ} g^{Ξ½Οƒ} k_{ab} F^a_{ΞΌΞ½} F^b_{ρσ},$$ where I also add in the space-time metric $g_{ΞΌΞ½}$, its dual $g^{ΞΌΞ½}$ and determinant $g$. The Maxwell Lagrangian density is also of Yang-Mills form, but with $k$ now only being a $1Γ—1$ metric. So ... what is it?

Adopt Cartesian coordinates $\left(x^1, x^2, x^3\right) = (x, y, z)$, with $x^0 = t$ and a metric given by the line element $$g_{ΞΌΞ½} dx^ΞΌ dx^Ξ½ = Ξ² dt^2 - Ξ± \left(dx^2 + dy^2 + dz^2\right),$$ where $Ξ±Ξ² > 0$ and light speed is given by $c = \sqrt{Ξ²/Ξ±}$. In this frame, we may write even the gauge field components in Maxwell form as: $$A^a_ΞΌ dx^ΞΌ = 𝐀^a Β· d𝐫 - Ο†^a dt \hspace 1em β‡’ \hspace 1em Ο†^a = -A^a_t, \hspace 1em 𝐀^a = \left(A^a_x, A^a_y, A^a_z\right), $$ for the components of the potential, using $d𝐫 = (dx, dy, dz)$, and $$Β½ F^a_{ΞΌΞ½} dx^ΞΌ ∧ dx^Ξ½ = 𝐁^aΒ·d𝐒 + 𝐄^aΒ·d𝐫∧dt \hspace 1em β‡’ \hspace 1em 𝐁^a = \left(F^a_{yz}, F^a_{zx}, F^a_{xy}\right), \hspace 1em 𝐄^a = \left(F^a_{xt}, F^a_{yt}, F^a_{zt}\right),$$ for the gauge field strengths, using $d𝐒 = (dy∧dz, dz∧dx, dx∧dy)$.

The Maxwell fields $Ο†$, $𝐀$, $𝐁$ and $𝐄$ (respectively: the electric potential, magnetic potential, magnetic induction and electric force) are Lie-vector valued. So, the Lie index $a$ is in the upper position.

The metric determinant is $g = -Ξ²Ξ±^3$, so $\sqrt{|g|} = \sqrt{Ξ²Ξ±^3}$.

The dual metric is diagonal, given by $g^{tt} = 1/Ξ²$, $g^{xx} = g^{yy} = g^{zz} = -1/Ξ±$.

Thus, the Lagrangian density comes out to: $$𝔏 = Β½ k_{ab} \left(\sqrt{\frac{Ξ±}{Ξ²}} 𝐄^a·𝐄^b - \sqrt{\frac{Ξ²}{Ξ±}} 𝐁^a·𝐁^b\right) = Β½ k_{ab} \left(\frac{1}{c} 𝐄^a·𝐄^b - c 𝐁^a·𝐁^b\right).$$

The dependence on the scaling of the metric coefficients $Ξ±$ and $Ξ²$ drops out. This scale-invariance property is specific to 4-D space-times.

The field equations derived from the Lagrangian involve the densities $$π”Š_a^{ΞΌΞ½} = -\frac{βˆ‚π”}{βˆ‚F^a_{ΞΌΞ½}}, \hspace 1em 𝔍_a^ΞΌ = \frac{βˆ‚π”}{βˆ‚A^a_ΞΌ},$$ respectively for the response field $π”Š$, and the charge-current density $𝔍$, given in component form by $$ 𝐃_a = \left(π”Š_a^{tx}, π”Š_a^{ty}, π”Š_a^{tz}\right), \hspace 1em 𝐇_a = \left(π”Š_a^{yz}, π”Š_a^{zx}, π”Š_a^{xy}\right),\\ ρ_a = 𝔍_a^t, \hspace 1em 𝐉_a = \left(𝔍_a^x, 𝔍_a^y, 𝔍_a^z\right). $$

The Maxwell fields $𝐃$, $𝐇$, $𝐉$ and $ρ$ (respectively: the electric displacement, magnetic force, electric current density and electric charge density) are Lie co-vector valued. In particular, $𝐃$ and $𝐄$ cannot be equated, nor can $𝐁$ and $𝐇$ be, because they're not even the same type of objects! In each pair, one is a Lie vector and the other is a Lie co-vector. Similarly, one of each pair go as components into a tensor density, and the other go as components into a tensor.

Applied to the Yang-Mills Lagrangian density, we have the following definitions and results: $$𝐃_a = \frac{βˆ‚π”}{βˆ‚π„^a}, \hspace 1em 𝐇_a = -\frac{βˆ‚π”}{βˆ‚π^a} \hspace 1em β‡’ \hspace 1em 𝐃_a = \frac{1}{c} k_{ab} 𝐄^b, \hspace 1em 𝐇_a = c k_{ab} 𝐁^b,$$ or defining the gauge-theoretic analogue of the permittivity and permeability respectively by $$Ξ΅_{ab} = \frac{1}{c} k_{ab}, \hspace 1em ΞΌ^{ab} = \frac{1}{c} k^{ab},$$ and inverting the $𝐁-𝐇$ relation, we have: $$𝐃_a = Ξ΅_{ab} 𝐄^b, \hspace 1em 𝐁^b = ΞΌ^{ab} 𝐇_a.$$

(We won't need the definitions $𝐉_a = βˆ‚π”/βˆ‚π€^a$ or $ρ_a = -βˆ‚π”/βˆ‚Ο†^a$ for any of what follows, here.)

The gauge group metric $k$ is the analogue of $Ξ΅_0 c$, where $Ξ΅_0$ is the free space permittivity, while the dual metric $k^{-1}$ is the analogue of $ΞΌ_0 c$, where $ΞΌ_0$ is the free-space permeability.

Correspondingly, the quadratic invariant for the gauge charge may be written as $$k^{-1}(e, e) = k^{ab} e_a e_b = ΞΌ^{ab} c e_a e_b,$$ which is the analogue of $ΞΌ_0ce^2 = e^2/{Ξ΅_0c} = 2hΞ±$, where (here) $Ξ±$ denotes the fine structure coefficient and $h$ Planck's coefficient.

The charge, itself, participates in the same kind of force and power laws as in the case of electromagnetism, with the relevant equations being: $$\frac{d𝐩}{dt} = e_a \left(𝐄^a + 𝐯×𝐁^a\right), \hspace 1em \frac{dE}{dt} = e_a \left(𝐯·𝐄^a\right)$$ for a body with gauge charge $e_a$ with velocity $𝐯$, momentum $𝐩$ and energy $E$. The dynamics may also be given, equivalently by $$\frac{d\left(𝐩 + e_a 𝐀^a\right)}{dt} = -βˆ‡U, \hspace 1em \frac{d\left(E + e_a Ο†^a\right)}{dt} = \frac{βˆ‚U}{βˆ‚t}, \hspace 1em U(t, 𝐫, 𝐯) = e_a \left(Ο†^a(t, 𝐫) - 𝐯·𝐀^a(t, 𝐫)\right),$$ but, in the case of non-Abelian fields, since the field-potential relations are non-linear $$𝐁^c = βˆ‡Γ—π€^c + f^c_{ab} 𝐀^a×𝐀^b, \hspace 1em 𝐄^c = -βˆ‡Ο†^c - \frac{βˆ‚π€^c}{βˆ‚t} + f^c_{ab} \left(Ο†^a 𝐀^b - 𝐀^a Ο†^b\right),$$ this can only be done consistently, if the charge also has a time-variability, as given by Wong's Equation: $$\frac{de_a}{dt} = f^c_{ab} \left(Ο†^b - 𝐯·𝐀^b\right) e_c.$$ The quadratic invariant $k^{ab} e_a e_b$ is preserved when $k$ is adjoint-invariant.

The precession may only show up between charge components in the same block diagonal. So, only $SU(3)$ precesses with itself (leading to color oscillation), and the $SU(2)$ part precesses with itself (leading to changing back and/or forth between electrons/tauons/muons and neutrinos; or between up/charm/top and down/strange/bottom quarks).

$\endgroup$

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.