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Unit-charge SU(2) Prasad–Sommerfield monopole profile

Physical scope

One explicit smooth positive-unit-flux field configuration in the fixed classical SU(2) adjoint-Higgs Prasad–Sommerfield model, verifying the first-order Bogomolny equation, finite energy, quantized surface charge, and energy saturation. It asserts no quantum state or empirical monopole.

Example

Assume ${A C}_{ω}$ and use natural units on oriented Euclidean $R^{3}$ with $ϵ_{123} = 1$ , the zero-potential static model with one real adjoint Higgs field, no other dynamical matter, and global group $S U (2)$ (not $S O (3)$ ). Let $A_{i}, Φ \in C^{\infty} (R^{3}, s u (2))$ have dimensions $[A_{i}] = [Φ] = {l e n g t h}^{- 1}$ . Let $T_{a} = - i σ_{a} / 2$ , $κ (X, Y) = - 2 {Tr}_{2} (X Y)$ with $κ (T_{a}, T_{b}) = δ_{a b}$ , $g > 0$ be dimensionless, and $v > 0$ have dimension ${l e n g t h}^{- 1}$ . The fields use $F_{i j} = \partial_{i} A_{j} - \partial_{j} A_{i} + [A_{i}, A_{j}]$ , $D_{i} Φ = \partial_{i} Φ + [A_{i}, Φ]$ , $B_{i} = \frac{1}{2} ϵ_{i j k} F_{j k}$ , and $E = (2 g^{2})^{- 1} \int \sum_{i} (κ (B_{i}, B_{i}) + κ (D_{i} Φ, D_{i} Φ)) d^{3} x$ . The allowed external electric probes include the fundamental doublet. For $x \in R^{3}$ with $r = ∣ x ∣ > 0$ , $n_{i} = x_{i} / r$ , and $u = v r$ , define

$H (u) = \coth u - \frac{1}{u}, K (u) = \frac{u}{\sinh u}, Φ^{a} (x) = - v H (u) n_{a}, A_{i}^{a} (x) = ϵ_{a i j} n_{j} \frac{1 - K (u)}{r} .$

Use the removable analytic limits $H (0) = 0$ and $K (0) = 1$ . At infinity $Φ \to - v {\hat{x}}^{a} T_{a}$ , the unbroken algebra is its pointwise stabilizer, and the equatorial transition $h (φ) = \exp (θ (φ) T_{3})$ has integer $ν = - (θ (2 π) - θ (0)) / (4 π)$ in the fundamental-doublet-compatible SU(2) convention. In these conventions the fields extend smoothly to $r = 0$ , satisfy the positive-charge Bogomolny equation $B_{i} = D_{i} Φ$ , have $m = 4 π v$ and $ν = 1$ , and have finite energy $E = 4 π v / g^{2}$ . They give an explicit smooth unit-flux solution of the first-order classical Bogomolny equation that saturates the classical energy bound. No quantum state, supersymmetry representation, or empirical observation is asserted.

Example references

David Tong, §2.1.3, equation (2.19), unit-charge SU(2) solution

Jeffrey A. Harvey, §§2.1–2.2, radial Bogomolny equation and Prasad–Sommerfield profiles

Static SU(2) adjoint-Higgs model

Normalized invariant form and SU(2) bracket

SU(2) surface-flux quantization

Classical monopole energy identity and bound

Facts & Assumptions

[A1]

${A C}_{ω}$ is the Axiom of Countable Choice, inherited through the model and surface-Stokes inputs. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

The fields are interpreted in the static zero-potential SU(2) adjoint-Higgs model with $F_{i j} = \partial_{i} A_{j} - \partial_{j} A_{i} + [A_{i}, A_{j}]$ , $D_{i} Φ = \partial_{i} Φ + [A_{i}, Φ]$ , $B_{i} = \frac{1}{2} ϵ_{i j k} F_{j k}$ , and energy coefficient $1 / (2 g^{2})$ . Static SU(2) adjoint-Higgs Prasad–Sommerfield monopole sector

[F2]

The basis has $[T_{a}, T_{b}] = ϵ_{a b c} T_{c}$ and the positive invariant form has $κ (T_{a}, T_{b}) = δ_{a b}$ . Compactness and normalized invariant form for SU(2)

[F3]

Flux quantization in this model gives $m = 4 π v ν$ and $T = m / g^{2}$ . Flux quantization in the specified SU(2) adjoint-Higgs sector

[F4]

The two square identities imply that a finite-energy positive-charge configuration with $B_{i} = D_{i} Φ$ has $E = T$ . Classical SU(2) monopole energy bound and Bogomolny equations

Verification

Given: [A1], [F1]–[F4], $g > 0$ , $v > 0$ , $u = v r$ , and the radial functions and fields in the Example.

1.1givenalgebra

The Taylor expansions $\sinh u = u + u^{3} / 6 + O (u^{5})$ and $\cosh u = 1 + u^{2} / 2 + O (u^{4})$ give $H (u) = u / 3 + O (u^{3})$ and $K (u) = 1 - u^{2} / 6 + O (u^{4})$ . Hence $H (u) / u$ and $(1 - K (u)) / u^{2}$ are analytic in $u^{2}$ near zero; writing $n_{a} = x_{a} / r$ shows $Φ^{a} = - v^{2} x_{a} / 3 + O (r^{2} x_{a})$ and $A_{i}^{a} = (v^{2} / 6 + O (r^{2})) ϵ_{a i j} x_{j}$ . Thus both fields extend smoothly through the origin.

1.2givenalgebra

For $u > 0$ , differentiation gives $H^{'} (u) = 1 / u^{2} - 1 / \sinh^{2} u = (1 - K (u)^{2}) / u^{2}$ and $K^{'} (u) = K (u) (1 / u - \coth u) = - K (u) H (u)$ . Also $\partial_{i} n_{a} = (δ_{i a} - n_{i} n_{a}) / r$ .

2.1F1F2step 1.2algebra

Substitution into $D_{i} Φ = \partial_{i} Φ + [A_{i}, Φ]$ gives the ordinary derivative $- v^{2} H^{'} (u) n_{i} n_{a} - v H (u) (δ_{i a} - n_{i} n_{a}) / r$ . The commutator term is $v H (u) (1 - K (u)) (δ_{i a} - n_{i} n_{a}) / r$ , using $ϵ_{a b c} ϵ_{b i j} n_{j} n_{c} = n_{i} n_{a} - δ_{i a}$ . Thus $D_{i} Φ^{a} = - v^{2} H^{'} (u) n_{i} n_{a} - v H (u) K (u) (δ_{i a} - n_{i} n_{a}) / r = - (1 - K (u)^{2}) n_{i} n_{a} / r^{2} + v K^{'} (u) (δ_{i a} - n_{i} n_{a}) / r$ , by step 1.2 and $u = v r$ .

3.1F1F2step 1.2step 2.1algebra

Write $A_{i}^{a} = f (r) ϵ_{a i j} x_{j}$ with $f (r) = (1 - K (v r)) / r^{2}$ . Direct differentiation gives $\partial_{i} A_{j}^{a} - \partial_{j} A_{i}^{a} = f^{'} (r) r (n_{i} ϵ_{a j ℓ} n_{ℓ} - n_{j} ϵ_{a i ℓ} n_{ℓ}) - 2 f (r) ϵ_{a i j}$ . The commutator is $[A_{i}, A_{j}]^{a} = f (r)^{2} r^{2} ϵ_{i j ℓ} n_{ℓ} n_{a}$ , since $(e_{i} \times n) \times (e_{j} \times n) = ϵ_{i j ℓ} n_{ℓ} n$ . Contracting their sum with $ϵ_{k i j} / 2$ and substituting $f = (1 - K (v r)) / r^{2}$ yields $B_{k}^{a} = - (1 - K (u)^{2}) n_{k} n_{a} / r^{2} + v K^{'} (u) (δ_{k a} - n_{k} n_{a}) / r$ . This equals step 2.1, so $B_{i} = D_{i} Φ$ for $r > 0$ , and smoothness extends the equality to $r = 0$ .

4.1F1F2step 1.1step 2.1step 3.1algebra

As $u \to \infty$ , $H (u) = 1 - u^{- 1} + O (e^{- 2 u})$ , $H^{'} (u) = O (u^{- 2})$ , and $K (u), K^{'} (u) = O (u e^{- u})$ . The formulas in steps 2.1 and 3.1 therefore give $B_{i}, D_{i} Φ = O (r^{- 2})$ at infinity, whose squared magnitude is integrable because $\int_{1}^{\infty} r^{2} r^{- 4} d r < \infty$ ; smoothness gives local integrability at the origin. Thus the energy is finite. The radial component is $B_{i}^{a} n_{i} = - (1 - K (u)^{2}) n_{a} / r^{2}$ , while $Φ^{a} = - v H (u) n_{a}$ , so $κ (Φ, B_{i}) n_{i} = v H (u) (1 - K (u)^{2}) / r^{2}$ . Taking the surface limit gives $m = 4 π v$ .

5.1F1F2step 4.1givenalgebra

By step 4.1 the asymptotic Higgs direction is $n_{\infty} (ω) = - ω_{a} T_{a}$ . For $ω \in S^{2}$ , put $w = - ω$ and $d_{\pm} = \sqrt{2 (1 \pm w_{3})}$ . On $U_{N} = {ω \in S^{2} : ω_{3} > - 1}$ and $U_{S} = {ω \in S^{2} : ω_{3} < 1}$ , respectively, define $u_{N} = \frac{1}{d_{-}} (\begin{array}{cc} w_{1} - i w_{2} & - (1 - w_{3}) \\ 1 - w_{3} & w_{1} + i w_{2} \end{array}), u_{S} = \frac{1}{d_{+}} (\begin{array}{cc} 1 + w_{3} & - w_{1} + i w_{2} \\ w_{1} + i w_{2} & 1 + w_{3} \end{array}) .$ The denominators are positive on the respective domains, which contain the closed northern and southern hemispheres. Since $∣ w ∣ = 1$ , multiplication gives $u_{α}^{*} u_{α} = I$ , $\det u_{α} = 1$ , and $u_{α} σ_{3} u_{α}^{- 1} = w_{a} σ_{a}$ ; these are smooth SU(2) lifts of $n_{\infty} = w_{a} T_{a}$ . On the positively oriented equator $ω = (\cos φ, \sin φ, 0)$ , multiplication gives $u_{N}^{- 1} u_{S} = diag (- e^{i φ}, - e^{- i φ}) = \exp ((- 2 φ - 2 π) T_{3})$ . Thus $θ (φ) = - 2 φ - 2 π$ is an explicit smooth real lift, with increment $- 4 π$ .

6.1F1F2step 3.1step 5.1givenalgebra

With $ι_{R} (ω) = R ω$ , the given gauge potential has angular pullback $ι_{R}^{*} A = - (1 - K (v R)) (ω \times d ω)_{a} T_{a}$ . Since $K (v R) \to 0$ and is independent of $ω$ , the adapted connections $A_{α, R} = {Ad}_{u_{α}^{- 1}} (ι_{R}^{*} A) + u_{α}^{- 1} d u_{α}$ converge in $C^{1}$ on each closed hemisphere to the $C^{1}$ form $A_{α, \infty} = - {Ad}_{u_{α}^{- 1}} ((ω \times d ω)_{a} T_{a}) + u_{α}^{- 1} d u_{α}$ . The radial magnetic field from step 3.1 gives $F_{α, R} = {Ad}_{u_{α}^{- 1}} (ι_{R}^{*} F) = (1 - K (v R)^{2}) T_{3} d Ω$ , where $d Ω$ is the outward area form of the unit sphere; hence these angular curvature forms converge uniformly to $T_{3} d Ω$ , the curvature of $A_{α, \infty}$ . In the same fixed frames, $q_{α, R} = {Ad}_{u_{α}^{- 1}} (Φ (R ω) / v) = H (v R) T_{3} \to T_{3}$ uniformly. In the original frame the limiting angular connection is $A_{\infty} = - (ω \times d ω)_{a} T_{a}$ and $n_{\infty} = - ω_{a} T_{a}$ , so $d n_{\infty} + [A_{\infty}, n_{\infty}] = 0$ by $(ω \times d ω) \times ω = d ω$ . Covariance gives $[A_{α, \infty}, T_{3}] = 0$ . The scaled radial limit is $B_{\infty} = - ω_{a} T_{a} = n_{\infty}$ , so its stabilizer condition holds. Since $H (v R)$ is constant in $ω$ , the Higgs limit holds in $C^{\infty}$ . Thus all the model's boundary conditions and the flux theorem's hypotheses hold.

7.1A1F3F4step 2.1step 3.1step 4.1step 5.1step 6.1algebra∎

By [F3] and step 4.1, $4 π v = m = 4 π v ν$ , so $ν = 1$ because $v > 0$ . Then $T = 4 π v / g^{2} > 0$ . Steps 2.1 and 3.1 give $B_{i} = D_{i} Φ$ , so [F4] gives $E = T = 4 π v / g^{2}$ . This verifies a classical unit-flux first-order solution and its saturation without asserting a quantum state.

Unit-charge SU(2) Prasad–Sommerfield monopole profile

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources