- #1
Charles_Henry
- 14
- 0
Homework Statement
1. Derive the Yang-Mills-Higgs equations of motion
[itex] (D_{v}F^{\mu, v})^{a} = - \epsilon^{abc}\phi^{b}(D^{\mu}\phi)^{c}[/itex] and [itex] (D_{\mu}D^{\mu}\phi)^{a} = - c(|\phi|^{2} - v^{2}) \phi^{a} [/itex]
from [itex] \mathcal{L} = - \frac{1}{4} F^{a}_{\mu v}F^{\mu v a} + \frac{1}{2}D_{\mu}\phi^{a}D^{\mu}\phi^{a} - U(\phi) [/itex]
where is c is a constant.
2. Show that in SU(2) Yang-Mills-Higgs theory the general solution to the equation [itex] D_{i}\hat{\phi}=0 [/itex] with [itex] |\hat{\phi}| = 1 [/itex] is
[itex] A^{a}_{i} = -\epsilon^{abc}\partial_{i}\hat{\phi}^{b}\hat{\phi}^{c} + k_{i}\hat{\phi}^{a} [/itex]
The attempt at a solution
1st question:
Assume finiteness of the energy by
[itex] \phi \rightarrow v [/itex] [itex] D_{mu} \phi \rightarrow 0 [/itex] [itex] F_{{\mu}v} \rightarrow 0 [/itex] as [itex] r \rightarrow \infty [/itex]
Let
[itex] F = \frac{1}{2} F_{{\mu},v}dx^{\mu} \wedge dx{v} = dA + A \wedge A [/itex] be
the gauge field of A.
with components
[itex] F_{{\mu}v} = \partial_{\mu}A_{v} - \partial_{v}A_{\mu} + [ A_{\mu}, A_{v}] = [D_{\mu}, D_{v}] [/itex]
We define gauge identities [itex] (A, A') [/itex] and [itex] (F, F') [/itex] where
[itex] A' = gAg^{-1} - dgg^{-1} [/itex] [itex] F' = gFg^{-1} [/itex] and [itex] g = g(x^{\mu}) \in G [/itex]
which satisfies the bianchi identity
[itex] DF = dF + [A,F] = d^2 A + dA \wedge A - A \wedge dA + A \wedge dA + A^3 - dA \wedge A - A^3 = 0 [/itex]
The field
[itex] \phi: \mathbb{R}^{D+1} \rightarrow \mathfrak{g} [/itex]
[itex] D\phi = d\phi + [A, \phi] [/itex] with gauge transformation,
[itex] \phi' = g \phi g^{-1} [/itex]
if [itex] G = SU(2) [/itex] choose a basis [itex] T_{a} , a =1,2,3 [/itex]
for the algebra of anti hermitian (2x2 matrices), such that
[itex] [T_{a},T_{b}] = - \epsilon_{abc}T_{c}
T_{a} = \frac{1}{2} i\sigma_{a} [/itex] where [itex] \sigma_{a} [/itex] are pauli matrices.
[itex] Tr (T_{a}T_{b}) = - \frac{1}{2} \delta_{ab} [/itex]
a general group element is [itex] g = exp (\alpha^{a}T_{a}) [/itex] with [itex] \alpha^{a} [/itex] being real. The components of [itex] D/phi [/itex] and [itex] F [/itex] with respect to the basis are given by
[itex] (D_{\mu} \phi)^{a} = \partial_{\mu} \phi^{a} - \epsilon^{abc} A^{b}_{\mu} \phi^{c} [/itex] and [itex] F^{a}_{{\mu}v} = \partial_{\mu}A^{a}_{v} - \partial_{v}A^{a}_{\mu} - \epsilon^{abc}A^{b}_{\mu}A^{c}_{v} [/itex]
when [itex] D+1 = 4 [/itex] the dual of the field tensor is
[itex] (\ast F)_{{\mu}v} = \frac{1}{2} \epsilon_{{\mu}v{\alpha}{\beta}}F^{{\alpha}{\beta}} [/itex]
we introduce a two form [itex] F = (\frac{1}{2}) F_{{\mu}v}dx^{\mu} \wedge dx^{v} [/itex] (Self-Dual)
or [itex] \ast F = F [/itex] [itex] \ast F = - F [/itex] (Anti-Self-Dual)
[itex] - Tr (F \wedge \ast F) = - \frac{1}{2} Tr (F_{{\mu}v} F^{{\mu}v})d^{4}x \frac{1}{4} F^{a}_{{\mu}v}F^{{\mu}va}d^{4}x [/itex]
where [itex] d^{4}x = \frac{1}{24}\epsilon_{{\mu} v {\alpha} {\beta} } dx^{\mu} \wedge dx^{v} \wedge dx^{\alpha} \wedge dx^{\beta} [/itex]
to be continued...