- #1
timb00
- 15
- 0
Hi,
I am interested in grand unification with extra dimensions. Especially the case when
extra dimensions are broken by orbifolding.
Now I am trying to understand how the Goldstonebosons appear in the spectrum of a
theory with global (for example SU(N)) symmetry. From the Goldstonetheorem it follows
that if this breaking is spontaneous then the particle spectrum has to include a set of
massless bosons with the same quantum numbers as the broken generators.
The model considered here consist of N_f complex bosons \phi_i. The lagrangian density is of the form
L = \frac{1}{2}\partial_M \Phi (\partial^M \Phi)^{\dagger}= \frac{1}{2}\sum_{i=1}^{N_f}\partial_{M} \phi_i (\partial^{M} \phi_i)^{\dagger}.
Where M=0,1,2,3,4,5 and the fifth dimension is compactified to
S^1 with Radius R. Breaking via Orbifold is incorporated if one
indentifies
y \sim -y .
This yields that the Space-time is M\times S^1/Z_2 and the field can be
written in as eigenstates of the Projection operator P with the eigenvalues
\pm1. These eigenstates are given by
\phi_+(x^{\mu},y) = \sum_n \phi^{(n)}(x^{\mu}) cos(\frac{ny}{R}),
\phi_-(x^{\mu},y) = \sum_n \phi^{(n)}(x^{\mu}) sin(\frac{ny}{R}).
We chose the parity of the fields to be
\underbrace{(-1,...,-1}_{M_f ~\text{times}},1,...,1).
This leads to a Lagrangian density of the form
L= \frac{1}{2}\sum_{i=1}^{M_f}\partial_{\mu} \phi_i^{(0)} (\partial^{\mu} \phi_i^{(0)})^{\dagger} + \frac{1}{2}\sum_{i=1}^{M_f}\sum_{n=-\infty,n\neq 0}^{\infty}\left(\partial_{\mu} \phi_i^{(n)} (\partial^{\mu} \phi_i^{(n)})^{\dagger} - \frac{n^2}{R^2} \phi_i^{(n)} \phi_i^{(n)\dagger} \right).
Most of the arguments given above a just summaries, and have to be discussed further.
But for my purpose it is enough. Form the Lagrangian given above one finds that Goldstone
bosons dose not appear in the spectrum.
Maybe you can explain to me why they do not occur in the spectrum?
I hope you understand my short review of the model, otherwise feel free to ask.
best regards,
Timb00
I am interested in grand unification with extra dimensions. Especially the case when
extra dimensions are broken by orbifolding.
Now I am trying to understand how the Goldstonebosons appear in the spectrum of a
theory with global (for example SU(N)) symmetry. From the Goldstonetheorem it follows
that if this breaking is spontaneous then the particle spectrum has to include a set of
massless bosons with the same quantum numbers as the broken generators.
The model considered here consist of N_f complex bosons \phi_i. The lagrangian density is of the form
L = \frac{1}{2}\partial_M \Phi (\partial^M \Phi)^{\dagger}= \frac{1}{2}\sum_{i=1}^{N_f}\partial_{M} \phi_i (\partial^{M} \phi_i)^{\dagger}.
Where M=0,1,2,3,4,5 and the fifth dimension is compactified to
S^1 with Radius R. Breaking via Orbifold is incorporated if one
indentifies
y \sim -y .
This yields that the Space-time is M\times S^1/Z_2 and the field can be
written in as eigenstates of the Projection operator P with the eigenvalues
\pm1. These eigenstates are given by
\phi_+(x^{\mu},y) = \sum_n \phi^{(n)}(x^{\mu}) cos(\frac{ny}{R}),
\phi_-(x^{\mu},y) = \sum_n \phi^{(n)}(x^{\mu}) sin(\frac{ny}{R}).
We chose the parity of the fields to be
\underbrace{(-1,...,-1}_{M_f ~\text{times}},1,...,1).
This leads to a Lagrangian density of the form
L= \frac{1}{2}\sum_{i=1}^{M_f}\partial_{\mu} \phi_i^{(0)} (\partial^{\mu} \phi_i^{(0)})^{\dagger} + \frac{1}{2}\sum_{i=1}^{M_f}\sum_{n=-\infty,n\neq 0}^{\infty}\left(\partial_{\mu} \phi_i^{(n)} (\partial^{\mu} \phi_i^{(n)})^{\dagger} - \frac{n^2}{R^2} \phi_i^{(n)} \phi_i^{(n)\dagger} \right).
Most of the arguments given above a just summaries, and have to be discussed further.
But for my purpose it is enough. Form the Lagrangian given above one finds that Goldstone
bosons dose not appear in the spectrum.
Maybe you can explain to me why they do not occur in the spectrum?
I hope you understand my short review of the model, otherwise feel free to ask.
best regards,
Timb00