| Topic: |
Science > Physics |
| User: |
"Barrow" |
| Date: |
03 Jun 2007 11:51:54 PM |
| Object: |
questions on simple SSB |
Dear all,
In QFT text book by Ryder, He showed the spontaneous symmetry
breaking(SSB) mechanism by the scalar potential lagrangian:
L = \partial_a\phi \partial^a\phi^* - m^2 \phi^* \phi - c(\phi^*
\phi)^2 ---(1)
There is a square term of the potential and a quartic term. Then He
put :
\phi = (\rho + a)\exp{ i \theta} ---(2)
where a is the minimum of \phi while m^2 < 0, it can be calculated to
be -m^2/2c.
Thus, the complex scalar potential \phi(x) is decomposed into two real
scalar fields. Now treat these real scalar fields, \rho and \theta as
physical ones.
Substitute the above form of \phi into the lagrangian, he then showed
that there is a \phi^2 term but no \theta^2 term. He concluded that
the field \rho is a massive field and the field \theta is a massive
field.
My problem:
(1) Why is the particle mass of a field determined by the square term
of the potential in the lagrangian?
(2) Ryder said : "As a result of SSB, what would otherwise be two
massive fields (the real parts of \phi), become one massive and one
massless field." I don't know how this conclusion is achieved. Because
one just substitute eq(2) into the lagrangian eq(1), one will find one
massive and one massless field. Thus, how could he conclude that if no
SSB, there will be two massive fields, and that come from the real
parts of \phi???
Any help will be appreciated. Sincerely Barrow
.
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| User: "Jim Black" |
|
| Title: Re: questions on simple SSB |
05 Jun 2007 11:30:54 AM |
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On Jun 3, 11:51 pm, Barrow <GRsemi...@gmail.com> wrote:
Dear all,
In QFT text book by Ryder, He showed the spontaneous symmetry
breaking(SSB) mechanism by the scalar potential lagrangian:
L = \partial_a\phi \partial^a\phi^* - m^2 \phi^* \phi - c(\phi^*
\phi)^2 ---(1)
There is a square term of the potential and a quartic term. Then He
put :
\phi = (\rho + a)\exp{ i \theta} ---(2)
where a is the minimum of \phi while m^2 < 0, it can be calculated to
be -m^2/2c.
Thus, the complex scalar potential \phi(x) is decomposed into two real
scalar fields. Now treat these real scalar fields, \rho and \theta as
physical ones.
Substitute the above form of \phi into the lagrangian, he then showed
that there is a \phi^2 term but no \theta^2 term. He concluded that
the field \rho is a massive field and the field \theta is a massive
field.
My problem:
(1) Why is the particle mass of a field determined by the square term
of the potential in the lagrangian?
Because when we apply the calculus of variations to get a wave
equation from the Lagrangian, that \phi^2 term in the Lagrangian gives
rise to a \phi term in the wave equation. This in turn gives rise to
the constant term in the equation \omega^2 - k^2 = const relating the
frequency of modes to their wavenumber. And the last step, of course,
is finding the energy / particle number eigenstates of each mode.
Then we find that the particles' energies and momenta are related to
the frequency and wavenumber of the mode, respectively. Thus our
\omega^2 - k^2 = const equation gets turned into the E^2 - p^2 = m^2
relationship.
(2) Ryder said : "As a result of SSB, what would otherwise be two
massive fields (the real parts of \phi), become one massive and one
massless field." I don't know how this conclusion is achieved. Because
one just substitute eq(2) into the lagrangian eq(1), one will find one
massive and one massless field. Thus, how could he conclude that if no
SSB, there will be two massive fields, and that come from the real
parts of \phi???
Most likely the two massive fields he is referring to in the first
clause are the real and imaginary parts of \phi.
--
Jim E. Black
.
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