Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 303 / 3 / d

Table of contents
- Solution d

Solution

For

d = 4 - ϵ

, define dimensionless variables

r = \frac{μ ^{2}}{Λ ^{2}}, u = g = Λ^{- ϵ} g .

To leading order, with

K_{4} = 1/ (8 π^{2})

,

\frac{d r}{d s} = 2 r + 4 (N + 2) K_{4} \frac{u}{1 + r},

\frac{d u}{d s} = ϵ u - 4 (N + 8) K_{4} \frac{u ^{2}}{( 1 + r ) ^{2}} .

The Gaussian fixed point is

(r_{*}, u_{*}) = (0, 0)

. Its thermal eigenvalue is

y_{t} = 2

, so

ν_{G} = \frac{1}{2} .

The interacting Wilson-Fisher fixed point is

u_{*} = \frac{2 π ^{2}}{N + 8} ϵ + O (ϵ^{2}), r_{*} = - \frac{N + 2}{2 ( N + 8 )} ϵ + O (ϵ^{2}) .

Linearizing the flow gives

y_{t} = 2 - \frac{N + 2}{N + 8} ϵ + O (ϵ^{2}) .

Since the correlation-length critical exponent is

ν = 1/ y_{t}

,

ν_{WF} = \frac{1}{2} + \frac{N + 2}{4 ( N + 8 )} ϵ + O (ϵ^{2}) .

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version