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Perturbative and non-perturbative infrared behavior of ...

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C.3. The bounce action for a triangular barrier in three dimensions<br />

It is then helpful to define the gradient <strong>of</strong> the potential V ′ (φ) in terms <strong>of</strong> the parameters at<br />

the extremal points,<br />

λF = Vmax−VF<br />

φmax−φF<br />

λT = − Vmax−VT<br />

φT −φmax<br />

≡ ∆VF<br />

∆φF<br />

≡ −∆VT<br />

∆φT<br />

(C.3.36)<br />

where the first has positive sign <strong>and</strong> the second has negative one.<br />

The solution <strong>of</strong> the equation <strong>of</strong> motion at the two side <strong>of</strong> the triangular barrier are solved by<br />

imposing the boundary conditions, <strong>and</strong> a matching condition at some radius r + Rmax, that has<br />

to be determinate.<br />

The choice <strong>of</strong> the boundary condition proceeds as follows. Firstly the field φ reaches the false<br />

vacuum φF at a finite radius RF <strong>and</strong> stays there. This imposes<br />

φ(RF) = φF<br />

˙φ(RF) = 0 (C.3.37)<br />

For the second boundary condition we work in the simplified situation such that the field has<br />

initial value φ0 < φT at radius r = 0. In this way we imposes the conditions<br />

φ(0) = φ0<br />

˙φ(0) = 0 (C.3.38)<br />

Solving the equations <strong>of</strong> motion we have two solution at the two sides <strong>of</strong> the barrier<br />

λTr2 φR(r) = φ0 −<br />

6<br />

φL(r) = φ F max − λFR2 F<br />

2 + λFR3 F<br />

3r<br />

+ λF<br />

6 r2<br />

for 0 < r < Rmax (C.3.39)<br />

for Rmax < r < RF<br />

By matching the derivatives at Rmax we are able to express RF as a function <strong>of</strong> Rmax<br />

R 3 F = (1 + c)R 3 max<br />

where c = −λT/λF. Out <strong>of</strong> the value <strong>of</strong> the field at Rmax we get two useful relations<br />

φ0 = φmax + λT<br />

6 R2 max<br />

∆φF = λF(3 + 2c − 3(1 + c) 2<br />

3)<br />

6<br />

R 2 max<br />

(C.3.40)<br />

(C.3.41)<br />

(C.3.42)<br />

The bounce action B can be evaluated by integrating the equation (C.3.35) from r = 0 to r = RF.<br />

We found<br />

B = 16√6π 1 + c<br />

5 (3 + 2c − 3(1 + c) 2/3 ) 3/2<br />

<br />

∆φ6 F<br />

(C.3.43)<br />

∆VF<br />

199

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