Energy of moving Sine-Gordon breather

  • #1
TOAsh2004
10
5
TL;DR Summary
I want to prove the formula ##E =\frac {E_0} {\sqrt {1 - v^2}} ## for the energy of the moving breather solution of the Sine-Gordon equation.
Hello everyone,

A few days ago I stumbled across the formula for the energy of a moving breather for the Sine-Gordon equation $$\Box^2 \phi = -Sin(\phi) $$ The energy in general is given by (c=1) $$ E = \int_{-\infty}^{\infty} \frac {1} {2} ((\frac {\partial \phi} {\partial x})^2+ (\frac {\partial \phi} {\partial t})^2) +1-Cos(\phi) \, dx ~~~~~~~~ (1) $$ and the moving breather solution in question is $$\phi(x,t) =4 arctan(\frac {\sqrt{1-w^2}} {w} \frac{Sin(w \frac{t-vx} {\sqrt{1-v^2}})} {Cosh(\sqrt{1-w^2} \frac {x-vt} {\sqrt{1-v^2}})}).$$ Here, v is the velocity of the breather, w is a parameter. Now it was claimed in different sources that the energy of this moving breather solution is $$E =\frac {E_0} {\sqrt {1 - v^2}}, ~~~~~~~~ (2) $$ where ##E_0## is the energy of the resting breather (v=0). I did try numerous attempts to derive this formula, by plugging in the breather solution into (1), but always ended up with integrals not even Mathematica was able to solve. I see, that (2) holds for travelling wave solutions ##\phi(x,t)=\phi(\frac {x-vt} {\sqrt{1-v^2}})##, but the breather solution does not have this symmetry. Can anyone provide me a hint on how one can derive (2)? I would very much appreciate it.

Greetings
 
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  • #2
Sources?
"I stumbled across" is really not a reference.......
 
  • #3
The equation is Lorentz invariant which means solutions for velocity, ##v##, may be written down given the ##v=0## solution. Doesn’t the energy expression just follow as the transform rest energy?
 
  • Informative
Likes hutchphd

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