# Energy of moving Sine-Gordon breather

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 ($$c=1$$) $$E = \int_{-\infty}^{\infty} \frac {1} {2} \left[\left(\frac {\partial \phi} {\partial x}\right)^2+ \left(\frac {\partial \phi} {\partial t}\right)^2\right] +1-\cos\phi \, dx, ~~~~~~~~ (1)$$ and the moving breather solution in question is $$\phi(x,t) =4 \arctan\left[\frac {\sqrt{1-w^2}} {w} \frac{\sin\left(w \frac{t-vx} {\sqrt{1-v^2}}\right)} {\cosh\left(\sqrt{1-w^2} \frac {x-vt} {\sqrt{1-v^2}}\right)}\right].$$ Here, $$v$$ is the velocity of the breather, and $$w$$ 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 $$\phi(x,t)=\phi\left(\frac {x-vt} {\sqrt{1-v^2}}\right) ,$$ 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.

• Which references? Which pages? Commented Jun 24 at 3:17
• The standing breather solution is given by: $\phi(x,t) =4 \arctan\left[\frac {\sqrt{1-w^2}} {w} \frac{\cos\left(w t\right)} {\cosh\left(\sqrt{1-w^2} \,x\right)}\right]$ ;see en.wikipedia.org/wiki/Sine-Gordon_equation Commented Jun 24 at 5:50
• Correction (of wiki): en.wikipedia.org/wiki/Frenkel%E2%80%93Kontorova_model Commented Jun 24 at 6:55

Since the sine-Gordon theory is $$SO(1,1)$$ invariant, if you can find the energy for the stationary breather, you can find it for a moving breather as well. All that is required is a Lorentz boost; note that $$E$$ is just the energy of a body with rest energy $$E_{0}$$, boosted by the Lorentz factor $$\gamma=(1-v^{2})^{-1/2}$$.
If you want to evaluate the integral expression $$(1)$$ directly, just perform a change of variables to the $$(x',t')$$ coordinates in which the breather is at rest. That will take the integral into the special form it takes when $$v=0$$, multiplied by an overall factor of $$\gamma$$.