Energy of moving Sine-Gordon breather

Question

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.

Answer 1

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 $\gamma$.