Quantum flux tubes possible shapes If given energy, can flux tubes be any shape as long as all quarks are connected and the amount of energy is insufficient to form a quark-antiquark pair, or will the shape of the flux tubes be a scaled up or elongated duplicate of the ground state? 
My current understanding of them is that they will stretch linearly unlike a normal string with the amount of energy applied to it. Is this understanding correct or false in any ways
 A: Think of the QCD vacuum inside the hadron as similar to a material with a dielectric constant. The electric field, here a QCD color charge field, as $\vec D = \vec E + 4\pi \vec P$. The self interaction of the gluons is similar to the polarization of a medium in classical electrodynamics. The chromo-electric displacement is then $\vec D = \epsilon\vec E$.  The difference is that the medium is antiscreening so that $\epsilon << 1$ and near zero. Thus the chromo-electric vector displacement is very small $\vec D \simeq 0$. Hence if we put a charge (a chromo-charge) $q$ in the medium this will punch a hole in the medium. Consequently the $\epsilon = 1$, the normalized vacuum state, in the hole, but with $\epsilon <<1$ n the medium there is an induced charge on the inner surface equal to the charge introduced. Hence the charge “punches open”a little bubble of antiscreening with the vacuum for $\epsilon << 0$
We then have the conditions on the electric field outside this hole as $E = q/\epsilon r^2$ and its interaction energy with the charge induced on the outside is ${\cal E}_q = q^2/\epsilon r$. The energy needed to generate the hole has two terms, one for the volume and the other for the area. We may think of this as a bit like a capacitor. So the energy needed to create this hole, which is a “QCD bag of gluons” is then ${\cal E}_{bag} = A\times vol + B\times area$. So the total area is then
$$
{\cal E} =  q^2/\epsilon r + A\times r^3 + B\times r^2
$$
I have graphed out the form of this as seen below.

The volume contribution is the main source of energy for large $r$ and for energy ${\cal E} = \int\vec F\cdot dr$ the force is approximately $F\propto r^2$. For intermediate energy the surface is the major contribution and one gets $F \propto r$, which is the harmonic oscillator. 
