$\kappa=8/3$ [@SchrammSAW; @Kennedy], we can use ${\cal P}(z,\bar z) ={\cal P}_0(\theta)$ as given by Schramm’s formula (for $\kappa=8/3$) ${\cal P}_0(\theta) = \cos^2(\theta/2)$, where $\theta$ is the angle with the real axis (see Fig. \[f:1\]). From this we obtain the force exerted by the polymer on the impenetrable object: $$\vec f = - \vec \nabla F=  kT \frac{\vec n_\theta} r \frac{\partial }{\partial \theta} \ln {\cal P}_0(\theta) =  - kT \frac{\vec n_\theta} r \tan \left(\frac{\theta }{2}\right) \label{single}$$ This result is valid in the (critical) limit $a ,\ell \ll r$, of object- and monomer-size small compared to $r$. Note that when approaching the boundary on the $x<0$ side, the object is repelled by a force diverging as $2 k T/y$, with $y$ the distance from the wall.

We can now use conformal invariance to obtain results in various geometries. The simplest one is the wedge geometry $\cal B$, see right of Fig. \[f:1\], with exterior angle $\phi$, the polymer being attached at the top of the wedge. Under the map $w=g(z)=z^{\pi/\phi}$ the wedge geometry (in coordinate $z=x+i y$) is mapped back to the half plane (in coordinate $w$). The case $\phi=2 \pi$ corresponds to the full plane with impenetrable positive real axis. Conformal 