Laser Pushing on Paraboloid

The tangent to the parabola $y = x^2$ makes an angle $\theta$ with the $x$ -axis at the point $(x,x^2)$ , where $\tan\theta = 2x$ . Thus the normal to the parabola at $(x,x^2)$ makes the same angle $\theta$ with the $y$ -axis. Thus a photon of momentum $p$ incident upon the paraboloid a distance $x$ from the axis is reflected back at an angle $2\theta$ to its original direction. Thus the $z$ -component of the change of momentum of that photon is $p(1 + \cos2\theta) = 2p(1 - \sin^2\theta) = \tfrac{2p}{4x^2 + 1}$ . Thus the $z$ -component of the force on the paraboloid due to a laser beam of power $P$ and radius $r$ is $F \; = \; \int_0^r \frac{P}{cp} \times \frac{2\pi x\,dx}{\pi r^2} \times \frac{2p}{4x^2+1} \; = \; \frac{4P}{cr^2}\int_0^r \frac{x\,dx}{4x^2 +1} \; = \; \frac{P}{2cr^2}\ln(4r^2+1)$ With $r=1$ we obtain $\alpha = \boxed{\tfrac12\ln5}$ .