# Lateral momentum of Gaussian beam

A beam of light carries momentum. What fraction of this is lateral rather along the propagation direction if we assume something like a Gaussian beam?

Wikipedia claims in the entry on Gaussian beams that the Poynting vector is entirely along the z-axis. But this expression is based on solving the paraxial approximation of the Helmholtz equation rather than Maxwell's equations, implicitly assuming that there is indeed no lateral energy or momentum flow.

(Simon, Sudarshan & Mukunda 1986) give an expression for the Poynting vector of an actual Maxwell gaussian beam as $$\mathbf{S}(x,y,z)=\frac{2}{\pi}\frac{1}{\sigma(z)^2}\left [\frac{x}{R(z)},\frac{y}{R(z)},1\right]$$ where $$R(z)=z [1+(k\sigma_0^2/2z)^2]$$ and $$\sigma(z)=\sigma_0 \sqrt{1+(2z/k\sigma_0^2)^2}$$. They find that the energy flow follows the geometrical optics rays that are normal to the phase curvature.

So by this account the fraction of energy going laterally would be $$E_{lat}/E_{total}=\frac{r/R(z)}{\sqrt{r^2/R(z)^2+1}},$$ which reasonably approaches 1 as $$r\rightarrow \infty$$. But there seems to be a Gaussian term missing from their expression since integrating the lateral energy flow across a plane diverges.

(Allen 2000) looks at Laguerre-Gaussian beams and gets a lateral Poynting component of $$p_r=\epsilon_0 \frac{\omega k r z}{z_R^2+z^2}|u|^2$$ and an axial component $$p_z = \epsilon_0 \omega k |u|^2$$ (assuming no azimuthal component). The fraction of lateral momentum is $$\frac{rz}{(z_R^2+z^2)\sqrt{\frac{r^2z^2}{(z_R^2+z^2)^2}+1}}$$. Integrating this times $$|u|^2=(C/w(z)^2)\exp(-2r^2/w(z)^2)$$ where $$w(z)^2=(2/k)(z_R^2+z^2)/z_R$$ gives $$P_r = \frac{Cz}{w(z)^2(z_R^2+z^2)}\int_0^\infty \frac{r}{\sqrt{\frac{r^2z^2}{(z_R^2+z^2)^2}+1}}\exp(-2r^2/w(z)^2) dr$$ which at least converges, but doesn't look like it has a neat analytic solution.

I assume there is a much simpler argument for how much of the momentum ends up sideways when we project a beam through a finite radius aperture or have a finite Gaussian beam waist width.

(My application is a consideration of the ultimate limitations of photon rockets; lateral momentum is wasted from a propulsion standpoint. Sure, there are other problems with photon rockets too, but one thing at a time.)

• "but doesn't look like it has a neat analytic solution" - why is that a Bad Thing? Why would one expect the existence of an analytical expression for a quantity like this? – Emilio Pisanty Oct 3 '18 at 11:55
• Gaussian beams imply the assumption of the paraxial approximation, which means $r\ll z_R$. As a result, you may approximate some of those quare roots with simpler expressions, which in turn may lead to analytic solutions. – flippiefanus Oct 3 '18 at 12:11
• Yes, one shouldn't fetishize analytic answers too much. Still, they tend to make scaling dependent on parameters clearer. Great point about using the paraxial approximation - that might give scalings too. – Anders Sandberg Oct 3 '18 at 15:36
• By "photon rocket", are you including vehicles with light sails driven by, e.g., Earth-based lasers? In that case, the main limitation is the aperture of the Earth-based laser or the aperture of an intermediary element along the lines of what Robert Forward proposed. Note that with an aperture of the size of the Sun, a laser beam could be focused to a spot a few hundred meters across at a distance of a few light-years. Lateral momentum wouldn't be an issue. – S. McGrew Jan 15 at 3:24

Firstly, Gaussian beam (and also Laguerre-Gauss beams) are solutions of the paraxial wave equation. Therefore, these beams are valid under the paraxial condition, which can be imposed by a requirement that the beam divergence angle is small: $$\theta_B = \frac{\lambda}{\pi w_0} \ll 1 ,$$ where $$\lambda$$ is the wave length and $$w_0$$ is the radius of the beam at its waist.
For a paraxial optical beam, the Poynting vector is given directly by the gradient of the phase. As such, it is equal to the optical current. For an optical field $$\psi(\mathbf{x})=|\psi(\mathbf{x})| \exp[ i\theta(\mathbf{x})]$$, the Poynting vector or optical current is given by $$\mathbf{S} = |\psi(\mathbf{x})|^2 \nabla \theta(\mathbf{x}) .$$