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In a chapter on the basics of laser physics, my textbook says the following:

The $c/2L$ spacing of resonance is typical of the so-called longitudinal modes, or $\text{TEM}_{00}$ modes, characterised by an electric field distribution given by: $$E(r) = E_0 \exp -r^2/w_0^2 \tag{A1.3}$$ In the Gaussian field distribution given by Eq.A1.3, the parameter $w_0$ has the meaning of a characteristic radius, called the spot size of the laser beam.
In particular, $w_0$ is the radius at which the field amplitude drops off to $1/e = 0.37$ and the power density (proportional to $E^2$) drops off to $1/e^2 = 0.13$ of the maximum value. In addition, by integrating $E^2(r)$ on $r$, we find that the relative power contained within $w_0$ is $1 - 1/e^2 = 0.86$ of the total beam power.

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Propagation of Gaussian modes is such that the mode distribution (Eq.A1.3) keeps itself unaltered in free propagation of the beam and in imaging of it by lenses and mirrors. Only the spot size $w(z)$ changes, and its relation to the beam waist or spot-size $w_0$ originated by the laser at $z = 0$ is: $$w^2(z) = w_0^2 + (\lambda z/\pi w_0)^2 \tag{A1.4}$$ In Eq.A1.4, when $z = 0$ we get $w = w_0$, the beam waist of the laser. For $z > 0$, $w$ increases as the quadratic sum of two terms: the beam waist $w_0$ and a propagation term proportional to distance $z$. Writing this term as $\theta_0 z$, we can see that $\theta_0 = \lambda / \pi w_0$, and this is the angular divergence of the beam, equal to the diffraction angle of the aperture $w_0$.
When $z$ is large enough ($z > \pi w_0^2 / \lambda$), the propagation term prevails and $w(z) = \theta_0 z$. Then, at the angle $\theta = r/z$ off the beam axis, the electric field given by Eq.A1.3 becomes $E(\theta) = E_0 \exp -\theta^2/\theta_0^2$. This tells us that $\theta_0$ duplicates in angle the spot-size meaning $w_0$ has in radius. In a real laser with spot size $w_0$, the $86\%$-power angle $\theta_\text{eff}$ is larger than $\theta_0 = \lambda / \pi w_0$, the ideal single mode value. Then we define an M-squared factor as $$M^2 = \theta_\text{eff}^2 / \theta_0^2 \tag{A1.4'}$$

I'm unsure about this part:

When $z$ is large enough ($z > \pi w_0^2 / \lambda$), the propagation term prevails and $w(z) = \theta_0 z$. Then, at the angle $\theta = r/z$ off the beam axis, the electric field given by Eq.A1.3 becomes $E(\theta) = E_0 \exp -\theta^2/\theta_0^2$.

I cannot find any mathematically coherent way in which $\theta = r/z$ with Eq.A1.3 produces $E(\theta) = E_0 \exp -\theta^2/\theta_0^2$. All of the sources that I have found online perform this derivation in a similar but different way, and most of the online sources seem to perform the derivation in the same way as each other, but none of them seem to perform the derivation in the exact same way as the author has in this textbook. Is there something wrong with the author's derivation, or am I missing something?

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1 Answer 1

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See Gaussian Beam. Note that you'd be better served considering equations for the intensity of a beam ($\propto |E|^2$) rather than the field so you don't have to worry as much about keeping track of various phase factors (which I neglect in this answer).

Equation A1.3 as written is only valid at $z=0$. For non-zero $z$ we should have

$$ E(r, z) = E_0 \exp(-r^2/w(z)^2) $$

Then for large $z$ we get $$ w(z) \approx z / (\pi w_0/\lambda) = \theta_0 z $$

So we get

$$ E(r, z) = E_0 \exp(-r^2/(\theta_0 z)^2) = E_0 \exp(-\theta^2/\theta_0^2) $$

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