# For a gaussian beam, how can I calculate $w_0$, $z$ and $z_R$ from knowledge of $w$, $R$ and $\lambda$?

I have a Gaussian beam somewhere away from the beam waist (also referred to as focus point). Then, according to text books (or wikipedia), the beam size $w$ is given by

$$w(z) = w_0 \sqrt{ 1+\left(\frac{z}{z_R}\right)^2 },$$

with $z$ the axial distance from the beam waist, $w_0$ the waist size and the Rayleigh length $z_R$ which is defined as

$$z_R=\frac{\pi w_0^2}{\lambda_0}.$$

The radius of curvature of the phase fronts are given by

$$R(z) = z \left( 1 + \left(\frac{z_R}{z}\right)^2 \right).$$

I have: $R$, $w$, and $\lambda_0$.

I am looking for: $w_0$, $z$, and $z_R$.

My idea was to solve the third equation for $z$,

$$z_{1,2} = \frac{R}{2} \pm \sqrt{\left( \frac{R}{2} \right)^2 - z_R^2},$$

and then inserting into the first equation

$$w = w_0 \sqrt{ 1 + \left( \frac{ \frac{R}{2} \pm \sqrt{\left( \frac{R}{2} \right)^2 - z_R^2} }{z_R} \right)^2 }.$$

Inserting $z_R$ yields

$$w = w_0 \sqrt{ 1 + \left( \frac{ \frac{R}{2} \pm \sqrt{\left( \frac{R}{2} \right)^2 - \left(\frac{\pi w_0^2}{\lambda_0}\right)^2} }{\frac{\pi w_0^2}{\lambda_0}} \right)^2 }$$

which is an equation for $w_0$. Since I am currently stuck solving it, my question is in principle twofold: what software/tool do you guys recommend me for solving such an equation for $w_0$ and if there is a different way determining the beam parameters (maybe I missed something) ?

• Wolfram Mathematica, maybe? There is a Mathematica SE... Commented Jul 27, 2016 at 20:47
• I don't know, but I do know that I would square the first equation, and I would not carry around square roots. Commented Jul 27, 2016 at 20:53
• @heather I was not aware that you can such ''complicated'' (not in the mathematical sense...) things with the free version... thanks!
– Alf
Commented Jul 28, 2016 at 5:46

When attacking a system of equations with square roots it is often helpful to get rid of the square roots until the very end (if possible). Good idea garyp.

$$w^2 = w_0^2 \left( 1 + \frac{z^2}{z_R^2} \right)$$ $$z_R = \frac{\pi w_0^2}{\lambda_0}$$ $$R = z \left( 1 + \frac{z_R^2}{z^2}\right)$$

When I look at this system of equations I notice that $z_R = \pi w_0^2 / \lambda_0$ is only relating $z_R$ and $w_0$, two unknowns, with $\lambda_0$, a known. This means I can easily use $z_R = \pi w_0^2 / \lambda_0$ to replace either the $w_0$ or the $z_R$ in all of the equations and I would be down to two equations with two unknowns. I recommend getting rid of the $w_0^2$ first as it only shows up in one place.*

$$w^2 = \frac{\lambda_0 z_R}{\pi}\left( 1 + \frac{z^2}{z_R^2} \right)$$ $$R = z \left( 1 + \frac{z_R^2}{z^2}\right)$$

Now move the $\lambda_0/\pi$ to the left hand side and expand the right hand sides to see that there is symmetry in the two remaining equations.

$$\frac{\pi w^2}{\lambda_0 } =z_R + \frac{z^2}{z_R}$$ $$R = z + \frac{z_R^2}{z}$$

Notice that if I make both right hand sides the same we can set them equal to each other.

$$\frac{\pi w^2}{\lambda_0 }z_R =z_R^2 + z^2$$ $$R z = z^2 + z_R^2$$

Now it is easy to use $z = (\pi w^2/(\lambda_0 R))z_R$ to plug into the earlier forms .

$$\frac{\pi w^2}{\lambda_0 } =z_R + \frac{z^2}{z_R} = z_R \left( 1 + \frac{\pi^2 w^4}{\lambda_0^2 R^2} \right)$$ $$R = z + \frac{z_R^2}{z} = z \left( 1 + \frac{ \lambda_0^2 R^2}{\pi^2w^4} \right)$$

The final results are

$$z_R = \frac{ \pi \lambda_0 R^2 w^2 } {\lambda_0^2 R^2 + \pi^2 w^4}$$ $$z =\frac{\pi^2 R w^4}{ \lambda_0^2 R^2 + \pi^2 w^4}$$

and we can use the relation for $z_R$ and $w_0$ to get

$$w_0 = \frac{ \lambda_0 \left|R \right| w } {\sqrt{\lambda_0^2 R^2 + \pi^2 w^4}}$$

Now $w_0$ must be greater than $0$ and so must $w$ but $R$ could be positive or negative. Therefore we need an absolute value on the $R$ in the above equation.

heather was right in that Mathematica would be very helpful.

Solve[{w^2 == w0^2 (1 + z^2/zr^2), zr == \[Pi] w0^2/lambda0, R == z (1 + zr^2/z^2)}, {w0, z, zr}] // FullSimplify


*I would like to mention that this step wasn't obvious to me from the start. I figured it out by realizing that $\pi/\lambda_0$ was not that important so I set $\lambda=\pi$. This way I could see the key features of the equations without the mess of coefficients.

The easiest way to do this is via the complex beam parameter $q$, which is given by $$\frac1q = \frac{1}{z+iz_R} = \frac{1}{R(z)}-i\frac{\lambda_0}{\pi w(z)^2}.$$ In your case you know all three parameters on the right hand side, so you know $q$, and therefore you can determine $z$ and $z_R$ as its real and imaginary parts. Doing this explicitly, \begin{align} q & = \frac{1}{\frac{1}{R(z)}-i\frac{\lambda_0}{\pi w(z)^2}} \\ & = \frac{ R(z)\pi w(z)^2}{\pi w(z)^2-i\lambda_0R(z)} \\ & = \frac{ R(z)\pi w(z)^2}{\pi w(z)^2-i\lambda_0R(z)} \frac{\pi w(z)^2+i\lambda_0R(z)}{\pi w(z)^2+i\lambda_0R(z)} \\ & = \frac{ R(z)\pi w(z)^2}{\pi^2 w(z)^4+\lambda_0^2R(z)^2}\left(\pi w(z)^2+i\lambda_0R(z)\right), \end{align} and from there you can read off $$z= \frac{R(z)\pi^2 w(z)^4}{\pi^2 w(z)^4+\lambda_0^2R(z)^2}$$ and $$z_R= \frac{\lambda_0R(z)^2\pi w(z)^2}{\pi^2 w(z)^4+\lambda_0^2R(z)^2}.$$ Finally, the beam waist is determined by the Rayleigh range: $$w_0^2 =\frac{\lambda_0}{\pi}z_R = \frac{\lambda_0^2 R(z)^2 w(z)^2}{\pi^2 w(z)^4+\lambda_0^2R(z)^2},$$ so $$w_0 = \frac{\lambda_0 R(z) w(z)}{\sqrt{\pi^2 w(z)^4+\lambda_0^2R(z)^2}}.$$ These agree with LasersMatter's results, but the $q$ route is a lot easier.

• wow, nice, I was not aware of the complex beam parameter! That is more elegant, I agree (sorry, I cannot accept 2 answers...). I definitely have to read up on Gaussian beams...
– Alf
Commented Jul 28, 2016 at 6:03