On beam models and their paraxial approximation

Starting with equation (2) we choose to work in the Lorentz gauge $\partial\cdot A=0$ and set $A^{0}=0$. We begin by considering linear polarisation. For this set-up our ansatz becomes:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\LCperp}{{\perp}} \newcommand{\LCp}{{+}} \newcommand{\w}{{\rm w}} \newcommand{\eps}{\varepsilon} \displaystyle \label{aperpansatz} \widetilde{A}^{\perp,\mu}(l) = -{\rm i} \epsilon^{\LCperp,\mu} E_0(2\pi)^4\, \delta(l\cdot l)\, \rho(l^{0}) \delta_{\epsilon}({l}^{\perp})\, l^3 \theta(l^3 l^0), \nonumber \end{align} \tag{ 4 }$$

where $\perp$ represents the combined transverse coordinate. From here on in we define the transverse component of the vector potential $A^{\perp}$, via $A^{\perp, \mu} = A^{\perp} \varepsilon^{\perp, \mu}$. Note that $\newcommand{\w}{{\rm w}} \widetilde{A}^\perp$ is supported only on the light cone $\delta(l\cdot l)$ with directionality enforced by the $\theta(l^{3}l^{0})$ term, so it is automatically a solution to the electromagnetic wave equation. The $\newcommand{\eps}{\varepsilon} \newcommand{\LCp}{{+}} \newcommand{\LCperp}{{\perp}} \newcommand{\e}{{\rm e}} \epsilon^\LCperp$ term is the real polarisation vector, $\rho(l^{0})$ represents the energy spectrum of the photon momenta and in order for the field to be real-valued must obey $\rho(l^{0}) = \rho^{\ast}(-l^0)$. Finally $\newcommand{\eps}{\varepsilon} \newcommand{\e}{{\rm e}} \delta_{\epsilon}({l}^{\perp})$ is the transverse distribution of the photon frequencies, representing the focusing of the beam. We maintain continuity with the plane wave limit by imposing the condition;

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{PWlimit} \lim_{\varepsilon\to\infty}\delta_{\varepsilon}(l^{\perp}) = \delta(l^{1})\,\delta(l^{2}). \nonumber \end{align} \tag{ 5 }$$

To recover the Gaussian beam result in the literature we use the 'nascent' delta function of the heat kernel:

$$\begin{align*} \newcommand{\e}{{\rm e}} \displaystyle \lim_{\varepsilon\to0}\frac{1}{\sqrt{2\pi\varepsilon}}\,\exp\left[{\frac{-x^{2}}{2\varepsilon}}\right] = \delta(x) \nonumber \end{align*}$$

to choose:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle \label{deltaregG} \delta_{\varepsilon}(l^{\perp}) = \frac{\w_{0}^{2}}{4\pi}\exp\left[-\frac{\left(l^{\perp}\w_{0}\right)^{2}}{4}\right]; \nonumber \end{align} \tag{ 6 }$$

where $\newcommand{\w}{{\rm w}} \varepsilon=2/\w_{0}^{2}$ has been chosen and $\newcommand{\w}{{\rm w}} \w_0$ is the beam waist.

To recover the four-potential $A^{\perp}(x)$ and show that this ansatz does indeed reproduce the field of a Gaussian beam as proposed, we must perform the momentum integrals in equation (2). We eliminate the l³ integral using:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{DLL} \delta(l\cdot l) =&\, \theta\big[(l^{0})^{2}-(l^\perp)^{2}\big]\nonumber \\ &\times\left[\frac{\delta\left(l^{3}-\sqrt{(l^{0})^{2}-(l^\perp)^{2}}\right)}{2\vert l^{3}\vert } +\left(l^{3}\to-l^{3}\right)\right], \nonumber \end{align} \tag{ 7 }$$

where $\theta(\cdot)$ is the Lorentz–Heaviside function.

Writing out $A^\perp(x)$ explicitly using equation (7) to replace $\delta(l\cdot l)$, we have:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \newcommand{\sig}{\varsigma} \displaystyle \label{Aperp} A^{\perp}(x)=&\,-{\rm i}E_{0}\frac{\w_{0}^{2}}{8\pi} \int {\rm d}l^{0}\,{\rm d}^{2}l^{\perp}~ {\rm{e}}^{-{\rm i}l^{0}t + il^{\perp}\cdot x^{\perp}-\frac{(l^{\perp}\w_{0})^{2}}{4}}\nonumber \\ &\rho(l^{0})\,\theta\big(l^{\sigma}\big)\nonumber \\ &\left[\theta(l^{0}){\rm{e}}^{{\rm i}z\sqrt{l^{\sigma}}} -\theta(-l^{0}){\rm{e}}^{-{\rm i}z\sqrt{l^{\sigma}}}\right],\nonumber \\ \nonumber \end{align} \tag{ 8 }$$

where $\newcommand{\sig}{\varsigma} l^{\sigma}=(l^{0}){\hspace{0pt}}^{2}-(l^{\perp}){\hspace{0pt}}^{2}$. As stated previously, we are looking to connect with the well known analytical result of the Gaussian beam in the paraxial limit. This is achieved using two approximations.

• (i)  
  Since we are not solving the paraxial wave equation directly, we must make assumptions about the photon momenta to ensure we recover the same result, namely that:

  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{paraxassump} l^\perp\ll l^0. \nonumber \end{align} \tag{ 9 }$$

  This allows us to simplify the integral by manipulating the square root term. Performing a Taylor expansion of $l^{3}$ and neglecting terms of order $\left(l^{\perp}/l^{0}\right){\hspace{0pt}}^4$ we recover a Gaussian term:

  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{sqrt} \sqrt{(l^{0})^{2}-(l^{\perp})^{2}} = \vert l^{0}\vert \sqrt{1-\left(\frac{l^{\perp}}{l^{0}}\right)^{2}} \approx \vert l^{0}\vert - \frac{(l^{\perp})^{2}}{2\vert l^{0}\vert }. \nonumber \end{align} \tag{ 10 }$$
• (ii)  
  The integration in $(l^{\perp}){\hspace{0pt}}^{2}$ is bounded by $(l^{0}){\hspace{0pt}}^{2}$. Since the Heaviside-Lorentz θ function only depends on $(l^{\perp}){\hspace{0pt}}^{2}$, we perform the integral over $l^\perp$ in polar co-ordinates and compute the integral over the angular dependence. Following this we may use the θ function to determine the limits for the remaining integration:

  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{Il0} I(l^{0},z) = 2\pi \int_{0}^{\vert l^{0}\vert } {\rm d}\rho~\rho ~{\rm{e}}^{-a\rho^{2}}\,J_{0}(\vert x^{\perp}\vert \rho); \nonumber \end{align} \tag{ 11 }$$

  where,

  $$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle a = \frac{\w_{0}^{2}}{4}+{\rm i}\frac{z}{2\vert l^{0}\vert }\nonumber \end{align*}$$

  and $J_{n}(x)$ is the nth order Bessel function of the first kind [58]. We find:

  $$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle \label{Aperp1} A^{\perp}(x) &= -{\rm i}E_{0}\frac{\w_{0}^{2}}{8\pi} \int {\rm d}l^{0}~ \rho(l^{0}){\rm{e}}^{-{\rm i}l^{0}t}\nonumber \\ &\left[\theta(l^{0}){\rm{e}}^{{\rm i}\vert l^{0}\vert z}I(l^{0},z) - \theta(-l^{0})(z\to-z)\right]. \nonumber \end{align} \tag{ 12 }$$

To perform the l⁰ integral we must state the form of the $\rho(l^0)$ function (frequency spectrum). To recover the Gaussian beam, we choose:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{RHO} \rho(l^{0}) = \frac{1}{\vert l^0\vert }\big[ \delta(l^{0}-\omega)+\delta(l^{0}+\omega) \big], \nonumber \end{align} \tag{ 13 }$$

where $\omega>0$ is the laser beam frequency.

Then we find:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\tr}{{\rm tr}} \newcommand{\w}{{\rm w}} \newcommand{\trm}[1]{{\rm #1}} \displaystyle A^{\perp}(x) = -{\rm i}E_{0}\frac{\w_{0}^{2}}{8\pi\omega}{\rm{e}}^{{\rm i}\omega(z-t)}I(\omega,z) + \trm{c.c.}. \nonumber \end{align} \tag{ 14 }$$

Now to deal with $I(\omega, z)$ we make the second assumption that $\newcommand{\w}{{\rm w}} \omega \w_0 \gg 1$ to perform the integral $I(\omega, z)$ analytically, giving:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\LCm}{{-}} \newcommand{\w}{{\rm w}} \displaystyle A^\perp = -\frac{E_{0}}{\omega}\frac{{\rm{e}}^{-\frac{\vert x^{\perp}\vert ^{2 } } {\w^{2} }}}{\sqrt{1+\varsigma^{2}}}\sin\left[\omega x^{\LCm}+\tan^{-1}\varsigma-\frac{\vert x^{\perp }\vert ^ {2}\varsigma}{\w^{2}}\right], \nonumber \end{align} \tag{ 15 }$$

where we have defined:

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\LCm}{{-}} \newcommand{\w}{{\rm w}} \newcommand{\sig}{\varsigma} \displaystyle \sig(z)&=\frac{z}{z_r};\nonumber \\ \w(z)&=\w_0\sqrt{1+\sig^2};\nonumber \\ x^\LCm&=t-z \nonumber \end{align*}$$

and $\newcommand{\w}{{\rm w}} z_{r}=\omega\w_{0}^{2}/2$ is the usual Rayleigh length. The well-documented [12, 31, 32] result for the electric field of a paraxial Gaussian beam using $\newcommand{\LCp}{{+}} \newcommand{\LCperp}{{\perp}} E^\LCperp = -\partial_t A^\LCperp$, is then

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\tr}{{\rm tr}} \newcommand{\LCm}{{-}} \newcommand{\w}{{\rm w}} \newcommand{\sig}{\varsigma} \newcommand{\trm}[1]{{\rm #1}} \displaystyle \label{EPERPG} E^{\perp}_{\trm{beam}}(x) = E_0\frac{{\rm e}^{-\frac{\vert x^{\perp}\vert ^2}{\w^2}}}{\sqrt{1+\sig^2}} \cos \bigg[\omega x^\LCm +\tan^{-1}\sig - \frac{\vert x^{\perp}\vert ^2\sig}{\w^2}\bigg]. \nonumber \end{align} \tag{ 16 }$$

Finally to achieve an expression for a Gaussian paraxial pulse simply adapt the frequency spectrum $\rho(l^{0})$ to the desired pulse profile of a Gaussian distribution:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{RHOP} \rho(l^{0})=\frac{\tau}{\vert l^{0}\vert \sqrt{4\pi }}\left(\exp\left[\frac{-\tau^{2}\left(l^{0}-\omega\right)^{2}}{4}\right]+(\omega\to-\omega)\right). \nonumber \end{align} \tag{ 17 }$$

Where τ is the pulse duration. On substituting into equation (12) and computing, we yield an expression for the Gaussian paraxial pulse which differs from the beam equation (16) only in the addition of a Gaussian temporal envelope:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\tr}{{\rm tr}} \newcommand{\LCm}{{-}} \newcommand{\trm}[1]{{\rm #1}} \displaystyle \label{EPERPGp} E_{\trm{pulse}}^{\perp}(x) = {\rm e}^{-\left(\frac{x^\LCm}{\tau}\right)^{2}}E^{\perp}_{\trm{beam}}(x). \nonumber \end{align} \tag{ 18 }$$

To demonstrate the flexibility of this method, we determine the field of an axicon beam. Taking our lead from [8] we consider the case where $A^{3}$ is the only non-zero spatial component of A and, choosing to work in the Lorentz gauge $\partial\cdot A = 0$ implies that $\partial_0 A^{0}=\partial_3 A^{3}$.

The method is the same as before, we evaluate the momenta integrals in equation (2) except in this case we have the ansatz:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle \widetilde{A}^{3} = -(2\pi)^{4}{\rm i}E_{0}\delta(l^{2})\rho(l^{0}) \delta_{\varepsilon}(l^{\perp})l^{3}\theta(l^{3}l^{0}). \label{a3ansatz} \nonumber \end{align} \tag{ 19 }$$

On performing the integrals we recover the expression for $A^{3}$:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle A^3(x) = -\frac{E_{0}}{\omega}\frac{{\rm{e}}^{-\frac{\vert x^{\perp}\vert ^{2 } } {\w^{2} }}}{\sqrt{1+\varsigma^{2}}}\sin\left[\omega x^{-}+\tan^{-1}\varsigma-\frac{\vert x^{\perp }\vert ^ {2}\varsigma}{\w^{2}}\right]. \nonumber \end{align} \tag{ 20 }$$

Now to establish the $E^{3}$ component of the electric field we recall that $A^{0}$ is non zero and hence has a contribution to $E^{3}$. Therefore differentiating $A^{3}$ with respect to z gives:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle \partial_{z}A^{3}(x) = E_{0}\frac{{\rm{e}}^{\frac{-\vert x^{\perp}\vert ^{2 } } {\w^{2} }}}{\sqrt{1+\varsigma^{2}}}\cos\left[\omega x^{-} + \tan^{-1}\varsigma-\frac{\vert x^{\perp }\vert ^ {2}\varsigma}{\w^{2}}\right]. \nonumber \end{align} \tag{ 21 }$$

We have neglected a term of order $\left(\omega z_{r}\right){\hspace{0pt}}^{-1}$ since $\newcommand{\w}{{\rm w}} \omega \w_{0}\gg 1$. This implies with the gauge condition:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle A^{0}(x) = \frac{E_{0}}{\omega}\frac{{\rm{e}}^{\frac{-\vert x^{\perp}\vert ^{2 } } {\w^{2} }}}{\sqrt{1+\varsigma^{2}}}\sin\left[\omega(t-z) + \tan^{-1}\varsigma-\frac{\vert x^{\perp }\vert ^ {2}\varsigma}{\w^{2}}\right]. \nonumber \end{align} \tag{ 22 }$$

Therefore $A^{0}(x) = -A^{3}(x)$. Calculating the electric field using $E^{3} = -\partial_{t}A^{3}-\partial_{z}A^{0}$ and $E^{\perp} = -\partial_{\perp} A^{0} = \partial_{\perp} A^{3}$, and neglecting terms of $O\left((\omega z_r){\hspace{0pt}}^{-1}\right)$ we find:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle E^{3} =-\frac{E_{0}}{\sqrt{1+\varsigma^{2}}}{\rm{e}}^{\frac{-\vert x^{\perp}\vert ^{2} } {\w^{2} }}\cos\left[\omega x^{-} + \tan^{-1}\varsigma-\frac{\vert x^{\perp }\vert ^ {2}\varsigma}{\w^{2}}\right], \label{eqn:E3a} \nonumber \end{align} \tag{ 23 }$$

and we note that, ignoring the different phase of individual terms, $E^{\perp}$ is a factor smaller than $E^{3}$ of the order:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\tr}{{\rm tr}} \newcommand{\trm}[1]{{\rm #1}} \displaystyle \label{eqn:Epa} E^{\perp} \sim \frac{x^{\perp}}{z_{r}} \frac{\trm{w}_{0}^{2}}{\trm{w}^{2}}\,E^{3}, \nonumber \end{align} \tag{ 24 }$$

and hence is next-to-leading-order. Equation (23) agrees with recognised [8] expressions for a Gaussian-focused axicon beam. We can see directly from equations (23) and (24) that they produce the characteristics associated with an axicon beam: maximum longitudinal E-field component in the centre and the absence of a transverse E-field on-axis.

We seek a beam description that is representative of observations in high intensity experiments showing that the transverse intensity profile is not always well-described by a Gaussian distribution but instead can have wide tails [6, 7]. The suggestion by [37] is that a Cauchy/Lorentzian distribution would be more representative. We note that the wide tails can be seen by calculating the average root mean square (RMS) width, which for a Gaussian $\newcommand{\w}{{\rm w}} \newcommand{\e}{{\rm e}} \exp\left[-(x/\w){\hspace{0pt}}^2\right]$, is $\newcommand{\w}{{\rm w}} (\sqrt{\langle x^2\rangle}=\w/\sqrt{2})$ whereas for a Lorentzian of the form $\newcommand{\w}{{\rm w}} 1/(1+(x/\w){\hspace{0pt}}^2)$ diverges $(\sqrt{\langle x^2\rangle}\to\infty)$.

We implement a simple variation into our ansatz, demonstrating the flexibility of the method. The function $\newcommand{\eps}{\varepsilon} \newcommand{\e}{{\rm e}} \delta_{\epsilon}(l^{\perp})$ represents the focusing as it did in the Gaussian case, but we alter its form from Gaussian to a distribution with wider tails. We seek a $\delta_{\varepsilon}(l^{\perp})$ function that produces an intensity profile with wide tails and satisfies the condition (5). We choose the Poisson kernel:

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\eps}{\varepsilon} \displaystyle \lim_{\epsilon\to 0}\frac{2\,\varepsilon}{x^{2}+\eps^{2}} =\lim_{\eps\to 0}\int_{-\infty}^{\infty} \frac{{\rm d}s}{2\pi}~{\rm e}^{{\rm i}xs -\vert s\,\eps\vert\,f[x;\eps]}=\delta(x). \nonumber \end{align*}$$

To acquire a beam that is symmetric under rotations about the propagation axis, we set x² to $l^\perp\cdot l^\perp$, square the Poisson kernel and choose $\newcommand{\w}{{\rm w}} \varepsilon=1/\w_0$. This leads to:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle \label{Ldeltareg} \delta_{\varepsilon}(l^\perp)=\frac{4\w_{0}^2}{\left[1+(\w_0 l^\perp)^2\right]^2}, \nonumber \end{align} \tag{ 25 }$$

which, unlike the Gaussian beam case, does not have a simple connection to the plane-wave limit as $\lim_{\varepsilon\to 0}$$\delta_{\varepsilon}(l^\perp) =\delta[(l^1){\hspace{0pt}}^2+(l^2){\hspace{0pt}}^2]$.

We use the same polarisation set-up and method used in the Gaussian beam derivation. Starting from equation (4) with equation (25), we proceed in the same manner, finding:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\tr}{{\rm tr}} \newcommand{\LCm}{{-}} \newcommand{\w}{{\rm w}} \newcommand{\trm}[1]{{\rm #1}} \displaystyle A^{\perp}(x)&{=}\LCm2{\rm i}E_{0}\w_{0}^{2}\int {\rm d}l^{0}~ \rho(l^{0})\theta(l^{0}){\rm{e}}^{-{\rm i}l^{0}t +{\rm i}\vert l^{0}\vert z}I_{L}(l^{0},z)&\nonumber \\ &\quad+\trm{c.c.},& \label{AperpL} \nonumber \end{align} \tag{ 26 }$$

where:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle \label{IL} I_{L}(l^{0},z) = 2\pi \int_{0}^{\vert l^{0}\vert } {\rm d}\rho~\frac{\rho{\rm{e}}^{-{\rm i}b\rho^{2}}}{\left[1+(\w_0\rho)^2\right]^2} ~\,J_{0}(\vert x^{\perp}\vert \rho), \nonumber \end{align} \tag{ 27 }$$

with $b=z/2\vert l^{0}\vert$.

Expanding the Gaussian or Lorentzian in equation (27) for small argument does not give a satisfactorily convergent expression in regions close to the beam axis. However, we can make use of the Bessel multiplication theorem ([59], page 142):

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{BesselMT} J_{\nu}(\lambda z) = \lambda^{\nu} \sum_{m=0}^{\infty} \frac{(-1)^{m}(\lambda^{2}-1)^{m}(z/2)^{m}}{m!}\,J_{\nu+m}(z). \nonumber \end{align} \tag{ 28 }$$

Setting $\newcommand{\w}{{\rm w}} z=\vert x^{\perp}\vert /\w_{0}$ so $\newcommand{\w}{{\rm w}} \lambda=\w_{0}\rho$:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle J_{0}(\vert x^{\perp}\vert \rho) &= \sum_{m=0}^{\infty} \frac{(-1)^{m}((\w_{0}\rho)^{2}-1)^{m}}{m!}\nonumber \\ &\quad\times\,\left(\frac{\vert x^{\perp}\vert }{\w_{0}}\right)^{m}\! J_{m}\left(\frac{\vert x^{\perp}\vert }{\w_{0}}\right). \nonumber \end{align} \tag{ 29 }$$

A benefit of this expansion is that only the $m=0$ term has a non-zero value at the origin. Therefore, we should expect the lowest orders of this expansion to already quite well approximate the paraxial case. Taking the 'paraxial' condition $\newcommand{\w}{{\rm w}} \vert x^{\perp}\vert \ll\w_{0}$, we acquire

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\tr}{{\rm tr}} \newcommand{\w}{{\rm w}} \newcommand{\trm}[1]{{\rm #1}} \displaystyle &A^{\perp}(x)=\frac{-4\pi {\rm i}\w_{0}^{2}}{\w_{0}^{2}\omega} E_{0}J_{0}\left(\frac{\vert x^{\perp}\vert }{\w_{0}}\right)&\nonumber \\ &\int {\rm d}l^{0}\rho(l^{0})\theta(l^{0})\,{\rm{e}}^{-{\rm i}l^{0}t +{\rm i}\vert l^{0}\vert z}G\left(\vert l^{0}\vert \w_{0},\frac{z}{\vert l^{0}\vert \w_{0}^{2}}\right)+\trm{c.c.}& \nonumber \end{align} \tag{ 30 }$$

where, after making the substitution $\newcommand{\w}{{\rm w}} u=\w_{0}\rho$:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle G(\mu,\nu)=\int_{0}^{\infty}{\rm d}u\,\frac{u}{\left[1+u^{2}\right]^{2}}\,{\rm e}^{\frac{-{\rm i}\nu u^{2}}{2}} \nonumber \end{align} \tag{ 31 }$$

and the assumption $\newcommand{\w}{{\rm w}} z_{r} \gg \w_{0}$ has been used to raise the upper limit of integration. Taking $\rho(l^{0})$ as in equation (13), gives for the beam result:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle &A^{\perp}(x)=\frac{-4\pi {\rm i}}{\omega} E_{0}J_{0}\left(\frac{\vert x^{\perp}\vert }{\w_{0}}\right)&\nonumber \\ &\left[{\rm e}^{-{\rm i}\omega x^{-}}G\left(\omega\w_{0},\frac{z}{\omega\w_{0}^{2}}\right)-{\rm e}^{{\rm i}\omega x^{-}}G\left(\omega\w_{0},\frac{-z}{\omega\w_{0}^{2}}\right)\right]. \nonumber \end{align} \tag{ 32 }$$

Given that the two occurrences of the G function occur as complex conjugates, we can write $\newcommand{\w}{{\rm w}} G\left(\omega\w_{0}, \frac{z}{\omega\w_{0}^{2}}\right)=g{\rm e}^{{\rm i}\gamma}$ and rewrite $A^{\perp}$ as:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle \label{AperpP} &A^{\perp}(x)=\frac{-4\pi {\rm i}}{\omega} E_{0}J_{0}\left(\frac{\vert x^{\perp}\vert }{\w_{0}}\right)g\left[{\rm e}^{{\rm i}(\gamma-\omega x^{-})}-{\rm e}^{-{\rm i}(\gamma-\omega x^{-})}\right],&\nonumber \\ &A^{\perp}(x)=\frac{8\pi}{\omega} E_{0}J_{0}\left(\frac{\vert x^{\perp}\vert }{\w_{0}}\right)g\sin\left[\gamma-\omega x^{-}\right].& \nonumber \end{align} \tag{ 33 }$$

Since we are working in the Lorentz gauge with $A^{0}=0$ we have that $E^{\perp}=-\partial_{t}A^{\perp}$. Suppose we choose, without loss of generality, a linearly-polarised background field with polarisation three-vector $\pmb{\varepsilon}_{x}=(1, 0, 0)$, then

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \newcommand{\mbf}[1]{{\bf #1}} \displaystyle \label{Eparax} \mbf{E}=8\pi \pmb{\varepsilon}_{x} E_{0}J_{0}\left(\frac{\vert x^{\perp}\vert }{\w_{0}}\right)g\cos\left[\gamma-\omega x^{-}\right], \nonumber \end{align} \tag{ 34 }$$

where $g=\vert G\vert$ and $\newcommand{\trm}[1]{{\rm #1}} \newcommand{\tr}{{\rm tr}} \gamma=\tan^{-1}\left(\trm{Im}(G)/\trm{Re}(G)\right)$. Using equation (1), the magnetic field becomes:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\mbf}[1]{{\bf #1}} \displaystyle \mbf{B}=\pmb{\varepsilon}_{y}\partial_{z}A^{\perp}, \nonumber \end{align} \tag{ 35 }$$

where $\pmb{\varepsilon}_{y} = (0, 1, 0)$, which due to the nature of the integral G and its dependence on z does not admit a neat solution but can be resolved from equation (33). The exact result can be written in a similar form:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\mbf}[1]{{\bf #1}} \displaystyle \label{Eexact} \mbf{E}_{e}=4\pi \pmb{\varepsilon}_{x} E_{0}g_{e}\cos\left[\gamma_{e}-\omega x^{-}\right], \nonumber \end{align} \tag{ 36 }$$

where $g_{e}$ and $\gamma_{e}$ represent the modulus and argument respectively of the integral $G_{e}$:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle G_{e}=\int_{0}^{\omega\w_{0}}{\rm d}u\,\frac{u}{\left[1+u^{2}\right]^{2}}\,{\rm e}^{\frac{-{\rm i} z u^{2}}{2\omega\w_{0}^{2}}}\,J_{0}\left(\frac{\vert x^{\perp}\vert }{\w_{0}}u\right). \nonumber \end{align} \tag{ 37 }$$

To compute the integral $G_{e}$ requires a numerical approach. Note: to consider a Lorentzian pulse we can again model the frequency distribution $\rho(l^{0})$ with a Gaussian distribution equation (17). The pulse calculation is not altogether straightforward and requires careful manipulation to produce an analytic paraxial result. By making the assumption that $\vert l^{0}\vert$ is peaked around the central frequency ω we can approximate $G(\vert l^{0}\vert, z)\to G(\omega, z)$, and perform the momentum integrals.

Here, we wish to analyse the transverse properties of the beam, and for this we calculate the energy density $T^{00}$ from the energy-momentum tensor [60]:

$$\begin{align*} \newcommand{\e}{{\rm e}} \displaystyle T^{\alpha\beta}=\frac{1}{4\pi}\left[g^{\alpha\mu}F_{\mu\lambda}F^{\lambda\beta}+\frac{1}{4}g^{\alpha\beta}F_{\mu\lambda}F^{\mu\lambda}\right]. \nonumber \end{align*}$$

To simplify calculations we consider a linear polarisation and determine whether the $A^{3}$ makes any significant contribution. Given that $\partial\cdot A=0$ that implies that:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \partial_{\perp}A^{\perp}=-\partial_{3}A^{3}. \nonumber \end{align} \tag{ 38 }$$

Taking the derivative of equation (33) with respect to $x^{\perp}$ gives for the paraxial case:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle \partial_{3}A^{3}=\frac{4\pi}{\omega\w_{0}} E_{0}J_{1}\left(\frac{\vert x^{\perp}\vert }{\w_{0}}\right)g\sin\left[\gamma-\omega x^{-}\right], \nonumber \end{align} \tag{ 39 }$$

since the dependency on the co-ordinate z in g and γ is much weaker than omega, we have that:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\w}{{\rm w}} \displaystyle A^{3}\sim \frac{1}{\omega\w_{0}}J_{1}\left(\frac{\vert x^{\perp}\vert }{\w_{0}}\right),\qquad A^{\perp}\sim J_{0}\left(\frac{\vert x^{\perp}\vert }{\w_{0}}\right), \nonumber \end{align} \tag{ 40 }$$

which for small arguments (e.g. when $\newcommand{\w}{{\rm w}} \vert x^{\perp}\vert \ll \w_{0}$), we can say:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle A^{3}\sim \frac{\vert x^{\perp}\vert }{z_{r}},\qquad A^{\perp}\sim 1, \nonumber \end{align} \tag{ 41 }$$

implying that $A^{3}$ is small compared to $A^{\perp}$ and given that $T^{00}$ is calculated using the square of the potential we suggest the contribution from $A^{3}$ is negligible. The energy density then becomes:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T^{00}=\frac{1}{8\pi}\left[\left(\partial_{0}A^{1}\right)^{2}+\left(\partial_{2}A^{1}\right)^{2}+\left(\partial_{3}A^{1}\right)^{2}\right], \nonumber \end{align} \tag{ 42 }$$

which can also be written as $T^{00}=\left(E^{2}+B^{2}\right)/8\pi$, which is the the mod-square of the Poynting vector.

In this section we analyse the properties of our Lorentzian beam, all plots show the energy density $T^{00}$ with minimum beam waist $\newcommand{\w}{{\rm w}} \w_{0}=\lambda$ (unless otherwise stated) to accentuate any focusing effects. We first confirm from figure 2 that our choice to use a Lorentzian focusing function does indeed produce a beam with wider tails than the Gaussian beam.

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We further compare with the Gaussian beam by evaluating our paraxial approximation, which we expect to be good on-axis. Since the only dependence on the transverse co-ordinate $x^{\perp}$ was found in the Bessel function equation (33) we find that the 'paraxial' result is very accurate on-axis, ($J_{0}(0)=1$), hence the second paraxial approximation using the Bessel multiplication theorem equation (28) does not apply. This is shown in figure 3.

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Off-axis however, there are significant variations between the paraxial and exact result. Firstly we observe from figure 4 that the paraxial transverse energy density profile has wide oscillating tails representative of the Bessel function $\newcommand{\w}{{\rm w}} J_{0}(x^{\perp}/\w_{0})$, whereas the exact beam result figure 5 exhibits a definite width and is non-oscillatory. This is to be expected since the paraxial approximation we made had the effect of removing the transverse momentum dependence from the integrand but gave the paraxial approximation with a $\newcommand{\w}{{\rm w}} J_{0}(x^{\perp}/\w_{0})$ envelope.

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The paraxial approximation presented is leading order, but including higher order terms in the Bessel approximation would have a damping effect on the oscillations off-axis. It is worth noting that the first paraxial approximation equation (10) has little effect for $\newcommand{\w}{{\rm w}} \w_{0}=\lambda$ as is the case for the Gaussian beam figure 6. Hence to improve the paraxial Lorentzian result significantly, higher order terms in the Bessel approximation should be included. This Bessel envelope has a significant effect on the beam shape. As observed from figure 7, the exact solution has the expected shape of a focused beam (a narrowing width towards the focus), however for the paraxial case we observe that $T^{00}$ seems to be bigger than it should be at an appreciable transverse distance from the focus, near the temporal peak. This has the effect of a beam broadening towards the centre, i.e. the beam will have a convex rather than concave shape. This brings into question the validity of using this leading-order paraxial approximation for describing any off-axis phenomena. Therefore, we consider more closely whether the paraxial approximation is representative within a certain regime. In figure 8 we plot how the field depends on transverse co-ordinate, in the plane of constant longitudinal co-ordinate, for three different cases. We see that the main peak at the centre of the paraxial profile is a good approximation within one Rayleigh length ($z_{r}$). However due to the absence of width broadening for the paraxial case, we see that the approximation becomes poorer for $z > z_{r}$. Figure 9 displays profiles of the Lorentzian beam for the same beam cross sections.

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