Thermal aberrations and structured light I: analytical model for structured pumps and probes

Abstract

Structured light, where optical beams are tailored in amplitude, phase and polarisation to some desired profile, has become topical of late, fuelled by the ease at which such fields can be created internal and external to the source. In this treatise, part I of a two part series, we consider the thermal effects (stress, lensing and phase aberrations) associated with high-power structured light, where the structure may be the pump, optically inducing the thermal effects in the medium, or the probe, experiencing thermally induced optical aberrations. We outline a general theory for arbitrary structured light pumps and probes, reducing to the prior studies of Gaussian and flat-top beams as special cases. We illustrate the power of the model using the structure of light as a new degree of freedom with which to mitigate thermally induced optical aberrations. Finally, in part II of this composite work (Scholes and Forbes, Appl Phys B, 2021. https://doi.org/10.1007/s00340-021-07656-z), we experimentally demonstrate the phase aberration predictions using a digital micro-mirror device for real-time simulation of such high-power thermal effects in a cheap, fast and versatile manner, without the need for elaborate high-power experiments. Our work brings together the disparate fields of thermal modelling and structured light, providing a framework for future work in the creation and delivery of high-power structured light fields.

Appendices

Appendix A Solution to the non-homogenous heat equation

For radially symmetric systems, we begin with

$$\begin{aligned} \begin{aligned} u(r,t)&= \sum _{n=1}^{\infty }p_n J_0(\gamma _n r)\exp (-\Delta t),\\ p_n&= \frac{2}{(RJ_1(\beta _n))^2}\int _0^R u(r,t') rJ_0(\gamma _n r) dr. \end{aligned} \end{aligned}$$

(40)

Assume the following form for the solution to Q(r, t).

$$\begin{aligned} \begin{aligned} Q(r,t) = \sum _{n=1}^{\infty }q_nR_n(r). \end{aligned} \end{aligned}$$

(41)

Replacing $R_n(r)$ with the Eigen-fuction

$$\begin{aligned} \begin{aligned} Q(r,t) = \sum _{n=1}^{\infty }q_n\textit{J}_0(\gamma _nr). \end{aligned} \end{aligned}$$

(42)

The Bessel functions have a known orthogonality relation given by

$$\begin{aligned} \begin{aligned} \int _{0}^{R}{} \textit{J}_0(\gamma _nr)r\textit{J}_0(\gamma _mr)dr = \frac{R^2}{2}\delta _{m,n}[\textit{J}_1(\beta _n)]^2. \end{aligned} \end{aligned}$$

(43)

Using Fourier’s trick we can exploit this orthogonality to isolate the function $q_n$

$$\begin{aligned} \int _{0}^{R}Q(r,t)r\textit{J}_0(\gamma _nr)dr&= \int _{0}^{R}\sum _{n=1}^{\infty }q_n\textit{J}_0(\gamma _nr)r\textit{J}_0(\gamma _mr)dr,\nonumber \\ \int _{0}^{R}Q(r,t)r\textit{J}_0(\gamma _nr)dr&= q_n\frac{R^2}{2}[\textit{J}_1(\beta _n)]^2, \end{aligned}$$

(44)

$$\begin{aligned} q_n = \frac{2}{(R\textit{J}_1(\beta _n))^2}\int _{0}^{R}Q(r,t)r\textit{J}_0(\gamma _nr)dr. \end{aligned}$$

(45)

The solution u(r, t) to the PDE is assumed to have the form

$$\begin{aligned} \begin{aligned} u(r,t) = \sum _{n=1}^{\infty }a_n(t)R_n(r). \end{aligned} \end{aligned}$$

(46)

Substituting this form back into the PDE gives

$$\begin{aligned} \begin{aligned}&\frac{\partial u}{\partial t} = \sum _{n=1}^{\infty } \frac{\partial a_n}{\partial t}R_n,\\ \end{aligned} \end{aligned}$$

(47)

and

$$\begin{aligned} \begin{aligned}&\nabla ^2u = \sum _{n=1}^{\infty }a_n\nabla ^2R_n. \end{aligned} \end{aligned}$$

(48)

It should be noted that Eq. 48 implicitly assumes that the derivative of an infinite sum is equivalent to the sum of its derivatives. This is generally true for PDEs with homogeneous boundary conditions,

$$\begin{aligned} \begin{aligned} \sum _{n=1}^{\infty } \frac{\partial a_n}{\partial t}R_n - \alpha \left(\sum _{n=1}^{\infty }a_n\nabla ^2R_n\right)&= Q(r,t),\\ \sum _{n=1}^{\infty } \frac{\partial a_n}{\partial t}R_n - \alpha \left(\sum _{n=1}^{\infty }a_n\nabla ^2R_n\right)&= \sum _{n=1}^{\infty }q_n R_n(r).\\ \end{aligned} \end{aligned}$$

(49)

To avoid calculating the Laplacian of the Eigen-function we can use the following relation. Since $R_n(r)$ is an Eigen-function it obeys the relationship

$$\begin{aligned} \begin{aligned} R_n^{''}(r) + \frac{1}{r}R_n^{'}(r) + \lambda _n^2R_n = 0. \end{aligned} \end{aligned}$$

(50)

Here the prime $'$ indicates the derivative. Substituting gives

$$\begin{aligned} \begin{aligned} \sum _{n=1}^{\infty } \frac{\partial a_n}{\partial t}R_n - \alpha \left\{ \sum _{n=1}^{\infty }a_n[-(\lambda _n)^2R_n]\right\}&= \sum _{n=1}^{\infty }q_nR_n,\\ \sum _{n=1}^{\infty }\left[ R_n\left( \frac{\partial a_n}{\partial t} + \alpha \lambda _n^2a_n - q_n\right) \right]&= 0,\\ \sum _{n=1}^{\infty }\left[ \textit{J}_0(\gamma _nr)\left( \frac{\partial a_n}{\partial t} + \Delta a_n - q_n\right) \right]&= 0.\\ \end{aligned} \end{aligned}$$

(51)

Since $\textit{J}_0(\gamma _nr) \ne 0$ for all r, $\frac{\partial a_n}{\partial t} + \Delta a_n - q_n =0$. This is now an ODE problem. Consider also the initial condition

$$\begin{aligned} \begin{aligned} u(r,0)&= 0,\\ \sum _{n=1}^{\infty }a_n(0)R_n(r)&= 0,\\ \sum _{n=1}^{\infty }a_n(0)\textit{J}_0(\gamma _nr)&= 0.\\ \end{aligned} \end{aligned}$$

(52)

thus

$$\begin{aligned} \begin{aligned} a_n(0)&= 0. \end{aligned} \end{aligned}$$

(53)

Hence we solve

$$\begin{aligned} \begin{aligned} \frac{\partial a_n}{\partial t} + \Delta a_n - q_n&=0,\\ \end{aligned} \end{aligned}$$

(54)

with

$$\begin{aligned} a_n(0)&= 0, \end{aligned}$$

(55)

$$\begin{aligned} \frac{{\partial a_{n} }}{{\partial t}} & = q_{n} - \Delta a_{n} , \\ \frac{{\frac{{\partial a_{n} }}{{\partial t}}}}{{q_{n} - \Delta a_{n} }} & = 1, \\ \int {\frac{{\frac{{\partial a_{n} }}{{\partial t}}}}{{q_{n} - \Delta a_{n} }}} dt & = \int 1 dt, \\ - \frac{{\log (q_{n} - \Delta a_{n} )}}{\Delta } & = t + C_{1} , \\ - \log (q_{n} - \Delta a_{n} ) & = (t + C_{1} )\Delta . \\ \end{aligned}$$

(56)

Therefore

$$\begin{aligned} \begin{aligned} a_n(t) = \frac{[-\exp (-\Delta t) +1]q_n}{\Delta }. \end{aligned} \end{aligned}$$

(57)

For non-zero initial conditions where $p_n$ is non-trivial, Eq. 57 becomes

$$\begin{aligned} \begin{aligned} a_n(t) = \frac{\exp (-\Delta t)\{[-1 + \exp (\Delta t)]q_n + \Delta p_n\}}{\Delta }. \end{aligned} \end{aligned}$$

(58)

Thus finally

$$\begin{aligned} \begin{aligned} u(r,t)&= \sum _{n=1}^{\infty }\frac{\exp (-\Delta t)\{[-1 + \exp (\Delta t)]q_n + \Delta p_n\}}{\Delta }{} \textit{J}_0(\gamma _nr).\\ \end{aligned} \end{aligned}$$

(59)

With

$$\begin{aligned} \begin{aligned} q_n(t)&= \frac{2}{(R\textit{J}_1(\beta _n))^2}\int _{0}^{R}Q(r,t)r\textit{J}_0(\gamma _nr)dr,\\ \end{aligned} \end{aligned}$$

(60)

and

$$\begin{aligned} \begin{aligned} p_n&= \frac{2}{(RJ_1(\beta _n))^2}\int _0^R u(r,t') rJ_0(\gamma _n r) dr,\\ \end{aligned} \end{aligned}$$

(61)

for $t'<t$.

For Cartesian symmetry, a procedurally identical process is used in two dimensions. Beginning with the solution to the homogenous heat equation in Cartesian coordinates

$$\begin{aligned} \begin{aligned} u(x,y,t)&= \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }p_{m,n}\sin (\gamma _n^\dagger x)\sin (\gamma _n^\dagger y)\exp (-\Delta ^\dagger t),\\ p_{m,n}&= \frac{4}{D^2}\int _0^D\int _0^Du(x,y,t')\sin (\gamma _n^\dagger x)\sin (\gamma _n^\dagger y)dxdy, \end{aligned} \end{aligned}$$

(62)

the $\sin ()$ functions are identified as the Eigen-functions for this system. Again by exploiting orthogonality the source Q(x, y, t) can be decomposed as

$$\begin{aligned} \begin{aligned} q_n = \frac{4}{D^2}\int _0^D\int _0^D Q(x,y,t)\sin (\gamma _n^\dagger x)\sin (\gamma _n^\dagger y)dxdy.\\ \end{aligned} \end{aligned}$$

(63)

Assuming

$$\begin{aligned} \begin{aligned} u(x,y,t) = \sum _{m=1}^{\infty }\sum _{n=1}^{\infty } a_{nm}S_nS_m, \end{aligned} \end{aligned}$$

(64)

leads to

$$\begin{aligned} \begin{aligned} \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\left\{ \sin (\gamma _n^\dagger x)\sin (\gamma _n^\dagger y)\left[ \frac{\partial a_{nm}}{\partial t}+\Delta ^\dagger a_{nm} - q_{nm} \right] \right\} = 0, \end{aligned} \end{aligned}$$

(65)

and so for

$$\begin{aligned} u(x,y,0)\,= 0,\nonumber \\ u(D,D,t)\,= 0,\nonumber \\ u(0,0,t)\,= 0,\end{aligned}$$

(66)

$$\begin{aligned} a_{nm} = \frac{q_{nm} - q_{nm}\exp (\Delta ^\dagger t)}{\Delta ^\dagger }. \end{aligned}$$

(67)

Generally speaking, the approach presented here may be expanded to n dimensions. To achieve this, the solution will comprise n summations with n Eigen functions (which are determined by the geometry and the boundary conditions) and n integrals in the decompositions of p and q.

Appendix B Derivation of the stress

The stress $\sigma (r,t)$ experienced by a material is found using the relationship between the radial stress $\sigma _r$ and the angular stress $\sigma _{\theta }$,

$$\begin{aligned} \begin{aligned} \sigma (t)&= |\sigma _{\theta } - \sigma _r|.\\ \end{aligned} \end{aligned}$$

(68)

With

$$\begin{aligned} \sigma _{\theta }&= \frac{\breve{\gamma } Y}{1-\breve{\nu }}\Bigl [\frac{1}{R^2}\int _{0}^{R}u(r)rdr + \frac{1}{r^2}\int _{0}^{r}u(\acute{r})\acute{r}d\acute{r} - u(r)\Bigr ],\nonumber \\ \sigma _{r}&= \frac{\breve{\gamma } Y}{1-\breve{\nu }}\Bigl [\frac{1}{R^2}\int _{0}^{R}u(r)rdr - \frac{1}{r^2}\int _{0}^{r}u(\acute{r})\acute{r}d\acute{r}\Bigr ], \end{aligned}$$

(69)

$$\begin{aligned} \int _{0}^{R}u(r)rdr&=\int _{0}^{R}\sum _{n=1}^{\infty }\Bigl [\frac{(-\exp (-\Delta t) + 1)q_n}{\Delta }{} \textit{J}_0(\gamma _n r) +\nonumber \\&\exp (-\Delta t)p_n\textit{J}_0(\gamma _n r)\Bigr ]rdr. \end{aligned}$$

(70)

Assuming the sum and the integral are interchangeable

$$\begin{aligned} \begin{aligned} \int _{0}^{R}u(r)rdr&=\sum _{n=1}^{\infty }\Bigl \{\frac{[-\exp (-\Delta t) + 1]q_n}{\Delta }\int _{0}^{R}{} \textit{J}_0(\gamma _n r)rdr +\\&\exp (-\Delta t)p_n\int _{0}^{R}{} \textit{J}_0(\gamma _n r)\Bigr \}rdr. \end{aligned} \end{aligned}$$

(71)

Using the integral property of Bessel Functions,

$$\begin{aligned} \begin{aligned} \int _{0}^{a}x\textit{J}_0(x)dx = a\textit{J}_1(a). \end{aligned} \end{aligned}$$

(72)

Using variable substitution. Let

$$\begin{aligned} \begin{aligned} \gamma _n r&= x,\\ dr&= \frac{dx}{\gamma _n}.\\ \end{aligned} \end{aligned}$$

(73)

For

$$\begin{aligned} \begin{aligned} r&= R,\\ x&=\gamma _n R,\\&=\beta _n,\\ \int _{0}^{R}u(r)rdr&=\sum _{n=1}^{\infty }\Bigl \{\frac{[-\exp (-\Delta t) + 1]q_n}{\Delta }\int _{0}^{R}{} \textit{J}_0(\gamma _n r)rdr \\+&\exp (-\Delta t)p_n\int _{0}^{R}{} \textit{J}_0(\gamma _n r)\Bigr \}rdr,\\ \end{aligned} \end{aligned}$$

(74)

becomes

$$\begin{aligned} \begin{aligned} \int _{0}^{R}u(r)rdr&=\sum _{n=1}^{\infty }\Bigl \{\frac{[-\exp (-\Delta t) + 1]q_n}{\Delta }\int _{0}^{\beta _n}{} \textit{J}_0(x)\frac{x}{\gamma _n}\frac{dx}{\gamma _n} \\+&\exp (-\Delta t)p_n\int _{0}^{\beta _n}{} \textit{J}_0(x)\frac{x}{\gamma _n}\frac{dx}{\gamma _n}\Bigr \}. \end{aligned} \end{aligned}$$

(75)

Using equation 72

$$\begin{aligned} \begin{aligned} \int _{0}^{R}u(r)rdr&=\sum _{n=1}^{\infty }\Bigl \{\frac{[-\exp (-\Delta t) + 1]q_n}{\Delta }\Bigl [\frac{1}{\gamma _n^2}\beta _n\textit{J}_1(\beta _n)\Bigr ] +\\&\exp (-\Delta t)p_n\Bigl [\frac{1}{\gamma _n^2}\beta _n\textit{J}_1(\beta _n)\Bigr ]\Bigr \},\\ \int _{0}^{R}u(r)rdr&=\sum _{n=1}^{\infty }\Bigl (\Bigl [\frac{R^2}{\beta _n}{} \textit{J}_1(\beta _n)\Bigr ]\Bigl \{\frac{[-\exp (-\Delta t) + 1]q_n}{\Delta } +\\&\exp (-\Delta t)p_n\Bigr \}\Bigr ). \end{aligned} \end{aligned}$$

(76)

Performing an identical operation with a dummy variable $\acute{r}$, it is found

$$\begin{aligned} \begin{aligned} \int _{0}^{r}u(\acute{r})\acute{r}d\acute{r}&=\sum _{n=1}^{\infty }\Bigl (\Bigl [\frac{R}{\beta _n}r\textit{J}_1(\gamma _nr)\Bigr ]\Bigl \{\frac{[-\exp (-\Delta t) + 1]q_n}{\Delta } \\+&\exp (-\Delta t)p_n\Bigr \}\Bigr ).\\ \end{aligned} \end{aligned}$$

(77)

For the temperature function u(r, t) where the source $Q(r,t) = 0$. Equations 76 and 77 reduce to,

$$\begin{aligned} \begin{aligned} \int _{0}^{R}u(r)rdr&=\sum _{n=1}^{\infty }\left[ \frac{R^2}{\beta _n}{} \textit{J}_1(\beta _n)\exp (-\Delta t)p_n\right] ,\\ \end{aligned} \end{aligned}$$

(78)

and

$$\begin{aligned} \begin{aligned} \int _{0}^{r}u(\acute{r})\acute{r}d\acute{r}&=\sum _{n=1}^{\infty }\left[ \frac{R}{\beta _n}r\textit{J}_1(\gamma _nr)\exp (-\Delta t)p_n\right] .\\ \end{aligned} \end{aligned}$$

(79)

Thus for $Q\ne 0$ the stress is

$$\begin{aligned} \begin{aligned} \sigma (t)&= \Bigl |\frac{\breve{\gamma } Y}{1 - \breve{\nu }}\Bigl \{\frac{2}{r}\sum _{n=1}^{\infty }\Bigl [\frac{-q_n\exp (-\Delta t) + q_n}{\Delta }\\& \quad +\frac{\Delta p_n(n)\exp (-\Delta t)}{\Delta }\Bigr ]\frac{R}{\beta _n}J_1(\gamma _nr) - u(r)\Bigr \}\Bigr |. \end{aligned} \end{aligned}$$

(80)

For $Q=0$

$$\begin{aligned} \begin{aligned} \sigma (t) = \left| \frac{\breve{\gamma } Y}{1 - \breve{\nu }}\left\{ \frac{2}{r}\sum _{n=1}^{\infty }\left[ p_n(n)\exp (-\Delta t)\right] \frac{R}{\beta _n}J_1(\gamma _nr) - u(r)\right\} \right| .\\ \end{aligned} \end{aligned}$$

(81)