%%\documentclass[lineno,pdflatex,sn-basic]{sn-jnl}% Basic Springer Nature Reference Style/Chemistry Reference Style
%%\documentclass[sn-basic]{sn-jnl}% Basic Springer Nature Reference Style/Chemistry Reference Style
 
%%\documentclass[pdflatex,sn-nature]{sn-jnl}% Style for submissions to Nature Portfolio journals
\documentclass[pdflatex,sn-basic]{sn-jnl}% Basic Springer Nature Reference Style/Chemistry Reference Style

\usepackage{graphicx}%
\usepackage{multirow}%
\usepackage{amsmath,amssymb,amsfonts}%
\usepackage{amsthm}%
\usepackage{mathrsfs}%
\usepackage[title]{appendix}%
\usepackage{xcolor}%
\usepackage{textcomp}%
\usepackage{manyfoot}%
\usepackage{booktabs}%
\usepackage{algorithm}%
\usepackage{algorithmicx}%
\usepackage{algpseudocode}%
\usepackage{listings}%
\usepackage{gensymb}%
\usepackage{pgfplots}%
\usepackage{hyperref}
\usepackage[nameinlink,capitalize]{cleveref}
% optional helper macros:
\newcommand{\Eq}[1]{Eq.~\eqref{#1}}
\newcommand{\Eqs}[1]{Eqs.~\cref{#1}}

\theoremstyle{thmstyleone}%
\newtheorem{theorem}{Theorem}%  meant for continuous numbers
%%\newtheorem{theorem}{Theorem}[section]% meant for sectionwise numbers
%% optional argument [theorem] produces theorem numbering sequence instead of independent numbers for Proposition
\newtheorem{proposition}[theorem]{Proposition}% 
%%\newtheorem{proposition}{Proposition}% to get separate numbers for theorem and proposition etc.

\theoremstyle{thmstyletwo}%
\newtheorem{example}{Example}%
\newtheorem{remark}{Remark}%

\theoremstyle{thmstylethree}%
\newtheorem{definition}{Definition}%

\raggedbottom
%%\unnumbered% uncomment this for unnumbered level heads

\begin{document}

\title[title]{Supplementary Materials
}

\author*[1]{\fnm{Aditya} \sur{Tummala}}\email{adityatummala@college.harvard.edu}

\author[2]{\fnm{Francesca} \sur{Marturano}}\email{fmarturano@mgh.harvard.edu}

\author[2]{\fnm{Giorgio} \sur{Bonmassar}}\email{giorgio.bonmassar@mgh.harvard.edu}

\affil*[1]{\orgdiv{Biomedical Engineering}, \orgname{Paulson School of Engineering and Applied Sciences, Harvard University}, \orgaddress{\city{Boston}, \state{MA} \postcode{02134}, \country{U.S.A}}}

\affil[2]{\orgdiv{Department of Radiology, Athinoula A. Martinos Center for Biomedical Imaging}, \orgname{Harvard Medical School/Massachusetts General Hospital}, \orgaddress{\city{Charlestown}, \state{MA} \postcode{02129}, \country{U.S.A}}}

\maketitle

\section{Complete Derivation}\label{completeDer}

First, we simplify the particle emission source to a point at $S=(0,0,0)$. We center a cylindrical substrate $W$ of radius \(R\) about the z-axis at $C=(0,D,0)$, where $D$ is defined as the distance between substrate center $C$ and the emission source $S$, and take a cross-section in the x-y plane (Figure 1). We parametrize the bottom hemisphere of the circular cross–section of the cylinder by
\begin{equation}\label{P_theta}
P(\theta)=(R\cos\theta,\;R\sin\theta + D),\qquad \theta\in[\pi,2\pi].
\end{equation}
as deposition only occurs on the bottom hemisphere of $W$ facing the substrate. $\hat{n_p}(\theta)$ is the outward unit normal at deposition point $P$ along the surface of the cross section and $\hat{n_{p0}}(\theta)$ is the same unit normal, but shifted to begin at $S$
\begin{equation}\label{n_p0}
\hat{n_{p0}}(\theta)= \boldsymbol{\hat{\imath}}\cos\theta + \boldsymbol{\hat{\jmath}}\sin\theta\
\end{equation} 
where $\boldsymbol{\hat{\imath}}$ and $\boldsymbol{\hat{\jmath}}$ are versors in the x-axis and y-axis directions respectively and $\hat{n_p}(\theta) \parallel \hat{n_{p0}}(\theta)$. Let the particle emission source be simplified to a point at \(S=(0,0)\) (a distance \(D>0\) below the center of the substrate cylinder) whereby the outward unit normal to the source is $\hat{n}_s = \boldsymbol{\hat{\jmath}}$. \footnote{Note: This geometric setup is defined so that film growth occurs in the negative $\hat{\jmath}$ direction, analogously to experimental PVD systems.}

\noindent A ray of particles from the source $S$ to the substrate at point $P$ is therefore

\begin{equation} \label{vec(r)}
\vec{r}(\theta) = \overrightarrow{SP(\theta)}
=\boldsymbol{\hat{\imath}}R\cos\theta+ \boldsymbol{\hat{\jmath}} (R\sin\theta + D), \qquad \theta \in [\pi, 2\pi]
\end{equation}

\noindent with magnitude

\begin{equation} \label{mag_vec(r)}
\|\vec{r}(\theta)\| = \sqrt{R^2 + D^2 + 2RD\sin{\theta}}
\end{equation}

\noindent and thus unit vector

\begin{equation} \label{unit_vec(r)}
\hat{r}(\theta) = \frac{\vec{r}(\theta)}{\|\vec{r}(\theta)\|} = \frac{\boldsymbol{\hat{\imath}}R\cos\theta+ \boldsymbol{\hat{\jmath}} (R\sin\theta + D)}{\sqrt{R^2 + D^2 + 2RD\sin{\theta}}}
\end{equation} 

\noindent Then, the cosine of the emission angle $\psi(\theta)$ between $\hat{n_s}$ and $\hat{r}(\theta)$ can be defined as 
\begin{equation} \label{cos(psi)}
\cos(\psi(\theta)) = \hat{r}(\theta) \cdot \hat{n}_s = \frac{(R\cos{\theta})(0) + (R\sin{\theta} + D)(1)}{{\|\vec{r}\|}}=\frac{R\sin\theta + D}{\|\vec{r}\|}
\end{equation}

\noindent and the incidence angle $\phi(\theta)$ between $\hat{n_p}$ and -$\hat{r}(\theta)$ can be defined as

\begin{equation} \label{cos(phi)}
\cos(\phi(\theta))
= -\hat{r}(\theta) \cdot \hat{n}_{p0}
= \frac{(-R\cos{\theta})(\cos{\theta})+(-R\sin{\theta}-D)(\sin{\theta})}{\|\vec{r}\|}
=\frac{-R-D\sin\theta}{\|\vec{r}\|}
\end{equation}

\noindent Since we seek to model a differential particle flux onto the differential area of deposition, we can use the property of solid angle to do so, as depicted in Figure \ref{solid_angle}. Treating the differential deposition area as $dA_p$ and the differential source area as $dA_s$, we have that the differential solid angle $d\omega$ $[sr]$, can be described as (Wang et. al. 2018)

\begin{equation} \label{solidangle}
d\omega = \frac{dA_p \cos(\phi(\theta))}{\|\vec{r}(\theta)\|^2}
\end{equation} 

\noindent Thus, in the cases when $\phi(\theta) = 0$, (by Figure 1, this occurs when $\theta = \frac{3\pi}{2}$), deposition is maximal, as $d\omega$ is at its maximum ($\cos{(0)}=1 \Rightarrow d\omega = \frac{dA_p}{\|\vec{r}(\frac{3\pi}{2})\|^2}$).

\subsection{Deposition Profile}\label{sec2.2}
With this geometric setup, we can describe $d\Phi_{emit}$ as the differential deposition rate at $dA_p$ and $\mathcal{L}_s$ [$N_p \space s^{-1} \space m^{-2} \space sr^{-1}]$ as the particle radiance where $N_p$ is defined as the number of particles. Then, as discussed in (Wang et. al. 2018), we have that
\begin{equation} \label{dPhi_emit1}
d\Phi_{emit}(\theta) = \mathcal{L}_s \cos^n(\psi(\theta)) \space dA_s \space d\omega, \qquad [\Phi_{emit}] = [N_p \space s^{-1}]
\end{equation} 
Introducing equation \Eq{solidangle} into \Eq{dPhi_emit1}, we attain
\begin{equation} \label{dPhi_emit2}
d\Phi_{emit}(\theta) = \mathcal{L}_s \frac{cos^n(\psi(\theta)) \space\cos(\phi(\theta))}{\|\vec{r}(\theta)\|^2}dA_s \space dA_p
\end{equation} 

\noindent To determine the local particle deposition density $dJ_p(\theta)$ $[N_p \space m^{-2} \space s^{-1}]$, we divide by $dA_p$:

\begin{equation} \label{dJ_p}
dJ_p(\theta) = \mathcal{L}_s \frac{cos^n(\psi(\theta)) \space \cos(\phi(\theta))}{\|\vec{r}(\theta)\|^2} \space dA_s 
\end{equation} 

\noindent Since we treat the source as a point emitter, integrating over its surface $A_s$ gives us
\begin{equation} \label{dJ_p2}
J_p(\theta) = \int_{A_s}\mathcal{L}_s \frac{cos^n(\psi(\theta)) \space \cos(\phi(\theta))}{\|\vec{r}(\theta)\|^2} dA_s
\end{equation} 

\noindent We collapse constants by defining constant $C=\mathcal{L}_s \space A_s$ $[N_p \space s^{-1}]$ and solve for $J_p(\theta)$:

\begin{equation} \label{dJ_p3}
J_p(\theta) = C \frac{cos^n(\psi(\theta)) \space \cos(\phi(\theta))}{\|\vec{r}(\theta)\|^2}
\end{equation} 

\noindent Next, we define $P_s(\theta) \in [0,1]$ as the sticking probability, i.e., the probability that a deposited particle adheres to those deposited before it. The rate of film growth $\dot{f}(t, \theta)$ $[m \space s^{-1}]$, then, is given by multiplying $J_p(\theta)$ by the particle volume $V_{p} [m^3 \space N_p^{-1}]$. Combining these factors, we define constant $K = C P_s V_{p}$ $[m^3 \space s^{-1}]$. Incorporating this into \Eq{dJ_p3}, we get
\begin{equation} \label{f_dot}
\dot{f}(t', \theta) = K\frac{cos^n(\psi(\theta)) \space \cos(\phi(\theta))}{\|\vec{r}(\theta)\|^2}
\end{equation} 
Integrating over time, we calculate $f(t, \theta)$ $[m]$,
\begin{equation} \label{f1}
f(t, \theta) = \int_{0}^{t}\dot{f}(t', \theta)dt'= Kt\frac{cos^n(\psi(\theta)) \space\cos(\phi(\theta))}{\|\vec{r}(\theta)\|^2}
\end{equation}
Substituting \Eq{cos(psi)} and \Eq{cos(phi)} into \Eq{f1},
\begin{equation*} \label{f2}
f(t, \theta) = Kt\frac{(\frac{(D+R\sin\theta)}{\|\vec{r}(\theta)\|})^n (\frac{(-R-D\sin\theta)}{\|\vec{r}(\theta)\|})}{\|\vec{r}(\theta)\|^2}
\end{equation*} 

\begin{equation} \label{f3}
f(t, \theta) = -Kt\frac{(D+R\sin\theta)^n (R+D\sin\theta)}{(R^2+D^2+2RD\sin\theta)^{\frac{n+3}{2}}}
\end{equation} 

\noindent However, note that $\lim_{\theta \to \pi, 2\pi} f(t, \theta) <0$. Specifically, this occurs where the term $R+D\sin(\theta) > 0$ (when $\sin(\theta)>-R/D$), i.e. near the boundaries approaching $\theta \to \pi, 2\pi$. In the raw analytic form, the model predicts small negative values near these edges because the cosine projection term allows for slight “overhang” where the surface normal begins to turn away from the source. However, in real physical vapor deposition, these regions simply receive negligible flux rather than negative deposition. Since we assume $D \gg R$, as is with almost all PVD systems, we can simplify the $R+D\sin(\theta)$ term to $D\sin(\theta)$, which is always negative across $\theta = \pi \to 2\pi$.
Therefore, we attain a final equation, modeling deposition at any point $P(\theta)$ along the circumference of the cylindrical substrate (fixed $R$) at time $t$:
\begin{equation} \label{f_final}
\boxed{
f(t, \theta) = -Kt\frac{(D+R\sin\theta)^n (D\sin\theta)}{(R^2+D^2+2RD\sin\theta)^{\frac{n+3}{2}}}
}
\end{equation} 

\noindent From above, we have that at some time $t$, substrate radius $R$, and angle around the substrate $\theta \in [\pi, 2\pi]$, the thickness of deposition at that point can be represented as \Eq{f_final}. To evaluate the total conductive cross-section formed by the deposited material $A_{cs}(t)$ $[m^2]$, we integrate \Eq{f_final} over angle $\theta$ from $\pi \to 2\pi$ (cross section in the x-y plane):

\begin{equation} \label{Acs}
A_{cs}(t) = \int_{\pi}^{2\pi}f(t, \theta) Rd\theta, \qquad \theta \in [\pi, 2\pi]
\end{equation}

\noindent For typical deposition conditions, the distance from the source to the substrate is significantly larger than the substrate radius ($R \ll D$). We therefore define the small parameter $\epsilon = \frac{R}{D} \ll 1$, and expand the above expression in powers of $\epsilon$ to isolate the leading-order dependence of the deposited cross-section on system geometry. Substituting \Eq{f_final} into \Eq{Acs}, setting $R=\epsilon D$:

\begin{equation*} \label{Acs1}
A_{cs}(t)=-Kt\int_{\pi}^{2\pi}\frac{D^n(1+\epsilon\sin\theta)^n(D)(\sin{\theta})}{(D^2(\epsilon^2 + 1 +2\epsilon\sin(\theta))^{\frac{n+3}{2}}}Rd\theta\\
\end{equation*}
\begin{equation*} \label{Acs2}
= -KtR\int_{\pi}^{2\pi}\frac{D^n(D)}{(D^2)^{\frac{n+3}{2}}} \frac{(\sin{\theta})(1+\epsilon\sin{\theta})^n}{(\epsilon^2 + 1+2\epsilon\sin{\theta})^{\frac{n+3}{2}}}d\theta\\
\end{equation*}
\begin{equation*} \label{Acs3}
= -KtR\frac{D^{n+1}}{D^{n+3}}\int_{\pi}^{2\pi}(\sin{\theta})\frac{(1+\epsilon\sin{\theta})^n}{(\epsilon^2+1+2\epsilon\sin{\theta})^{\frac{n+3}{2}}}d\theta\\
\end{equation*}
\begin{equation} \label{Acs4}
= -\frac{KtR}{D^2}\int_{\pi}^{2\pi} (\sin{\theta})(1+\epsilon\sin{\theta})^n(\epsilon^2+1+2\epsilon\sin{\theta})^{-\frac{n+3}{2}}d\theta\\
\end{equation}

\noindent Expanding to first order in $\epsilon = \frac{R}{D}$, dropping $O(\epsilon^2)$, by the binomial theorem, we have that $(1+x)^a = 1 + ax + O(x^2) + O(x^3)+...$. Applying this to \Eq{Acs4}, we expand each term in the integrand:

\begin{equation} \label{Acs5}
A_{cs}(t)= -\frac{KtR}{D^2}\int_{\pi}^{2\pi}(\sin{\theta})(1+n\epsilon\sin{\theta}+O(\epsilon^2))(-(\frac{n+3}{2})(2\epsilon\sin{\theta})+1+O(\epsilon^2))d\theta\\
\end{equation}
Dropping second-order error terms ($O(\epsilon^2))$ at each step,
\begin{equation*} \label{Acs6}
A_{cs}(t)\approx -\frac{KtR}{D^2}\int_{\pi}^{2\pi}(\sin{\theta})(1+n\epsilon\sin{\theta})(1-(n+3)\epsilon\sin{\theta})d\theta\\
\end{equation*}
\begin{equation*} \label{Acs7}
\approx -\frac{KtR}{D^2}\int_{\pi}^{2\pi}(\sin{\theta})(1-(n\epsilon\sin{\theta})-(3\epsilon\sin{\theta})+(n\epsilon\sin{\theta}))d\theta\\
\end{equation*}
\begin{equation*} \label{Acs8}
\approx -\frac{KtR}{D^2}\int_{\pi}^{2\pi}(\sin{\theta})(1-3\epsilon\sin{\theta})d\theta\\
\end{equation*}
\begin{equation*} \label{Acs9}
\approx -\frac{KtR}{D^2}\int_{\pi}^{2\pi}(\sin{\theta}-3\epsilon\sin^2{\theta})d\theta\\
\end{equation*}
\begin{equation} \label{Acs10}
\approx -\frac{KtR}{D^2}(-2-3\epsilon(\frac{\pi}{2})).
\end{equation}

\noindent Adding the error term $O(\epsilon^2)$ back in,

\begin{equation} \label{Acs11}
A_{cs}(t)\approx -\frac{KtR}{D^2}(-2-\frac{3\epsilon\pi}{2})+ O(\epsilon^2)\\
\end{equation}
Since $\epsilon=\frac{R}{D}$, substituting and distributing the \textbf{above expression (need to change to number}, we get our final expression for the cross-sectional area of deposition ($A_{cs}(t)$).

\begin{equation} \label{Acs12}
A_{cs}(t)\approx \frac{2KtR}{D^2}+\frac{3\pi KtR^2}{2D^3}+O(\frac{R^2}{D^2}), \qquad [A_{cs}(t)] = [m^2]
\end{equation}

\noindent Here, the first term ($\frac{2KtR}{D^2}$) is dominant, the second term ($\frac{3\pi KtR^2}{2D^3}$) is a small correction ($\propto \frac{R}{D}$), and the rest is negligible by the aforementioned condition of $R \ll D$.\\\\
\noindent Now that we have the cross-sectional area of deposition $A_{cs}$, we can proceed with calculating the predicted resistance of the cylindrical substrate after time $t$ seconds of deposition and a constant volumetric "flow-rate" of material $K$ $[m^3 \space s^{-1}]$. First, we note that simple trace resistance is given by
\begin{equation} \label{traceR}
    \Omega = \frac{\rho L}{A}
\end{equation}
where $\Omega$ $[\Omega = ohms]$ is resistance, $\rho$ $[\Omega \space m]$ is the resistivity of deposited material, $L$ $[m]$ is the length of the substrate, and $A$ $[m^2]$ is the cross-sectional area of the substrate. 
\\\\
\noindent The resistivity ratio, using the correction for thin metal layers based on the Fuchs-Sondheimer (FS) and Mayadas-Shatzkes (MS) models (\cite{Jeong2023}) to account for boundary and scattering effects in thin films, is given by:

\begin{equation} \label{FSMS}
    \frac{\rho_0}{\rho_{\text{eff}}} \simeq 1 - \frac{3}{2}\alpha + 3\alpha^2 - 3\alpha^3 \log(1 + \frac{1}{\alpha}) - \frac{3(1 - P)}{8k}
\end{equation}

\noindent where $\rho_{\text{eff}}$ $[\Omega \space m]$ depends on the thickness $f(t,\theta)$, $\rho_0$ $[\Omega \space m]$ is the bulk resistivity of Au, $k = \frac{f(t,\theta)}{l_0}$ $[m \space m ^{-1}]$ (unitless), and $P$ is the fraction of electrons specularly scattered at the external surfaces (\cite{Jeong2023}; \cite{Mayadas1968}; \cite{Tellier1977}). Additionally, $\alpha = \frac{l_0r_{gb}}{a_g(1-r_{gb})}$ $[\Omega \space m]$ whereby $l_0$ $[m]$ is the mean free path of electrons, $r_{gb}$ $[m]$ is the grain boundary reflection constant, and $a_g$ $[m]$ is average material crystalline diameter. \\

\noindent Let $F(\alpha) := 1 - \frac{3}{2}\alpha + 3\alpha^2 - 3\alpha^3 \log(1 + \frac{1}{\alpha})$. Then, by the FS-MS model, we have that 
\begin{equation} \label{FSMSsubk}
\frac{\rho_0}{\rho_{\text{eff}}} \cong F(\alpha) - \frac{3(1 - P)}{8k} = F(\alpha) - \frac{3(1 - P)}{8}\frac{l_0}{f(t,\theta)}
\end{equation}

\noindent To calculate the resistance across the entire cylindrical substrate, we can find longitudinal conductance per unit length $G(t)$ as the parallel sum over the cross-section of the cylindrical substrate
\begin{equation} \label{G}
G(t) = \int_{\pi}^{2\pi}\sigma_{eff}(f(t,\theta)) \space f(t,\theta) \space Rd\theta,
\end{equation}
where effective conductance $\sigma_{eff}(f(t,\theta)) =(\rho_{eff}(f(t,\theta)))^{-1}$. Thus substituting into \Eq{FSMSsubk},
\begin{equation*} \label{rhosig_eff}
\rho_0\sigma_{eff}(f(t,\theta)) \simeq F(\alpha) - \frac{3-3P}{8}\frac{l_o}{f(t,\theta)}\\
\end{equation*}
\begin{equation} \label{sig_eff}
\sigma_{eff}(f(t,\theta)) \simeq \frac{F(\alpha)}{\rho_0} - \frac{3-3P}{8}\frac{l_o}{\rho_0f(t,\theta)}
\end{equation}
Substituting \Eq{sig_eff} into \Eq{G} to find $G(t)$,
\begin{equation*} \label{G2}
G(t) = \int_{\pi}^{2\pi}[\frac{F(\alpha)}{\rho_0} - \frac{3-3P}{8}\frac{l_o}{\rho_0f(t,\theta)}]  \space f(t,\theta) \space Rd\theta \\
\end{equation*}
\begin{equation*} \label{G3}
=\frac{F(\alpha)}{\rho_0}\int_{\pi}^{2\pi} \underbrace{f(t,\theta) \space Rd\theta}_{A_{cs}} - \int_{\pi}^{2\pi}\frac{3-3P}{8}\frac{l_o}{\rho_0} \space Rd\theta \\
\end{equation*}
\begin{equation*} \label{G4}
=\frac{F(\alpha)}{\rho_0}A_{cs}(t) - \frac{3-3P}{8}\frac{l_o}{\rho_0} \space\int_{\pi}^{2\pi} Rd\theta \\
\end{equation*}
\begin{equation*} \label{G5}
=\frac{F(\alpha)}{\rho_0}A_{cs}(t) - \frac{3-3P}{8}\frac{l_o}{\rho_0} \space\pi R\\
\end{equation*}
\begin{equation} \label{G6}
=\frac{1}{\rho_0}[F(\alpha)A_{cs}(t)-\frac{3-3P}{8}\pi Rl_0]
\end{equation}
Therefore, since resistance $R_{cyl}=G(t)^{-1}$, 
\begin{equation*} \label{Omega1}
R_{cyl}(t) \cong \frac{L_{cyl}\rho_0}{F(\alpha)A_{cs}(t)-\frac{3-3P}{8}\pi R l_0} \\
\end{equation*}
\begin{equation*} \label{Omega2}
= \frac{L_{cyl}\rho_0}{F(\alpha)(\frac{2KtR}{D^2}+\frac{3\pi KtR^2}{2D^3}+O(\frac{R^3}{D^4}))-\frac{3-3P}{8}\pi R l_0} \\
\end{equation*}
\begin{equation} \label{Omega 3}
= \frac{L_{cyl}\rho_0}{F(\alpha)(\frac{2KtR}{D^2})(1+\frac{3\pi R}{4D}+O(\frac{R^2}{D^2}))-\frac{3-3P}{8}\pi R l_0}
\end{equation}

\noindent To account for ambient deposition temperatures $T_{amb} \neq 25 \degree C$, in addition to modifications to the temperature-dependent constants listed above, we adjust $R_{cyl}$ by temperature-dependent constant $T_{adj} := (1+\alpha(T_{amb}-25\degree C))$(Jeong et. al. 2023):
\begin{equation} \label{Omega 4}
= \frac{L_{cyl}\rho_0}{F(\alpha)(\frac{2KtR}{D^2})(1+\frac{3\pi R}{4D}+O(\frac{R^2}{D^2}))-\frac{3-3P}{8}\pi R l_0}T_{adj}
\end{equation}
Simplifying further,
\begin{equation} \label{OmegaFinal}
\boxed{
R_{cyl}(t) = \frac{L_{cyl}\rho_0T_{adj}}{R[F(\alpha)(\frac{2Kt}{D^2})(1+\frac{3\pi R}{4D}+O(\frac{R^2}{D^2}))-\frac{3-3P}{8}\pi l_0]}, \qquad R \ll D}
\end{equation}
where $F(\alpha) = 1 - \frac{3}{2}(\frac{l_0r_{gb}}{a_g(1-r_{gb})}) + 3(\frac{l_0r_{gb}}{a_g(1-r_{gb})})^2 - 3(\frac{l_0r_{gb}}{a_g(1-r_{gb})})^3 \log(1 + \frac{a_g(1-r_{gb})}{l_0r_{gb}})$ and $T_{adj} = (1+\frac{l_0r_{gb}}{a_g(1-r_{gb})}(T_{amb}-25\degree C))$.

\section{Additional Characterization}
As discussed in Section 5.2.3 of the main article, we see island formation, flaking, and cracking on wire substrates post-deposition (See Supplementary Figure \ref{sFigSEM}). 

\begin{figure} [h!]
    \centering
    \includegraphics[width=0.75\linewidth]{Deposition Profile Paper/Supp Figures/SuppFig SEM Images.png}
    \caption{SEM images of wire substrate cross section after 1 $\mu$m Au deposition, noting the island formation and flaking of the trace.}
    \label{sFigSEM}
\end{figure}

\section{Code}
All code developed for this study is open-access, and is available on \href{https://github.com/adityatummala5/Thin-Film-Modeling.git}{GitHub}.

\bibliographystyle{sn-basic}
\bibliography{sn-bibliography}

\end{document}