# Real-Time Tomography of Gas-Jets with a Wollaston Interferometer
## 3 Data Analysis for Interferometry and Real-Time Tomography
### Numerical Tools
As long as cylindrical symmetry of the gas distribution is assumed, obtaining the radial density distribution \(\rho(r)\) from \(\phi(y)\) is possible. The corresponding integral relation is called inverse Abel transformation or Abel inversion. The phase shift \(\phi\), extracted with the method explained before, is proportional to a projection of the gas jet along the propagation direction of the laser. Here, cylindrical symmetry of the gas distribution is assumed. This assumption enables one to reconstruct the radial density distribution from \(\phi\). This is achieved by the inverse Abel transform [16; 17], which reads:  
\[\phi(y)=\frac{\Delta s}{\lambda}2\pi=\frac{2\pi}{\lambda}2\int_{0}^{\infty} \left[\eta(y,z)-1\right]\;\mathrm{d}z=\frac{4\pi}{\lambda}\int_{y}^{\infty} \left[\eta(r)-1\right]\frac{r}{\sqrt{r^{2}-y^{2}}}\mathrm{d}r,\]  
\[F(y)=\phi(y)\cdot\frac{\lambda}{2\pi}=2\int_{y}^{\infty}\left[\eta(r)-1 \right]\frac{r}{\sqrt{r^{2}-y^{2}}}\mathrm{d}r,\]  
\[\stackrel{{\mathrm{inv.\ Abel}}}{{\Longrightarrow}}\eta(r)-1=- \frac{1}{\pi}\int_{r}^{\infty}\frac{\mathrm{d}F(y)}{\mathrm{d}y}\frac{ \mathrm{d}y}{\sqrt{y^{2}-r^{2}}}. \tag{5}\]  
The derivative and the integral occurring in Equation (5) are calculated following [18]. The key points are summarized here. It is not suitable to approximate the derivative of \(F\) by the discrete differential quotient due to noise, since this would amplify noise drastically. A more sophisticated method based on Gaussian filters is needed. Consider the following Fourier identity for \(f\), \(g\) functions with compact support. \[F(f)\cdot\mathcal{F}\left(\frac{\mathrm{d}g}{dx}\right) =\int_{-\infty}^{+\infty}\!\!\mathrm{d}x\,e^{-\mathrm{i}2\pi x \omega}f(x)\cdot\int_{-\infty}^{+\infty}\mathrm{d}x\,e^{-\mathrm{i}2\pi x \omega}\frac{\mathrm{d}g(x)}{dx} \tag{6}\] \[=\int_{-\infty}^{+\infty}\!\!\mathrm{d}x\,e^{-\mathrm{i}2\pi x \omega}f(x)\cdot\int_{-\infty}^{+\infty}\mathrm{d}x\,e^{-\mathrm{i}2\pi x \omega}\left(\mathrm{i}2\pi\omega\right)\,g(x)\] \[\mathcal{F}(g)(\omega)=\int_{-\infty}^{+\infty}e^{-i2\pi\alpha y}\cdot b(y) \cdot e^{i[2\pi fy+\alpha(y)]}\;\mathrm{d}y.\]0 \[\mathcal{F}(g)(\omega)=\int_{-\infty}^{+\infty}e^{-i2\pi\alpha y}\cdot b(y) \cdot e^{i[2\pi fy+\alpha(y)]}\;\mathrm{d}y.\]1 \[\mathcal{F}(g)(\omega)=\int_{-\infty}^{+\infty}e^{-i2\pi\alpha y}\cdot b(y) \cdot e^{i[2\pi fy+\alpha(y)]}\;\mathrm{d}y.\]2  
When \(g\) is set to a Gaussian distribution, rearranging this identity yields the smoothed derivative of a noisy signal \(f\) by using only the derivative of \(g\) instead of \(f\) (Equation (6)). The derivative of the Gaussian function \(g\) can be easily evaluated from its analytical expression. Another issue is the singularity inside the integral of Equation (5), when the integral is approximated on a discrete domain. This is solved by setting the first value of the integral (\(y=r\)) to the second value (\(y=r+\Delta y\)). For the limit of many points (i.e., infinitesimal grid spacing \(\Delta y\)), the numerical value of the integral will converge to the analytical value [18].