Dual degrees of freedom diagnosis with high energy electron lens radiography

Jiahao Xiao; Yingchao Du; Haoqing Li; Yongtao Zhao; Liang Sheng

doi:10.11884/HPLPB202234.210548

Journals >High Power Laser and Particle Beams >Volume 34 >Issue 6 >Page 064010 > Article

High Power Laser and Particle Beams
Vol. 34, Issue 6, 064010 (2022)

Dual degrees of freedom diagnosis with high energy electron lens radiography

Jiahao Xiao^1、2, Yingchao Du^1、2、*, Haoqing Li^1、3, Yongtao Zhao⁴, and Liang Sheng³

Author Affiliations

¹Department of Engineering Physics, Tsinghua University, Beijing 100084, China

²Key Laboratory of Particle and Radiation Imaging (Tsinghua University), Ministry of Education, Beijing 100084, China

³State Key Laboratory of Intense Pulsed Radiation Simulation and Effect, Northwest Institute of Nuclear Technology, Xi’an 710024, China

⁴School of Physics, Xi’an Jiaotong University, Xi’an 710049, China

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DOI: 10.11884/HPLPB202234.210548 Cite this Article

Jiahao Xiao, Yingchao Du, Haoqing Li, Yongtao Zhao, Liang Sheng. Dual degrees of freedom diagnosis with high energy electron lens radiography[J]. High Power Laser and Particle Beams, 2022, 34(6): 064010 Copy Citation Text

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Abstract

The evolution of electromagnetic field and fluid is important in the research of high energy density physics, controlled nuclear fusion, and laboratory astrophysics. But in experiment, it is difficult to get the density and electromagnetic field distribution simultaneously. Based on high energy electron lens radiography, this paper proposes dual degrees of freedom diagnosis (DDFD) by constructing areal density difference. Combining the Monte-Carlo simulation and beam optics analysis, the feasibility of this method when the diagnosed system includes relatively strong electromagnetic field has been validated. Besides, changing the aperture as a ring can effectively improve the resolution in low E/B field and low areal density situation. The simulation results indicate that this method works well. Considering the characters of electron beams, this method is quite suitable for the electromagnetic fluid diagnosis.

Keywords

areal density measuring dual degrees of freedom diagnosis E/B field diagnosis high energy electron lens radiography ultrafast image

$ f\left(\varphi ,{t}\right)=\dfrac{N(t,\varphi )}{{N}_{0}}=\dfrac{1}{{\phi }_{0}\left(t\right)\sqrt{2\mathrm{\pi }}}{\mathrm{e}}^{-\tfrac{1}{2}{\left(\tfrac{\varphi }{{\varphi }_{0}\left(t\right)}\right)}^{2}} $

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$ {\varphi }_{0}\left(t\right)=\dfrac{13.6\mathrm{M}\mathrm{e}\mathrm{V}}{\beta cp}\sqrt{t}(1+0.038\mathrm{l}\mathrm{n}(t\left)\right) $

()

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$ \boldsymbol{\theta }=\displaystyle\int \dfrac{e{\boldsymbol{B}}_{\perp }(x,y,{\textit{z}})}{p}\mathrm{d}{{\boldsymbol{z}}}+\displaystyle\int \dfrac{{\boldsymbol{E}}_{\perp }(x,y,{\textit{z}})e}{cp}\mathrm{d}{{\boldsymbol{z}}} $

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$ \left[\begin{array}{c}{x}_{i}\\ {x}_{i}'\\ {y}_{i}\\ {y}_{i}'\end{array}\right]=\left[\begin{array}{cccc}{R}_{11}& {R}_{12}& 0& 0\\ {R}_{21}& {R}_{22}& 0& 0\\ 0& 0& {R}_{33}& {R}_{34}\\ 0& 0& {R}_{43}& {R}_{44}\end{array}\right]\left[\begin{array}{c}{x}_{0}\\ {x}_{0}'\\ {y}_{0}\\ {y}_{0}'\end{array}\right] $

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$ \left[\begin{array}{c}{x}_{\mathrm{F}}\\ {y}_{\mathrm{F}}\end{array}\right]=\left[\begin{array}{cc}{R}_{12\mathrm{F}}& 0\\ 0& {R}_{34\mathrm{F}}\end{array}\right]\left[\begin{array}{c}{x}_{0}'\\ {y}_{0}'\end{array}\right] $

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$ j({x}_{0}',{y}_{0}',t)=f({\rho }',{\varphi }',t)=f\Bigg(\sqrt{{x}_{0}'^{2}+{y}_{0}'^{2}},{\varphi }',t\Bigg) $

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$ {T}_{{\rm{r}}}(t,\boldsymbol{\theta })=\iint\nolimits_{({\rho }',{\varphi }')\in {S}'} f({\rho }',{\varphi }',t){\rho }'\mathrm{d}{\rho }'\mathrm{d}{\varphi }' $

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$ \left\{ \begin{array}{l} {T}_{{\rm{r}}}(t,\boldsymbol{\theta })={\displaystyle\int }_{0}^{\varepsilon }{\displaystyle\int }_{0}^{2\mathrm{\pi }}{f}_{t}\left({\rho }_{t}'\right){\rho }'\mathrm{d}{\rho }'\mathrm{d}{\varphi }'\\ \dfrac{{l}_{x}}{{R}_{12\mathrm{F}}}=\dfrac{{l}_{y}}{{R}_{34\mathrm{F}}}=\varepsilon \end{array} \right.$

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$ {\rho }_{{\rm{t}}}'=\sqrt{{({\rho }'\mathrm{c}\mathrm{o}\mathrm{s}{\varphi }'-{\rho }_{\boldsymbol{\theta }}')}^{2}+{\left({\rho }'\mathrm{s}\mathrm{i}\mathrm{n}{\varphi }'\right)}^{2}} $

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$ {T}_{{\rm{r}}}(t,{\boldsymbol{\theta}} )={\displaystyle\int }_{{\varepsilon }_{1}}^{{\varepsilon }_{2}}{\displaystyle\int }_{0}^{2\mathrm{\pi }}{{{f}}}_{{\rm{t}}}\left({\rho }_{{\rm{t}}}'\right){\rho }'\mathrm{d}{\rho }'\mathrm{d}{\varphi }' $

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$ \left\{\begin{array}{c}{T}_{{\rm{r}}1}={T}_{{\rm{r}}}(t+{t}_{1},\theta )\\ \begin{array}{c}{T}_{{\rm{r}}2}={T}_{{\rm{r}}}(t+{t}_{1},\theta )\\ {T}_{{\rm{r}}3}={T}_{{\rm{r}}}(t+{t}_{1},\theta )\\ {T}_{{\rm{r}}4}={T}_{{\rm{r}}}(t+{t}_{1},\theta )\end{array}\end{array}\right. $

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