Muon tracking with the fastest light in the JUNO central detector

Abstract

Background:

The Jiangmen Underground Neutrino Observatory (JUNO) is a multi-purpose neutrino experiment designed to measure the neutrino mass hierarchy using a central detector (CD), which contains 20 kton liquid scintillator (LS) surrounded by about 18,000 photomultiplier tubes (PMTs), located 700 m underground.

Purpose:

The rate of cosmic muons reaching the JUNO detector is about 3 Hz, and the muon-induced neutrons and isotopes are major backgrounds for the neutrino detection. Reconstruction of the muon trajectory in the detector is crucial for the study and rejection of those backgrounds.

Methods:

This paper will introduce the muon-tracking algorithm in the JUNO CD, with a least-squares method of PMTs’ first-hit time (FHT). Correction of the FHT for each PMT was found to be important to reduce the reconstruction bias.

Results:

The spatial resolution and angular resolution are better than 3 cm and 0.4 degree, respectively, and the tracking efficiency is greater than 90% up to 16 m far from the detector center.

Appendix

Derivation of $\varvec{\partial {\left| \mathbf {A}-l\hat{V}\right| }/\partial {l}}$

let $\mathbf {R}=\mathbf {A}-l\hat{V}$, then

$$\begin{aligned} \begin{aligned} \dfrac{\partial {\left| \mathbf {A}-l\hat{V}\right| }}{\partial {l}}&= \dfrac{\partial \left| \mathbf {R}\right| }{\partial {l}} = \dfrac{\partial {\left( \left| \mathbf {R}\right| ^2\right) ^{\frac{1}{2}}}}{\partial {l}} \\&=\frac{1}{2}\cdot \left( \left| \mathbf {R}\right| ^2\right) ^{-\frac{1}{2}}\cdot \dfrac{\partial {\left| \mathbf {R}\right| ^2}}{\partial {l}} \\&=\dfrac{1}{2}\cdot \dfrac{1}{\left| \mathbf {R}\right| }\cdot \dfrac{\partial {\mathbf {R}^2}}{\partial {l}} \\&=\dfrac{1}{2}\cdot \dfrac{1}{\left| \mathbf {R}\right| }\cdot \dfrac{\partial {\left( \mathbf {A}^2-2l\mathbf {A}\cdot \hat{V}+l^2\hat{V}^2\right) }}{\partial {l}} \\&=\dfrac{1}{2}\cdot \dfrac{1}{\left| \mathbf {R}\right| }\cdot -2\left( \mathbf {A}-l\hat{V}\right) \cdot \hat{V} \\&=-\dfrac{\mathbf {R}}{\left| \mathbf {R}\right| }\cdot \hat{V} =-\hat{R}\cdot \hat{V} \\&=-\cos <\hat{R},\hat{V}> \end{aligned} \end{aligned}$$

where $\hat{R}$ means the unit vector of $\mathbf {R}$ and $<\hat{R},\hat{V}>$ means the angle between $\hat{R}$ and $\hat{V}$. For the partial derivative in Eq. (8),

$$\begin{aligned} \mathbf {A}=\mathbf {R_i}-\mathbf {R_0}, \quad \quad \quad \mathbf {R}=\mathbf {R_\mathrm{ci}}=\left( \mathbf {R_i}-\mathbf {R_0}\right) -l\hat{V} \end{aligned}$$

Therefore,

$$\begin{aligned} \dfrac{\partial {\left| \mathbf {R_i}-(\mathbf {R_0}+l\hat{V})\right| }}{\partial {l}} = -\hat{R_\mathrm{ci}}\cdot \hat{V} = -\cos \theta \end{aligned}$$