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An unstable high-energy particle enters a detector and leaves a track 1.05 mm long before it decays. Its speed relative to the detector was 0.992c. What is its proper lifetime? That is, how long would it have lasted before decay had it been at rest with respect to the detector?

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mary24457181

The particle's lifetime relative to the detector is 0.00106 s.

The particle's rest frame lifetime, t_0, can be calculated using the Lorentz time dilation equation:

t_0 = t/(γ) = (1.05 mm)/(0.992c) = 0.00106 s.

The Lorentz time dilation equation:

The photon rest lifetime, t_0, can be calculated using the Lorentz time dilation equation: t_0 = t/(γ) = (1.05 mm)/(0.992c) = 0.00106 s.

The γ is calculated using the Lorentz transformation equation:

γ = 1/√(1 - (v/c)^2) = 1/√(1 - (0.992c)^2) = 7.5.

Therefore, the photon lifetime with respect to the detector is

t_0 / γ = 0.00106 s/7.5 = 0.00106 s.

Learn more about the Lorentz: law :

https://brainly.com/question/28546820

https://brainly.com/question/14121617

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