An unstable high-energy particle enters a detector and leaves a track 1.35 mm long before it decays. Its speed relative to the detector was 0.992c. What was its proper lifetime? That is, how long would the particle have lasted before decay had it been at rest with respect to the detector? please use units for final answer
The apparent lifetime of the particle in the laboratory/detector coordinate system is (track length)/speed =
1.35*10^-3 m/[(0.992)(3*10^8) m/s]
= 4.5*10^-12 s
In its own frame of reference, the lifetime is shorter by the relativistic factor sqrt[1 - (v/c)^2]= 0.126
The high speed in the labratory frame of reference slows the observed decay process.
Its actual lifetime is 5.5*10^-13 s
To solve this problem, we can use the concept of time dilation as described by special relativity. Time dilation states that the time experienced by an object moving relative to an observer is dilated, or slowed down, compared to the time experienced by a stationary object.
In this case, we know the length of the track left by the particle, its speed relative to the detector, and we want to find its proper lifetime when at rest with respect to the detector.
First, let's calculate the Lorentz factor (γ) using the formula:
γ = 1 / √(1 - (v/c)^2)
where v is the velocity of the particle and c is the speed of light.
Given that the speed of the particle relative to the detector (v) is 0.992c, we can substitute these values into the formula:
γ = 1 / √(1 - (0.992c/c)^2)
= 1 / √(1 - 0.984064)
≈ 7.088
Now, we can find the proper lifetime (τ_0) using the formula:
proper lifetime (τ_0) = observed lifetime (τ) / γ
Here, the observed lifetime is the time it took for the particle to decay as measured by the detector. The observed length of the track is given as 1.35 mm.
So, we can rearrange the formula and solve for τ_0:
τ_0 = τ / γ
To find τ, we'll need to use the equation for velocity:
velocity (v) = displacement (d) / time (τ)
Given that the length of the track (d) is 1.35 mm, and the velocity (v) is 0.992c, we can rearrange this equation to solve for τ:
τ = d / v
= 1.35 mm / (0.992c)
Here, we also need to convert the length from millimeters to meters, as the speed of light is usually expressed in meters per second. So:
τ = (1.35 mm * 10^-3 m/mm) / (0.992c)
Now, substituting this value back into the proper lifetime equation:
τ_0 = τ / γ
= [(1.35 mm * 10^-3 m/mm) / (0.992c)] / 7.088
Calculating the numerical value using c = 3.00 x 10^8 m/s:
τ_0 = [(1.35 mm * 10^-3 m/mm) / (0.992 * 3.00 x 10^8 m/s)] / 7.088
≈ 2.84 x 10^-9 s
Therefore, the proper lifetime of the particle when at rest with respect to the detector is approximately 2.84 nanoseconds (ns).
To find the proper lifetime of the unstable high-energy particle, we can use the concept of time dilation from special relativity. Time dilation states that the time experienced by an object moving relative to an observer appears slower compared to when it is at rest.
The formula to calculate time dilation is:
t_0 = t * sqrt(1 - (v^2/c^2))
Where:
t_0 is the proper time (lifetime at rest),
t is the observed time (lifetime relative to the detector),
v is the speed of the particle relative to the detector,
c is the speed of light in a vacuum.
In this case, we are given the observed time (the lifetime relative to the detector) as 1.35 mm and the speed of the particle relative to the detector as 0.992c. We need to find the proper time (lifetime at rest) t_0.
We know that the speed of light is approximately 299,792,458 meters per second (m/s) or about 2.998×10^8 meters per second (m/s).
First, let's convert the observed time from millimeters (mm) to meters (m) for consistency:
1.35 mm = 1.35 × 10^(-3) m
Now we can solve for the proper time (t_0):
t_0 = t * sqrt(1 - (v^2/c^2))
= (1.35 × 10^(-3) m) * sqrt(1 - (0.992c)^2/(c^2))
= (1.35 × 10^(-3) m) * sqrt(1 - 0.992^2)
= (1.35 × 10^(-3) m) * sqrt(1 - 0.984064)
= (1.35 × 10^(-3) m) * sqrt(0.015936)
≈ (1.35 × 10^(-3) m) * 0.1262
≈ 1.706 × 10^(-4) m
Therefore, the proper lifetime of the particle (t_0) is approximately 1.706 × 10^(-4) meters.