Problem 82 - Fundamentals of Physics

Question

An elementary particle produced in a laboratory experiment travels \(0.230 \mathrm{~mm}\) through the lab at a relative speed of \(0.960 \mathrm{c}\) before it decays (becomes another particle). (a) What is the proper lifetime of the particle? (b) What is the distance the particle travels as measured from its rest frame?

Step-by-Step Solution

Answer

(a) Proper lifetime \(\tau \approx 8.6 \times 10^{-13}\) seconds. (b) Rest frame distance \(d_0 \approx 2.9 \times 10^{-4}\) meters.

Step 1: Identify Known Variables

We are given the particle's travel distance in the lab frame as \(0.230\, \mathrm{mm}\) and its speed as \(0.960c\). We are asked to find the proper lifetime (\( \tau \)) and the rest frame distance (\( d_0 \)).

Step 2: Use Time Dilation Formula

In the lab frame, the time \( t \) measured is related to the proper time \( \tau \) by the time dilation formula, \( t = \gamma \tau \), where \( \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \). First, compute \( \gamma \) for \( v = 0.960c \).

Step 3: Calculate Lorentz Factor \(\gamma\)

The Lorentz factor is \( \gamma = \frac{1}{\sqrt{1-(0.960)^2}} \). Evaluate this to find \( \gamma \approx 3.57 \).

Step 4: Calculate Time in Lab Frame

The distance traveled in the lab frame is \(0.230\, \mathrm{mm} = 0.230 \times 10^{-3} \mathrm{m}\). The time in lab frame is \( t = \frac{d}{v} = \frac{0.230 \times 10^{-3}}{0.960c} \).

Step 5: Convert Lab Time to Proper Time

Using the time dilation formula, obtain \( \tau = \frac{t}{\gamma} \). Substitute the value of \( t \) from Step 4 and \( \gamma \).

Step 6: Verify Proper Time

Calculate \( \tau \) using the values from previous steps. This gives the proper lifetime of the particle.

Step 7: Calculate Rest Frame Distance

The rest frame distance \( d_0 \) is calculated using \( d_0 = v \tau \). Substitute \( \tau \) from Step 6 and compute \( d_0 \) to find the distance in the particle's rest frame.

Key Concepts

Proper Time

Proper time, denoted as \(\tau\), is a fundamental concept in relativity. It represents the time interval experienced by an object moving along with the particle, meaning it is the time measured in the particle's own rest frame. In our exercise, the proper lifetime of a particle is asked, which is the time that would be measured by an observer moving with the particle, not affected by the effects of relative motion.

To calculate the proper time, we need to use the time dilation formula, which relates the time measured in an observer's frame (lab frame, in this case) with the proper time. This relationship highlights how clocks moving relative to an observer run more slowly, a core phenomenon of special relativity. Understanding this enables us to grasp why proper time is 'shorter' compared to the dilated time measured by a stationary observer.

Lorentz Factor

The Lorentz factor, symbolized by \(\gamma\), is central to calculations in special relativity because it quantifies the amount of time dilation and length contraction. It is calculated using the formula: \(\gamma = \frac{1}{\sqrt{1-v^2/c^2}}\), where \(v\) is the velocity of the object and \(c\) is the speed of light.

In our problem, with the particle traveling at \(0.960c\), the Lorentz factor is computed to be approximately 3.57. This large value indicates significant relativistic effects due to the high velocity, close to the speed of light. As \(v\) approaches \(c\), \(\gamma\) increases dramatically, showcasing how relativistic effects become pronounced at high speeds.

Time Dilation

Time dilation is a fascinating consequence of Einstein's theory of special relativity. It implies that time, as measured in a moving frame of reference, will appear to be slower compared to a stationary observer's frame. In simpler terms, if you're moving fast enough, time literally slows down for you relative to someone who is at rest.

In the problem, time dilation tells us that while the lab observer sees the particle as having a longer lifetime, the proper time (lifetime measured in the particle's rest frame) is actually shorter. This is represented mathematically as \(t = \gamma \tau\), linking time \(t\) in the lab's reference frame to the proper time \(\tau\). Here, \(\gamma\) being greater than 1 means \(t\) is expanded compared to \(\tau\).

Rest Frame Distance

The rest frame distance refers to the distance as measured by an observer who is at rest relative to the moving object. In the problem, this would be the distance the particle thinks it travels, versus what an external observer might measure.

To determine this distance, denoted \(d_0\), we use its relationship with the proper time: \(d_0 = v \tau\). Here, \(v\) is the velocity, and \(\tau\) is the proper time calculated earlier. By substituting the known values, we can find how far the particle travels in its own frame of reference before decaying. This concept elegantly ties together velocity, time dilation, and the particle's lifetime in understanding motion in relativity.