Suppose a cosmic ray colliding with a nucleus in Earth’s upper atmosphere produces a muon that has a velocity
The muon then travels at constant velocity and lives 2.20 μs as measured in the muon’s frame of reference. (You can imagine this as the muon’s internal clock.) How long does the muon live as measured by an earthbound observer (
[link] )?
A muon in Earth’s atmosphere lives longer as measured by an earthbound observer than as measured by the muon’s internal clock.
As we will discuss later, in the muon’s reference frame, it travels a shorter distance than measured in Earth’s reference frame.
Strategy
A clock moving with the muon measures the proper time of its decay process, so the time we are given is
The earthbound observer measures
as given by the equation
Because the velocity is given, we can calculate the time in Earth’s frame of reference.
Solution
Identify the knowns:
Identify the unknown:
Express the answer as an equation. Use:
with
Do the calculation. Use the expression for
to determine
from
:
Remember to keep extra significant figures until the final answer.
Significance
One implication of this example is that because
at 95.0% of the speed of light
the relativistic effects are significant. The two time intervals differ by a factor of 3.20, when classically they would be the same. Something moving at 0.950
c is said to be highly relativistic.
A non-flat screen, older-style television display (
[link] ) works by accelerating electrons over a short distance to relativistic speed, and then using electromagnetic fields to control where the electron beam strikes a fluorescent layer at the front of the tube. Suppose the electrons travel at
through a distance of
from the start of the beam to the screen. (a) What is the time of travel of an electron in the rest frame of the television set? (b) What is the electron’s time of travel in its own rest frame?
The electron beam in a cathode ray tube television display.
Strategy for (a)
(a) Calculate the time from
Even though the speed is relativistic, the calculation is entirely in one frame of reference, and relativity is therefore not involved.
Solution
Identify the knowns:
Identify the unknown: the time of travel
Express the answer as an equation:
Do the calculation:
Significance
The time of travel is extremely short, as expected. Because the calculation is entirely within a single frame of reference, relativity is not involved, even though the electron speed is close to
c .
Strategy for (b)
(b) In the frame of reference of the electron, the vacuum tube is moving and the electron is stationary. The electron-emitting cathode leaves the electron and the front of the vacuum tube strikes the electron with the electron at the same location. Therefore we use the time dilation formula to relate the proper time in the electron rest frame to the time in the television frame.