Solutions to Special Relativity Problems (3)

Recall the problems set at the end of (3):

1) With what speed would a clock have to be moving to run at a rate that is one half the rate of a clock at rest?

Solution

We have: t = t’/ [1 - v^2/c^2]^ ½

But the proper time is defined such that:

t' = t/2 or t'/ t = ½

Then:

[1 - v^2/c^2] = (t'/ t)^2

and:

v^2/c^2 = 1 - (t'/ t)^2

v^2 = c^2[1 - (t'/ t)^2]

so:

v = c[1 - (t'/ t)^2]^½ = c[1 - 0.5^2]^½ = c[0.75]^½ = 0.866c


2) An atomic clock is placed on a Jumbo Jet. The clock measures a time interval of 3600 s when the jet is oving at v = 300 m/s. What corresponding time would an identical clock left on the ground measure? (Hint: whenever v << c (e.g. v/c << 1), we have 1 + v^2/2c^2 and not [1 - v^2/c^2]^1/2)

Solution

The proper time t' = 3600 s

Since v = 300 m/s = (10^-6) c and hence v/c << 1 we need the form:

t = t’/ [1 + v^2/2c^2]


t = 3600s/ [1 + (10^-12)c^2/2c^2]

Since the numerator is only slightly larger than 1, then the time t will be:

3600 s/(1.000000000001)= 3600.0000000018

3600 + 1.8 x 10^-9 s

or slightly longer than one hour.


3) A muon formed high in the Earth's atmosphere travels at v = 0.99c for a distance of 4.6 km before it decays into an electron, a neutrino and an anti-neutrino.

a) How long does the muon survive as measured in its rest frame?

Solution

The proper time t' applies to the muon's reference frame.

So:

t = t’/ [1 - v^2/c^2]^ ½


and t' = t [1 - v^2/c^2]^ ½

v = 0.99 c and v^2 = (0.99c)^2 = 0.98c^2

Then: t' = t [1 - 0.98c^2/c^2]^ ½ = t [0.02]^ ½ = t(0.14)

recall distance travelled = 4.6 km = 4600 m

To get t' we need to find t first, e.g.

t = 4600m/ (2.97 x 10^8 m/s) = 1.55 x 10^-5 s

Then: t' = (1.55 x 10^-5 s) (0.14) = 2.1 x 10^-6 s


b) How far does the muon travel as measured in its frame?

Solution

The distance traveled in its frame is just the proper length, L'

so:

L' = 4600 m [[1 - v^2/c^2]^ ½ = 4600m (0.02)^ ½

L' = 4600 m (0.14) = 644 m


4) The average lifetime of a pi meson in its own frame of reference is 2.6 x 10^-8 s. If the meson moves with v = 0.95c, what is its mean lifetime as measured by an observer on Earth?

Solution

The proper time t' = 2.6 x 10^-8 s

t = t’/ [1 - v^2/c^2]^ ½

and v = 0.95c


so:

t = (2.6 x 10^-8 s)/ [1 - (0.95c)^2/c^2]^ ½

t = (2.6 x 10^-8 s)/ [0.0975]^ ½ = (2.6 x 10^-8 s)/ 0.312

t = 8.3 x 10^-8 s