1)
We have: t = t’/ [1 - v

But the proper time is defined such that:

t' = t/2 or t'/ t = ½

Then:

[1 - v

and:

v

v

so:

v = c[1 - (t'/ t)

2) The proper time t' = 3600 s

Since v = 300 m/s = (10

^{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}= 0.866c2) 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, 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) 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}

where 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 = 4600 m/ (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) The distance traveled by the muon as measured in its frame is just the

proper length, L' so:

L' = 4600 m [1 - v

L' = 4600 m (0.14) = 644 m

4) The proper time t' = 2.6 x 10

t = t’/ [1 - v

so:

t = (2.6 x 10

t = (2.6 x 10

t = 8.3 x 10

L' = 4600 m [1 - v

^{2}/c^{2}]^{½}= 4600 m (0.02)^{½ }L' = 4600 m (0.14) = 644 m

4) The proper time t' = 2.6 x 10

^{-8}st = t’/ [1 - v

^{2}/c^{2}]^{½}and v = 0.95cso:

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)/ 0.312t = 8.3 x 10

^{-8}s
Which is the mean lifetime of the muon as measured by an observer on Earth.

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