Pulmano, Ray Angelo E.
Sacro, John Rich
PHYS 34.01: Problem Set 1
Q1. Consider a ball fired from a stationary frame S.
(a.) What is the velocity of the ball relative to you?
Figure 1. Ball and Rocket in Frame S
Using Newtonian or Galilean Relativity,
𝑣𝐵/𝑅 = 𝑣𝐵/𝑆 − 𝑣𝑅/𝑆
^
^
^
^
^
= 3 𝑚/𝑠 𝑥 + 2 𝑚/𝑠 𝑦 − (− 3 𝑚/𝑠 𝑥)
= 6 𝑚/𝑠 𝑥 + 2 𝑚/𝑠 𝑦
(b.) What should your velocity be relative to S so that the ball will appear stationary to you?
To make the ball appear stationary to me relative to frame S, the rocketship (and I) should
have the same velocity as the ball.
To show this mathematically,
𝑣𝐵/𝑅 = 𝑣𝐵/𝑆 − 𝑣𝑅/𝑆 = 0
^
^
⇒ 𝑣𝐵/𝑆 = 𝑣𝑅/𝑆 = 3 𝑚/𝑠 𝑥 + 2 𝑚/𝑠 𝑦
Q2. Tabulate the dilated lifetime of the muon for different speeds.
0. 999𝑐, 0. 9999𝑐, ... , 0. 99999999𝑐
−6
Take the lifetime in the muon’s frame of reference to be ∆𝑡0 = 2. 1969811 × 10
𝑠. At what
speed will the time be 1000 times longer than the rest frame lifetime?
To get the dilated time ∆𝑡 for each relative speed of the muon, we use the following formula:
∆𝑡0
∆𝑡 =
2
1−(𝑢/𝑐)
Muon’s Relative Speed (𝑢)
Dilated Time (∆𝑡)
∆𝑡/∆𝑡0
−5
0. 999𝑐
4. 913447245 × 10
0. 9999𝑐
1. 553418791 × 10
0. 99999𝑐
4. 91223101 × 10
0. 999999𝑐
1. 553380343 × 10
0. 9999999𝑐
4. 912218854 × 10
0. 99999999𝑐
1. 5533799555 × 10
𝑠
22. 36454
𝑠
70. 70697
𝑠
222. 5900
−4
−4
−3
−3
𝑠
707. 0522
𝑠
2235. 895
−2
𝑠
7070. 5203
Based from the table, the time will be dilated by a factor of 1000 when the speed is
0. 9999990𝑐 < 𝑢 < 0. 9999999𝑐. To compute this exactly, we let ∆𝑡 = 1000(∆𝑡0) and isolate
𝑢.
1000(∆𝑡0) =
∆𝑡0
2
1−(𝑢/𝑐)
2
⇒ 1000 1 − (𝑢/𝑐) = 1
2
⇒ 1 − (𝑢/𝑐) =
2
⇒ (𝑢/𝑐) = 1 −
⇒𝑢 = 𝑐 1 −
(
2
1
1000
)
1
1000000
1
1000000
= 299999850 𝑚/𝑠
In terms of c, 𝑢 = 0. 9999995𝑐, which agrees with the interval above.
( 0) created in your laboratory at time 𝑡𝑙𝑎𝑏 = 0. A rocket ship move past your
Q3. Consider a pion π
−4
lab from point A to point B in a span of 1 × 10
𝑠.
Figure 3. Pion in the lab and a rocketship.
0
(a.) What is the lifetime of π according to an observer inside the rocketship?
Let the lab be in frame S and the rocketship in frame S’. To get how much time is dilated for
the observer in the rocketship, we must get the velocity 𝑢 of the rocketship relative to the frame S.
This is given by the simple speed-distance-time formula:
𝑢 = 𝑑/𝑡
=
7
1000 𝑚
−4
1×10
= 10 𝑚/𝑠
𝑠
−17
The lifetime of a pion is ∆𝑡0 = 8. 43 × 10
. Using the time-dilation formula to obtain
the lifetime of the pion as observed from the rocketship, we get
∆𝑡 =
∆𝑡0
2
1−(𝑢/𝑐)
−17
=
8.43×10
(
1−
7
10 𝑚/𝑠
8
3×10 𝑚/𝑠
= 8. 43468724 × 10
−17
𝑠
2
)
(b.) How long did it take to travel from point A to point B from the perspective of the rocketship?
Viewing this from frame S’, we introduce ∆𝑡0' the proper time of how long it took for the
rocketship to get from point A to point B. |𝑢| = |𝑢'| as the two frames move relative to each other.
Noting that the time it took for the rocketship to move from point A to point B as measured by the
−4
lab is ∆𝑡' = 1 × 10
∆𝑡0' =
(
𝑠, we rearrange the time-dilation formula:
)
2
1 − (𝑢'/𝑐) ∆𝑡'
2
=
1 −
[(107 𝑚/𝑠)/(3 × 108 𝑚/𝑠)] (1 × 10−4 𝑠)
−5
= 9. 9944429 × 10
𝑠 = 0. 9944429 × 10
−4
𝑠
Q4. An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight
down toward the surface of the earth with a speed of 0.99540c relative to the earth. A scientist at
rest on the earth’s surface measures that the particle is created at an altitude of 45.0 km.
Figure 4. Unstable particle from the atmosphere moving towards the Earth's surface
(a.) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the
surface of the earth?
In the scientist’s reference frame, we use the speed-time-distance formula to obtain the time
he measured:
∆𝑡 =
=
𝑑
𝑢
45,000 𝑚
8
0.99540(3×10 𝑚/𝑠)
−4
= 1. 506931887 × 10
𝑠
(b.) Use the length-contraction formula to calculate the distance from where the particle is created to
the surface of the earth as measured in the particle’s frame.
In the Earth’s (scientist’s) reference frame, 𝑑 = 𝑙0 = 45. 0 𝑘𝑚. Using the contracted length
formula,
𝑙 = 𝑙0
1 −
(
0.99540𝑐 2
𝑐
)
2
= (45, 000 𝑚) 1 − 0. 99540 = 4, 311. 28 𝑚 = 4. 31128 𝑘𝑚
(c.) In the particle’s frame, how much time does it take the particle to travel from where it is created
to the surface of the earth? Calculate this time both by time dilation formula and from the distance
calculated in part (b). Do the two results agree?
Using the time-dilation formula:
∆𝑡0 =
(
)
2
1 − (𝑢/𝑐) ∆𝑡
2
−4
(
−5
)
= 1 − (0. 99540𝑐/𝑐) 1. 506931887 × 10 𝑠 = 1. 443735124 × 10 𝑠
Using the speed-time-distance formula, while noting that 𝑑 = 𝑙 = 𝑙0' in the particle’s reference
frame:
𝑑
𝑢
∆𝑡 =
−5
4,311.28 𝑚
=
8
(
)
0.99540 3×10 𝑚/𝑠
= 1. 443735124 × 10
𝑠.
Therefore, the two results agree.
Q5. An unstable particle is created at 𝑥 = 5. 0 𝑚 and decayed upon reaching 𝑥 = 15. 0 𝑚 as seen
−5
in frame S. Frame S measured the lifetime of the particles to be 2 × 10 𝑠. What is the lifetime of
the particle in frame S’ if the particle traversed a distance of ∆𝑥' = 20. 0 𝑚 as measured by S’?
To get ∆𝑡', we can use the spacetime interval and input the given, shown by:
2
2
2
2
(𝑐∆𝑡) − (∆𝑥) = (𝑐∆𝑡') − (∆𝑥') .
Isolating ∆𝑡'
2
2
2
2
⇒ 𝑐 (∆𝑡') = (𝑐∆𝑡) − (∆𝑥) + (∆𝑥')
2
2
2
2
[ 2
] 2
[(𝑐∆𝑡)2 − (∆𝑥)2 + (∆𝑥')2]/𝑐2
⇒ (∆𝑡') = (𝑐∆𝑡) − (∆𝑥) + (∆𝑥') /𝑐
⇒ ∆𝑡' =
2
2
2
(∆𝑥) +(∆𝑥')
=
(∆𝑡) −
=
(2 × 10
2
𝑐
−5
2
2
2
(15.0 𝑚−5.0 𝑚) +(20.0 𝑚)
𝑠) −
2
(3×108 𝑚/𝑠)
−5
= 2. 000008333 × 10
𝑠
Q6. Show that
𝑣𝑥'+𝑢
(a.) 𝑣𝑥 =
𝑢𝑣𝑥'
1+
𝑐
2
We know that
𝑣𝑥−𝑢
𝑣𝑥' =
1−
𝑢𝑣𝑥
𝑐
2
Isolating 𝑣𝑥,
(
)= 𝑣 − 𝑢
𝑢𝑣𝑥
𝑣𝑥' 1 −
2
𝑐
𝑢𝑣𝑥𝑣𝑥'
⇒ 𝑣𝑥' −
2
𝑐
𝑥
= 𝑣𝑥 − 𝑢
𝑢𝑣𝑥𝑣𝑥'
⇒ 𝑣𝑥' + 𝑢 = 𝑣𝑥 +
2
𝑐
(
⇒ 𝑣𝑥' + 𝑢 = 𝑣𝑥 1 +
𝑣𝑥'+𝑢
⇒ 𝑣𝑥 =
1+
𝑢𝑣𝑥'
𝑢𝑣𝑥'
2
𝑐
)
∎
2
𝑐
(b.) the Galilean relationship 𝑣𝑥 = 𝑣𝑥' + 𝑢 is only an approximation in the limit 𝑣𝑥' << 𝑐 and
𝑢 << 𝑐. That, is 𝑣𝑥 ≃ 𝑣𝑥' + 𝑢 + ອ(α)
Recall,
𝑣𝑥 = (𝑣𝑥' + 𝑢)
1
1+
𝑢𝑣𝑥'
𝑐
2
Note that a geometric series can be evaluated using the equation below:
+∞
𝑛−1
∑ 𝑎𝑟
=
𝑛=1
𝑎
1−𝑟
, if |𝑟| < 1.
𝑢𝑣𝑥'
If 𝑣𝑥' << 𝑐 and 𝑢 << 𝑐,
2
𝑐
< 1 and is a very small number. We let 𝑟 =−
𝑎 = (𝑣𝑥' + 𝑢) and plug them into the summation since |𝑟| < 1, such that
+∞
)(
(
𝑣𝑥 = ∑ 𝑣𝑥' + 𝑢 −
𝑛=1
⎡
= (𝑣𝑥' + 𝑢)⎢1 −
⎢
⎣
(
𝑢𝑣𝑥'
2
𝑐
)
𝑛−1
(
𝑢𝑣𝑥'
+ −
2
𝑐
)(
≅𝑣𝑥' + 𝑢 − 𝑣𝑥' + 𝑢
𝑢𝑣𝑥'
2
𝑐
𝑢𝑣𝑥'
2
𝑐
2
3
) + (− ) +...⎤⎥⎥⎦
𝑢𝑣𝑥'
2
𝑐
)= 𝑣 ' + 𝑢 −
𝑥
2
2
𝑢𝑣𝑥' +𝑢 𝑣𝑥'
2
𝑐
𝑢𝑣𝑥'
2
𝑐
and
Since the terms from
2
( ) and so on are very small, they can be neglected.
𝑢𝑣𝑥'
2
𝑐
2
Thus, the leading correction ອ(α) =−
2
𝑢𝑣𝑥' +𝑢 𝑣𝑥'
2
𝑐