Phys 4271 - Exercises - 16 January 2013

These are some simple exercises in special relativity calculations. They will not be collected forcredit. They are mainly to help you remember some of the basics. Solutions will be posted in afew days.

1. A missile is fired from New York to Los Angeles (don’t ask why), a distance of about 5000km. How fast must the missile go so that an observer on the ground in Omaha perceives itslength to be shorter than its length at rest by 1% ? (Ans.: 4.2× 107m/s)

Solution: With a contraction of 1%, L/L◦ = 0.99 =√

1− β2. Thus β = 0.14 orv = βc = 4.2× 107m/s.

2. A clock in a moving spaceship is observed to run at a speed of only 3/5 that of when it is atrest on earth. How fast is the spaceship moving? (Ans.: 2.4× 108m/s)

Solution: The observed time-dilation gives the Lorentz factor as γ = 5/3, so thenβ2 = 1− 1/γ2 = 16/25, and the speed is then v = βc = (4/5)c = 2.4× 108m/s.

3. What speed will a clock have to be moving in order to run at a rate that is one-half the rateof the clock at rest? (Ans.: 2.6× 108m/s)

Solution: This means that any given time interval ∆t measured by an observer who watchesthe clock go past will be twice as long as the interval ∆t′ measured in the rest frame of theclock. The Lorentz time-dilation is given by ∆t = γ∆t′, so we can rearrange things to obtainthe speed or β:

γ =1√

(1− β2)⇒ β2 = 1− 1

γ2

In our case, γ = ∆t/∆t′ = 2 so

β =

√1− 1

22= 0.866

giving v = 0.866c = 2.6× 108m/s.

4. How fast must a meter stick be moving if its length is observed to shrink to 0.5 m ? (Ans.:2.6× 108m/s)

Solution: An observer measures the moving stick to be L, while someone moving with thestick measures its rest length L′. In our case L = 0.5L′. The Lorentz length contractionexpression tells us that γ = L′/L = 2. As in the last problem,

β =

√1− 1

22= 0.866

so again v = 0.866c = 2.6× 108m/s.

5. A clock on a spacecraft runs 1 second slower per day when compared to an identical clock onearth. What is the speed of the spacecraft relative to the earth? (Hint: it is helpful to makean approximation using β2 � 1) (Ans.: 1.44× 106m/s)

Solution: Use the Lorentz time dilation relation between two frames ∆t = γ∆t′ to determinethe Lorentz factor γ: In the at-rest earth frame S, let ∆t = 1 day = 86400 sec. In themoving spacecraft frame S ′, the clock is one second per day slow, so in that same period∆t′ = 86399 sec. Thus γ = 86400/86399 = 1 + 1.16× 10−5. Since γ differs little from 1.0,the speed is well under the speed of light, β � 1.0. Use the approximate expression:

γ ≈ 1 +1

2β2 ⇒ β ≈

√2(γ − 1) =

√2(1.16× 10−5) = 0.0048

to obtain v = βc = .0048c = 1.44× 106 m/s.

6. The pion is a subatomic particle which has an average lifetime of 2.6× 10−8 sec in its ownframe of reference (i.e., when it is at rest). If the particle is observed moving at a speed of0.95c, what is (a) its mean lifetime and (b) the average distance it travels before decaying?(Ans.: (a) 8.33× 10−8s, (b) 24 m)

Solution: (a) τ = γτ ′ with β = 0.95 so

τ = τ ′(1− β2)−1/2 = 2.6× 10−8sec(1− 0.952)−1/2 = 8.33× 10−8sec

(b) d = vτ = 0.95(3× 108m/s)(8.33× 10−8sec) = 24m.

7. If the pion above is seen to travel 10 m before it decays, how fast must it be moving? (Ans.:2.37× 108m/s)

Solution: If a pion travels 10m in our frame, v∆t = 10m. In the pion’s rest frame,∆t′ = 2.6× 10−8sec. Lorentz time dilation tells us that ∆t = γ∆t′, so

10m = v∆t = vγ∆t′ = v(1− v2/c2)−1/2(2.6× 10−8sec)

Squaring and rearranging,

(3.85× 108m/sec)2(1− v2/c2) = v2

(3.85× 108m/sec)2 = v2(1 + (3.85× 108m/sec)2/c2) = v2(1 + 1.64)

v = 2.37× 108m/sec = 0.789c

8. An astronaut at rest on Earth has a heartbeat rate of 70 beats/min. What will this rate bewhen she is traveling in a spaceship at 0.90c as measured by (a) another astronaut in theship with her and (b) by an observer at rest on Earth? (Ans.: (b) 30.5 beats/min)

Solution:(a) To this astronaut, the subject is at rest so they will measure 70 beats/min.(b) The time between beats (when observed at rest) is ∆t′ = (1/70) min. When she is seento be moving, the time interval is

∆t = γ∆t′ =1√

1− (0.9)2∆t′ = 2.29(1/70) = 1/(30.5) min

or 30.5 beats/min.

9. A particle having a speed of 0.92c has a momentum of 10−16kg ·m/s. What is its mass?(Ans.: 1.42× 10−25kg)

Solution:

γ =1√

1− 0.922= 2.5516

m =p

γv=

10−16kg ·m/s(2.5516)(0.92)(3× 108m/s)

= 1.42× 10−25kg

10. Calculate the momentum, kinetic energy, and total energy of an electron traveling at a speedof (a) 0.01c, (b) 0.1c, and (c) 0.9c. (Ans.: for part (a), 5.11 keV/c, 26 eV, 511.026 keV)

Solution: (a)

p = γmv =(511 keV/c2)(0.01c)√

1− 0.012= 5.11 keV/c

E = γmc2 =(511 keV/c2)(c2)√

1− 0.012= 511.026 keV

K = E −mc2 = 511.026 keV − 511.00 keV = 26eV(≈ 30eV)

Similarly for parts (b) and (c):

(b) β = 0.1: p = 51.4 keV/c, E = 513.6 keV, K = 2.6 keV.

(c) β = 0.9: p = 1055 keV/c, E = 1172 keV, K = 661 keV.

11. What is the speed of an electron when its kinetic energy is: (a) 10% of its rest energy, (b)equal to its rest energy, and (c) 10 times the rest energy? (Ans.: (a) .417c, (b) .866c, (c).996c)

Solution:

(a) If K = 0.1mc2 then E = K +mc2 = 1.1mc2 so γ = 1.1. Then

β =

√1− 1

γ2=

√1− 1

1.12= 0.417

and v = 0.417c.

(b) As above, γ = 2.0 and v =√

3c/2 = 0.866c.

(c) As above, γ = 11.0 and v = 0.996c.

12. Across what potential difference does an electron have to be accelerated in order to reach aspeed of v = 2× 107 m/s? Comment on whether you need to use relativistic calculationshere.

Solution: Assume the speed is exact. Relativistically, put the electron through potential ∆Vand find the kinetic energy is K = q∆V = e∆V (where “e” is the charge of an electron).Relativistically K = mc2(γ − 1). In this problem,

γ = 1/√

1− ((2× 107)/(2.9979× 108))2

and for an electron with mc2 = 511 keV, ∆V = 1141.0 volts.

Nonrelativistically K = e∆V = 12mv2. so

∆V =mv2

2e=

(9.1094× 10−31 kg)(2× 107m/s)2

2(1.6022× 10−19 C)= 1137.1 volts

These results differ by about 4 volts or 0.34%. Use of relativistic techniques is neccessaryonly if this level of precision is required.

(Note: I used a more precise expression for the speed of light c than we normally do. As youmay have noticed, the answer is slightly different that way. But we’re looking for rathersmall differences, so it was probably important to use the more precise value.)

13. Protons are accelerated at Fermi National Accelerator Laboratory near Chicago to a (total)energy equal to 400 times their rest energy. (a) What is the speed of these protons? (b)What is their kinetic energy in MeV? (Ans.: 0.999997c, 3.74× 105 MeV = 374 GeV)

Solution: (a) E = 400mc2 = γmc2, so γ = 400. Then β =√

1− 1/γ2 ⇒ v = 0.999997c. (b)

K = (γ − 1)mc2 = 399 mc2 = 399× 938.3 MeV = 3.74× 105 MeV