Desai_12010-12050

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Section A HW 1 [12010-12050]Ch. 1: 9,10,12; Ch. 3: 4,8,12

Nikhil Desai

Chapter 1

Problem 9

a: Let the spatial separation between Earth and Andromeda, in the rest frame of the

Earth, be ∆xr, and the time in which the spaceship is observed to make the trip in thisframe be ∆tr. The invariant spacetime interval ∆s corresponding to the trip is specifiedby the equation ∆s2 = (c∆tr)2 − (∆xr)2. Since ∆s is invariant under the change of reference frame, we know that if ∆xs and ∆ts signify spatial and temporal separation,respectively, in the spaceship frame, then ∆s2 = (c∆ts)2−∆x2

s. But ∆xs is clearly zero,

so ∆s = c∆ts, and so

ts =∆s

c=

 c2∆t2r −∆x2r

c=

 ∆t2r −

∆x2rc2

.

Substituting, we have that ∆ts = 2.00

×105 years.

b: We observe that vr = ∆xr∆tr

= 0.995c.

c: We substitute into the formula from part (a), to find that ∆ts = 6.33× 104 years.

Additionally, β = ∆xr

c∆tr= 0.9995.

d: We are given ∆tr and wish to calculate the speed necessary to achieve this for thetrip. Using the invariance of the interval, we once again have that ∆s2 = c2∆t2

r−∆x2

r=

c2∆t2s. Consequently,c∆tr =

 ∆x2r + c2∆t2s.

As a result,

β =∆xr

c∆tr=

∆xr

 ∆x2r + c2∆t2s.

To simplify this calculation, we observe that

∆xr ∆x2r + c2∆t2s

=

1 +

c2∆t2s

∆x2r

−1

2

≈ 1− c2∆t2s

2∆x2r.

Substituting, we get that β ≈ 1 − 5 × 10−11.

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Problem 10

a: We know that Samantha’s information is carried at the speed of light from here

to Zircon, as viewed in the Eart-Zircon reference frame, which we shall call the ”restframe.” Let the distance Samantha traveled in the rest frame be ∆xr, and the timethis travel takes to be ∆tr. It is clear that on Samantha’s outbound trip, βr = 1,since she travels at the speed of light. Thus, ∆xr = c∆tr. The spacetime intervalcorresponding to the outbound trip is ∆s =

 (c∆tr)2 − ∆x2r =

 (c∆tr)2 − (c∆tr)2 =

0. In Samantha’s reference frame, the distance between start and destination is zero,and so she does not age.

b: Samantha ages for one year during her stay on Zircon, then again instantaneously(for her) transports herself back to Earth. She has aged for a total of one year.

c: In the rest frame, Samantha’s information takes two million years to reach Zirconand two million years to return. She also spends one year on Zircon, for a total of 

4,000,001 years.d: Assuming that the conquest of Zircon is instantaneous, it should take at least4 million years for the dictator to hear back.

e: Samantha should still age just one year; the travel to and from the galaxy is, forthe astronaut, instantaneous (as ascertained previously), while the astronaut will ageone year with relation to the time spent in the galaxy’s reference frame. This is a totalof one year.

Problem 12

a: Ignoring time dilation, we see that half of the mesons would remain after thegroup has traveled a distance δ = cτ, where τ  is their half-life. By the charts, we then

have that δ = 5.4 meters.b: The ”characteristic distance” at which half of the particles are undecayed was

(erroneously) calculated to be δ = cτ  above. The proper distance for the mesons shouldin reality be δ = vτ , where τ  is the half-life of the particles as measured  in the Earth’srest frame, and v is the observed speed of the particles. The spacetime interval cor-responding to a particle traveling this distance is s =

 (cτ )2 − δ2 =

√c2τ 2 − v2τ 2.

Letting β denote the ratio of particle speed to c, we have that v = βc, so

s = 

c2τ 2 − v2τ 2 = τ  

c2 − β2c2 = cτ  

1− β2.

In the reference frame of a traveling particle, we know this corresponds to an intervalof time τ  and a spatial interval of zero, so we have that cτ  = s = cτ  1− β2, and so

τ  =τ  

1 − β2= γτ .

Consequently,δ

δ=

vτ 

cτ =

(βc)(γτ )

cτ = βγ  =

β 1 − β2

.

Substituting for β, we find this to be approximately 15.1 .

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Chapter 3

Problem 4

a: yes. b: yes. c: no. d: no. e: no. f: yes. g: no.

Problem 8

a: The particle radiates light isotropically at each point in its trajectory. Let theparticle travel at v, the light at c = c

n, where n is the medium’s index of refraction.

Observe that in a time t, the particle travels vt while the light travels ct. In order for asingle cone to form, the wavefront formed by the traveling light must be perpendicularto the direction of propagation. Consequently, the angle φ as given in the figure is givenby

cos φ =ct

vt=

c

n

v=

1v

c n=

1

βn,

which was to be shown.b: To produce Cerenkov radiation, a particle moving in Lucite must have a speed

greater than 23

c, as n = 32

. A particle moving in Lucite has a maximum velocity of just

less than c, for which the angle of the shock cone is

arccos

1

(1)32

= arccos

2

3

≈ 48.19 degrees.

Problem 12

a: The two legs of the trip take different times. In the first leg, from A to B, thewind blows against the plane and it travels with an actual velocity of (c − v), taking atime ∆x

c−v(where ∆x is the distance between A and B.) In the second leg, from B to A,

it travels with an actual velocity of  c + v and takes time ∆x

c+v. Thus, the total time taken

is

∆t =∆x

c− v+

∆x

c + v= ∆x

(c + v) + (c− v)

(c − v)(c + v)

=

(∆x)(2c)

c2 − v2

We can rearrange this to obtain

2∆x

c ·

1c2

−v2

c2 =

2∆x

c ·1

1 − vc 2 ,

The original round trip would have taken a total time of 2∆x

c

, so the round-trip is

longer by a factor of  1

1−v2

c2

. The reason that equivalent ”boosts” and ”drags” given by

the wind on the different legs do not cancel out is because the relationship of velocity totime taken to travel a given distance is not linear (in fact, it is an inverse relationship).

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b: The plane flies at speed c such that its motion in the direction of the wind isexactly countered by the wind itself, traveling at speed v. This means that for both legs,the plane has a lateral speed of 

√c2

−v2, and consequently the round-trip of the plane

is2∆x√c2 − v2

=

2∆x

c

c√

c2 − v2

=

2∆x

c

1 1 −

v

c

2 ,

which is a factor of  1 1−( v

c)2

greater than the travel time in still air, as was to be shown.

c: It is clear that1 

1 −v

c

2 <1

1 −v

c

2when v < c, since if  v < c, v

c< 1, so 0 < 1− v

2

c2< 1 and so

1−v

c

2<

 1 −

v

c

2.

Consequently, the plane flying across the wind returns first, and we can infer that theplane flying in the direction of the wind will return last. The difference between thearrival times of these two flights will be

∆t =2∆x

c

1

1−v

c

2 − 1 1 −

v

c

2

= 2∆xc

1− v

2

c2−1

−

1 − v2

c2−

1

2

≈ 2 · ∆x

c·

1 +v2

c2

−

1 +v2

2c2

= 2 · ∆x

c·

v2

2c2

=

∆x · v2

c3

d: We know that the first plane to arrive back will be the one flying perpendicu-lar to the direction of the wind, while the last to arrive will be parallel to the wind.Consequently, the wind velocity can be calculated by using the formula from part c toget

v =  c3∆t

∆x

.

Substituting, we get for the problem that v = 2.78 m/s.

e: If there were no reflection, then the approximate time lag of the light beams wouldbe, as calculated before,

∆t =v2∆x

c3= 7.3× 10−16 s.

This is too small to detect with current measuring tools.

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If there were reflection, then ∆t would be in the form

i= v

2∆xi

c3= v2

c3

i

∆xi,where

i

∆xi = ∆x. Thus, the same time delay would be observed.f: We know that for any frequency of light, f λ = c, and f  = T −1, so

T  =1

f =

1

c/λ=

λ

c.

Substituting, we have that T ≈ 2.0 × 10−25 seconds.Michelson and Morley’s observations indicated that the total change in delay between

the two configurations is less than T 

100, and consequently 2∆t < T 

100. Replacing the

expression for ∆t, we get

2

∆x

c3· v2

<T 

100

and so

v < 

c3T 200∆x

= 3.5× 103 m/s.

This is less than one-ninth the speed of Earth in its orbit.g: Not necessarily. Perhaps the ether is “locally” stationary - a small clump of ether

might surround the Earth and make the speed of the ether locally zero. We could testthis by performing the experiment on some body moving relative to the Earth, such asthe Moon, and seeing if the same results are obtained there.

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