Solutions to Special Relativity (4)

We now go through the solutions at the end of instalment (4):

Problems:

1) Assume two astronauts are traveling at v = 0.95c on a journey to the system of Alpha Centauri. We on Earth would say that it takes 4.2 / 0.95c = 4.4 years to reach the system 4.2 light years distant. But the astronauts dispute this.

(a) How much time passes on the astronauts' clocks?

Solution

Let t(A) be the astronauts' clock and t(E) be the time recorded on an Earth clock.

Then, we have t(E) = 4.4 yrs.

And:

t(A) = t(E)[1 - v^2/c^2]^½

t(A) = (4.4 yrs.) [1 - (0.95c)^2/c^2]^½ = 1.37 yrs.


(b) What is the distance to Alpha Centauri as measured by the astronauts? (Hint: this is an exact analog of the muon path length problem (#3) from the previous problem set)

Solution

Since we know: t(A) = 1.37 yrs.

then the distance D(A) = (0.95c) (1.37 yrs) = 1.31 Ly


2) According to Hubble's law, the distant galaxies are receding from us at speeds proportional to their distances, d, e.g. v = Hd. Where H = 2.26 x 10^-18 s^-1, currently).

a) How far away would a galaxy be in light years whose velocity relative to the Earth is c?

Solutions

In this case, v = c = 3 x 10^8 m/s

d = v/H = (3 x 10^8 m/s)/ (2.26 x 10^-18 s^-1)

d = 1.32 x 10^26 m

Converting to light years:

d = (1.32 x 10^26 m)/ (9.5 x 10^15 m /Ly) = 1.4 x 10^10 Ly


b) Would it be observable from Earth?

Given that modern telescopes can penetrate to about 1.8 x 10^10 Ly, the galaxy should easily be observable to the Hubble but might be more problematical for land-based scopes.


3) A galaxy in Hydra emits light with a red shift corresponding to a recessional velocity of 6 x 10^4 km/s.

a) What is its distance according to Hubble's law?

b) What is the value of z?

c) Assume this galaxy passed Earth T years ago and has moved with constant velocity ever since, what is the value of T?

Solutions

We know the recessional velocity v = 6 x 10^4 km/s

By Hubble's law: v = Hd so the distance d = v/H

Then, attending to the proper units for v, H:

d = (6 x 10^7 m/s)/(2.26 x 10^-18 s^-1)= 2.6 x 10^25 m

and d = (2.6 x 10^25 m)/(9.5 x 10^15 m /Ly) = 2.8 x 10^9 Ly


(b) z = v/c = (6 x 10^7 m/s)/(3 x 10^8 m/s) = 0.2

(c) T = d/v = (2.6 x 10^25 m)/(6 x 10^7 m/s) = 4.3 x 10^17 s

But 1 yr. = 3.15 x 10^7 s

so T = (4.3 x 10^17 s)/(3.15 x 10^7 s/ yr)

T = 1.36 x 10^10 years, or 13.6 billion years


4) Some observations reported on the quasar 3C-9 suggest that when it emitted the light that just reached Earth it was receding at a velocity of 0.8c. One of the lines identified in its spectrum has a wavelength of 1200 Å (angstroms) when emitted from a stationary source.

a) At what wavelength must this spectral line have appeared in the observed spectrum of the quasar?

b) What is its red shift, z?

Solutions

Let L(o) be the normal wavelength = 1200 Å

and L be the red-shifted value.

We know v = 0.8c so we must use the modified Doppler version, viz.

L/L(o) = (1 - v/c)^ ½ /(1 + v/c)^ ½]

L/L(o) = (1 + 0.8)^ ½/ (1 = 0.8)^ ½ = (1.8/0.2)^ ½

L/L(o) = (9)^ ½ = 3

then:

L = 3 L(o) = 3 (1200 Å) = 3600 Å


(b) The red shift of the quasar is found from:

1 + z = (1 + v/c)^ ½]/(1 - v/c)^ ½

1 + z = (1.8/0.2)^ ½ = (9)^ ½ = 3

Then: z = 3 - 1 = 2