## Thursday, October 25, 2018

### Solutions To Special Relativity Problems (3)

Solutions to the problems from the Oct. 17 instalment:

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

Then, we have t E = 4.4 yrs.

And:

t A  = t E  [1 - v2/c2]½

t A = (4.4 yrs.) [1 - (0.95c)2/c2]½ = 1.37 yrs.

(b)  Since we know: t A  = 1.37 yrs.

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

2) In this case, v = c = 3 x 108 m/s

d = v/H = (3 x 108 m/s)/ (2.26 x 10-18 s-1)

d = 1.32 x 1026 m

Converting to light years:

d = (1.32 x 1026 m)/ (9.5 x 1015 m /Ly) = 1.4 x 1010 Ly

b) Would it be observable from Earth?

Given that modern telescopes can penetrate to about 1.8 x 1010 Ly, the galaxy should easily be observable to the Hubble but might be more problematical for land-based scopes.

3) We know the recessional velocity v = 6 x 104 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 107 m/s)/(2.26 x 10-18 s-1)= 2.6 x 1025 m

and d = (2.6 x 1025 m)/(9.5 x 1015 m /Ly) = 2.8 x 109 Ly

(b) z = v/c = (6 x 107 m/s)/(3 x 108 m/s) = 0.2

(c) T = d/v = (2.6 x 1025 m)/(6 x 107 m/s)

= 4.3 x 1017 s

But 1 yr. = 3.15 x 107 s

so T = (4.3 x 1017 s)/(3.15 x 107 s/ yr)

T = 1.36 x 1010 years, or 13.6 billion years

4) Let
lo  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/lo = (1 - v/c) ½ /(1 + v/c) ½

l/lo = (1 + 0.8) ½/ (1 - 0.8) ½ = (1.8/0.2) ½

l/lo = Ö(9) = 3

then:

l = 3 lo = 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 =
Ö(9)  = 3

Then: z = 3 - 1 = 2

c)  The corrected velocity, v
=  c [(z 2 + 2z) / (z 2 + 2z + 2)]

Then:  v =  c[4 + 4]/ [4 + 4  + 2]  =   8c/10  = 4c/5 = 0.8c