HOMEWORK SOLUTIONS

WEEK 1 PROBLEMS

#17.) Calculate, for the judge, how fast were you going in miles per hour when you ran the red light because it appeared Doppler-shifted green to you. Take red light to have a wavelength of 650 nm and green to have a wavelength of 550 nm.

This is a pretty straightforward Doppler-shift problem... the driver is approaching the red light at a high rate of speed, which causes the light that they sees to be shifted in wavelength (or frequency) such that it appears to be green to them. So all you need to do is apply the Doppler-shift formula for light (eq. 1.13):

f_obs = Ö(1+v/c) f_source (1.13)

Ö(1-v/c)

Since we’re given wavelengths instead of frequencies, using

c = lf

Þf = c

So that eq. 1.13 looks like

c = Ö(1+v/c) c__

l_obs Ö(1-v/c) l_source

Þ l_obs = Ö(1-v/c) l_source

Ö(1+v/c)

We know that the wavelength seen by the person in the car (l_obs) is green, which is 550 nm, and that the wavelength given off by the light (l_source) is red, which is 650 nm. What we are interested in is the speed required to accomplish this, which is v in that equation. So solve for it to get:

v = (l_source² - l_obs²) c (1)

(l_source² + l_obs²)

Plug and chug from here to get

v = 0.166c = 1.11 x 10⁸ mi/hr

#18.) (a)How fast and in what direction must galaxy A be moving if an absorption line found at 550 nm (green) for a stationary galaxy is shifted to 450 nm (blue) for A?

Ok, first of all, if you look at the Doppler-shift equation

l_obs = Ö(1-v/c) l_source

Ö(1+v/c)

as given, this is for the observer approaching the source. You can see that l_obs is gonna be smaller than l_source. Colors with smaller wavelengths are more towards the blue end of the spectrum, so if observer and source are closing on one another, the light is said to be blue shifted. If you look at the corresponding equation for the observer moving away from the source:

l_obs = Ö(1+v/c) l_source

Ö(1-v/c)

you can see that l_obs is gonna be larger than l_source. Light with larger wavelengths are towards the red end of the spectrum, so if observer and source are moving away from one another, the light is said to be red shifted. So, since the light seen from galaxy A has a smaller wavelength than that ordinarily seen, that means that the light has been blue shifted and the galaxy is therefore moving towards the observer. To find out how fast it must be moving to achieve this blue shift, use eq (1) from problem #17:

v = (l_source² - l_obs²) c

(l_source² + l_obs²)

which I calculate to be

v = 0.2c

(b) How fast and in what direction is galaxy B moving if it shows the same line shifted to 700 nm (red)?

From part (a), we know that since the wavelength gets larger, that means that the line is red shifted, and therefore galaxy B is moving away from the observer. To get the speed with which it is moving, apply the analogue of eq (1) for redshifts:

v = (l_obs² - l_source²) c

(l_obs² + l_source²)

which I get as

v = 0.2c

#23.) Speed of light in a moving medium. The motion of a medium such as water influences the speed of light. This effect was first observed by Fizeau in 1851. Consider a light beam passing through a horizontal column of water moving with speed v.

(a) Show that if the beam travels in the same direction as the flow of water, the speed of light measured in the laboratory frame is given by

u = c(1+nv/c)

n(1+v/nc)

where n is the index of refraction of the water.

First of all, you need to remember what the index of refraction of a medium is… it is defined to be the speed of light in vacuum, c, divided by the speed of light in the medium as measured in a reference frame stationary with respect to the medium, v,:

n º c

Therefore, you can express the speed of light in the rest frame of some medium, v, as

v = c

In other words, an observer outside of the medium will measure the speed of a beam of light passing through the stationary medium to be v. However, if the medium is flowing with respect to the lab, the rest frame of the medium is also moving in the same direction and speed as the fluid. So to get the velocity of the beam of light passing through the medium in the same direction as the flow as measured from outside the medium, we must use the relativistic velocity addition formulas to transform from the rest frame of the fluid, which is moving with speed v with respect to the lab, to the lab frame. Call the direction of flow x. In the rest frame of the fluid, which I will call the stationary frame S, it appears that the lab frame S’, is moving with speed –v. So use eq 1.32 to transform to the lab frame:

u_x’ = u_x – v (1.32)

1-(u_xv/c²)

v is the speed of the moving lab frame, S’, so

v = -v

and the thing we are watching is the beam of light which, in the medium, has speed c/n. Therefore

u_x = c

Substitution into (1.32) yields

u_x’ = c/n + v

1-(c/n)(-v/c²)

Pull a factor of c/n out of the numerator and simplify the denominator to get

u_x’ = c(1+nv/c)

n(1+v/nc)

(b)Show that for v<<c, the expression above is in good agreement with Fizeau’s experimental result

u » c + v – v

n n²

This proves that the Lorentz velocity transformation and not the Galilean velocity transformation is correct for light.

v<<c

Þ v<<1

So we can drop terms of order (v/c)² and higher. Take the result from part (a)

u_x’ = c(1+nv/c)

n(1+v/nc)

and write this as

u_x’ = c(1+nv/c)(1+v/nc)^-1

Look at the term

(1+v/nc)^-1

To simplify things, call

v = x

So that

(1+v/nc)^-1 = (1+x)^-1

Now, Taylor expand this term:

(1+x)^-1 = 1 – x + x² – x³ + x⁴ - …

Since x~(v/c), we can drop terms of order x² and higher. This makes

(1+x)^-1 » 1 – x

Substitute this approximation into the formula for u_x’:

u_x’ » c(1+nv/c)(1-v/nc)

= c + v – v

n n²

Woo hoo!