Spaceship Doppler frequency

Question

A spaceship starts falling under gravity with an acceleration $g$ as measured by an observer Barry at rest on Earth. At the instant that the ship starts to fall, an astronaut Harry at the base of the rocket ship sends a light signal of frequency $w$ vertically upward to another astronaut Sally a distance $h$ above.

Barry argues that the light signal reaching Sally ought to be Doppler shifted toward the blue. This Doppler shift $Δw$ is given by $\frac{Δw}{w_{Doppler}} = \frac{Δu}{c} $

where $Δu$ is the velocity of the rocket ship after a time $Δt = \frac{h}{c}$

What I want to know is how that Doppler shift equation came about mathematically? The text said they used the formula for the low velocity approximation to the relativistic Doppler shift which is

$w' = w \sqrt{\frac{c+v}{c-v}} $

But I just don't see how that happened. Maybe I'm missing something...

Answer 1

I sure hope my assumption is correct that you're just needing help understanding where an equation that you encountered in your book came from, rather than this having been a homework problem that you were supposed to do. I fully support the policy here of not doing people's homework problems for them.

From the starting point of

$$w'=w\sqrt{\frac{c+v}{c-v}}\ ,$$

divide both the numerator and the denominator by $c$, giving

$$w'=w\sqrt{\frac{1+v/c}{1-v/c}}\ .$$

As a general rule, as can be verified by a Taylor expansion, if $\epsilon \ll 1$, then

$$\frac{1}{1-\epsilon}\approx 1+\epsilon\ .$$

Noting that $v/c \ll 1$, we apply that approximation to the expression for $w'$ to give

$$w'\approx w\sqrt{(1+v/c)(1+v/c)}\ ,$$

which, keeping only terms under the square root that are first order in $v/c$, gives

$$w'\approx w\sqrt{1+2\frac{v}{c}}\ .$$

Another general rule which can be verified with a Taylor expansion is that if $\epsilon\ll 1$, then

$$\sqrt{1+\epsilon}\approx 1+\frac{\epsilon}{2}\ .$$

Applying that to the above equation gives

$$w'\approx w\left(1+\frac{v}{c}\right)\ .$$

Multiplying both sides of that equation by $1-v/c$ and again keeping only terms that are first order in $v/c$ gives

$$w'\left(1-\frac{v}{c}\right)\approx w\ .$$

A little bit of simple algebra rearranges that equation into

$$\frac{w'-w}{w'}\approx \frac{v}{c}\ ,$$

which says the same thing except for using different symbols as

$$\frac{\Delta w}{w_{Doppler}}=\frac{\Delta u}{c}\ .$$