How Does the Relativistic Doppler Shift Equation Simplify for Low Velocities?

[gnome]

relativistic doppler shift

I'm trying to show that this equation for the doppler shift for light:
f' = [√(1+(v/c))/√(1-(v/c))] * f
reduces to
&Delta;f/f = -v/c for v<<c

So I expanded (1+v/c)^(1/2) = 1 + v/(2c)
and (1-v/c)^(1/2) = 1 - v/(2c)
dropping higher order terms on the assumption that they are negligible.
Now I have
f' = [(1+v/(2c))/(1-(v/(2c))] * f
&Delta;f = f' - f = f * [(1+v/(2c))/(1-(v/(2c)) - 1]
&Delta;f/f = (2c + v)/(2c - v)

That doesn't look like -v/c.

On the other hand, if I multiply the original equation by &radic;(1-(v/c))/&radic;(1-(v/c)) I get
f' = [&radic;(1-(v^2/c^2))/(1-(v/c))] * f
and then expand &radic;(1-(v^2/c^2)) = 1-v^2/(2c^2)
then
f' = [(1-v^2/(2c^2))/(1-(v/c))] * f
&Delta;f/f = (1-v^2/(2c^2))/(1-(v/c)) - 1
= (2c^2 - v^2)/(2c^2 - 2cv) - 1
= (2cv - v^2)/(2c^2 - 2cv)
= v(2c - v)/(c(2c - 2v))

Close but no cigar.

Or, can I just claim that (2c-v)/(2c-2v) &sim; 1?

(The minus sign can be attributed to how we define &Delta;f.)

 

[jamesrc]

Starting here:

Originally posted by gnome

&Delta;f = f' - f = f * [(1+v/(2c))/(1-(v/(2c)) - 1]
&Delta;f/f = (2c + v)/(2c - v)

Can't you say:

$$\frac{\Delta f}{f} = \frac{1+\frac{v}{2c}}{1-\frac{v}{2c}} - 1$$
$$\frac{\Delta f}{f} = \frac{2\frac{v}{2c}}{1-\frac{v}{2c}}$$

and since v<<c, 1>>v/2c --> 1-v/2c ~ 1, and then you've proven it.

Or can you expand again at that point, so that: 1/(1+x) ~ 1 - x (for x close to zero), then:


$$\frac{\Delta f}{f} = \frac{v}{c}(1-\frac{v}{2c})$$

drop the second order terms and you've got the answer.

I'm not sure if that's all right, but that'd be my guess; hope it helps.

 

[gnome]

I guess that's it. I'm always uncertain as to which terms I can legitimately "drop", but I don't see any other argument that works here.

Thanks, James.