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gnome

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**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

Δf/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

Δf = f' - f = f * [(1+v/(2c))/(1-(v/(2c)) - 1]

Δf/f = (2c + v)/(2c - v)

That doesn't look like -v/c.

On the other hand, if I multiply the original equation by √(1-(v/c))/√(1-(v/c)) I get

f' = [√(1-(v^2/c^2))/(1-(v/c))] * f

and then expand √(1-(v^2/c^2)) = 1-v^2/(2c^2)

then

f' = [(1-v^2/(2c^2))/(1-(v/c))] * f

Δf/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) ∼ 1?

(The minus sign can be attributed to how we define Δf.)

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