- #1
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.)
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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