Your responses:
> The Doppler shift can only convey information about the relative speeds if the light itself actually had those speeds.
> If the frequency and wavelength canceled each other out, no Doppler effect would be encoded. The existence of the Doppler shift proves light speed is variable.
> Whether it is a wavelength or frequency, the difference in distance between crests and troughs vs. time is identical.
> It would either be a compression wave or frequency difference and not both combined. Your math is divorced from physics, so it is a lie.
> As I was pointing out to Valev, it is the wavelength and not the frequency that changes.
Let's go through it step by step.
Please respond to my questions below.
The medium is vacuum all the way.
If the frequency of a wave is f and its wavelength is λ
and the speed of the wave is c', then the equation:
f⋅λ = c' (1)
must be fulfilled BY DEFINITION!
Question #1: Do you understand and accept this?
__________________
A star is emitting a H-alpha spectral line with
frequency f₀ and wavelength λ₀. The speed of
light relative to the star is c, so:
f₀⋅λ₀ = c (2)
Question #2: Do you understand and accept this?
_____________________________
The star is receding from the Earth with the speed v.
We are measuring the frequency of the H-alpha line
to be f and the wavelength to be λ.
According to you, the speed of the light from the star is c-v.
So we have:
f⋅λ = c-v (3)
Question #3: Do you understand and accept this?
_____________________________
Let's find the Doppler shift of the light from the star.
Let the star be stationary in frame K'(t, x')
and let the Earth be stationary in frame K(t, x)
The wave can be expressed as:
In the rest frame of the star K': A⋅cos(Φ'(t, x'))
In the rest frame of the Earth: A⋅cos(Φ(t,x))
where:
Φ'(t,x') = 2πf₀t - (2π/λ₀)x' (4)
Φ(t,x) = 2πft - (2π/λ)x (5)
According to equation (2) we have:
2π/λ₀ = 2πf₀/c so (4) can be written:
Φ'(t,x') = 2πf₀t - (2πf₀/c)x' (6)
We transform Φ'(t,x') to Φ(t,x) by using the Galilean transform:
x' = x + vt
Φ(t,x) = 2πf₀t - (2πf₀/c)(x + vt) = 2πf₀(1-v/c)t + (2πf₀/c)x
f₀/c = 1/λ₀ so:
Φ(t,x) = 2πf₀(1-v/c)t + (2π/λ₀)x
by comparison with (5) we see that:
f = f₀(1-v/c) = f₀((c-v)/c) (7)
λ = λ₀
The frequency is Doppler shifted, but there is no Doppler shift of
the wavelength.
Question #4: Do you understand and accept this?
_____________________________
We can now insert (7) in (3)
f⋅λ = f₀(1-v/c)λ = f₀λ(c-v)/c = (c-v)
so f₀λ/c = 1 => λ = λ₀
Inevitable conclusion:
======================
If the speed of light from the receding star is c-v
then the Doppler shift of the frequency is (1-v/c)
and there is no Doppler shift of the wavelength.
Question #5: Do you understand and accept this?
--
Paul
https://paulba.no/