Eric Gisse says...

On Mar 23, 10:11=A0pm, Koobee Wublee wrote:

[...]

You've just confessed that you don't understand where Doppler shift
comes from, even in sound waves, despite your floundering attempts to
understand it using either Galilean or Lorentz transforms. This utter
incompetence on YOUR part you blame on others.

More cheap shots.

The bottom line is that using the same method the Galilean transform
predicts no Doppler shift even for sound waves.

Yes, double down on your stupidity.

It's really hard for me to guess why Koobee thinks that
the method used by Einstein predicts no Doppler shift
in the nonrelativistic case. I cannot imagine what
Koobee might have done wrong. It's possible that
he's still confused about the difference between
Doppler shift and time dilation. He thinks
that since there is no time dilation in the
nonrelativistic case, that there is no Doppler
shift.

But Doppler shift doesn't come from the
transformation for time, it comes from
the transformation for the *spatial*
coordinate.

The phase for a traveling wave is
given (both relativistically and
non-relativistically) by:

Phi = kx - wt

To compute w' in a new coordinate
system, you rewrite x and t in terms
of x' and t':

x = gamma (x' + vt')
t = gamma (t' + v/c^2 x')

So in the frame F', we have:

Phi' = gamma k (x'+vt') - gamma w (t'+v/c^2 x')
= gamma (k-vw/c^2) x' - gamma (w-vk) t'

So
k' = gamma (k-vw/c^2)
w' = gamma (w-vk)

To get the nonrelativistic limit, you just
take the limit in which v/c is small, so
gamma is approximately 1, and v/c^2 is approximately
zero. This produces:

k' = k
w' = w-vk

In the Galilean case, there is no shift for *wave-length*
(k = 2pi/L, where L is the wavelength), but there is still
a shift for frequency (w is actually 2pi f, where f is
the frequency).

--
Daryl McCullough
Ithaca, NY