In sci.physics, Greg Neill

wrote
on Sat, 11 Feb 2006 13:15:00 -0500
:
"The Ghost In The Machine" wrote in message
...
In sci.physics, Greg Neill

wrote
on Fri, 10 Feb 2006 09:25:57 -0500
:
"The Ghost In The Machine" wrote in message
...

The spacecraft will pass by 161080.7 lines per second.
(93000 * sqrt(1+.5) / sqrt(1-.5) )

Hey Ghost, you appear to be using the dilation factor
for relativistic Doppler shift rather than the
gamma factor 1/sgrt(1-(.5)^2) for coordinate
transforms . Any particular reason?


Because it's the correct one. :-) (At least, it's correct
from a mathematical standpoint; so far no experiments
have falsified it, either, at least AFAIK.)

[snip nice derivation]

I'm curious as to why you went to the trouble of
using light pulses rather than just considering the
line separations as fixed lengths in one frame which
are contracted when viewed in the other.

A naive approach would be to simply assume that the
distance between lines has been shortened from the
point of view of the moving observer, so that:

L' = L*sqrt(1 - v^2/c^2)

and the observer is passing the lines with a relative
velocity of v from his point of view, so the frequency
of line passing becomes:

f = v/L' = v/(L*sqrt(1 - v^2/c^2)

For the given example v = 0.5c and L = 1 mile, so

f = 0.5c/(1mile*sqrt(1 - 0.5^2))

= 107387.15 Hz


As you can see, the results are different. In any event, you
probably didn't see (it's at the very end) an alternative
derivation focusing on a rod in O's space, which A is trying
to measure. (Replication of this rod yields markers or lines.)

It turns out to also have sqrt(1-v/c)/sqrt(1+v/c) as
a ratio, rather than the expected 1/sqrt(1-(v/c)^2).

A bit surprising to me, but logical enough. I'll reprise it,
in case you didn't see it the first time:

O has a rod of length L (in his space); A passes over it.
A hits the first endpoint at x_A = 0, x_O = 0, t_A = 0, t_O = 0.

The second endpoint is hit when x_A = 0 (as A is the one
observing it) and x_O = L (since O has the rod). We don't
particularly care about t_O. What is t_A?

g = 1/sqrt(1-v^2/c^2)
x_A = (x_O - v * t_O) * g
t_A = (t_O - v * x_O/c^2) * g

0 = (L - v * t_O) * g
L = v * t_O
t_O = L/v

t_A = (L/v - v * L/c^2) * g
= (L/v) * (1 - v^2/c^2) * g
= (L/v) * sqrt(1-v/c) / sqrt(1+v/c)

As you can see, the Lorentz can be tricky.

--
#191,
It's still legal to go .sigless.

"Spaceman" wrote in message ...

"Sam Wormley" wrote in message
news:xdrHf.761197$x96.326443@attbi_s72...
| Spit****, you don't understand Doppler shift... it has to
| do with measured shifts in frequency and wavelength of
| the EM radiation and is observer dependent. The speed of
| light is a fundamental constant of nature and is observer
| independent.

Sam,
You don't understand that a "fundamental constant" should not be
affected by doppler shift, yet for light to doppler shift
without a medium as a cause, it would have to have
relative speed differences to occur at all.
You truly are brainwashed beyond help it seems.

You are obviously wrong because we measure Doppler
shifts for light and we measure constant speed for
light from both moving and stationary sources. So
you're just engaging in one of your favorite
pastimes: arguing with the universe.

"Greg Neill" wrote in message
. ..
| You are obviously wrong because we measure Doppler
| shifts for light and we measure constant speed for
| light from both moving and stationary sources.

You can't measure doppler shifts of light
without a relative speed change of the light.
You are full of crap.
No relative speed change = No doppler shift.
Sheesh!

"The Ghost In The Machine" wrote in message
...
In sci.physics, Greg Neill


[snip]

L' = L*sqrt(1 - v^2/c^2)

and the observer is passing the lines with a relative
velocity of v from his point of view, so the frequency
of line passing becomes:

f = v/L' = v/(L*sqrt(1 - v^2/c^2)

For the given example v = 0.5c and L = 1 mile, so

f = 0.5c/(1mile*sqrt(1 - 0.5^2))

= 107387.15 Hz


[snip]

g = 1/sqrt(1-v^2/c^2)

[snip]


t_A = (L/v - v * L/c^2) * g
= (L/v) * (1 - v^2/c^2) * g
= (L/v) * sqrt(1-v/c) / sqrt(1+v/c)

As you can see, the Lorentz can be tricky.

Indeed. Your final two lines are not the same expression!

The first of the pair yields my answer (when the time
period is inverted to yield frequency), while the second
gives your answer. In particular, the first is:

T = (L/v)*(1 - v^2/c^2)*(1 - v^2/c^2)^(-1/2)
= (L/v)&(1 - v^2/c^2)^(1/2)

Inverting to yield frequncy:

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

Which is the expression that I derived above.