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Natural Science Forum / Physics / Relativity / March 2005



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Doppler Shift Implies EM speed is NOT c.

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Henri Wilson - 30 Mar 2005 05:10 GMT
An RF signal is emitted by an antenna. It's frequency is nu and its speed
relative to the antenna is c.

The 'frequency' refers to the 'number of wavecrests' emitted per second.

If an observer who is moving wrt the source intercepts the signal, any change
in the arrival frequency of wavecrests can only indicate that the relative
speed of the wave is NOT c wrt the observer.

How could it be otherwise?
If the arrival speed was c wrt the observer, the arrival frequency WOULD REMAIN
AT nu.

HW.
www.users.bigpond.com/hewn/index.htm

Sometimes I feel like a complete failure.
The most useful thing I have ever done is prove Einstein wrong.
Tom Capizzi - 30 Mar 2005 05:39 GMT
> An RF signal is emitted by an antenna. It's frequency is nu and its speed
> relative to the antenna is c.
[quoted text clipped - 10 lines]
> REMAIN
> AT nu.

Seems like you never heard of something called wavelength. c = lambda * nu.
Since
c is constant, if frequency changes then wavelength changes inversely.
But you know this already. Why are you trolling this group?

. > HW.
> www.users.bigpond.com/hewn/index.htm
>
> Sometimes I feel like a complete failure.
> The most useful thing I have never done is prove Einstein wrong.
The Ghost In The Machine - 30 Mar 2005 18:00 GMT
In sci.physics.relativity, H@..(Henri Wilson)
<H@>
wrote
on Wed, 30 Mar 2005 04:10:35 GMT
<799k41l6dp6bmt2uefpcr3anfb80o3kl00@4ax.com>:
> An RF signal is emitted by an antenna. It's frequency is nu and its speed
> relative to the antenna is c.

In its reference frame, yes.

> The 'frequency' refers to the 'number of wavecrests' emitted per second.

Or the number of wavecrests received by a receiver.

> If an observer who is moving wrt the source intercepts
> the signal, any change in the arrival frequency of
> wavecrests can only indicate that the relative
> speed of the wave is NOT c wrt the observer.

False.

> How could it be otherwise?
> If the arrival speed was c wrt the observer, the
> arrival frequency WOULD REMAIN AT nu.

No, it wouldn't.  You need to follow the bouncing Lorentz transform.
:-)

Let's give it specific figures here.

An antenna at 10.24 MHz (I pick this because AIUI it's a frequency
of a certain system that a lot of people depend on :-) ) is
centered on a coordinate system.  The moving observer has
velocity v relative to this coordinate system, and has his *own*
coordinate system.

For purposes of clarity the antenna's coordinate system I will
designate as x_A, t_A and the receiver's as x_R, t_R.  Then
Lorentz specifies the transformation between these two pairs
of coordinates as

(x_R,t_R) = ( (x_A - v * t_A) * g, (t_A - v * x_A /c^2) * g)

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

Since x_A is identically zero (the antenna is not moving in its space)
we get

(x_R,t_R) = ( ( - v * t_A) * g, t_A * g)

Since the observer is not at -v * t_A, he has to wait
an additional v * t_A * g / c seconds to get the wavecrest;
the time observed is therefore t_A * g * (1 + v/c) and
the frequency is then f_R = f_A / (g * (1 + v/c)).

At v = 3 km/s = 10^-5c and f_A = 10.24 MHz we get
f_R = 10.2398976005 MHz.  In a pure emissive theory
scenario one would get f_REmissive = f_A * (1 - v/c)
= 10.2398976000 MHz.

Note that this is a pure SR effect; GR also is a factor in
any realistic calculation involving orbiting satellites (hint hint). :-)

The result is small, but measurable.

[.sigsnip]

Signature

#191, ewill3@earthlink.net
It's still legal to go .sigless.

John C. Polasek - 31 Mar 2005 02:54 GMT
>An RF signal is emitted by an antenna. It's frequency is nu and its speed
>relative to the antenna is c.
[quoted text clipped - 6 lines]
>
>How could it be otherwise?
As I pointed out in sci.physics.relativity:

No, c + v and c - v are not valid equations. c is at right angles to
any v you can name.

c + v and c - v can be used for the Doppler effect, but there is never
linear addition of velocities because
Doppler is a *positonal* effect: with each cycle the source is closer
by z*lambda and therefore has that much less distance to go which
*simulates* a velocity change allowing you to write c + v.

Once you understand that, you will see why the Michelson Morley
experiment was a null: v + ic = v - ic. See Fig. 7 in my gravity paper
at dualspace.net.

>If the arrival speed was c wrt the observer, the arrival frequency WOULD REMAIN
>AT nu.
[quoted text clipped - 4 lines]
>Sometimes I feel like a complete failure.
>The most useful thing I have ever done is prove Einstein wrong.
Mr. Dual Space
 
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