On Mon, 12 May 2008 08:45:09 -0700 (PDT), Dono <sa... at (no spam) comcast.net
wrote:
On May 8, 8:58 am, Surfer <n... at (no spam) spam.net> wrote:
That was back in 2003. In 2008 it was discovered that 3-space speeds
in the range 420..450 km/s, perfectly resolved hitherto unresolved
spacecraft earth flyby anomalies
http://www.scieng.flinders.edu.au/cpes/people/cahill_r/Cahill_flyby.pdf
snip
Peter, check out the relativistic Doppler formula and its experimental
verifications (many)
Contrast with Cahill's formula pulled out of his butt.
I have been doing that.
From:http://en.wikipedia.org/wiki/Relativistic_Doppler_effect
For an observer moving TOWARD the source
where
fs = source frequency
fo = observed frequency
V = relative speed
the above page would give the relativistic Doppler formula as:
fo = fs Sqrt[(1+V/c)/(1-V/c)]
When a radar signal is reflected from an incoming spacecraft, the
above formula is applied twice.
Let
f = frequency of transmitted radar signal
fi = frequency of reflected radar signal received back at ground
station
V = speed of spacecraft relative to ground station
Then
fi = f (1+V/c)/(1-V/c)
= f (c+V)/(c-V) (2)
That is identical to equation (2) in Cahills paper.
[So here Cahill is using the relativistic Doppler formula as used by
space agencies]
However, he notes:
"The use of (2) instead of (1) is the origin of the
putative anomalies."
So what is (1)?
As before let:
f = frequency of transmitted radar signal
fi = frequency of reflected radar signal received back at ground
station
V = speed of spacecraft relative to ground station
Also let:
v = velocity of 3-space in ground frame
theta_i be an angle between v and direction of space craft velocity
vi = component of v in same direction as space craft velocity
= v cos(theta_i)
Then (see paper for derivation)
fi = f (c+vi)/(c+vi-V) * (c-vi+V)/(c-vi) (1)
Since (1) and (2) are different they will cause different velocities
to be calculated for a given Doppler shift.
This difference is small, however in the case of an inertial flyby,
the incoming speed Vi of a spacecraft is predicted to be EXACTLY the
same as the outgoing speed Vo, so if Vi as measured by doppler shift
is not exactly the same as the Vo as measured by doppler shift, we
have an anomaly.
Such are being repeatedly observed for spacecraft earth flybys, when
velocities are calculated using equation (2).
So if equation (1) was used instead of (2) what would the difference
be?
In the paper it is the value Delta_V_infinity, given by equation (7)
(see the paper for derivation)
Let
v = speed of 3-space in ground frame
V_infinity = asymptotic speed of spacecraft
theta_i = angle between incoming velocity and 3-space velocity
theta_f = angle between outgoing velocity and 3-space velocity
Then (7) is
Delta_V_infinity
= (v^2/c^2) ((cos(theta_f))^2 - (cos(theta_i))^2) V_infinity
It turns out that values of v in the range 420..450 km/s give values
for Delta_V_infinity that are equal to the observed anomalies.
In other words, Cahills formula (1) gives more accurate results in
such cases than (2).
NOTE:
The relativistic Doppler formula incorporates the effects of time
dilation in the source. This seems to be confirmed by relativistic
Doppler experiments.
However, this means that double application of the formula to doppler
radar measurement of spacecraft speed incorporates:
- time dilation of the transmitter relative to the spacecraft
- time dilation of the spacecraft relative to the receiver
But if the receiver and transmitter are in the SAME frame, there
should be NO relative time dilation between them.
Hence this circular inclusion of time dilation effects, where none are
required, would seem to be an error.
Is there an implication here that the relativistic Doppler formula
should not be used for Doppler radar?
-- Surfer
