index $ \def \hieruit {\quad \Longrightarrow \quad} \def \slechts {\quad \Longleftrightarrow \quad} \def \SP {\quad ; \quad} \def \MET {\quad \mbox{with} \quad} \def \half {\frac{1}{2}} $
Classical | UAC |
---|---|
$$ v^2 = \frac{2GM}{r} + C \\ V^2 = \frac{2GM}{r_0} + C \\ v^2 - V^2 = 2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) \\ v^2 = V^2 + 2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) $$ | $$ e^{-H_i\,t} u^2 = \frac{2GM}{r} + C \\ e^{-H_i\,0} V^2 = \frac{2GM}{r_0} + C \\ e^{-H_i\,t} u^2 - V^2 = 2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) \\ u^2 = e^{H_i\,t} \left[V^2 + 2GM\left(\frac{1}{r}-\frac{1}{r_0}\right)\right] $$ |
> f(x) := 1/sqrt(sqrt(1-x^2)); > series(f(x),x); 2 4 6 1 + 1/4 x + 5/32 x + O(x )Brain off, computer on! Can't we do this by hand anymore? $$ f(x) = (1-x)^{-1/4} \approx f(0)+f'(0)\,x = 1-\frac{1}{4}(-1)\,x \quad \mbox{with} \quad x=\left(\frac{v}{c}\right)^2 $$ Alas. Little hope is offered by the (Wikipedia) Observations:
Galileo I | NEAR | Rosetta-I | |
---|---|---|---|
$v$ at $\infty$ (km/s) | $8.949$ | $6.851$ | $3.863$ |
$u-v$ at $\infty$ (mm/s) | $3.92\pm 0.08$ | $13.46\pm 0.13$ | $1.82\pm 0.05$ |
$(v/c)^2/4$ at $\infty$ | $0.2227653395\times 10^{-9}$ | $0.1305589167\times 10^{-9}$ | $0.4150954942\times 10^{-10}$ |
$(v/c)/2$ at $\infty$ | $1.49254368272984\times 10^{-5}$ | $1.14263033627086\times 10^{-5}$ | $6.44281125392813\times 10^{-6}$ |
$(u-v)/v$ at $\infty$ | $0.4380377696\times 10^{-6}$ | $0.1964676690\times 10^{-5}$ | $0.4711364225\times 10^{-6}$ |
$v$ at perigee (km/s) | $13.738$ | $12.739$ | $10.517$ |
$u-v$ at perigee (mm/s) | $2.560\pm 0.050$ | $7.210\pm 0.0700$ | $0.670\pm 0.0200$ |
$(v/c)^2/4$ at perigee | $0.5249834672\times 10^{-9}$ | $0.4514080275\times 10^{-9}$ | $0.3076680158\times 10^{-9}$ |
$(v/c)/2$ at perigee | $2.29127802089382\times 10^{-5}$ | $2.12465907620465\times 10^{-5}$ | $1.75406217880978\times 10^{-5}$ |
$(u-v)/v$ at perigee | $0.1863444461\times 10^{-6}$ | $0.5659784912\times 10^{-6}$ | $0.6370638015\times 10^{-7}$ |
The quantity $\,(v/c)^2/4\,$ is also called second order Doppler shift in some circles (e.g. manufacturers of atomic clocks).
It is expected that there must also exist sort of a first order Doppler shift associated with varying particle mass. Have we
encountered such a thing? In Relativity Special we find something like particle mass varying
in the line of sight:
$$
1+z_i = \sqrt{\frac{1+\beta}{1-\beta}} \MET \beta = \frac{v_\parallel}{c} \MET \ln(1+z_i)= \ln\left(\frac{m_0}{m}\right)
$$
It is assumed that the VPM is increasing instead of decreasing with $\,v\,$:
$$
\frac{m}{m_0} = \sqrt{\frac{1+(v/c)}{1-(v/c)}} \\
\hieruit \frac{u-v}{v} = \sqrt{\sqrt{\frac{1+(v/c)}{1-(v/c)}}} - 1 \approx \frac{1}{2}\frac{v}{c}
$$
This explains the entries with $\,(v/c)/2\,$ in the table. It may be concluded that the second order Doppler shift is orders of
magnitude to small and that the first order Doppler shift is orders of magnitude too large to explain the Flyby anomaly.
Perhaps the true mass in Special Relativity, when properly re-defined, is something in between
Transverse and longitudinal mass.
The Earth's Rotation Retardation