Latest revision 20-12-2022

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}} $

Flyby Anomaly

A Basic Equation as derived at the end of that section is repeated here for convenience. $$ e^{-H_i/c_0\,\cdot\, r}\;\vec{a}_T = \vec{a}_t - \half H_i\,\vec{v}_t $$ Another formulation of the same equation has been $$ \vec{a}_T\,e^{H_i(t-t_0)} = \frac{d^2\vec{r}}{dt^2} - \half H_i \frac{d\vec{r}}{dt} $$ Without loss of generality we can take $\,t_0=0\,$, resulting in: $$ e^{H_i\,t} \frac{d^2\vec{r}}{dT^2} = \frac{d^2\vec{r}}{dt^2} - \half H_i \frac{d\vec{r}}{dt} $$ Let Newton's law of gravity come into play. $$ m\,\vec{a}_T = m\frac{d^2\vec{r}}{dT^2} = - G\frac{M\,m}{r^3}\vec{r} \hieruit - \frac{GM}{r^3}\vec{r} = \frac{d^2\vec{r}}{dT^2} $$ $G=$ the Gravitational constant $=6.674 \times 10^{-11}\,m^3\,kg^{-1}\,s^{-2}$. $M=$ mass of central body, for example the Earth mass $=5,9722 \times 10^{24}\,kg$.
The left hand side can be modified as follows. Physics students will recognize the derivation of Potential energy. $$ \mbox{LHS} = - e^{H_i\,t} \frac{GM}{r^3}\left(\vec{r}\cdot\frac{d\vec{r}}{dt}\right) = e^{H_i\,t} \frac{d}{dt} \left(\frac{GM}{r}\right) $$ The right hand side can be modified as follows. Physics students will recognize the derivation of Kinetic energy. $$ \mbox{RHS} = \left(\left[\frac{d^2\vec{r}}{dt^2} - \half H_i \frac{d\vec{r}}{dt}\right]\cdot\frac{d\vec{r}}{dt}\right) \\ = \half \frac{d}{dt}\left|\frac{d\vec{r}}{dt}\right|^2 - \half H_i \left|\frac{d\vec{r}}{dt}\right|^2 \\ = \half \left[\frac{d}{dt} - H_i\right]\left|\frac{d\vec{r}}{dt}\right|^2 $$ Progress can be made with help of Operator Calculus: $$ \left[ \frac{d}{dt} - H_i \right] = e^{H_i\,t} \frac{d}{dt} e^{-H_i\,t} $$ By putting RHS = LHS and taking care of the factor $1/2$ we get: $$ e^{H_i\,t} \frac{d}{dt} e^{-H_i\,t} \left|\frac{d\vec{r}}{dt}\right|^2 = e^{H_i\,t} \frac{d}{dt} \left(\frac{2GM}{r}\right) $$ It's sort of a miracle that exponential factors at both sides cancel, thus making further integration quite easy: $$ \require{cancel} \cancel{e^{H_i\,t}} \frac{d}{dt} e^{-H_i\,t} \left|\frac{d\vec{r}}{dt}\right|^2 = \cancel{e^{H_i\,t}} \frac{d}{dt} \left(\frac{2GM}{r}\right) \\ e^{-H_i\,t} \left|\frac{d\vec{r}}{dt}\right|^2 = \frac{2GM}{r} + C \\ \left|\frac{d\vec{r}}{dt}\right|^2 = e^{H_i\,t} \left(\frac{2GM}{r} + C\right) $$ The intrinsic Hubble parameter $H_i$ may be considered as being very small. An obvious simplification is $H_i=0$, which shall be called the Classical approach. With that approach, let us finally replace the absolute value of the derivative of the position to time by what it is commonly called. $$ \mbox{speed} = v = \left|\frac{d\vec{r}}{dt}\right| \hieruit v^2 = \frac{2GM}{r} + C $$ Still having $C$ as an integration constant. It can be determined by assuming that speed $v=V$ at infinity $r\to\infty$ is given: $$ v^2 = \frac{2GM}{r} + V^2 = GM \left( \frac{2}{r} - \frac{1}{a} \right) $$ The result is written in the latter form because it resembles (not at all by coincidence) the Vis-viva equation, with $a = -GM/V^2 \lt 0$ for a hyperbola.

Flyby anomaly

What we have found for the Flyby anomaly speed squared - with UAC, so call it $\,u^2\,$ - is a slight modification of the common Vis-viva equation: $$ u^2 = e^{H_i\,t} GM \left(\frac{2}{r} - \frac{1}{a}\right) = e^{H_i\,t} \left(\frac{2GM}{r} + C\right) $$ Assume that the central body (planet) is at rest. Then $\,u=$ speed of the satellite, $\,r=$ distance between the two bodies centers of mass, $\,t=$ (atomic) time with that distance $\,r(t)\,$, $\,GM=$ Gravitational constant times Mass of the central body, $\,H_i=$ Hubble parameter. Due to the anomaly, determining an integration constant like $V^2$ has become more problematic, because we would have to assume that $\,r = \infty\,$ at $\,t = 0\,$, which is impossible. Let $\,r_0 = r(t=0)\,$ and $\,V\,$ be the speed at that place $\,r_0\,$ at timestamp $\,t=0\,$. Then it's not hard to see that the following more general expressions can be proposed.
ClassicalUAC
$$ v^2 = \frac{2GM}{r} + C \\ V^2 = \frac{2GM}{r_0} + C \\ v^2 - V^2 = 2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) \\ v^2 = V^2 + 2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) $$ $$ e^{-H_i\,t} u^2 = \frac{2GM}{r} + C \\ e^{-H_i\,0} V^2 = \frac{2GM}{r_0} + C \\ e^{-H_i\,t} u^2 - V^2 = 2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) \\ u^2 = e^{H_i\,t} \left[V^2 + 2GM\left(\frac{1}{r}-\frac{1}{r_0}\right)\right] $$
Conclusion: $$ u^2 = e^{H_i\,t}\,v^2 = \frac{m}{m_0}\,v^2 $$ To be compared with an analogous outcome in our Galaxy project: $$ \vec{u} = \frac{d\vec{s}}{dt} = \frac{d\vec{s}}{dT}\frac{dT}{dt} = \vec{v} \frac{\sqrt{m_0/m}}{m_0/m} = \vec{v} \sqrt{m/m_0} $$ Thus, according to UAC, the quotient of the anomalous and the classical speed depends only on our Hubble parameter $\times$ (atomic) time and it shows very slow exponential growth. $$ \frac{u^2}{v^2} = e^{H_i\,t} \hieruit \frac{u-v}{v} = e^{\half H_i\,t} - 1 \approx \half H_i\,t $$ Not approximately but exactly an analogous result can be obtained with orbital time $\,T\,$ instead of atomic time $\,t\,$: $$ \frac{u^2}{v^2} = \frac{m}{m_0} = \left(\frac{T-A}{T_0-A}\right)^2 \hieruit \frac{u}{v}=1+\frac{T-T_0}{T_0-A} \\ \hieruit \frac{u-v}{v} = \frac{T-T_0}{T_0-A} = \half H_i(T-T_0) $$ But I am stuck with the next step, because nobody knows how to extract time from Flyby anomaly empirical data.
Perhaps there is a way out, however. Let's take a fresh look at this equality only: $$ \frac{u^2}{v^2} = \frac{m}{m_0} $$ In the booklet by Brillouin (page 87) varying mass is promoted as A New Approach to Special Relativity. Could it be that the Flyby Anomaly is just an effect of Special Relativity? A first order approximation should be sufficient if such is the case. $$ m = \frac{m_0}{\sqrt{1-(v/c)^2}} \hieruit \frac{u^2}{v^2} = \frac{1}{\sqrt{1-(v/c)^2}} \\ \hieruit \frac{u-v}{v} = \frac{1}{\sqrt{\sqrt{1-(v/c)^2}}} - 1 \approx \frac{1}{4}\left(\frac{v}{c}\right)^2 $$ The latter as verified with MAPLE:
> 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