In our last blog, we observed that Newton's slope,
z=\equiv r'/r= \pm \tan\theta (1)
where \theta is the complementary angle to angle between the tangent to the curve and the direction of the force to the source at the focus. Integrating (1) gives a straight line, however,
r=C\sec\theta (2)
where C is an arbitrary constant of integration.
The metric corresponding to one is completely flat,
ds^2= E dr^2 + G d\theta^2 (3)
where the metric coefficients are E=1 and G=r^2. The Binet equation, which determines the trajectories, is
u" + G'/2G = F/(h^2u^2) (4)
where the central force, F, has to be "written in by hand", and h is the conserved angular momentum with u=1/r. This was Newton's inverse problem where given the central force determine the trajectory. Obviously, Newton, basing his ideas on Kepler's laws, knew that the trajectory was an ellipse
1/r=u= A + B\cos\theta (5)
where A and B are integration constants obtained by integrating the Binet equation, (4). However, since Newton's slope is given by Eq. (2), it requires that A\equiv 0, reaffirming that the trajectory is a straight line.
On the basis of the metric (3) when the metric coefficients are functions of r, assuming spherical symmetry, Newton's slope (1) generalizes to
\sqrt{E/G}r' = \pm \tan\theta (6)
We are interested in non-Euclidean geometries of constant curvature, and this limits the possibilities on the metric coefficients. For a pseudo-sphere, which is the archtype of hyperbolic geometry of constant negative curvature, E=p^2/r^2, with p a constant, and G=r^2.
Integrating (6) gives
1/r=A \pm B\ln(\cos\theta) (7)
where A and B are constants. As shown in the figure
If A=0 there results a parabola, opening to the right if the - sign is chosen, The negative sign gives closed trajectories, while the + sign can result in hyperbolas, or, in general, open trajectories. As we will dedicate a separate blog to this case, his will not concern us here though.
Rather, considering E=1 and G=\sinh(r), for hyperbolic geometry and G=\sin(r) for elliptic geometry we get the equation for the slope of constant negative curvature
u'=-\csch^2(r) r'= \mp \tan\theta. (8)
Integrating the first equation gives
u=\coth(r) (9)
while integrating the second equation results in
\cos\theta= u -\sqrt{u^2-1) =\coth(r)-\csch(r) =\tanh(r/2) (10)
Since \sec\theta=u+\sqrt{u^2-1}, we can write
u=(\cos\theta+\sec\theta)/2 (11)
coressponding to
r= \pm\coth^{-1}((\cos\theta+\sec\theta)/2)). (12)
The plus and minus signs refer to the Joukowski airfoil opening to the right or left respectively as shown in the figure.
The corresponding Binet equation is
2u"+u=\sec^3\theta. (13)
The rhs is obtained in Euclidean space by introducing a central force F, according to (4). Since \sec\theta=u+\sqrt}u^2-1}, and r=\coth(u) in hperbolic geometry of constant negative curvature. In the near Euclidean approximation u>1 so that r=\coth^(-1)(u)~1/u, which is the Euclidean result. If there were a force present, the approximation \sec^3\theta~u^3, would lead to an inverse-fifth central force which traces out a circular trajectory passing through the origin. The inclusion of higher order terms would lead to a distortion of the circular orbit.
For large values of ( |u| > 1 ), the function ( \text{coth}^{-1}(u) ) approaches ( 1/u ). This simplification aligns with Euclidean geometry and supports the approximation you’re considering for large ( u ).
In this case, using ( r = 1/u ) for large ( |u| ) simplifies the expression and makes it easier to work with, especially if you’re looking for a Euclidean result.
That’s an interesting point. In classical mechanics, the Binet equation does indeed require the explicit introduction of the force function to describe the motion under a central force. In this case, ( \sec^3\theta ) arises naturally from the geometry of the problem without needing to introduce a force term by "hand". This suggests that the geometry itself is dictating the dynamics, which is quite elegant.
This geometric approach can sometimes reveal deeper insights into the nature of the system under study. It’s always fascinating when the mathematics aligns so neatly with physical phenomena, allowing for a more intuitive understanding of the forces and motions involved.
The Joukoswski transform (12) converts a circle into an ellipse. If this were Euclidean space the ellipse would be centered at the origin. The square of a Joukowski transform is also a Joukowski transform, as shown in the figure below.
When applied to the centered ellipse it shifts the origin to the focus. This is true in Euclidean space. It is significant because the Hookean law of attraction at the center, has no become Newton's inverse square law at the focus. Hooke's and Newton's laws are duals.
However, in non-Euclidean spaces, the center and focus are ill defined. However, what we have shown is that the inverse-fifth law is valid in the near Euclidean limit. And it so happens that the inverse-fifth law is its own dual!
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