# Can Kepler's 3rd Law be applied in higher dimensions to describe orbits (quadric surfaces)?

The 2-D orbital path of a celestial body revolving around another celestial body due to the force of gravity can be described by a conic section; this is the curve formed by intersecting a plane with the surface of a cone. This generalized conic section is given by:

$$A x^2 + B xy + C y^2 + D x + E y + F = 0$$

The ellipse is one example of a conic section (others being parabola and hyperbola); this happens when $$B^2 - 4 A C < 0$$.

Kepler's 3rd Law can be applied to elliptical orbits; this relates the orbital period $$T$$ and the semi-major axis $$a$$ of the orbit such that:

$$\frac{a^3}{T^2}$$ $$=$$ $$\frac{G (M + m)}{4 \pi^2}$$

As I understand it, Kepler's Law holds when the orbital path is 2-D. What happens when the orbital path is 3-D? I am aware that the conic section in 3-D is referred to as a quadric surface, which is given by:

$$A x^2 + B y^2 + C z^2 + D xy + E xz + F yz + Gx + Hy + Jz + K = 0$$

Say one were to rotate the 2-D orbit by its polar angle in spherical coordinates - that is, the altitude (as opposed to azimuth) - by a non-zero amount such that the orbit was in 3-D (changing z-coordinates as a function of time). Doing this, we can generate an ellipsoid, which is given by: $$(\frac{x}{a})^2$$ + $$(\frac{y}{b})^2$$ + $$(\frac{z}{c})^2$$ = 1.

Question:

Could the logic behind Kepler's 3rd Law be used to relate the orbital period $$T$$ of a body on an ellipsoidal path to any of the ellipsoids parameters $$a, b, c$$? What if the path weren't ellipsoidal, but some other quadric? Can the quadric surface always be reduced to a rotate 2-D conic section? Or are ALL orbits considered Keplerian?

• You may enjoy this article by John Baez: Planets in the Fourth Dimension. – PM 2Ring Dec 17 '19 at 10:56
• +1 your question is a bit long but it's a good question and should certainly have a well-founded mathematical answer. I'm a little confused by the 3D rotation part though, I can't visualize around what axis exactly you are rotating. You would get an ellipsoid, paraboloid or hyperboloid if you rotated a conic section orbit around its line of apses; would that be a more suitable term than "its polar angle in spherical coordinates"? – uhoh Dec 19 '19 at 2:40
• @uhoh This is the first time I’ve heard the term “line of apses”. Basically, the the xy-plane can be rotated towards/away from 3rd dimension (z). The “polar angle” that links to the wikipedia page is theta for physicists and phi for mathematicians. – user24627 Dec 19 '19 at 4:16
• @uhoh The polar angle convention is described in the top-right figure of the wiki page in the link. – user24627 Dec 19 '19 at 4:28
• I forgot to mention Cris Moore (who I learned about from a John Baez post). His figure-eight orbit is relatively well-known, but he also found a bunch of interesting 3D orbits in rotating frames, like this. See his gallery for more examples. – PM 2Ring Dec 19 '19 at 9:10

Trajectories of a system through some solution space, are in physics always solutions of some set of differential equations. However using the right formalism, one can often derive properties of the solution, without actually knowing it.

For the Kepler-problem, there is such a trick.

Let us consider the Lagrangian $$L=E_{\rm kin} - E_{\rm pot}$$ of the Kepler-problem: $$L = \frac{1}{2} m (v^2_x + v^2_y + v^2_z) - \frac{\alpha}{\sqrt{x^2+y^2+z^2}},$$ but to simplify the notation we shall abbreviate this to $$L = \frac{1}{2}mv^2-\frac{\alpha}{r}$$ The fact that this simplifies to a 1-D problem is of importance later. The equation of motion of the Kepler-problem can be derived from the corresponding Euler-Lagrange equation and results in Newton's law plus $$E_{\rm pot}$$.

Now we can multiply the Lagrangian by a constant $$L'= c L$$, and the resulting equation of motion, hence the solutions of the equation of motion, will remain identical. If we choose $$c$$ carefully, then we get the desired information about the solution of the equation of motion, by only touching $$L$$. This principle is called the principle of scale-invariance, and it doesn't apply to all Lagrangians or all systems in physics.

Now introduce a different idea, that of rescaling of space and time: Imagine a 'zoom' in space and a 'fast-forward' or 'slow-down' in time, such that $$x$$ and $$t$$ are rescaled according to $$x'\rightarrow c_x x$$, $$t'\rightarrow c_t t$$, where $$c_x$$ and $$c_t$$ are some new constants, that are arbitrary initially. From those rescalings follow now, for the scaling of v that $$v' \rightarrow c_x/c_t v$$ and the potential, because it only depends on one length, scales as $$\alpha /r' \rightarrow \alpha / (c_x r)$$.

Let us plug-in all those results from the scalings into $$L'$$ and relate this to the unprimed, original quantities, $$L' = \frac{1}{2}mv'^2 - \frac{\alpha}{r'} = \frac{1}{2}m v^2 \cdot \frac{c_x^2}{c_t^2} - \frac{\alpha}{c_x \cdot r}$$.

Now the latter looks a little like a mess. But we want to write this as $$L'=cL$$, which is only possible, if we can factor out those factors containing all the $$c_x$$ and $$c_t$$s. The condition for this is $$\frac{c_x^2}{c_t^2} = c \cdot \frac{1}{c_x}$$ or $$c_x^3/c_t^2 =c$$, which is one statement of Kepler's third law, namely that the spatial-scale determining and the temporal-scale determining constants have the above relation w.r.t each other. Note that as long as the potential remains a central potential, it doesn't matter how many coordinates we have, i.e. if $$r=\sqrt{x^2+y^2+z^2 + a^2+b^2+...}$$ this statement retains validity.

• I am still trying to fully understand the last part of your answer, but the question is technically answered. Regarding the scale factor $c_x$ in the lagrangian $L'$, why are $c_y$ and $c_z$ not included (the rest makes sense)? And given $c$ in terms of $c_x$ and $c_t$ is constant, how can this account for 3d- parabolic and hyperbolic orbits? – user24627 Dec 20 '19 at 11:12
• @allthemikeysaretaken $c_x$ and all the other spatial constants, are the same when all spatial dimensions are isotropic w.r.t each other, i.e. there is rotational symmetry between them. This can be broken in general relativity, where also Kepler 3 doesn't hold anymore, and no closed loop orbit solutions exist. As to the hyperbolic and parabolic solutions, they will always exist, because they're the result of effective centrifugal potential plus gravitational potential. The latter happens in the orbital plane, which is conserved no matter how many dimensions there are. – AtmosphericPrisonEscape Dec 20 '19 at 16:08
• The Lagrangian is $L = T - U$, whereas the total energy $<E>$ is given by the hamiltonian $H = T + U$. To my understanding, the orbit is elliptical if $E < 0$, hyperbolic if $E > 0$, and parabolic for $E = 0$. Does it follow that the constants $c_x, c_y, ...$ determine the sign of E? – user24627 Dec 21 '19 at 6:54
• @allthemikeysaretaken As long as you can factor them out, it doesn't matter if you're rescaling $T-U$ or $T+U$ – AtmosphericPrisonEscape Dec 21 '19 at 8:24
• Very interesting answer! +1 Slightly related in that the orbit solution is 1d : Can the radial oscillations of an elliptical orbit be solved using a fictitious centrifugal potential? – uhoh Jan 14 at 5:19