The Kepler Problem (Part 2)

I’m working on a math project involving the periodic table of elements and the Kepler problem—that is, the problem of a particle moving in an inverse square force law. That’s one reason I’ve been blogging about chemistry lately! I hope to tell you all about this project sometime—but right now I just want to say some very basic stuff about the ‘eccentricity vector’.

This vector is a conserved quantity for the Kepler problem. It was named the ‘Runge–Lenz vector’ after Lenz used it in 1924 to study the hydrogen atom in the framework of the ‘old quantum mechanics’ of Bohr and Sommerfeld: Lenz cite Runge’s popular German textbook on vector analysis from 1919, which explains this vector. But Runge never claimed any originality: he attributed this vector to Gibbs, who wrote about it in his book on vector analysis in 1901!

Nowadays many people call it the ‘Laplace–Runge–Lenz vector’, honoring Laplace’s discussion of it in his famous treatise on celestial mechaics in 1799. But in fact this vector goes back at least to Jakob Hermann, who wrote about it in 1710, triggering further work on this topic by Johann Bernoulli in the same year.

Nobody has seen signs of this vector in work before Hermann. So, we might call it the Hermann–Bernoulli–Laplace–Gibbs–Runge–Lenz vector, or just the Hermann vector. But I prefer to call it the eccentricity vector, because for a particle in an inverse square law its magnitude is the eccentricity of that orbit!

Let’s suppose we have a particle whose position \vec q \in \mathbb{R}^3 obeys this version of the inverse square force law:

\ddot{\vec q} = - \frac{\vec q}{q^3}

where I remove the arrow from a vector when I want to talk about its magnitude. So, I’m setting the mass of this particle equal to 1, along with the constant saying the strength of the force. That’s because I want to keep the formulas clean! With these conventions, the momentum of the particle is

\vec p = \dot{\vec q}

For this system it’s well-known that the following energy is conserved:

H = \frac{1}{2} p^2 - \frac{1}{q}

as well as the angular momentum vector:

\vec L = \vec q \times \vec p

But the interesting thing for me today is the eccentricity vector:

\vec e = \vec p \times \vec L - \frac{\vec q}{q}

Let’s check that it’s conserved! Taking its time derivative,

\dot{\vec e} = \dot{\vec p} \times \vec L + \vec p \times \dot{\vec L} - \frac{\vec p}{q} + \frac{\dot q}{q^2} \,\vec q

But angular momentum is conserved so the second term vanishes, and

\dot q = \frac{d}{dt} \sqrt{\vec q \cdot \vec q} = \frac{\vec p \cdot \vec q}{\sqrt{\vec q \cdot \vec q}} = \frac{\vec p \cdot \vec q}{q}

so we get

\dot{\vec e} = \dot{\vec p} \times \vec L - \frac{\vec p}{q} + \frac{\vec p \cdot \vec q}{q^2}\, \vec q

But the inverse square force law says

\dot{\vec p} = - \frac{\vec q}{q^3}

so

\dot{\vec e} = - \frac{1}{q^3} \, \vec q \times \vec L - \frac{\vec p}{q} + \frac{\vec p \cdot \vec q}{q^2}\, \vec q

How can we see that this vanishes? Mind you, there are various geometrical ways to think about this, but today I’m in the mood for checking that my skills in vector algebra are sufficient for a brute-force proof—and I want to record this proof so I can see it later!

To get anywhere we need to deal with the cross product in the above formula:

\vec q \times \vec L = \vec q \times (\vec q \times \vec p)

There’s a nice identity for the vector triple product:

\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c) \vec b - (\vec a \cdot \vec b) \vec c

I could have fun talking about why this is true, but I won’t now! I’ll just use it:

\vec q \times \vec L = \vec q \times (\vec q \times \vec p) = (\vec q \cdot \vec p) \vec q - q^2 \, \vec p

and plug this into our formula

\dot{\vec e} = - \frac{1}{q^3} \, \vec q \times \vec L - \frac{\vec p}{q} + \frac{\vec p \cdot \vec q}{q^2}\, \vec q

getting

\dot{\vec e} = -\frac{1}{q^3} \Big((\vec q \cdot \vec p) \vec q - q^2 \vec p \Big) - \frac{\vec p}{q} + \frac{\vec p \cdot \vec q}{q^3}\, \vec q

But look—everything cancels! So

\dot{\vec e} = 0

and the eccentricity vector is conserved!

So, it seems that the inverse square force law has 7 conserved quantities: the energy H, the 3 components of the angular momentum \vec L, and the 3 components of the eccentricity vector \vec e. But they can’t all be independent, since the particle only has 6 degrees of freedom: 3 for position and 3 for momentum. There can be at most 5 independent conserved quantities, since something has to change. So there have to be at least two relations betwen the conserved quantities we’ve found.

The first of these relations is pretty obvious: \vec e and \vec L are at right angles, so

\vec e \cdot \vec L = 0

But wait, why are they at right angles? Because

\vec e = \vec p \times \vec L - \frac{\vec q}{q}

The first term is orthogonal to \vec L because it’s a cross product of \vec p and \vec L; the second is orthogonal to \vec L because \vec L is a cross product of \vec q and \vec p.

The second relation is a lot less obvious, but also more interesting. Let’s take the dot product of \vec e with itself:

e^2 = \left(\vec p \times \vec L - \frac{\vec q}{q}\right) \cdot \left(\vec p \times \vec L - \frac{\vec q}{q}\right)

or in other words,

e^2 = (\vec p \times \vec L) \cdot (\vec p \times \vec L) - \frac{2}{q} \vec q \cdot (\vec p \times \vec L) + 1

But remember this nice cross product identity:

(\vec a \times \vec b) \cdot (\vec a \times \vec b) + (\vec a \cdot \vec b)^2 = a^2 b^2

Since \vec p and L are at right angles this gives

(\vec p \times \vec L) \cdot (\vec p \times \vec L) = p^2 L^2

so

e^2 = p^2 L^2 - \frac{2}{q} \vec q \cdot (\vec p \times \vec L) + 1

Then we can use the cyclic identity for the scalar triple product:

\vec a \cdot (\vec b \times \vec c) = \vec c \cdot (\vec a \times \vec b)

to rewrite this as

e^2 = p^2 L^2 - \frac{2}{q} \vec L \cdot (\vec q \times \vec p) + 1

or simply

e^2 = p^2 L^2 - \frac{2}{q} L^2 + 1

or even better,

e^2 = 2 \left(\frac{1}{2} p^2 - \frac{1}{q}\right) L^2 + 1

But this means that

e^2 = 2HL^2 + 1

which is our second relation between conserved quantities for the Kepler problem!

This relation makes a lot of sense if you know that e is the eccentricity of the orbit. Then it implies:

• if H > 0 then e > 1 and the orbit is a hyperbola.

• if H = 0 then e = 1 and the orbit is a parabola.

• if H < 0 then 0 < e < 1 and the orbit is an ellipse (or circle).

But why is e the eccentricity? And why does the particle move in a hyperbola, parabola or ellipse in the first place? We can show both of these things by taking the dot product of \vec q and \vec e:

\begin{array}{ccl} \vec q \cdot \vec e &=& \vec q \cdot \left(\vec p \times \vec L - \frac{\vec q}{q} \right) \\ \\ &=& \vec q \cdot (\vec p \times \vec L) - q \end{array}

Using the cyclic property of the scalar triple product we can rewrite this as

\begin{array}{ccl} \vec q \cdot \vec e &=& \vec L \cdot (\vec q \times \vec p) - q \\ \\ &=& L^2 - q \end{array}

Now, we know that \vec q moves in the plane orthogonal to \vec L. In this plane, which contains the vector \vec e, the equation \vec q \cdot \vec e = L^2 - q defines a conic of eccentricity e. I won’t show this from scratch, but it may seem more familiar if we rotate the whole situation so this plane is the xy plane and \vec e points in the x direction. Then in polar coordinates this equation says

er \cos \theta = L^2 - r

or

r = \frac{L^2}{1 + e \cos \theta}

This is well-known, at least among students of physics who have solved the Kepler problem, to be the equation of a conic of eccentricity e.

Another thing that’s good to do is define a rescaled eccentricity vector. In the case of elliptical orbits, where H < 0, we define this by

\vec M = \frac{\vec e}{\sqrt{-2H}}

Then we can take our relation

e^2 = 2HL^2 + 1

and rewrite it as

1 = e^2 - 2H L^2

and then divide by -2H getting

- \frac{1}{2H} = \frac{e^2}{-2H} + L^2

or

- \frac{1}{2H} = L^2 + M^2

This suggests an interesting similarity between \vec L and \vec M, which turns out to be very important in a deeper understanding of the Kepler problem. And with more work, you can use this idea to show that -1/4H is the Hamiltonian for a free particle on the 3-sphere. But more about that some other time, I hope!

For now, you might try this:

• Wikipedia, Laplace–Runge–Lenz vector.

and of course this:

• The Kepler problem (part 1).

This entry was posted on Friday, December 31st, 2021 at 2:56 am and is filed under physics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

15 Responses to The Kepler Problem (Part 2)

  1. Toby Bartels says:
    31 December, 2021 at 3:24 am

    When you write down the ‘nice cross product identity’, you stop writing the arrows too early; everything on the left-hand side needs one.

    Incidentally, while this is a useful identity, the purpose to which you put it might be overkill; a lot more people will probably recognize the famous fact that if two vectors are orthogonal, then the magnitude of their cross product is the product of the their magnitudes. (For that matter, the full identity is secretly the more general famous fact that the magnitude of the cross product of two vectors is the product of their magnitudes and the sine of the angle between them, and it's good for people to look at that identity until they realize that that's indeed what it says.)

    Reply
    • John Baez says:
      31 December, 2021 at 6:19 am

      Thanks for catching that mistake! I fixed it.

      Yes, I was having fun noticing proving the basic facts the eccentricity vector use most of the basic identities involving the cross product. When I teach vectors I sometimes show

      (\vec a \times \vec b)\cdot (\vec a \times \vec b) + (\vec a \cdot \vec b)^2 = a^2 b^2

      by brute force—a real workout for the students, but fundamentally straightforward—and then use the previously shown fact that

      \vec a \cdot \vec b = ab \cos \theta

      where \theta is the angle between the two vectors to conclude that

      \|\vec a \times \vec b\| = ab |\sin \theta|

      Is there a better way to get this last formula starting from the component definition of the cross product?

      Reply
      • Toby Bartels says:
        31 December, 2021 at 6:32 am

        A better way from the component-wise definition? Not that I know of, only that it seems backwards to try to do that at all. Because how do you motivate that definition in the first place? So I start with the geometric definition, prove geometrically that it’s bilinear, and then the component-wise formula comes very quickly after that; not as a definition, but as a theorem to aid in calculation.

        Reply
  2. ron davison says:
    31 December, 2021 at 3:52 am

    can this math be used for orbital eccentricity vectors of earth and moon for Lagrange zone calculations for Artemis?

    Reply
    • John Baez says:
      31 December, 2021 at 7:48 am

      One needs much more fancy math to study the 3-body problem, but I believe all this math about the 2-body problem is a prerequisite for that.

      Thanks for bringing my attention to ARTEMIS, NASA’s mission to study Acceleration, Reconnection, Turbulence and Electrodynamics of the Moon’s Interaction with the Sun.

      Reply
  3. Frederic Barbaresco says:
    31 December, 2021 at 7:20 am

    For Kepler problem, more general than Laplace-Runge-Lenz vector, it was shown by Souriau [1] that every three dimensional dynamical systems involving central potentials do admit a conserved vector and this general vector has been constructed and analyzed by Bacry, Ruegg and Souriau. Shortly afterwards, it was shown that this generalized conserved vector is multi-valued. Bacry, Ruegg and Souriau showed that such a vector is exceptionally one-valued, in the Kepler case, and corresponds generally to a piecewise conserved quantity, the Fradkin-Bacry-Ruegg-Souriau (FBRS) perihelion vector.
    Jean-Marie Souriau applied developments in symplectic geometry to the Kepler problem in papers [2,3,4,5], where Kepler Manifold was shown to be a coadjoint orbit of SO(4, 2), and a complex structure of Kepler Manifold was revealed: Kepler Manifold is diffeomorphic to the conifold in C4 with the conifold point removed.

    On Kepler problem, I invite you to read the following papers:
    [1] H. Bacry, H. Ruegg and J.M. Souriau, Dynamical groups and spherical potentials in Classical Mechanics, Comm. Math. Phys. 3 (1966) 323
    [2] J.-M. Souriau, Sur la variété de Kepler. Symp. Math., 14, pp. 343-360, (1974). Translated by Jamie B. Jorgensen
    [3] J.-M. Souriau, Géometrie globale du problème à deux corps. Proc. IUTAMISSIM Symp. on Mod. Devl. Anal. Mech., Atti Acad.Sci. Torino, Suppl. 117, pp. 369-418, (1983).
    [4] Jean-Marie Souriau, Sur la variété de Kepler, Teoria Geometrica Dell’Integrazione E Varieta Minimali, 23-25 May 1973
    [5] Jean-Marie Souriau, Géométrie Globale du Problème à 2 Corps, Proceedings of the IUTAMISSIM, Symposium on “Modern Developments in Analytical Machanics”, Academy of Sciences of Turin, Turin, June 7-11th 1982
    [6] Alain Guichardet, Le problème de Kepler: Histoire et théorie, Editions Ellipse, Ecole Polytechnique, 2012 – https://www.amazon.fr/Probleme-Kepler-Histoire-Theorie/dp/2730215964
    |7] Alain Guichardet, Sur le problème de Kepler, HAL Id: hal-00576029, https://hal.archives-ouvertes.fr/hal-00576029, submitted on 11 March 2011
    [8] Alain Albouy, Le rôle de la structure projective sous-jacente de l’espace dans les particularités de la gravitation newtonienne. Dominique Flament (dir), Série Documents de travail (Équipe F2DS), Histoires de géométries : textes du séminaire de l’année 2003, Paris, Fondation Maison des Sciences de l’Homme, 2004
    [9] Y. Grandati, A. Berard and H. Mohrbach, On Peres approach to Fradkin-Bacry-Ruegg-Souriau’s perihelion vector, Cent. Eur. J. Phys., 9 (2011) 88.
    [10] Y. Grandati, A. Berard and H. Mohrbach, Fradkin-Bacry-Ruegg-Souriau perihelion vector for Gorringe-Leach equations, Celest. Mech. Dyn. Astr. 106 (2010) 109.
    [11] Partha Guha, E. Harikumar, N.S. Zuhair, Fradkin-Bacry-Ruegg-Souriau vector in kappa-deformed space-time, April 2015European Physical Journal Plus 130(10)
    [12] Jian Zhou, On Geometry and Symmetry of Kepler Systems. I., arXiv:1708.05504v1 [math-ph], 18 August 2017

    Reply
    • John Baez says:
      31 December, 2021 at 10:18 pm

      Thanks! I know some of this work because it’s summarized in Guillemin and Sternberg’s elegant book Variations on a Theme by Kepler, and also Cordani’s large book The Kepler Problem. But I really should read more of Souriau’s original work, and other work, because I’m including a bit of a historical review of the Kepler problem as part of a paper I’m writing. It’s a really fascinating subject!

      In my blog article here I was just wanting to record some basic calculations without getting into the more sophisticated aspects. Everything I did here is completely well-known, but I want to be able to see every step spelled out in my own notation, so I transcribed the calculations I did in my notebook to this blog.

      Reply
  4. Raphael says:
    31 December, 2021 at 8:14 am

    Would be interesting to see how it pans out with the Levi-Civita-Tensor+Einstein summation formalism rather than utilizing the vector algebra.

    Reply
    • John Baez says:
      31 December, 2021 at 10:19 pm

      I don’t think it dramatically changes things here. When we get to more fancy aspects of the Kepler problem, we start doing calculations on the 3-sphere, and then tensors and quaternions can be helpful.

      Reply
  5. amarashiki says:
    1 January, 2022 at 5:07 am

    What about the Kepler problem in higher dimensions where the 3d product is not available? Is everything OK if we substitute it by the wedge product? I guess there is a quaternionic/octonionic version of this, isn’t it?

    Reply
    • Toby Bartels says:
      1 January, 2022 at 5:33 am

      The whole thing is essentially a two-dimensional problem. The cross products here are conventional, but it's useful to think of the angular momentum as a bivector all along, and then it looks the same in any number of dimensions.

      Reply
    • John Baez says:
      1 January, 2022 at 4:40 pm

      Yes, if I were only interested in the classical Kepler problem it would have made more sense to work in 2 dimensions, since as Toby hints the orbit of a given particle always stays in a 2-dimensional plane, no matter what dimension we’re in, thanks to conservation of angular momentum.

      The quantum Kepler problem is more seriously dimension-dependent, and that’s what I’m actually interested in… the spectrum of the Hamiltonian

      H = -\nabla^2 - \frac{1}{r}

      depends on the dimension of space. You can see the spectrum worked out in this paper my friend Guowu Meng:

      • Guowu Meng, The MICZ-Kepler problems in all dimensions.

      In fact the main point of this paper is that he handles the case where the point charge at the origin is also a magnetic monopole. But you can set the magnetic charge to zero.

      But the n-dimensional Kepler problem is closely connected to the free particle on S^n, and one reason I’m specially interested in the n = 3 case is that the 3-sphere is the Lie group SU(2). (Another reason is that we live in 3 dimensions.)

      Of course, arguably it’s weird to study a 1/r potential in dimensions other than 3, since that’s not how electrostatics works in other dimensions.

      Reply
      • amarashiki says:
        1 January, 2022 at 4:51 pm

        Great. But the Kepler problem in HIGHER DIMENSIONS should be connected to the spectrum of the interacting laplacian:
        H=-\nabla^2-\dfrac{1}{r^{D-3}}, for D=d+1\geq 3. Interestingly, some time ago, I wondered what happens with the Kepler problem in 2d AND lower dimensions. In fact, with D=2, \;\; d=1 we get something like the harmonic potential, in D=3, \;\; d=2 we have a logarithmic potential (similar to some potentials we see in conformal mechanics). Also, what about 0D laplacian (the identity) and negative dimensions? I wondered about the Kepler problem in negative dimensions too (sorry if I seem too crazy and mad with this, John). What is a negative dimensional laplacian? I guess maybe it is an integral operator, but I am not sure about that. Moreover, there are several different definitions of negative dimensional derivatives (Caputo’s, Riemann-Liouville,…), and thus, I am not sure if we can define negative dimensional operator calculus and the Kepler problem in negative dimensions. I don’t know also if there is some work about this thing…I apologize if I am a mathematical ignorant, but negative dimensional derivatives (related to absement and other derivatives) is something I sometimes mention to my high-school students (please, don’t tell me I am a mad man now…).

        Reply
        • John Baez says:
          1 January, 2022 at 5:13 pm

          Amarshiki wrote:

          Great. But the Kepler problem in HIGHER DIMENSIONS should be connected to the spectrum of the interacting laplacian:
          H=-\nabla^2-\dfrac{1}{r^{D-3}}, for D=d+1\geq 3

          That’s an interesting problem. It becomes extremely complicated for d = 4 because the Hamiltonian becomes ill-defined: more precisely, there are infinitely many different ways to define it. For d = 5 the problem may become undefined.

          But by the way, this is not called the Kepler problem. The name ‘Kepler problem’ refers to

          H=-\nabla^2-\dfrac{1}{r}

          no matter what dimension you’re in. This problem has special features (see the paper by Guowu Meng above).

          (You seem to be mixing up negative dimensions with negative powers of the derivative operator; I’m not interested in talking about this stuff.)

        • amarashiki says:
          1 January, 2022 at 8:51 pm

          Yeah! I know I mixed stuff… True, but I was searching for investigations with the operator D^\alpha-\dfrac{1}{r^\beta}, fractional derivatives (even negative) with higher dimensions because of fractional quantum mechanics and a possible relation with the spectrum E(d)=-const\dfrac{1}{n^d}, I suppose you imagine why given the functional dependence of that energy spectrum. I am not saying the link is obvious, but It came to my mind years ago… Surely you caught it…Indeed I posted about it in physics stack exchange long ago. I know the relation with the riemannium and also non-extensive entropies with that spectrum, but nobody knows if it helps at all with the Riemann hypothesis…

          I have read the Meng paper several times these years, since I am interested in hidden symmetries and asymptotic symmetries now (beyond the fact that BH solutions with hidden symmetries are interesting for astrophysics in the next years to test the nature of astrophysical BH-likely Kerr, but they could have another richer structure), there is another paper about MICZ kepler problem I can not remember, in which it is sketched a possible connection also with black hole physics with the Kepler problem. Or I think so … Maybe, I am confusing things…I will seek my database if I find out that paper…I know I was not dreaming this time…

          It is curious that the Kepler problem (in higher dimensions!) imposes the 1/r potential. I had not thought too much about that. Maybe that was the reason because of my negative dimension affair/confusion too. Surely you know this recent paper too: https://arxiv.org/abs/2106.02313, related to octonions and the SU(8) monopole! Nersessian papers about MICZ problems are interested too, e.g., https://arxiv.org/abs/0711.1037

          Remark (for readers): I was confusing the fractionalized negative dimensional vector operators ^n\nabla^\alpha, where alpha is the fractional power and n is the dimension of space in nabla. Also, D^\alpha\equiv D^{\alpha_1\alpha_2\ldots\alpha_n} can have different definitions if extended to real or complex values of \alpha_i (not easy, not trivial). I confused the fact that is not in general true that the negative dimension matches (in absolute value) the components of the nabla vector. Sorry about my mistake John.

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