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We are supposed to show that orbits in 4D are not closed. Therefore I derived a Lagrangian in hyperspherical coordinates $$L=\frac{m}{2}(\dot{r}^2+\sin^2(\gamma)(\sin^2(\theta)r^2 \dot{\phi}^2+r^2 \dot{\theta}^2)+r^2 \dot{\gamma}^2)-V(r).$$

But we are supposed to express the Lagrangian in terms of constant generalized momenta and the variables $r,\dot{r}$. But as $\phi$ is the only cyclic coordinate after what I derived there, this seems to be fairly impossible. Does anybody of you know to calculate further constant momenta?

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    $\begingroup$ Hint: Bertrand's theorem. See also this Phys.SE post. FYI Bertrand's theorem on mathoverflow here. $\endgroup$ – Qmechanic May 29 '13 at 23:38
  • $\begingroup$ @Lipschitz, I believe that for a given trajectory you can always choose the coordinates so that all angles but $\phi$ are equal to zero. This is because the force is central, so the trajectories are flat (no force drives them out of the plane). $\endgroup$ – Peter Kravchuk May 30 '13 at 9:25
  • $\begingroup$ @Lipschitz, not zero, just constant, of course. $\endgroup$ – Peter Kravchuk May 30 '13 at 11:11
  • $\begingroup$ Vaguely related: physics.stackexchange.com/q/55638 $\endgroup$ – dmckee May 30 '13 at 15:04
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Hints:

  1. Prove that the angular momentum $L^{ij}:=x^ip^j-x^jp^i$ is conserved for a central force law in $d$ spatial dimensions, $i,j\in\{1,2,\ldots ,d\}.$

  2. Since the concept of closed orbits does not make sense for $d\leq 1$, let us assume from now on that $d\geq 2$.

  3. Choose a 2D plane $\pi$ through the origin that is parallel to the initial position and momentum vectors. Deduce (from the equations of motion $\dot{\bf x} \parallel {\bf p}$ and $\dot{\bf p} \parallel {\bf x}$) that the point mass continues to be confined to this 2D plane $\pi$ (known as the orbit plane) for all time $t$. Thus the problem is essentially 2+1 dimensional with radial coordinates $(r,\theta)$ and time $t$. [In other words, the ambient $d-2$ spatial dimensions are reduced to passive spectators. Interestingly, this argument essentially shows that the conclusion of Bertrand's theorem are independent of the total number $d\geq 2$ of spatial dimensions; namely the conclusion that only central potentials of the form $V(r) \propto 1/r$ or $V(r) \propto r^2$ have closed stable orbits.]

  4. Deduce that the Lagrangian is $L=\frac{1}{2}m(\dot{r}^2 +r^2\dot{\theta}^2) -V(r)$.

  5. The momenta are $$p_{r}~=~\frac{\partial L}{\partial \dot{r}}~=~m\dot{r}$$ and $$p_{\theta}~=~\frac{\partial L}{\partial \dot{\theta}}~=~mr^2\dot{\theta}.$$

  6. Note that $\theta$ is a cyclic variable, so the corresponding momentum $p_{\theta}$ (which is the angular momentum) is conserved.

  7. Deduce that the Hamiltonian is $H=\frac{p_{r}^2}{2m}+\frac{p_{\theta}^2}{2mr^2}+ V(r)$.

  8. Interpret the angular kinetic energy term $$\frac{p_{\theta}^2}{2mr^2}~=:~V_{\rm cf}(r)$$ as a centrifugal potential term in a 1D radial world. See also this Phys.SE post. Hence the problem is essentially 1+1 dimensional $H=\frac{p_{r}^2}{2m}+V_{\rm cf}(r)+V(r)$.

  9. From now on we assume that the central force $F(r)$ is Newtonian gravity. Show via a $d$-dimensional Gauss' law that a Newtonian gravitational force in $d$ spatial dimensions depends on distance $r$ as $F(r)\propto r^{1-d}$. (See also e.g. the www.superstringtheory.com webpage, or B. Zwiebach, A First course in String Theory, Section 3.7.) Equivalently, the Newtonian gravitational potential is $$V(r)~\propto~\left\{\begin{array}{rcl} r^{2-d} &\text{for}& d~\neq~ 2, \\ \ln(r)&\text{for}& d~=~2. \end{array}\right. $$

  10. So from Bertrand's theorem, candidate dimensions $d$ for closed stable orbits with Newtonian gravity are:

    • $d=0$: Hooke's law (which we have already excluded via the assumption $d\geq 2$).
    • $d=3$: $1/r$ potential (the standard case).
    • $d=4$: $1/r^2$ potential (suitably re-interpreted as part of a centrifugal potential).

    We would like to show that the last possibility $d=4$ does not lead to closed stable orbits after all.

  11. Assume from now on that $d=4$. Notice the simplifying fact that in $d=4$, the centrifugal potential $V_{\rm cf}(r)$ and the gravitational potential $V(r)$ have precisely the same $1/r^2$ dependence!

  12. Thus if one of the repulsive centrifugal potential $V_{\rm cf}(r)$ and the attractive gravitational potential $V(r)$ dominates, it will continue to dominate, and hence closed orbits are impossible. The radial coordinate $r$ would either go monotonically to $0$ or $\infty$, depending on which potential dominates.

  13. However, if the repulsive centrifugal potential $V_{\rm cf}(r)$ and the attractive gravitational potential $V(r)$ just happen to cancel for one distance $r$, they would continue to cancel for all distances $r$. Newton's second law becomes $\ddot{r}=0$. Hence a closed circular orbit $\dot{r}=0$ is possible. However, this closed circular orbit is not stable against perturbations in the radial velocity $\dot{r}$, in accordance with Bertrand's theorem.

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  • $\begingroup$ Following your reasoning, in four dimensions or higher the potential $1/r$ has always closed stable orbits, but I think I proved the contrary, please correct me if I'm wrong. $\endgroup$ – Ikiperu May 31 '13 at 16:18
  • $\begingroup$ Yes, an attractive $1/r$ potential in $d\geq 2$ spatial dimensions has closed stable orbits. $\endgroup$ – Qmechanic May 31 '13 at 18:31
  • $\begingroup$ Then where is the error in my answer? (it's up yours). I can't find it, I will be grateful if you could help. $\endgroup$ – Ikiperu May 31 '13 at 19:20
  • $\begingroup$ @Ikiperu gravity is not a $1/r$ potential in $d\ne 3$. You can see this just from dimensional analysis in Gauss's law, since the dimensions of density change with the number of dimensions of space, and therefore the dimensions of $G$ change (the units of the potential and the laplacian, of course, stay the same). This is the eighth point in the above answer. $\endgroup$ – zkf May 31 '13 at 20:59
  • $\begingroup$ @zkf It's clear, and from that it's obvious that exist orbits that are non-closed (I would like to see a detailed proof that all orbits are non-closed, which is the OP's question). I'm not convinced by the argument of Qmechanich (how 1=>2), in fact I think I found a counterexample for the potential $1/r$ (see my answer upthere), if my solution is correct there are non-closed orbits, contrary to his statement, I hope someone could check the calculation. $\endgroup$ – Ikiperu May 31 '13 at 21:26
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The 4 dimensional case contains the 3d, so closed orbits exist if exist in three dimensions. On contrary, if non-closed orbits exist in 3d, there exist obviusly also in 4d. According to Bertrand theorem, the two only potentials that in three dimensions admit stable closed orbits are the the harmonic and the newtonian, so it remains to prove that in dimension 4 these two potential admit non-closed orbits, for simplicity I will do the calculation for the newtonian but the proof for the harmonic is equally simple.

Introducing cylindrical coordinates for $x,y$ and $z,w$ separately the lagrangian becomes: $$ Lag=m(\dot{r_1}^2+\dot{r_2}^2+r_1^2\dot{\theta_1}^2+r_2^2\dot{\theta_2}^2)/2-V(\sqrt{r_1^2+r_2^2}) $$ or, introducing the angular momenta: $$ Lag=m(\dot{r_1}^2+\dot{r_2}^2+L_1^2/(m r_1)^2+L_2^2/(m r_2)^2)/2-V(\sqrt{r_1^2+r_2^2}) $$ changin again to cylindrical coordinates ($(r_1,r_2)\to(\rho,\phi),\phi \in (0,\pi/2))$ we have ($L_{complex}=L_1+i L_2=Le^{i \Phi}$): $$ Lag=m(\dot{\rho}^2+\rho^2\dot{\phi}^2+(\cos^2\Phi/\cos^2\phi+\sin^2\Phi/\sin^2\phi)(L/(m\rho))^2)/2-m k/\rho $$ by choosing

$$\phi(0)=\arctan{\tan^{2/3}\Phi}=\phi_0,\,\dot{\phi}(0)=0, $$$$ \rho(0)=(1+\tan^{2/3}\Phi)(\cos^2\Phi+\sin^2\Phi/\tan^{4/3}\Phi)L^2/(2 k m^2)=\rho_0,\,\dot{\rho}(0)=0,\,\theta_1(0)=0,\,\theta_2(0)=0$$ we obtains: $$ x(t)=\rho_0\cos\phi_0\cos(\omega_1 t) $$ $$ y(t)=\rho_0\cos\phi_0\sin(\omega_1 t) $$ $$ z(t)=\rho_0\sin\phi_0\cos(\omega_2 t) $$ $$ w(t)=\rho_0\sin\phi_0\sin(\omega_2 t) $$ where $\omega_1=L\cos\Phi/(m\cos\phi_0\rho_0^2),\omega_2=L\sin\Phi/(m\sin\phi_0\rho_0^2)$, now if we choose $\Phi$ such that: $$ \tan \Phi=q\tan\phi_0=q\tan^{2/3}\Phi,\;q\notin \mathbb{Q} $$ condition that is even less stricter than $\tan\Phi$ irrational. This is an example of non closed bound orbit. The harmonic case is analog with a different $\rho_0$. This construction is possible for $d\ge 4$ by simply ignoring coordinates. The question at this point is: for non-harmonic or non-newtonian ($1/r$ to be clear) are the stable orbits non-closed for every $d$?

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Assuming the the OP is asking whether the orbits are closed for a $\frac 1 {r^3}$-law, then it is surely relevant to point out that this is one of the three cases where the explicit form of the orbits are known. They are the so-called Cotes spirals---typical examples are curves of the form $r \cos ((A \theta+\epsilon)=1$, $r \cosh(A \theta+\epsilon)=1$ and hyperbolic spirals (for which, see Wikipedia). This is in Newton's Principia.

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