homework and exercises - No stable closed orbits for a Newtonian gravitational field in $dneq 3$ spatial dimensions

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?

Answer

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.