Friday, January 3, 2014

n-dimensional force, d <−3 implies not stable.) n≥4 is unstable, while n=2 doesn't have closed orbits but is stable. –

http://physics.stackexchange.com/questions/50142/gravity-in-other-dimensions-than-3-and-stable-orbits

     只有平方反比連心勢 (\beta =1) 與徑向諧振子勢 (\beta =2)能夠造成穩定的，閉合的，非圓形的公轉軌道。

牛頓旋轉軌道定理[编辑]

牛頓旋轉軌道定理表明，對於一個感受到線性作用力或平方反比作用力的移動中的粒子，假設再增添立方反比力於此粒子，只要因子 \alpha 是有理數，則粒子的軌道仍舊是閉合軌道。根據牛頓旋轉軌道定理的方程式，增添的立方反比力 \Delta F(r)=\frac{k}{r^3} 為
\Delta F(r)=\frac{L_{1}^{2}}{mr^{3}} \left( 1 - \alpha^{2} \right) ；
其中，\ell_{1} 是粒子原本的角動量，m 是粒子的質量。
所以， \alpha^2=1 - \frac{mk}{\ell_1^2} 。
由於 \alpha 是有理數，\alpha 可以寫為分數 m/n ；其中，m 和 n 都是整數。對於這案例，增添立方反比力使得粒子完成 m 圈公轉的時間等於原本完成 n 圈公轉的時間。這種產生閉合軌道的方法不違背伯特蘭定理，因為，增添的立方反比力與粒子的原本角動量有關。

Explore our sites

Stack Exchange
tour help


Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

up vote9down votefavorite
4
I have heard from here that stable orbits (ones that require a large amount of force to push it significantly out of it's elliptical path) can only exist in a three spatial dimensions because gravity would operate differently in a two or four dimensional space. Why is this?
share|improve this question
add comment

3 Answers


up vote13down voteaccepted
Specifically what that is referring to is the 'inverse-square law', nature of the gravitational force, i.e. the force of gravity is inversely proportional to the square of the distance:
F g ∝1 d 2     .
If you expand this concept to that of general power-law forces (e.g. when you're thinking about the virial theorem), you can write:
F∝d a   ,
Stable orbits are only possible for a few, special values of the exponent 'a  '---in particular, and more specifically 'closed1', stable orbits only occur for a=−2  (the inverse-square law) and a=1  (Hooke's law). This is called 'Bertrand's Theorem'.
Now, what does that have to do with spatial dimensions? Well, it turns out that in a more accurate description of gravity (in particular, general relativity) the exponent of the power-law ends up being one-less than the dimension of the space. For example, if space were 2-dimensional, then the force would look like F∝1 d    , and there would be no closed orbits.
Note also that a<−3  (and thus 4 or more spatial dimensions) is unconditionally unstable, as per @nervxxx's answer below.
1: A 'closed' orbit is one in which the particle returns to its previous position in phase space (i.e. its orbit repeats itself).
share|improve this answer
1  
You don't need general relativity, you just need Gauss's Law and stoke's theorem to hold to derive the d-1 rule. –  Jerry Schirmer Jan 13 '13 at 23:26
 
@JerrySchirmer thanks, good point --- but isn't Stoke's theorem requisite on the force being expressed as the divergence of a field---which itself is unique to a = -2, and 1? –  zhermes Jan 14 '13 at 0:00
 
+1. Also, so this is actually a question about 2+1  right, not 1+1  ? –  kηives Jan 14 '13 at 4:55
 
The problem is, there could still be stable orbits with this. The initial velocity would just have to me equal to rad(Gm) instead of rad(Gm/r). The required orbital velocity would just be independent of distance. I suppose the only thing is that elliptical orbits would be impossible, which is probably what the video I linked was talking about. –  Garan Jan 14 '13 at 5:15
1  
@zhermes While everything you say about closed orbits is correct, there is indeed a true sense of instability that sets in in 4+ dimensions: any perturbation to any orbit will send the separation either to infinity or to 0. See nervxxx's answer, which basically follows the method in, e.g., Goldstein. –  Chris White Jan 14 '13 at 18:34
show 10 more comments

up vote6down vote
I'll try to answer it by considering radial deviations from a circular orbit. First we have to assume two things about our n-dimensional universe: Newton's second law still holds, that is,
for a particle's position vector in n-dimensions x ⃗ =(x 1 ,x 2 ,⋯x n )  ,
mx ⃗  ¨ =F ⃗ ,   
where F ⃗   is some n-dimensional force,
and also that the law of gravity is given by Gauss' law:
∇⋅g ⃗ =−4πGρ,   
where g ⃗   is the gravitational force field. (See wikipedia for more information).
The solution to that pde is
g ⃗ ∼=−r 1−n e r  ^ ,   
for n≥2  . (For n=1  the motion is on a line and because it's always attractive the 'orbit' will still remain an 'orbit')
Since the motion will always be constrained to move in the 2-plane spanned by the initial radial vector r ⃗  0   and the initial velocity vector v ⃗  0   , it is easiest to analyze the motion in cylindrical coordinates. That is, Newton's second law becomes
m(r ¨ −θ ˙  2 r) m(rθ ¨ +2r ˙ θ ˙ ) mx 3  ¨  mx 4  ¨   mx n  ¨   =F r  =F θ  =F x 3   =F x 4   ⋯ =F x n  ,   
where x 1   and x 2   are coordinates of the plane spanned by v ⃗  0   and r ⃗  0   . Here r  really means x 2 1 +x 2 2  − − − − − −  √   , but it turns out that because the motion is just 2-D i.e. x 3 =x 4 =⋯x n =0  , we can say r=x 2 1 +⋯+x 2 n  − − − − − − − − − − −  √   .
Now we make use of the fact that gravity is always radial, so F θ =0  and we can combine the first two equations to get
r ¨ −L 2  r 3   =F r =f(r),   
where L  is a constant of motion (in 3D this is the angular momentum).
For a circular orbit at r=r c   , r ¨ =0  , so we are left with
−L 2  r 3   =f(r).   
Consider small deviations from r c   : x=r−r c   . Plugging this into newton's law and expanding to first order, one gets
x ¨ +[−3f(r c )/r c −f ′ (r c )]x=0.   
This is a simple harmonic equation if the stuff in the parenthesis is positive. So we obtain a stability condition
[−3f(r c )/r c −f ′ (r c )]>0.   

Let's check this on a radial force f(r)=−kr d   . The stability condition gives
−kr d c −kd 3  r d c <0,   
which implies d>−3  . So if the force law goes as r d   where d>−3  , then the orbit is not stable. One can, with a bit more work, show that d=−3  is also unstable.
So for dimensions n≥4  , the orbit is unstable. It appears, however, that for d=−1  or −2  , the orbit is stable, so this gives us the result that orbits in 3-dimensions (our world) and also that of 2-dimensions are stable, in disagreement with the video's statement. I might be wrong, though.
cheers.
share|improve this answer
 
Agreed. (Except I think you meant to say d<−3  implies not stable.) n≥4  is unstable, while n=2  doesn't have closed orbits but is stable. –  Chris White Jan 14 '13 at 7:52
add comment

up vote0down vote
I assume this is talking about Newtonian gravity (i.e., not relativity). Let's consider the effective potential:

V eff (r)=L 2  2mr 2   +V(r) 

where V  is the ordinary potential energy, and L  is the angular momentum. First, you may ask why the effective potential has this form. Remember that for a single particle, L=mr 2 ω  , so this is equivalently,

V eff (r)=ω 2 r 2  2m  +V(r) 

This first term appears from the equations of motion for a free particle. Phrasing it in terms of angular momentum is convenient because under central forces, angular momentum is a conserved quantity.
Why do we use the effective potential? Because it helps us talk solely about the radial motions of a particle, lumping the angular motions in with the real potential. A local extremum of the effective potential tells us about an equilibrium distance.
Now, in 3d, the potential V(r)  for gravity is −GMm/r  . What this means is that, as r→0  , the effective potential will eventually blow up, thanks to the angular momentum part, overcoming the gravitational part and forcing the particle outward again unless it lies on a direct infall trajectory.
In 2d, the potential is different. Why is this? Newtonian gravity deals with differential equations of the form ∇ 2 V∝ρ  . The point-source solution to this equation (the Green's function) is proportional to lnr  --compare, for example, the electric potential of an infinite line charge. This is exactly the same geometry and differential equation, at least in structure.
Let's check for a second that this is the case. Let V=Clnr  in 2d for some constant C  . Then the gravitational force is

F=−∂V ∂r  =−C/r 

which is inward for all positive C  . This is important. In 2d, then, our effective potential looks like,

V eff =Kr −2 +Clnr 

for two constants K,C  . The force is

F eff =2Kr −3 −Cr −1 =−r −1 (−2Kr −2 +C) 

So r eq =2K/C − − − − −  √   . But is this equilibrium stable?

∂F eff  ∂r  =−6Kr −4 +Cr −2  

At r eq   , this evaluates to −6C 2 /4K+C 2 /2K=−C 2 /K  .
Hm. That would suggest the equilibrium point is stable. So, perhaps someone has a reference to suggest this. I'm stuck.
share|improve this answer
add comment


Your Answer

 
discard
By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.


Posted by at

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)