Potential of particles moving on a circle attracted by elastic force

  • #1
l4teLearner
7
3
Homework Statement
I would like to derive the potential energy for ##l## equal point particles ##P_1, P_2, . . . , P_l## ##(l > 2)## on a circle of radius ##R## and centre ##O##. The costraint is smooth and the particle ##P_i## is attracted to ##P_{i−1}, P_{i+1}## with an elastic force (##P_0= P_l##).
Relevant Equations
Assuming the elastic potential energy between each couple of adjacent particles as ##1/2kd^2## where ##d## is the distance between the particles.
I use ##l-1## lagrangian coordinates ##\alpha_1,...,\alpha_{l-1}## . ##\alpha_i## is the angle between ##OP_{i-1}## and ##OP_{i}##.
As the length of a chord between two rays with angle ##\alpha## is ##d=2Rsin(\alpha/2)##, I write the potential energy of the system as
$$V=2kR^2\{\sum_{1}^{l-1}sin^2(\frac{\alpha_i}{2})+sin^2(\frac{2\pi-\sum_{1}^{l-1}\alpha_i}{2})\}$$
I am not fully convinced of this solution because when I evaluate the equilibrium position of 3 particles (placed at the vertices of an equilateral triangle) and 4 particles (forming a square) I find they are not stable ( hessian of V is not positive definite around the equilibrium position).
Any hint?
thanks
 
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  • #2
Hi, :welcome:

l4teLearner said:
Homework Statement: I would like to derive the potential energy for ##l## equal point particles ##P_1, P_2, . . . , P_l## ##(l > 2)## on a circle of radius ##R## and centre ##O##. The constraint is smooth and the particle ##P_i## is attracted to ##P_{i−1}, P_{i+1}## with an elastic force (##P_0= P_l##).

I find they are not stable

Why are you expecting stability ?


[edit] However,

l4teLearner said:
the elastic potential energy between each couple of adjacent particles as ##1\over 2kd^2## where ##d## is the distance between the particles.
seems to me a repulsive potential ?

##\ ##
 
Last edited:
  • #3
BvU said:
Hi, :welcome:



Why are you expecting stability ?


[edit] However,


seems to me a repulsive potential ?

##\ ##
thanks BvU for your reply.
Actually the way in which I wrote the formula was a bit ambigous, I will re-edit the post with
##\frac{1}{2}kd^2##.
I am expecting stability because of
1) for both cases ##l=3## and ##l=4## if I displace one point mass by ##\epsilon## with respect to its equilibrium position (assuming the other are held in place, respectively on the other 2 vertices of the equilateral triangle for ##l=3## or on the other 3 vertices of the square for ##l=4##), the resulting forces should push the displaced particle in the other direction
2) because a further question asks to calculate the fundamental frequencies for ##l=3## :)

[edit] Note: it seems like I cannot edit my original post, probably my edit time expired.
 
Last edited:
  • #4
l4teLearner said:
...##\frac{1}{2}kd^2##.
formally ##1/2kd^2## is the same relationship; I was just challenging a bit.


l4teLearner said:
I am expecting stability because of
1) for both cases ##l=3## and ##l=4## if I displace one point mass by ##\epsilon## with respect to its equilibrium position (assuming the other are held in place, respectively on the other 2 vertices of the equilateral triangle for ##l=3## or on the other 3 vertices of the square for ##l=4##), the resulting forces should push the displaced particle in the other direction
2) because a further question asks to calculate the fundamental frequencies for ##l=3## :)


##{1\over 2} kd^2## is stable at ##d=0##. I can imagine ##V=\sum ## to be stable with all ##P## in the same position..

Can you post the complete problem statement verbatim ? If we are working towards something like three identical springs for ##l=3##, I would expect ##{1\over 2} k(d-d_0)^2## with ##d_0=R\sqrt 3## to do the trick.

1709722334307.png
:smile:

##\ ##
 
  • #5
BvU said:
formally ##1/2kd^2## is the same relationship; I was just challenging a bit.





##{1\over 2} kd^2## is stable at ##d=0##. I can imagine ##V=\sum ## to be stable with all ##P## in the same position..

Can you post the complete problem statement verbatim ? If we are working towards something like three identical springs for ##l=3##, I would expect ##{1\over 2} k(d-d_0)^2## with ##d_0=R\sqrt 3## to do the trick.

:smile:

##\ ##
Here is the complete statement (I will omit the book it comes from as I dont't know the policy of this forum):

Consider ##l## equal point particles ##P_1, P_2, . . . , P_l(l > 2)## on a circle of radius #R$# and centre ##O##. All particles move without friction and the point ##P_i## is attracted by its neighbouring points ##P_{i−1}, P_{i+1}## with an elastic force (set ##P0= Pl##). Write down the potential of the system and prove that the configurations in which neighbouring rays form equal angles are equilibrium configurations. Study its stability (up to rotations). Compute the fundamental frequencies for ##l = 3##. What is the general procedure?

The picture you did matches my understanding of the problem. In my solution I have assumed ##d_0=0##, would it make any difference from the equilibrium standpoint? Until ##d_0## stays ##\leq R\sqrt{3}## I would say this may change the oscillating frequency but not the fact that an arrangement is stable or not (am I wrong?).

Indeed, the configuration with all point particles in the same place is a stable equilibrium configuration. To this extent I have calculated ##\nabla V=0##. As ##\frac{\partial V}{\partial \alpha_i}## is $$\frac{\partial V}{\partial \alpha_i}=2kR^2\{
sin(\frac{\alpha_i}{2})cos(\frac{\alpha_i}{2})
-
sin(\frac{2\pi-\sum_{1}^{l-1}\alpha_i}{2})
cos(\frac{2\pi-\sum_{1}^{l-1}\alpha_i}{2})
\}$$ the following configurations are equilibrium ones:

$$\alpha_i=2\pi/l \text{ } \forall i$$
or
$$\alpha_0...\alpha_i=0,\alpha_{i+1}...\alpha_{j}=\pi,\alpha_{j+1}...\alpha_{l-1}=0\text{ for }0\leq i\leq j \leq l-1$$

This matches my understanding of the problem. However, when I try to evaluate the stability of the equilibrium and calculate the Hessian matrix of V, I find it not to be positive definite for ##l=3## and ##0## for ##l=4##.
 
Last edited:
  • #6
l4teLearner said:
I have assumed d0=0, would it make any difference from the equilibrium standpoint?
Yes ! A huge difference:
1709744802756.png

(figure for ##l=3, k=R=1##)

But I grant you that ##d_0=0## is more in agreement with the problem statement.


##\ ##
 
  • #7
BvU said:
Yes ! A huge difference:

(figure for ##l=3, k=R=1##)

But I grant you that ##d_0=0## is more in agreement with the problem statement.


##\ ##
Thank you! I think you plotted my formula for ##V## when varying just one of the ##\alpha_i## right? Thinking about this, I think that the reason behind this, at least with ##l=3##, is that the constraint reaction component projected on the spring is stronger when the spring is extended, so the "strength" with which the spring pulls back (projected on the allowed trajectory) when extended actually tends to ##0## when the spring is a diameter of the circle, as it is fully neutralized by the costraint reaction...

that said can we then say that ##l=3## and ##l=4## are not stable?
if so, what is the meaning of the exercise asking for the fundamental frequencies when ##l=3##?

thanks
 
  • #8
l4teLearner said:
Thank you! I think you plotted my formula for ##V## when varying just one of the ##\alpha_i## right?
Yes.

l4teLearner said:
Thinking about this, I think that the reason behind this, at least with ##l=3##, is that the constraint reaction component projected on the spring is stronger when the spring is extended, so the "strength" with which the spring pulls back (projected on the allowed trajectory)
I think it indeed has to do with the interplay with the constraint forces

l4teLearner said:
when extended actually tends to ##0## when the spring is a diameter of the circle, as it is fully neutralized by the costraint reaction...
Things get complicated long before then.$$V=2kR^2\{\sum_{1}^{l-1}sin^2(\frac{\alpha_i}{2})+sin^2(\frac{2\pi-\sum_{1}^{l-1}\alpha_i}{2})\}$$ for ##l=3## and even distribution, with ##k=R=1## this simplifies to$$V=2 \sin^2\left (\frac{\alpha}{2}\right )+2\sin^2\left (\frac{\alpha}{2}\right )+2\sin^2\left (\frac{\alpha}{2}\right ) $$where ##\alpha = {2\over 3}\pi##.

a disturbance ##\varepsilon## of the even distribution gives $$\begin{align*}V&=2 \sin^2\left (\frac{\alpha+\varepsilon}{2}\right )+2\sin^2\left (\frac{\alpha}{2}\right )+2\sin^2\left (\frac{\alpha-\varepsilon}{2}\right ) \\ \ \\
&=\left(1-\cos\left(\alpha+\varepsilon\right) \right )
+\left(1- \cos\left(\alpha \right) \right )
+\left(1- \cos\left(\alpha-\varepsilon\right) \right ) \\ \ \\
&=3-\cos\alpha\left (1+2\cos\varepsilon \right ) = 3-3\cos\alpha + \varepsilon^2\cos\alpha +\mathcal{O} (\varepsilon^4)
\end{align*}
$$With ##\cos\alpha=-0.5## the blue parabola in the figure. Unstable.

Things get even more complicated when potential is not ##{1\over 2} kd^2## but ##{1\over 2} k(d-d_0)^2##. With a devious choice for ##d_0## the ##\varepsilon^2## term vanishes and a fourth power pops up !


l4teLearner said:
that said can we then say that ##l=3## and ##l=4## are not stable?
3 definitely. Haven't looked at four or more yet. (well, almost falling asleep ...) here the above ##V## would look like ##V=4-4\cos\alpha - \varepsilon^2\cos\alpha + \mathcal{O} (\varepsilon^4) ## but now ##\cos\alpha = 0##. I think the higher order terms all cancel too, so: not stable, but :sleep:

[edit] Haha, ##V=4\ \forall \varepsilon## -- cost me precious sleep :smile:

l4teLearner said:
if so, what is the meaning of the exercise asking for the fundamental frequencies when ##l=3##?
Here at PF we are not good at all in mind reading :smile:

##\ ##
 
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  • #9
BvU said:
Yes.


I think it indeed has to do with the interplay with the constraint forces


Things get complicated long before then.$$V=2kR^2\{\sum_{1}^{l-1}sin^2(\frac{\alpha_i}{2})+sin^2(\frac{2\pi-\sum_{1}^{l-1}\alpha_i}{2})\}$$ for ##l=3## and even distribution, with ##k=R=1## this simplifies to$$V=2 \sin^2\left (\frac{\alpha}{2}\right )+2\sin^2\left (\frac{\alpha}{2}\right )+2\sin^2\left (\frac{\alpha}{2}\right ) $$where ##\alpha = {2\over 3}\pi##.

a disturbance ##\varepsilon## of the even distribution gives $$\begin{align*}V&=2 \sin^2\left (\frac{\alpha+\varepsilon}{2}\right )+2\sin^2\left (\frac{\alpha}{2}\right )+2\sin^2\left (\frac{\alpha-\varepsilon}{2}\right ) \\ \ \\
&=\left(1-\cos\left(\alpha+\varepsilon\right) \right )
+\left(1- \cos\left(\alpha \right) \right )
+\left(1- \cos\left(\alpha-\varepsilon\right) \right ) \\ \ \\
&=3-\cos\alpha\left (1+2\cos\varepsilon \right ) = 3-3\cos\alpha + \varepsilon^2\cos\alpha +\mathcal{O} (\varepsilon^4)
\end{align*}
$$With ##\cos\alpha=-0.5## the blue parabola in the figure. Unstable.

Things get even more complicated when potential is not ##{1\over 2} kd^2## but ##{1\over 2} k(d-d_0)^2##. With a devious choice for ##d_0## the ##\varepsilon^2## term vanishes and a fourth power pops up !



3 definitely. Haven't looked at four or more yet. (well, almost falling asleep ...) here the above ##V## would look like ##V=4-4\cos\alpha - \varepsilon^2\cos\alpha + \mathcal{O} (\varepsilon^4) ## but now ##\cos\alpha = 0##. I think the higher order terms all cancel too, so: not stable, but :sleep:

[edit] Haha, ##V=4\ \forall \varepsilon## -- cost me precious sleep :smile:


Here at PF we are not good at all in mind reading :smile:

##\ ##
Thanks a ton for your help. Do I have to mark this as solved, or close it, or something?

thanks
 
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  • #10
Curious how the exercise continues and what comes out..
 
  • #11
l4teLearner said:
Thanks a ton for your help. Do I have to mark this as solved, or close it, or something?

thanks
No.

Do not mark as solved.
 
  • #12
BvU said:
Curious how the exercise continues and what comes out..
So my current status is the following. I am evaluating the Hessian ##H## of ##V## to look at the eigenvalues for the equilibrium position ##\alpha=\alpha_i=\frac{2\pi}{l}## for each ##i##.

$$\frac{\partial^2 V}{\partial^2 \alpha_i}=2kR^2\{
\frac{1}{2}(cos^2(\frac{\alpha_i}{2})-sin^2(\frac{\alpha_i}{2}))
+
\frac{1}{2}(cos^2(\frac{2\pi-\sum_{1}^{l-1}\alpha_m}{2})-sin^2(\frac{2\pi-\sum_{1}^{l-1}\alpha_m}{2}))\}$$
$$\frac{\partial^2 V}{\partial\alpha_i\partial\alpha_j}=2kR^2\{
\frac{1}{2}(cos^2(\frac{2\pi-\sum_{1}^{l-1}\alpha_m}{2})-sin^2(\frac{2\pi-\sum_{1}^{l-1}\alpha_m}{2}))\}$$
where ##i\neq j##.

Hence, $$H=2kR^2 cos(2\pi/l) M$$ where ##M## is an ##(l-1) \times (l-1)## matrix with ##M_{ij}= 1## for ##i=j## and ##M_{ij}=1/2## for ##i \neq j##.
I am currently trying to derive a generic recurrent polynomial formula for the eigenvalues (or at least for their sign) of $M$ to classify the stability for each ##l##.
For ##l=4## $$H=2kR^2 cos(\pi/2) \begin{bmatrix}1 & 1/2 &1/2\\1/2 & 1 & 1/2\\1/2 & 1/2 &1\end{bmatrix}=0$$
and for ##l=3## $$M=\begin{bmatrix}1 & 1/2\\1/2 & 1\end{bmatrix}$$ is positive definite, but ##cos(2\pi/3) < 0## hence the Hessian ##H## is not positive definite as discussed.

Maybe there is some quick way to see the sign of the eigenvalues of ##M## and I expect them to be positive, and ##l>4## to be stable.
 
  • #13
I think we can prove that ##M## is positive definite because it corresponds to the following quadratic form:
$$\sum_{1}^{l}x_i^2+\sum_{1\leq i<j \leq l}x_i x_j $$

It can be proved by contradiction that
##\sum_{1}^{l}x_i^2+\sum_{1\leq i<j \leq l}x_i x_j > 0## if some ##x_i \neq 0##.

In fact, the contrary would imply ##\sum_{1\leq i<j \leq l}x_i x_j \leq 0##.

But on the other hand, as ##\sum_{1\leq i<j \leq l}x_i x_j = \sum_{1}^{l}x_i^2+2\sum_{1\leq i<j \leq l}x_i x_j## is a perfect square, we have:

$$\sum_{1\leq i<j \leq l}x_i x_j = \sum_{1}^{l}x_i^2+2\sum_{1\leq i<j \leq l}x_i x_j
-
(\sum_{1}^{l}x_i^2+\sum_{1\leq i<j \leq l}x_i x_j)
> 0 $$

Which is a contradiction.

As ##\sum_{1}^{l}x_i^2+\sum_{1\leq i<j \leq l}x_i x_j \geq 0## is always positive if some ##x_i \neq 0##, it means that ##M## is positive definite.
 

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