Physics Challenge by QuantumQuest #1

[Greg Bernhardt]

Submitted and Judged by @QuantumQuest
Solution credited to: @TSny

RULES:
1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.

CHALLENGE:
Three point masses $m_1, m_2$ and $m_3$ which are located at the non-collinear points $P_1, P_2$ and $P_3$ respectively can interact only through gravitational attractions. The masses are isolated in space and they have no interaction with other objects. We suppose that an axis $\sigma$ is passing through the center of mass of the system of the three given masses and is perpendicular to the plane of the triangle $P_1P_2P_3$. Which conditions must angular velocity of the system (regarding given axis) and distances $P_1P_2 = d_{12}$,$P_2P_3 = d_{23}$ and $P_1P_3 = d_{13}$ must satisfy in order the shape and the size of the triangle $P_1P_2P_3$ stay constant, as the system is rotating?

 

[zwierz]

Let $O$ stand for the center of mass to the system. Introduce a coordinate frame that rotates about the axis $\sigma$ with angular velocity $\omega=const$.
With respect to this moving frame the potential energy of the system is
$$W(\overline {OP}_1,\overline {OP}_2,\overline {OP}_3)=-\gamma\sum_{1\le i<j\le 3}\frac{m_im_j}{|\overline {OP}_i-\overline {OP}_j|}+\frac{\omega^2}{2}\sum_{i=1}^3|\overline {OP}_i|^2$$
This is a function of six variables -- coordinates of the vectors $\overline {OP}_i$.
We are given with a configuration of particles such that the particles rest with respect to the rotating system.
To solve the problem one should find conditional extrema of the function $W$ under the following constrain
$$\sum_{i=1}^3m_i \overline {OP}_i=0.\qquad (*)$$
These are three scalar equations.
Using the Lagrange multipliers
we have
$$\frac{\partial W}{\partial \overline {OP}_i}+\overline\lambda m_i=0,\quad i=1,2,3.\qquad (**)$$
Therefore we have a system of 4 algebraic vector equations (*)-(**) with respect to unknowns $$\overline {OP}_i,\quad i=1,2,3,\quad \overline\lambda,\quad\omega.$$
One also should take into account the additional three scalar equations $|\overline {OP}_i-\overline {OP}_j|=d_{ij}$.I just wonder if why does the author think that such an overdetermined system must have a solution :)

 

[Charles Link]

@zwierz Please compute the frequency along with any other additional necessary constraints. Before posting, we looked over this problem quite carefully and also solved it. I think you will find the solution quite interesting.

 

[mfb]

I just wonder if why does the author think that such an overdetermined system must have a solution :)

It has solutions, and it has useful relations you did not find so far.

 

[zwierz]

Very well, it would be interesting to look at your solution once. For now I suspect that you just implicitly assume that the particles are connected by weightless rods to provide the conditions $|\overline{OP}_i-\overline{OP}_j|=d_{ij}$. In other case the problem is hardly solvable for arbitrary $d_{ij}$

 

[mfb]

There are no rods.
The problem is solvable.

Did you see the condition given in the last sentence of the problem?

 

[zwierz]

ok. I await with great interest how will you solve this problem for arbitrary $d_{ij}$

 

[Charles Link]

ok. I await with great interest how you solve this problem for arbitrary $d_{ij}$

You are allowed to assign constraints, if necessary, to the $d_{ij}$.

 

[zwierz]

that's quite another matter.
thus it is a linear system relative to $\overline {OP}_i$

 

[QuantumQuest]

I just wonder if why does the author think that such an overdetermined system must have a solution :)

It does have a solution and I really fail to see how it is overdetermined. I think that it is rather a well determined system but you see it through an overwhelming way.

I suspect that you just implicitly assume that the particles are connected by weightless rods

Did you see anything that even remotely implies a such thing in the wording of the problem?

thus it is a linear system relative to$\overline {OP}_I$

Again, the wording of the problem is exact. There is absolutely nothing missing.

 

[zwierz]

there are misprints in my first post, it should be written as follows
$$W(\overline{OP}_1,\overline{OP}_2,\overline{OP}_3)=-\gamma\sum_{1\le i<j\le 3}\frac{m_im_j}{|\overline{OP}_i-\overline{OP}_j|}-\omega^2\sum_{i=1}^3\frac{m_i}{2}|\overline{OP}_i|^2$$

If the particles do not lie on the same line then from equations (*)-(**) we get

$$\omega^2=\frac{\gamma(m_1+m_2+m_3)}{d^3_{13}}$$
and the
equilateral
triangle is solely possible $d_{12}=d_{13}=d_{23}$

The case when all the particles lie on the same line is boring

 

[QuantumQuest]

@zwierz

I really don't want to be any discouraging and I see that you put good efforts in an (unnecessarily in my opinion) cumbersome way. Now, while this is perfectly acceptable, you have to also abide by the rules in order to get credit. We all abide by these rules when we participate in a challenge here at PF. So, I want you to give a full proof including the necessary math, as of rule #1. I cannot really see how you get to $\omega$ and that the triangle is equilateral.

If the particles do not lie on the same line then from equations (*)-(**) we get

$\omega^2=\frac{\gamma(m_1+m_2+m_3)}{d^3_{13}}$
and the
equilateral
triangle is solely possible $d_{12}=d_{13}=d_{23}$

 

[zwierz]

I don't care about credits. I think about problem if find it interesting.
If you wish the details they are as follows. Equations (**) give
$$(a_{12}+a_{13}-\omega^2 m_1)\overline{OP}_1-a_{12}\overline{OP}_2-a_{13}\overline{OP}_3=-\overline{\lambda}m_1,\qquad (***)$$
$$(a_{21}+a_{23}-\omega^2 m_2)\overline{OP}_2-a_{21}\overline{OP}_1-a_{23}\overline{OP}_3=-\overline{\lambda}m_2,\qquad (!)$$
$$(a_{31}+a_{32}-\omega^2 m_3)\overline{OP}_3-a_{31}\overline{OP}_1-a_{32}\overline{OP}_2=-\overline{\lambda}m_3,\qquad (!)$$
here
$$a_{ij}=\frac{\gamma m_im_j}{d^3_{ij}}$$
If you sum these three equations and use (*) you obtain $\overline{\lambda}=0$.
Now express $\overline{OP}_3$ from (*) and put it into (***) this gives
$$(a_{12}+a_{13}-\omega^2 m_1)\overline{OP}_1-a_{12}\overline{OP}_2+a_{13}(\frac{m_1}{m_3}\overline{OP}_1+\frac{m_2}{m_3}\overline{OP}_2)=0$$
Since $\overline{OP}_1$ and $\overline{OP}_2$ are linearly independent you get two scalar equations to express $\omega$.
Then do the same for (!) (!)

 

[Charles Link]

@zwierz I have a couple of questions about your solution=and yes, you did get the correct answer. $\\$ 1) In your first posting, you describe the energy in a rotating coordinate system, so that apparently there is no kinetic energy term. Is that why the term that looks like the kinetic energy has a "-" sign in your second posting? I am familiar with $[\frac{d \vec{r}}{dt}]_o=\frac{d \vec{r}}{dt}+\omega \times \vec{r}$, but have never seen a potential energy term transformed or introduced in a rotating system due to what is kinetic energy in the stationary frame. $\\$ 2) Your energy term $W$ is a scalar=I found it difficult to take a vector derivative (gradient) on the vector-type Lagrange multiplier constraint term. Instead, I squared each vector component and set the sum of the squares equal to zero as the constraint to try and check your solution. $\\$ 3) The resulting set of equations does not appear to be simple. If you assume $d_{12}=d_{13}=d_{23}$, the result follows for $\omega^2$, but reaching that result appears non-trivial. Did you see an easy shortcut to this result? $\\$ Additional item you may find of interest, as was pointed out by @mfb in discussions regarding the equilateral triangle solution, is that for $m_3$ negligible, the system describes the Lagrangian points L4 and L5 for the sun, earth, and satellite system. $\\$ Anyway, I'm glad you got the correct result.

 

[zwierz]

In your first posting, you describe the energy in a rotating coordinate system, so that apparently there is no kinetic energy term. Is that why the term that looks like the kinetic energy has a "-" sign in your second posting?

this term is a potential energy of the inertial force.

If you assume d12=d13=d23 d_{12}=d_{13}=d_{23}

I have proved that but not assumed

Additional item you may find of interest, as was pointed out by @mfb in discussions regarding the equilateral triangle solution, is that for m3 m_3 negligible, the system describes the Lagrangian points L4 and L5 for the sun, earth, and satellite syste

Yes, I also recollected that but unfortunately after I had solved the problem

The resulting set of equations does not appear to be simple. If you assume d12=d13=d23 d_{12}=d_{13}=d_{23} , the result follows for ω2 \omega^2 , but reaching that result appears non-trivial.

Actually I do not understand which nontriviality you mean. A term expressed from one equation is substituted to another equation that are absolutely standard things.
I did not check that but I almost sure that equations (***),(!),(!) with $\lambda=0$ are the second Newton laws for each particle relative to an inertial frame

 

[Charles Link]

Actually I do not understand which nontriviality you mean. A term expressed from one equation is substituted to another equation that are absolutely standard things.

@zwierz Thank you. I'm going to need to take a second look at it. The result may follow directly upon setting equal the results for $\omega^2$.

 

[zwierz]

Exactly. It is time to go to bed for me 00.20 :)

 

[mfb]

Here is an alternative approach that might be easier to follow (I saw the problem before, don't count that as new solution):

Let the center of mass be the origin of the coordinate frame: $m_1 r_1 + m_2 r_2 + m_3 r_3 = 0$
To keep shape and size, the masses have to rotate at constant $\omega$ and constant radius. This leads to three equations for the three masses:

$$\omega^2 \vec r_1 = \frac {Gm_2 \vec d_{12}}{|d_{12}|^3} + \frac {Gm_3 \vec d_{13}}{|d_{13}|^3}$$
$$\omega^2 \vec r_2 = \frac {Gm_1 \vec d_{21}}{|d_{12}|^3} + \frac {Gm_3 \vec d_{23}}{|d_{23}|^3}$$
$$\omega^2 \vec r_3 = \frac {Gm_1 \vec d_{31}}{|d_{13}|^3} + \frac {Gm_2 \vec d_{32}}{|d_{23}|^3}$$
Here $d_{12} = r_1 - r_2 = -d_{21}$ is pointing from mass 2 to mass 1. We can see $\omega \propto \sqrt{G}$ already.

By taking differences, we can get rid of the bare vectors r. Equation 1 minus equation 2 and dividing by G:
$$\frac{\omega^2}{G} \vec d_{12} = \frac{(m_1+m_2) \vec d_{12}}{|d_{12}|^3} + \frac {m_3 \vec d_{13}}{|d_{13}|^3} + \frac {-m_3 \vec d_{23}}{|d_{23}|^3}$$
We can investigate this along the d₁₂ vector and perpendicular to it. Take the cross product with d₁₂ to study the perpendicular direction. The first two terms vanish.
$$0 = \frac {m_3 \vec d_{13} \times \vec d_{12}}{|d_{13}|^3} + \frac {-m_3 \vec d_{23} \times \vec d_{12}}{|d_{23}|^3}$$
As $\vec d_{13} = \vec d_{12}+\vec d_{23}$, we have $\vec d_{13} \times \vec d_{12} = (\vec d_{12}+\vec d_{23}) \times \vec d_{12} = \vec d_{23} \times \vec d_{12}$. Use this:
$$ \frac {m_3 \vec d_{23} \times \vec d_{12}}{|d_{13}|^3} = \frac {m_3 \vec d_{23} \times \vec d_{12}}{|d_{23}|^3}$$
Simplify:
$$|d_{23}|= |d_{13}|$$
We can repeat this with other differences, but we can also use that the problem is invariant under permutations of the masses:
$$ |d_{23}| = |d_{13}| = |d_{12}| = d$$
We get an equilateral triangle of arbitrary size. To determine $\omega$, take the scalar product with d₁₂ where we took the cross product before, and use $\vec d_{12}\vec d_{12}=d^2$ and $\vec d_{12}\vec d_{13}=\frac{d^2}{2}$ and $\vec d_{12}\vec d_{23}=\frac{-d^2}{2}$:
$$\frac{\omega^2}{G} d^2 = \frac{(m_1+m_2) d^2}{d^3} + \frac {m_3}{2d^3} + \frac {m_3 d^2}{2d^3}$$
Simplify: $$\omega = \sqrt{ \frac{G(m_1+m_2+m_3)}{d^3}}$$

If one mass is negligible, this is a well-known result from the Kepler problem, and the small mass is in one Lagrangian point.

 

[QuantumQuest]

@zwierz

I want to be fair regarding challenge so I want to ask you some things about your solution overall, in order to have a complete picture

To solve the problem one should find conditional extrema of the function $W$ under the following constrain
$\sum_{i=1}^3m_i \overline {OP}_i=0.\qquad (*)$

What kind of extrema you look for and why? What does this constraint represent? (I take $W$ in the corrected form you provided in #11)

Using the Lagrange multipliers
we have

​

$\frac{\partial W}{\partial \overline {OP}_i}+\overline\lambda m_i=0,\quad i=1,2,3.\qquad (**)$

How did you arrive there?

Also, your math in #13 are severely shortcut but I see your point. What I don't see presented in a clear way is how do you conclude that distances are equal.

 

[Charles Link]

@zwierz I successfully worked through the equations of your solution: Using the first gradient equation, and a substitution (as you had mentioned) for $\vec{r}_3$ into that equation from the center of mass equation resulted in an equation of the form $A \vec{r}_1+B \vec{r}_2=0$. This gives the result that $A=0$ and $B=0$ if $\vec{r}_1$ and $\vec{r}_2$ are not colinear. Setting $B=0$ gives $m_2( \frac{1}{d_{12}^3}-\frac{1}{d_{13}^3})=0$ so that $d_{12}=d_{13}$ Setting $A=0$ gives the equation for $\omega^2$. By symmetry of the equations, the result $d_{12}=d_{13}=d_{23}$ follows without further processing. $\\$ An additional comment is I think the minus sign that you used for the kinetic energy term is basically because in the Lagrangian formalism of mechanics, $T$ and $V$ have opposite sign=the Lagrangian $L=T-V$. $\\$ In any case, a very interesting method that you used.

 

[zwierz]

An additional comment is I think the minus sign that you used for the kinetic energy term is basically because in the Lagrangian formalism of mechanics, T T and V V have opposite sign=the Lagrangian L=T−V L=T-V .

The equations are the same ,terminology can be different. My argument is as follows. The particle $P_i$ experiences an inertial force
$\overline F_i=m_i\omega^2 \overline{OP}_i$ its potential energy is
$$\overline F_i=-\frac{\partial W_i}{\partial \overline{OP}_i},\quad W_i=-\frac{\omega^2 m_i}{2}|\overline{OP}_i|^2.$$ There is also Coriolis force but it does not play a role since we consider relative equilibria but the Coriolis force depends linearly on relative velocity. Formal fact is
if a Lagrangian has the form
$$L=\frac{1}{2}g_{ij}(q)\dot q^i\dot q^j+a_i(q)\dot q^i-W(q)$$ then $q(t)=q_*$ is an equilibrium iff $\frac{\partial W}{\partial q}(q_*)=0$.

What kind of extrema you look for and why? What does this constraint represent?

read somewher what a center of mass is

How did you arrive there?

read a math. analysis textbook

 

[QuantumQuest]

read somewher what a center of mass is

I did it more than thirty years ago, so I don't think that is my understanding at issue here.

read a math. analysis textbook

I also did this more than thirty years ago. Again the issue here is that you explain nothing.

The purpose of my questions was for some points that you make fast and without any justification whatsoever to get explained, so other members here who will read your solution will understand, appreciate your solution and learn something from it. That is the purpose. I didn't ask for myself. I already know all these I asked very well.

 

[zwierz]

It is impossible that this equilateral-triangle solution has never been noted somewhere in a textbook. I remember once I heard that the planar three body problem has a set of solutions such that all the particles belong to a line for all time. It would be much more interesting task to describe such solutions independently.

 

[TSny]

Here's a solution that I think is similar to zwierz's and mfb's, but hopefully different enough to be of interest.

Go to the rotating frame so that there is a fictitious centrifugal force on each particle. Let the origin be at the center of mass. $\mathbf{r}_1$ is the position vector of particle 1 relative to the center of mass, etc.

The net force on each particle must be zero in this frame since the particles remain at rest. For particle 3:

$\frac{Gm_1m_3}{d_{13}^3}\left(\mathbf{r}_1 - \mathbf{r}_3 \right) + \frac{Gm_2m_3}{d_{23}^3}\left(\mathbf{r}_2 - \mathbf{r}_3 \right) + m_3\omega^2 \mathbf{r}_3 = 0$

Since the origin is at the center of mass, we can substitute $m_1 \mathbf{r}_1 =-m_2\mathbf{r}_2 -m_3\mathbf{r}_3$ and collect terms in $\mathbf{r}_2$ and $\mathbf{r}_3$ to get (after cancelling a common factor of $m_3$):

$Gm_2\left(\frac{1}{d_{23}^3} -\frac{1}{d_{13}^3}\right) \mathbf{r}_2 + \left(\omega^2 – \frac{Gm_1}{d_{13}^3} - \frac{Gm_2}{d_{23}^3} - \frac{Gm_3}{d_{13}^3}\right)\mathbf{r}_3 = 0$.

Since the particles are non-collinear, the center of mass cannot lie on the line connecting particles 2 and 3. Hence, the vectors $\mathbf{r}_2$ and $\mathbf{r}_3$ cannot be parallel. Therefore, the coefficients of $\mathbf{r}_2$ and $\mathbf{r}_3$ in the above equation must be zero separately.

This yields $d_{13} = d_{23}$ and $\omega^2 = G(m_1+m_2+m_3)/d_{13}^3$.

Repeating the whole argument for particle 2 would clearly yield $d_{12} = d_{23}$ and $\omega^2 = G(m_1+m_2+m_3)/d_{12}^3$.

Thus, conclude that we must have $d_{12} = d_{23} = d_{13} \equiv d$ and $\omega^2 = G(m_1+m_2+m_3)/d^3$.

 

[QuantumQuest]

@TSny

Very well done!

 

[TSny]

Very well done!

Thank you.