Mass points in equilibrium

In 3D Euclidian space, find a collection of mass points lying on the line [0,1]
such that:

1) There is more than 1 point.
2) Each point has non-zero mass.
3) The total mass is 1.
4) Each point attracts each other point with inverse square gravitational
force.
5) The entire system is in static equilibrium - that is, the total force on
each point exists and is zero.

Quote:

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Originally Posted by bobbyk

In 3D Euclidian space, find a collection of mass points lying on the line [0,1]
such that:

1) There is more than 1 point.
2) Each point has non-zero mass.
3) The total mass is 1.
4) Each point attracts each other point with inverse square gravitational
force.
5) The entire system is in static equilibrium - that is, the total force on
each point exists and is zero.

---

May I be the first to say....

Huh?

We can certainly put a point mass on a line and add two other masses on either side of it to cause the original mass to be in static equilibrium, but then the two added masses are attracting each other.

I have no clue how to construct this and I would be very interested in knowing how it is done. Personally I don't think it's possible to construct this situation based on how the question is worded.

-Dan

Quote:

---

Originally Posted by bobbyk

In 3D Euclidian space, find a collection of mass points lying on the line [0,1]
such that:

1) There is more than 1 point.
2) Each point has non-zero mass.
3) The total mass is 1.
4) Each point attracts each other point with inverse square gravitational
force.
5) The entire system is in static equilibrium - that is, the total force on
each point exists and is zero.

---

1. What line?

2. No such arrangement of masses on a finite line segment is possible.
(consider the forces acting on the left most particle - it is not zero so this
point cannot be in static equilibrium) (if your line is of doubly infinite extent of
course equal masses equaly spaced will be in such an equilibrium, but its
unstable)

RonL

topsquark and Captain Black

I didn't say the number of points was finite - in fact, your two responses
showed that it couldn't be. For an infinite number of points, of course,
there doesn't have to be a left-most point.

I'm not trying to trick anybody - this is a serious question.

I don't know how to place the points, but there is a way to do so and this fact was stated by Ulam in one of his books. I think it was "A Collection of
Mathematical Problems". I can't find my copy right now, or I would give you
the page number.

I posted this to see if anybody knows how to do it.

Thanks for your responses.

Bob

Quote:

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Originally Posted by bobbyk

topsquark and Captain Black

I didn't say the number of points was finite - in fact, your two responses
showed that it couldn't be. For an infinite number of points, of course,
there doesn't have to be a left-most point.

I'm not trying to trick anybody - this is a serious question.

I don't know how to place the points, but there is a way to do so and this fact was stated by Ulam in one of his books. I think it was "A Collection of
Mathematical Problems". I can't find my copy right now, or I would give you
the page number.

I posted this to see if anybody knows how to do it.

Thanks for your responses.

Bob

---

Well, then any infinite series of points, equally spaced and extending to infinity in both directions, with positions symmetric about the origin should do the trick.

However, in your original post you said:

Quote:

---

Originally Posted by bobbyk

In 3D Euclidian space, find a collection of mass points lying on the line [0,1]
such that:

1) There is more than 1 point.
2) Each point has non-zero mass.
3) The total mass is 1.

---

[0, 1] is a line segment of length 1 along (presumably) some axis. This doesn't imply an infinite line. Also since the total mass of the system is 1 and we have an infinite number of points, the mass of each point is effectively 0, violating condition #2. I'm afraid that I don't see the infinite line to be a feasible answer given your conditions.

Edit: In fact given that the mass of each point is effectively 0 the points don't actually attract one another. So the placement of the points is irrelevant, the construction will be static anyway. We still need the infinite line, though. If we place an infinite number of even (effectively) 0 mass points on a finite line segment it forms a line of charge in which we do have internal gravitational forces (since the total mass of the line segment is non-zero.)

-Dan

It's not an infinite line, it is a line of length 1.

And don't you really think that the sum of an iinfinite set of non-zero masses
can add up to 1?

Bob

Quote:

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Originally Posted by bobbyk

It's not an infinite line, it is a line of length 1.

And don't you really think that the sum of an iinfinite set of non-zero masses
can add up to 1?

Bob

---

Let the line segment be [0,1] (three dimensions are irrelevant)

Divide the segment into n sub intervals I_i, of length 1/n, then each of
these subintervals contains mass m_i, with centre of mass at x_i.

Now to me it looks like if the sysyem is in equilibrium the gravitational force
due to all the other sub intervals acting on the i'th sub interval must be zero,
and we have reduced the problem to one of equilibrium of a finite number of
particles on a finite interval, so we are back to the problem that both
topsquark and I beleive can have no solution.

Now are there any get out clauses left?

A: Definition of line - not necessarily straight nor static (rotating string of pearls comes to mind).
B: Definition of equilibrium - not necessarily static (ie Lagrange point like arrangement).

RonL

Quote:

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Originally Posted by CaptainBlack

Now are there any get out clauses left?

A: Definition of line - not necessarily straight nor static (rotating string of pearls comes to mind).
B: Definition of equilibrium - not necessarily static (ie Lagrange point like arrangement).

RonL

---

Take three masses two of them equal, with the two equal masses either
side of the other on a straigt line segment at a distance $\displaystyle R$ from the
central mass. Then if the central mass is $\displaystyle m_1$ and the other two are
both of mass $\displaystyle m_2$ and (if I have done my sums correctly) the line segment
rotates about the central mass at an angular velocity of:

$\displaystyle
\dot{\theta}=\sqrt{\frac{G}{R^3}\left(m_1+\frac{m_ 2}{4}\right)}
$

we will have a stable configuration.

(there should be a simpler configuration with just two masses but I
can't be bothered to look for it just now - I take that back, just let
$\displaystyle m_1=0$. This is the simplest version of the rotating string of pearls
like configuration)

RonL

Quote:

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Originally Posted by bobbyk

And don't you really think that the sum of an iinfinite set of non-zero masses
can add up to 1?

Bob

---

Let the mass of each point be the same. Then the mass of N particles will be $\displaystyle m = \frac{1}{N}$. As we take the limit of N as N goes to infinity,
$\displaystyle m \to \lim_{N \to \infty}\frac{1}{N} \to 0$

We may presume that the mass of each particle is infinitesimal, but that still means we can't measure a non-zero mass for any one of them, only for an infinite collection of them.

So my answer is, yes you can have an infinite set of masses that add up to 1. But the mass of each particle is effectively 0, which violates one of your stated conditions.

-Dan

But what if the masses are NOT all the same - like 1/2+1/4+1/8+ ... =1?
I'm sure Ulam's solution didn't have all the masses equal!
Bob

Quote:

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Originally Posted by bobbyk

But what if the masses are NOT all the same - like 1/2+1/4+1/8+ ... =1?
I'm sure Ulam's solution didn't have all the masses equal!
Bob

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Possibly I'm missing a geometry here, but if your proposed construction were made then the (1/2) mass would be attracted to the rest of the collection, thus it would not be static.

-Dan

This was NOT a proposed constructoin, but a trivial example of how an infinite number of masses can add up to 1. If you're not aware of this,
then we have nothing more to say! You are kidding aren't you?
Please say you are!
Bob

Quote:

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Originally Posted by CaptainBlack

Take three masses two of them equal, with the two equal masses either
side of the other on a straigt line segment at a distance $\displaystyle R$ from the
central mass. Then if the central mass is $\displaystyle m_1$ and the other two are
both of mass $\displaystyle m_2$ and (if I have done my sums correctly) the line segment
rotates about the central mass at an angular velocity of:

$\displaystyle
\dot{\theta}=\sqrt{\frac{G}{R^3}\left(m_1+\frac{m_ 2}{4}\right)}
$

we will have a stable configuration.

(there should be a simpler configuration with just two masses but I
can't be bothered to look for it just now - I take that back, just let
$\displaystyle m_1=0$. This is the simplest version of the rotating string of pearls
like configuration)

RonL

---

Is this supposed to be a solution to the problem I posed? If so, congratulations! You did it!

Quote:

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Originally Posted by bobbyk

Is this supposed to be a solution to the problem I posed? If so, congratulations! You did it!

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But its not static equilibrium! (but then I do know its not possible for static
equilibrium)

RonL

But it is! Why don't you try to find the solution instead of saying it's
impossible?

Bob