|Nov9-10, 07:40 PM||#1|
Two separate barycenters of a binary system in a small closed universe?
I decided to move the thought experiment that led me to this question to the bottom of this thread. I will first state what my question is.
Suppose that we live in a closed universe of three spatial dimensions and this universe is in a state of rapid collapse. Now suppose that all of the matter in universe was now concentrated into two separate super massive objects during the final moments of our universe. There would exist a barycenter between these objects.
The only problem is that the word "between" has a different sense in a very small closed universe. For if I am located at one of these objects and I travel to the other, I could also travel in the opposite direction and still arrive at the other object. Thus there are two distinct paths that I could take to the other object by traveling in a straight line, and thus there are two barycenters? Or am I wrong? The original content of this thread starts below.
Suppose that we had a small closed one-dimensional universe, in which there only existed two masses. Let the first mass be m and the second mass be equal to 4m. As we are supposing a closed one-dimensional universe, we can represent such a universe as the boundary of a circle. Let us place the first mass at 0 degrees along this circle and call this point A, and the second mass at 90 degrees along this circle and call this point B.
Now let us find the barycenter of these two masses. Using the simple two-body formula, we see that the distance, s1, from the center of the more massive body (B) to the barycenter is given by:
s1 = k(4m)/(4m + m), where k is the distance between A and B. Let k = 1, then s1 = 1/5, such that the barycenter is located at a distance of one-fifth k from B towards A. Now we recall that the separation between B and A is also 90 degrees in this closed system, therefore the barycenter is located 18 degrees away from B towards A, such that the barycenter is located at 72 degrees. Call this point C.
However, we are in a small closed universe, such that we must now consider the fact that the distance between B and A is also represented by the reflex angle 270, which is a distance of 3k, or simply 3. We cannot ignore the fact that there in fact exists a second barycenter, as the distance between A and B is both 1 and 3, or the separation is both 90 and 270 degrees. The gravitational force of the objects would be exerted upon each other from both directions, and significantly from both directions in a small universe. The second barycenter is located at a distance of 0.6 from B away from A, which is the same as 144 degrees. Call this point C'.
Now there also exists a point F such that the gravitational force between them is equal. To solve for this we will consider the equation Gmxmy/d^2. A third mass m3 will be introduced as an object of negligible mass. Let us call the distance between A and F r1 and the distance between B and F r2. Let the ratio between the masses m1 and m2 be represented as β, then we are trying to solve for this system.
Gm1m3/r12 = Gβm1m3/r22, which simplifies to 1/r12 = β/r22. Recall that r1 + r2 = k = 1
Then we see that r2 = r1(β)1/2, such that 1 = (r1 + r1(β)1/2) therefore r1 = 1/(1 + (β)1/2). You can then deduce that r2 = (β)1/2/(1 + (β)1/2).
From these equations, substituting β for 4, we can determine that F is located at a distance of 41/2/ (1 + 41/2) from B towards A. This evaluates to a distance of 2/3, so r2 = 2/3 and r1 = 1/3. Therefore F is located at 30 degrees.
There is also another point F2 which is located at a distance of 2 from B towards A, or a distance of 1 from A away from B, where the force of gravity is equal. This point is located at 270 degrees. In general the location of this point is given by r1 = (β1/2 + 1)/(β - 1), with r2 remaining equal to r1β1/2, or r2 = (β + β1/2)/(β - 1)
However, we cannot ignore the fact that again the forces of gravity are being exerted in both directions of this small closed universe. Therefore if we consider the "other" distance between A and B, which is 3, that the force of gravity would be equal at a distance of two units from B away from A, which is the same as one unit from A away from B. This point is also located at 270 degrees and we shall call it point F'.
Now just as we used r1 = (β^1/2 + 1)/(β - 1) to determine the location of F2, we shall use it to determine the location of F2' in respect to the reflex arc. This tells us that r1 is a distance of 3 units from A towards B, or equivalently that r2 6 units from B towards A, which is again at 270 degrees! So F2' = 270.
Here is a list of the points that we have determined in this small and closed one-dimensional universe. See first attachment.
A = 0, B = 90, C = 72, C' = 144, F = 30, F2 = 270, F' = 270, F2' = 270
Now suppose we a had a closed two-dimensional universe represented by the surface area of a sphere. Suppose we first wanted to find the points at which the force of gravity was equally exerted by these two masses (the scalar force of gravity, thus we are ignoring the direction of these forces).
Let us first imagine this in flat space. We have two masses m1 and m2 separated by some distance k, in which k will be equal to 1. We want to know where the scalar force of gravity between both of these objects will be equal. We already know that there exists a point F at a distance of β1/2/(1+β1/2) from B towards A and a second point F2 located at a distance of (β + β1/2)/(β - 1) from B towards A.
However, what if m3 was at some point P, such that the locations of B, A and P formed a triangle with θ being the measure of the angle BAP. How far would an object have to travel from A towards P until the force of gravity from A and B were being equally exerted on this object?
Well we already know that the length of one side of this triangle is equal to BA, which equals 1. However the other sides BP and AP (or r2 and r1 respectively) remain unknown. All that we do know is that the ratio between r1 and r2 must always be equal to β1/2, such that r2 = r1(β)1/2, in order for the force of gravity to be equally exerted by m1 and m2.
Using the law of cosines, we can swiftly solve this problem.
c2 = a2 + b2 -2ab(CosC)
Then we substitute c for r2, a for k (which is equal to 1) and b for r1, such that we get:
r22 = 1 + r12 - 2r1(Cosθ). Now we substitute r2 with r1(β)1/2 to result with:
[r1(β)1/2]2 = 1 + r12 - 2r1(Cosθ), which becomes βr12 = 1 + r12 -2r1(Cosθ).
We can turn this into 0 = (β-1)r12 + 2(Cosθ)r1 -1
Now we can use the quadratic formula to solve for r1. Here is the quadratic formula.
x = [-b +/- (b2 - 4ac)1/2]/2a
Here we substitute x for r1, a for (β-1), b for 2(Cosθ) and c for (-1). Then we have (accepting only the positive square root):
r1 = [-2cos(θ) + [4cos2(θ) - 4(β-1)(-1)]1/2]/(2(β-1)), which simplifies to the final answer:
r1 = (-cos(θ) + [cos2(θ) + β - 1]1/2)/(β-1).
It just so happens that this formula is a polar equation for a perfect circle centered at the midpoint of F and F2. See the second attachment. In three dimensions, you need simply rotate curve about the x-axis to obtain a sphere. Every point on this sphere represents where the scalar force exerted by gravity is equal.
Now we can take this formula and apply it to our small closed two dimensional universe and realize that if we find the unique Great Circle that passes through A and B, and place F and F2 along their respective places on the Orthodrome, we can quickly construct the unique Small Circle that has F and F2 as its diameter that represents all of the points where the force of gravity is being exerted equally.
Conversely, if we look at the reflex of the orthodrome, we can draw another unique Small Circle using F' and F2' as its diameter to represent all points of equal gravitational pull in respect to the reflex arc BC'A. In the case of β = 4, this second small circle is simply a point, or a circle of zero radius.
The formula for finding the barycenter remains the same. We will again result with two distinct barycenters located at C and C' long the unique great circle that contains the orthodrome. If β equals 4, they will still be located at 30 degrees and 144 degrees, just like in our one-dimensional universe. This is interesting because as where the first barycenter is located along the shortest path between points on the face of a sphere (the orthodrome), the second barycenter is located along the furthest distance between two points on the face of a sphere (the reflex of the orthodrome).
It should be noted that there exists a special case in a two dimensional universe where there are infinitely many barycenters. This occurs when the masses A and B are antipodal points. What happens here is that there is no unique great circle that passes through A and B, because A and B are the diameter of the sphere. Thus any arc extending from A to B, in any direction, will be a Great Circle, thus there are infinitely many Great Circles that encompass A and B. Now to find where these barycenters are, one must draw just one Great Circle through A and B and find the location of C and C'. The unique small circle that encompasses C and C' as its diameter and is perpendicular to arc AB represents all of the barycenters between mass A and mass B. This scenario applies in a three dimensional universe as well.
Now for my question: In a very small closed three-dimensional universe, in which there only existed two masses, could there indeed exist two barycenters in a binary system? If so, what kind of path would these bodies take when they tried to orbit about C and C' spontaneously?
|Nov11-10, 12:33 AM||#2|
I'll rephrase this entire thread, since I apparently did not make sense to anybody.
In a small closed universe, could two massive objects have two distinct barycenters?
|Nov11-10, 03:10 AM||#3|
Firstly, if you have a closed universe according to General Relativity, and matter is unevenly distributed, then the space time curvature won't be the same everywhere, so you can't model it as a circle or sphere or whatever.
Secondly, why do you think the barycentre is important? In a three body system with two massive bodies and one light body, the light body won't orbit around the barycentre of the two massive bodies, but will follow some complicated path instead.
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