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Photon, Lagrange Point, Binary Black Hole

by Edward Solomo
Tags: binary, black, hole, lagrange, photon, point
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Edward Solomo
#1
Jun17-10, 12:15 AM
P: 72
Hello,

I am interested in what would happen if a photon became nested inside the Lagrangian point of a binary black hole system that was already far into the process of merging. It seems that the photon would be "frozen."
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nicksauce
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Jun17-10, 08:47 AM
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Do Lagrange points even exist though, when general relativistic effects are important? My guess is that the answer is unknown, since no analytic solutions to the two-body problem are known.
Edward Solomo
#3
Jun17-10, 07:31 PM
P: 72
Quote Quote by nicksauce View Post
Do Lagrange points even exist though, when general relativistic effects are important? My guess is that the answer is unknown, since no analytic solutions to the two-body problem are known.
It is my understanding that the singularities are the only places where time and space truly cease to exist. This means that the L1 point must exist somewhere between the singularities, I am not sure if any other Lagrangian point (L2 - L5) exist when the black holes are close to merging.

We can make an attempt using Newtonian Mechanics that suggests there exists an L1 point by showing there exists an "equi-gravitational sphere" that encompasses the smaller of two colliding masses. These equations assume that time is static, so they have no real use outside of extremely small time intervals (although every small lapse of time is critical during a black hole merger, so this argument may be valid).

Let the mass of the larger object be equal to M1
Let the mass of the smaller object be equal to M2
Let the mass of negligible object be equal to M3

Fg1 = GM1M3/d2
Fg2 = GM2M3/d2

M1 > M2
M1 = αM2
The distance between the center of masses for M1 and M2 is equal to d.

READ: The alpha symbol, α, is quantifying how much more massive M1 is than M2. This is the most important variable in the equations to come.

Our goal is to solve for at what distances from M1 and M2 is the acceleration due to gravity equal applied upon the negligible mass M3.

Suppose that we are considering the line drawn directly between their centers of gravity. Given that M1 = αM2 and that the strength of the gravitational field changes linearly in respect to mass, then we can conclude that the strength of gravity due to both masses is equal when we are a distance of (1/α + 1) from the center of M2. This is because (α/α+1) / (1/α +1) = α


Now suppose we are considering a right angle from the center of M2. We can now extend slightly longer from the center of M2 before the strengths become equal. However we'll need to use the Law of Cosines to solve this problem.

c2 = a2 + b2 -2ab*Cos(C)

Now we are solving for this right triangle. (Do not confuse letter "a" with alpha "α")
a = d
b = k
c = αk
Angle C = pi/2

As Cos(pi/2) = 0, we can ignore this for right triangles. Substitute for the variables.

(αk)2 = 12 + k2 - 0

Then put into quadratic form:

0 = α2k2 - k2 - 1
0 = (α2 -1)k2 - 1

Now use the quadratic equation. x = [-b +/- (b2 - 4ac)1/2]/2a
k = [0 +/- ((0 - 4(α2 -1)(-1))1/2]/2(α2 -1)
k = 2(α2 - 1)1/2/2(α2 -1)
k = [α2 -1]1/22-1

Now suppose we are considering any angle from the center of M2. We must now incorporate the Law of Cosines fully into our quadratic equation.

a = d
b = k
c = αk
Angle C = θ

(αk)2 = 12 + k2 - 2(1)(k)(Cos θ)

0 = α2k2 - k2 - 1 + 2k(Cos θ)
0 = (α2 -1)k2 + 2k(Cos θ) - 1

k = [-2Cosθ +/- (((2Cosθ)2) - 4(α2 -1)(-1))1/2]/2(α2 -1)

We can simplify this to the following (and we only accept the positive square root).

0 = [-cosθ + (cos2θ + α2 - 1)1/2]/(α2 -1)

Now it turns out that this equation describes a perfect circle about an imaginary center (just like the planets have an imaginary focus in their elliptical orbits). To extend this into three dimensions, we rotate this circle about the axis that is drawn between the centers of M1 and M2, giving us a sphere.

The L1 point is found at θ = 0 and the L2 point is found at θ = pi. The center of this circle, C, is found at the midpoint between L1 and L2. Knowing that Cos(0) = 1 and Cos(pi) = -1, we can quickly solve the quadratic equations to find how far the L1 and L2 points are from the center of M2, and use this fact to solve where C is located. Remember that d is the distance between the centers of M1 and M2.

The distance between L1 and M2 = (1/α+1)d
The distance between L2 and M2 = (1/α-1)d
The distance between C and M1 = (1 + distance from C to M2)d

The distance between C and M2 is equal to the distance between L2 and M2 subtracted by the distance between L1 and M2, divided by 2, we subtract because L1 and L2 are on opposite sides of M2. Then we get the following (after simplification)
The distance between C and M2 = (α + 2)/(α2 -1)d

Other distances:
The distance between L1 and M1 = (α/α+1)d
The distance between L2 and M1 = (α/α-1)d

So we can see that we have an equi-gravitational sphere around this imaginary center C (unless M1 = M2, in which case we have a plane that is perpendicular to the axis between their centers). Using basic vector algebra, we realize that the L1 point (in static time) is the only stable point on this sphere.

Now let M1 be a supermassive black hole and M2 be a stellar black hole. This suggests that for some very small lapses of time during a black hole merger, there are points in space where photons may become "frozen" between them, the warping of space at this point is also non-differentiable, as it appears as a very thin spike emanating from the space-time voids on either side.

nicksauce
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Jun17-10, 07:38 PM
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Photon, Lagrange Point, Binary Black Hole


Fg1 = GM1M3/d2
Fg2 = GM2M3/d2
Well, the point is that these equations won't be valid in a strongly curved spacetime such as that of a merging black holes.
Edward Solomo
#5
Jun17-10, 07:53 PM
P: 72
Quote Quote by nicksauce View Post
Well, the point is that these equations won't be valid in a strongly curved spacetime such as that of a merging black holes.
Well if you'd like to rewrite those equations using the Riemann curvature tensor in a field of increasing positive curvature you'd probably arrive at the same or a similar conclusion. I think you'd need a supercomputer for that as well though.

However, when you view the black hole merger video, you do see a point between them on the manifold that appears to be spiking upwards. This is the point that I'm referring to.
Edward Solomo
#6
Jun18-10, 08:56 PM
P: 72
Is this question stupid or too hard to answer?
Thinkor
#7
Nov1-11, 02:56 PM
P: 36
I think GR may have a problem with Lagrange points because the action in GR is constructed so that (for the vacuum solution) the field equations minimize the curvature of spacetime weighted by another factor, but I don't see how curvature can be defined at Lagrange points.
pervect
#8
Nov1-11, 05:56 PM
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Even Newtonian Lagrange points can only stably capture objects if a few conditions are met

a) when the mass ratio of the two objects which form the Lagrange point is large enough
b) when the velocity of the trapped object relative to the Lagrange point is small enough.

For instance, if someone tried to park a spaceship in the Lagrange point of a binary star system with a 1:1 mass ratio, it wouldn't work, the Lagrange orbit wouldn't be stable.

(Interestingly enough, there's an old and rather good short SF story based on this, called "Trojan Fall").

If you formed your black hole system with a large enough mass ratio so it had a stable Newtonian Lagrange point, it could still only capture slow moving objects. And light isn't a slow moving object, in fact you could say it moves as fast as anything possibly can.


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