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what is the observed distance range of gravity?
14 years 9 months ago #23549
by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
A caution to anyone who plays with the
free gravity simulator
: You can run the simulations as fast as you like. You can watch planets zipping around like a swarm of bee's. Unfortunately, if you run it too fast, the accuracy goes out the window. If you're simulating our solar system, you can eject Mercury and Venus from the galaxy by running the simulation too fast.
Fractal Foam Model of Universes: Creator
Fractal Foam Model of Universes: Creator
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14 years 9 months ago #23552
by Stoat
Replied by Stoat on topic Reply from Robert Turner
I never did like Mercury and Venus anyway[][8D] I've spent the last hour going crazy looking at a couple of exponential graphs. The main reason why I always try to get Joe Keller to do all the work.
I was looking at MOND. Say we've got c^2 / b^2 b being the speed of gravity. Our answer will be a tiny number. I reckon that it will be 6.626E-34 but we should try as many numbers as we like.
We can write that as 1/eta^2 which will be the reciprocal of 6.626E-34 that's 1.509E 33
With MOND we have F = GMm / r^2 = m* mu(a/a_0)a
We can lose that lower case m which is the mass of a star. The upper case M is the mass of the galaxy.
Milgrom takes a_0 as being 1.2E-10 which he says is the acceleration
a = v^2 / r (here v is the speed of light and r is the age of the universe) Yeah, the universe has to be about four and a half times bigger with this but we'll leave that for now.
We do the maths for the acceleration of a star at the edge of our galaxy find out what a / a_0 is, that comes out at 2.718 as near as damn it. Then we need to look at mu(x) = 1 - e^-x which is
1 - 1 / e^2.718
The answer is, 0.934
Now if we do the same for my speed of gravity the answer comes out at 0.9869
A Newtonian infinite speed of gravity will get closer to one.
Now I think that there's a phase change at the speed of light, which means we flip the sign in the Lorentzian but I'll leave it at that for now.
I was looking at MOND. Say we've got c^2 / b^2 b being the speed of gravity. Our answer will be a tiny number. I reckon that it will be 6.626E-34 but we should try as many numbers as we like.
We can write that as 1/eta^2 which will be the reciprocal of 6.626E-34 that's 1.509E 33
With MOND we have F = GMm / r^2 = m* mu(a/a_0)a
We can lose that lower case m which is the mass of a star. The upper case M is the mass of the galaxy.
Milgrom takes a_0 as being 1.2E-10 which he says is the acceleration
a = v^2 / r (here v is the speed of light and r is the age of the universe) Yeah, the universe has to be about four and a half times bigger with this but we'll leave that for now.
We do the maths for the acceleration of a star at the edge of our galaxy find out what a / a_0 is, that comes out at 2.718 as near as damn it. Then we need to look at mu(x) = 1 - e^-x which is
1 - 1 / e^2.718
The answer is, 0.934
Now if we do the same for my speed of gravity the answer comes out at 0.9869
A Newtonian infinite speed of gravity will get closer to one.
Now I think that there's a phase change at the speed of light, which means we flip the sign in the Lorentzian but I'll leave it at that for now.
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14 years 9 months ago #23553
by Stoat
Replied by Stoat on topic Reply from Robert Turner
My latest problem with this. How to account for this number 1.2E-10 that Milgrom has to introduce. With my speed of gravity, if we plug in about 300km per second and square it, then divide by the speed of gravity squared, we end up with rougly Milgrom's number to the fourth power. Then Milgrom's number is going to have its fourth root taken.
At the moment I'm toying with an argument of Robert Carroll's, that all matter has a space around it which is half of its energy. He worked out the fall of to infinity and found it to be a fourth power fall off.
With that idea I took the fourth root of my answer and it still came out as a big bang universe having to be just over four times larger than it's supposed to be. Milgrom doesn't seem bothered by this, and I'm not either. I suspect that Milgrom accepts the idea of a faster than light speed of gravity but isn't going to say that.
That number though, 1.2E-10 is still a loose end.
I suggest a look at the wikipedia article on MOND, as the equations are set out nicely, and equations on this board are often difficult to read. i was going to put in Carroll's maths but I haven't a clue how to write a long s for this board.
At the moment I'm toying with an argument of Robert Carroll's, that all matter has a space around it which is half of its energy. He worked out the fall of to infinity and found it to be a fourth power fall off.
With that idea I took the fourth root of my answer and it still came out as a big bang universe having to be just over four times larger than it's supposed to be. Milgrom doesn't seem bothered by this, and I'm not either. I suspect that Milgrom accepts the idea of a faster than light speed of gravity but isn't going to say that.
That number though, 1.2E-10 is still a loose end.
I suggest a look at the wikipedia article on MOND, as the equations are set out nicely, and equations on this board are often difficult to read. i was going to put in Carroll's maths but I haven't a clue how to write a long s for this board.
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14 years 9 months ago #23554
by Stoat
Replied by Stoat on topic Reply from Robert Turner
I suppose with this Milgrom thing, we're got to look at the curvature of curves. We've got a huge radius for the speed of gravity, and a small radius for the speed of light. Actually we can just swop them over, as informationally the speed of gravity radius is smaller than the speed of light radius.
The radius of the larger circle is so big that we may as well draw a straight line along the x axis. We draw a circle with its centre at zero on the y axis, resting on the x axis. A wheel on a road. To get Milgrom's number to come out right we'd have to move that wheel a little forward, or back. From Milgrom we've got
a = v^2/r where a = 1.2E-10
And he takes this to be the speed of light squared divided by the age of the universe. To get his age to come out right, we have to stick sine theta in.
a = v^2 sin theta / r theta
I make that about twelve and a half degrees.
The radius of the larger circle is so big that we may as well draw a straight line along the x axis. We draw a circle with its centre at zero on the y axis, resting on the x axis. A wheel on a road. To get Milgrom's number to come out right we'd have to move that wheel a little forward, or back. From Milgrom we've got
a = v^2/r where a = 1.2E-10
And he takes this to be the speed of light squared divided by the age of the universe. To get his age to come out right, we have to stick sine theta in.
a = v^2 sin theta / r theta
I make that about twelve and a half degrees.
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14 years 9 months ago #15208
by Stoat
Replied by Stoat on topic Reply from Robert Turner
The other Milgrom options for mu.
mu(x) = x(1 + x^2)^-0.5 or written the other way, mu(x) = x / sqrt(1 + x^2)
mu(x) = x / 1 + x
mu(x) = x(1 + x^2)^-0.5 or written the other way, mu(x) = x / sqrt(1 + x^2)
mu(x) = x / 1 + x
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14 years 8 months ago #23872
by Gregg
Replied by Gregg on topic Reply from Gregg Wilson
Putting theories and mathematics to the side, for a moment. The statement was made that gravitons cannot be measured. One might consider what magnetism is. If atomic structure is not "wide open" as proposed by Rutherford, Bohr, Pauling - then perhaps nuclear structure causes a structural change in the gravitational flux.
Gregg Wilson
Gregg Wilson
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