Solution to the Galaxy Rotation Problem, without Dark Matter

Discussion in 'Alternative Theories' started by Scott Myers, Feb 2, 2013.

1. Scott MyersNewbieRegistered Senior Member

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290
This is not to say I am estimating the mass OF the Black Hole in Andromeda, but by using (V=1,000,000 m/squared) X (R=5.271ly) divided by (G=6.67398 x10 -11th power) I am concluding. 1.021083770 Billion Solar masses. This is without any GTD equation adjustment,as yet for our first radius.

Anyone have time to check my calcs?

Here were the conversions
R= (5.271 ly= 4.98664452 X 10^16 meters)
V= (1000 km/s = 1000000 m/s)
G= (6.67398 x 10^-11)

Answer (747177024803790242104411460627691422509.kg =1.02108377 x 10^9 Solar Masses)

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3. I'm a bit burnt out today. I got up during the night and wrote up my insight about hidden velocity and hidden acceleration. Thanks those who replied for I do think we need to account for the energy loss as the stars migrate in toward the central bulge (there is going to be this visible kinetic energy and hidden kinetic energy.

Have we got a mechanism to account for the "hidden mass"? Here is my very tentative thoughts.
Mass plus "hidden mass" =>
Gravity plus "hidden gravity" =>
time dilation plus "hidden time dilation" =>
Velocity plus "hidden velocity" of the inner stars

But I really want to know how to account for it, for else how are we going to plan out the macros?
How does GTD hide this inner mass? Have you thought of a possible mechanism?

For if we say all this extra mass is in the central super-massive BH the velocity of the inner stars would need to be super-super fast. OK I say they are going super fast but only because there is two components to their velocity, the velocity that is measured plus the "hidden velocity" (hidden in their time dilation).
This extra mass in the central super-massive BH means it is easier to account for the speed of the outer most stars but not the inner ones. And then what happens to the ones in the middle region? Will the time dilation and the extra mass is in the central super-massive BH account for it?

How does this "Answer (747177024803790242104411460627691422509.kg =1.02108377 x 10^9 Solar Masses)" compare to the current estimate of "mass OF the Black Hole in Andromeda"? Were you close?

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5. No, the curve he drew in the diagram is pretty close to what we actually expect. If anything, it was the curve he drew for the sun and solar system that was a bit off. It should have had a sharper bend and dropped off faster. In both cases he was doing a rough sketch of the curves and not an actual plot. He was simply showing the general resemblance of the pattern, not making a detailed side by side comparison.

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7. Scott MyersNewbieRegistered Senior Member

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290
Yes, I agree OK, it's more of a representation to easily understand the discrepancies we need to account for with Dark Matter, or my initial postulate.

8. Your math works out in Kg but not solar masses.

It comes out to 3.75635727e 8 Solar masses.

9. Scott MyersNewbieRegistered Senior Member

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290
Well, I have 1.02 billion Solar Masses, but it seems we have nothing to compare it too. I solved for the entire mass within the radius remeber, of 5.27ly, so that is not in any way reflect what mass would be in the very central mass of the actual Black Hole. the current estimates I find published is 140 million Solar Masses, of the Black Hole. I have no idea what the total mass is, except that it is nearly the same as what I have come up with. We will need better data the farther we go here.

Again, there is no hidden, or apparent mass to account for, it is only that our mass estimates (I claim) are not right because we have not corrected the velocity, and it will increase with our GTD corrections. It's not missing, just not calculated right, with the proper speeds used in the equation, the proper mass will be deduced. You'll see that better if I can complete the next calc shortly. ASAP, I am populating the next equation and can compare it to the published numbers used for GTD on the Earth to be sure I am doing them properly.

10. So I have not looked specifically at Andromeda as yet, for I wanted to understand what we were dealing with. There is no question there is unaccounted for mass (I term this "missing mass"), and that is why Zwicky came up with Dark Matter. They have put that matter around the galaxy as a global halo, not just in the core. If you found by your calculations all this extra unaccounted for mass was in the core you might still have issues. (For your core looks 10 times more massive than the current estimate).

A statement like, quote Scott Myers: "Again, there is no hidden, or apparent mass to account for, it is only that our mass estimates (I claim) are not right"; to me this is the very definition of "missing mass".
Whatever the reasons are the current calculations rely on DM. If you get rid of the DM, you then have "unaccounted for mass" or "missing mass".

11. Scott MyersNewbieRegistered Senior Member

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290
So it that just 376 million Solar masses withing the 5.27 Light Year radius of Andromeda, correct? The Black hole iteself, is going to be extremely tiny by radius. Not anywhere near 1 light year, so I have no way of checking this. The nearest velocities we have observed are these blue stars as 1000 km/s.

Thanks much!

12. Where are you getting your data from? What device do you use to get so many figures in your answer? Is that 39 - 40 digits is it? Don't expect that from a macros. Read about the rotational speed of the stars around the core on Andromeda they aren't that flat. Is it a good example to work with? When they describe the flat rotational speed of galaxies which one do they cite as a good example?

13. Scott MyersNewbieRegistered Senior Member

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290

This publication has some nice comparison charts and good data, that we may be able to make good use of. That's just one source, there are others of course.
Calculator here: http://web2.0calc.com/

14. Where is "This publication has some nice comparison charts and good data, that we may be able to make good use of"?

15. These are the variables I need defining: from http://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere

Dim T0 As Double '*is the proper time between events A and B for a slow-ticking observer within the gravitational field,
[This will be the one we could find???]

Dim Tf As Double '*is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object (this assumes the fast-ticking observer is using*Schwarzschild coordinates, a coordinate
[What do you use here?]

Dim M As Double '*is the*mass*of the object creating the gravitational field,
[What do you use here?]

Dim G As Double '*is the*gravitational constant,
[You have already used this constant] G = 6.67398 * 10 ^ -11 [where did you get this from?]

Dim r As Double '*is the radial coordinate of the observer (which is analogous to the classical distance from the center of the object, but is actually a Schwarzschild coordinate),
[What do you use here?]

Dim c As Double 'speed of light
[You have already used this constant] c = 299792458 m/sec [same as you?]

Dim r0 As Double '*is the*Schwarzschild radius*of M. r0=2GM/c^2

Which formula are you using to need V ? "V= (1000 km/s = 1000000 m/s)"

16. Have you been able to check your calculations against these 2 examples and their answers. I did and my answer for the Earth was correct but the Sun was out by a little but not much.
"To illustrate then, a clock on the surface of the Earth (assuming it does not rotate) will accumulate around 0.0219 seconds less than a distant observer over a period of one year. In comparison, a clock on the surface of the sun will accumulate around 66.4 seconds less in a year."

17. How much time will be lost in a year standing on a planet in the middle of Andromeda?

It seems odd that the Earth and the Sun are not severely affected by time dilation as we are also inside a huge massive galaxy? Somehow I think we are treating the Solar System as the zero point rather than some really distant point in the Universe.
Or if we are really considering Tf * to be a far distant point there is virtually no effect from gravitational time dilation, in the last 1/3 of the outer parts of the galaxy. Not enough to account for a flat rotation speed.
* Tf as in "Dim Tf As Double '*is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object (this assumes the fast-ticking observer is using*Schwarzschild coordinates, a coordinate "

18. Scott MyersNewbieRegistered Senior Member

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290
If you resolved properly for Earth What did you usfor T? For the examples. That was my difficluly...
Plug in the R/G/M/C(2) which 89875517873681764 km/s and then the correct unit of time, and we have our first correction for Time, that we can apply to our intitial velocity.

so: USE

R= (5.271 ly= 4.98664452 X 10^16 meters)
G= (6.67398 x 10^-11)
C= (299 792 458 m / s )
M=(7.47177024803790242104411460627691422509.kg X 10^38

Drop as many characters as you need to fit the digits in to your calculator, for mass, but the closer the better of course.

We have a good starting point, but I will find better data as we continue outward. This first one is only a good test of the overall effects we might see.

19. Using the upper estimate for our galaxy's mass and the lower estimate for the Sun's orbital distance, and assuming that all of that mass was inside the Sun's orbit (this fudges things to give the maximum time dilation), it still only works out to a time dilation factor for the the solar system that works out to 0.99999644. That's the equivalent of under 2 min a year.

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21. Scott MyersNewbieRegistered Senior Member

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290
The formula I'm using V in is just the initial mass calc... using thee nearest published speeds I could find, closest to the center, of about 5ly.

22. Scott MyersNewbieRegistered Senior Member

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290
From 2005, here is some of the data gleened to determine some of the initial data points. You'll see that we use Parsecs, so we will have to get used to translating parsecs to meters, to use for our mass calcs, but we don't need to build the entire galaxy in a day...

http://iopscience.iop.org/0004-637X/631/1/280/pdf/61873.web.pdf

What we need is to do thye Time Dilation in an agreeable way, for this first initial dataset, so we can all be on the same page with the process I am trying to use here.

Our initial parameters for this first 5ly is near enough for us to determine how to proceed, or not to proceed at all.

We have our first mass, radius, G/ C and the Time dilation equation that needs to be solved by us to check our math. We will then take our next step. We will then adjust our new proper time to correct the velocity, then calc the mass one more time using our initial mass equation. Like I did the first time. "(V=1,000,000 m/squared) X (R=5.271ly) divided by (G=6.67398 x10 -11th power) "

Then we stop there for a moment, because there is an error that could be made, putting our calcs into an infinite loop. I'll explain once we get that far.

Time Dilation solution for the parameters given is the next step. I'll try and solve, but as stated, I can't figure out the correct time unit i.e. (between events Av & B) to use. If you guys used, and solved the Earth, or Sun examples given, what did you use? That way I can also verify, and be sure I am usiong the equation correctly as well.

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Cheers and happy mathing

23. I have begun writing the macro to ultimately so we can find out where the adjustments need to be made (non-looping yet) At the moment I'm using
T0 = Tf * (1 - (2 * G * M / (r * c ^ 2))) ^ 0.5 as my only formula.
I set tf = 1 the imaginary place where 1 sec passing in 1 sec for it seems that any calculation of time is going to be affect by the time dilation in our own region of the Milky Way Galaxy.

The output is a number like 0.99999644 so I would say you and I are using the same formula.
Can you share with us the values (5 significant figures only please) that you used for the Sun in your example please?