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mercredi 17 septembre 2014

"Why do mainstream astrophysicists refuse to admit that dark matter does not exist and that Vera Rubin used the wrong curve according to Newton's equations?" An enemy of materialism Tony Nguyen tried to call me a liar


by Yanick Toutain
17/9/14


"you're obviously lying or mistaken about your simulation as my previous reply to another comment of yours explained." Tony Nguyen

Dialogue of the deaf with an anti-materialist in bad faith


On Quora I asked

Why do mainstream astrophysicists refuse to admit that dark matter does not exist and that Vera Rubin used the wrong curve according to Newton's equations?

When I understood that Vera Rubin was wrong and that the curve of the speeds of the stars of a galaxy she expected to find was a absurd mistake, when I checked the speed curve in a galaxy (NGC 1566 and many others ) was ALWAYS an increasing curves and I published two texts and YouTube video, tell me for what mysterious reasons, in March-April 2012 and later, supporters of the dark matter acted ...
 (more) 
Write your answer, or answer later.
1 ANSWER
Thanks for the ask to answer, Panagiotes Koutelidakes. For anyone wondering, I accidentally declined.

You do realize that Vera observed the galaxies with a new sensitive spectrograph and found out that and announced that the curve wasincreasing. You're not in disagreement with her, you're in disagreement with astrophysics that use dark matter in their theoretical framework.

F=G*m1*m2/(r^2)

F stands for force. Force can be equal to m*a or mass times acceleration. 

G stands for the gravitational constant which is about 6.67×10^−11 N·(m/kg)^2

N stands for Newtons which is equal to 1kg·m/(s^2)

Acceleration is equal to distance divided by time squared. Meter (m) is a measurement for distance. Seconds (s) is a measurement for time.

Here's simple inverse equation.

Let x = 5/t
If t = 1, then x = 5
If t = 5, then x = 1

If you were to increase r (the distance between the centres of the masses), then F would be smaller, not bigger.

In case if you haven't figured it out already, the equation at the top of this answer is known as Newton's Law of Universal Gravitation.

If objects really do orbit faster the farther they are from each other, then how come F is inverse to r? You claim to use Newton's calculus, but if you did, you would've encountered his law of gravitation.

Edit: Forgot to address his simulation.

Not true at all by Tony Nguyen on Referential

Second Edit: He's not willing to admit that he's wrong about something, but I am. We're both wrong, just in different ways. We were also almost right, but in different ways.

I manually counted the number of all connections in a network with 4 nodes. 

A-B, A-C, A-D, B-C, B-D, C-D

Multiply 6 by two (for every action, there's an equal and opposite reaction) and you get 12.

Mine: 4! = 24
His: 2*4*(4-1) = 24

Neither of these are 12.

However.

4!/(4-2)! = 12
4*(4-1) = 12

There are 20 force interactions with five objects with mass. Add E for A, B, C, D multiply by two, then add that to 12.

5!/(5-2)! = 20
5 * (5-1) = 20

However, would it still be possible to simulate a 200 body problem?

200!/(200-2)! = 39800

That's the number of operations required to render a single frame of a hypothetical spiral galaxy. Another frame requires another 39800 operations. So it seems that it could be simulated given our current computational capacity.

Does that mean he's right about dark matter not existing? He could be, or not. I think we're gonna need more math proofing, simulations, and observations of actual galaxies to verify his findings. Since it isn't science if it isn't reproducible and verifiable. Thanks for the experience, Yanick. I found it enlightening.

This reminds me about superstring theory. There are many different approaches that can be simultaneously valid. They're just different ways of looking at things. That's to be expected given that there's always uncertainty. It is possible to unify theories just as M-theory has done with all accepted superstring theories.
  
Yanick Toutain
By following the link you would have seen this 
++++
The following animation 
has been realized with
the monochromic 
photography of a galaxy.
The source of the colored image
is Wikipedia's propandist entry
in favor of the swindle 
of dark matter
It's sufficient to use 
Newton's basis equations
GM X / R^3 et GM Y / R^3 
to obtain 
the gravitational deviation
received by each star 
from the totality of 
the others stars of the galaxy.

Then, the equality 
speed = SQR (acceleration * radius) 
enabled to give a good evaluation
of the revolution speed of 
all the stars (high speeds symbolized
in red then yellow , 
middle speeds in green, 
small in blue then violet)
It's easy to see that the 
most red zones - quickest - 
are in the periphery
Tony Nguyen You expect me to believe you managed to simulate something with about a BILLION objects at the same time all influencing each other which implies a FACTORIAL of a billion force interactions (A number with 8.5 billion DIGITS). IBM100, a super computer, can perform about 1 thousand trillion operations per second. It would still take ages for a modern supercomputer to simulate what you supposedly claimed to simulate.
Yanick Toutain
Start with a small galaxy (200 stars). Increase the number of stars. Gradually, nothing really changes.
A small galaxy acts as a very large: the SPEED BIG are ALWAYS the periphery !!!
Tony Nguyen 200! = 7.8865786736479050355236321393218506229513597768 × 10^374

30 quadrillion is a number with 17 digits. That doesn't come anywhere near 374 digits.

200! / 3*10^17 = 2.6288595578826350118412107131072835409837865922 × 10^358

It would take our supercomputers that many seconds to render a single frame of your simulation with 200 stars and all 200! force interactions.

200! / 3*10^17 / 60/24/365/1000 = 5.0016353840993816815852563034765668587971586608 × 10^349

That's the same duration except it's measured in millenniums.
Yanick Toutain
II do not see that the factorials to come here!
Do you understand that ONE STAR in a galaxy undergoes gravitational attraction of ALL others. And so it is for a galaxy formed "N stars" to find the value of "N-1 vectors centripetal acceleration."
So twice since I need accelerations abcisses and ordinates to make the sum then.
If I want to study the value of the gravitational acceleration exerted by 199 stars on one .... it 199 calculations for X and 199 for Y calculations I need (if this galaxy is kind enough to stay fit almost flat disc)
I will calculate 199 times -GM * X / R ^ 3
(with X (or Y) of the relative value of each star coordinate (relative to the other) and the distance R between them)
As the same calculation, I'll do it for 199 other stars, so it will take me 200 times 199 calculations (2 times for X and Y)
200 * 199 * 2
That is to say 2 * N * (N-1)
or 2 * (N ^ 2 - N)
Je ne vois pas ce que les factorielles viennent faire ici !
Avez-vous bien compris que UNE ETOILE dans une galaxie subit l'attraction gravitationnelle de TOUTES les autres. Et donc qu'il s'agit pour une galaxie formée de "N étoiles" de trouver la valeur de "N-1 vecteurs accélération centripète".
Et donc le double puisque j'ai besoin des accélérations en abcisses et en ordonnées pour en faire la somme ensuite.
Si je veux étudier la valeur de l'accélération gravitationnelle exercée par 199 étoiles sur une seule.... il me faut 199 calculs pour X et 199 calculs pour Y (si cette galaxie a la gentillesse de rester en forme de disque quasi plat)
Je vais calculer 199 fois -GM *X / R^3
(avec X (ou Y) la valeur de la coordonnée relative de chaque étoile (par rapport à l'autre) et R la distance qui les sépare)
Comme ce même calcul, je vais le faire pour les 199 autres étoiles, il me faudra donc 200 fois 199 calculs (fois 2 pour les X et Y)
200*199*2
C'est à dire 2 * N*(N-1)
ou encore 2* (N^2 - N)
Tony Nguyen Here's why they are relevant. There are two force interactions between  the Earth and the moon, not one. The moon and the Earth orbit around a  barycenter. Since Earth has greater mass, the barycenter is closer to  the centre of the Earth. 2 factorial (also written as 2!) is equal to 2*1.  Add the sun and you have 3 objects with 3! (3*2*1=6) force interactions. 6 if you count the two between Earth and moon, two more between Earth and sun, and two more between moon and sun.
Yanick Toutain
The trajectory of the Earth is modified by two bodies (Sun and Moon) 
Both X and Y components are calculated 
The trajectory of the Moon is modified by two bodies (the Sun and Earth) 
Both X and Y components are calculated 
The trajectory of the Sun (a little) modified by 2 bodies (Moon and Earth) 
Both X and Y components are calculated 
There is therefore 6 calculations in X and Y 

With body 3 must therefore 
2 * N * (N-1) calculations 
2 * 3 * 2 = 12 calculations 
This is exactly what I explained above! 
79600 calculations for 200 stars in a mini-galaxy! 
For 10,000 stars that amounts to almost 200 million calculations. 
(199980000)
Tony Nguyen Actually, 2 * 3 * (3-1) = 12. 12 is not 6.

It doesn't even calculate the correct number of force interactions between objects.

Your equation isn't even a calculus equation, it's basic arithmetic.

4! = 4*3*2*1 = 24
5! = 5*(4!) = 120

The number of force interactions increase exponentially the more objects there are.
Yanick Toutain
There is therefore 6 calculations in X and Y 
6 FOR X
6 FOR Y
EQUAL 12
Yanick Toutain
Your answer - which is not a - explains why it seems that no one has bothered to actually VERIFY by modeling a galaxy whose stars each suffer a gravitational pull from all others. (- GM * X / R ^ 3 and - GM * Y / R ^ 3) 
Modesty is the door to wisdom. 
So write a program to do a galaxy and THEN turn we can see WHAT LINE PROGRAMMING we disagree
Tony Nguyen I don't even need to bother verifying the calculus since you're obviously lying or mistaken about your simulation as my previous reply to another comment of yours explained.
Yanick Toutain
" since you're obviously lying or mistaken about your simulation "
YOU ACCUSE ME TO BE A LIAR ?
Tony Nguyen "...or mistaken..." Please don't cherry pick.
Yanick Toutain
You assumed I could lie ..... because you persist in using factorial in a case where it is totally irrelevant! 
I tried to explain in detail ... but you do not seem to really read my sentences .... 
Just one thing 
1° I know programming since 1981 
2 The program to calculate the acceleration vectors I made it ​​in 2012 and I saw it run at idle for debugging and fast .... 
3 I saw the stars light up one by one in different colors depending on the value of the acceleration vector. 
IF you knew programming, long ago you would have realized it was two loops nested FOR NEXT.
Vous avez supposé que je pourrais mentir parce que..... vous vous acharnez à utiliser des factorielles dans un cas où elle sont totalement hors sujet !
J'ai tenté de vous expliquer en détail... mais vous semblez ne pas vraiment lire mes phrases....
Juste une chose
1° Je sais programmer depuis 1981
2° Le programme pour calculer les vecteurs accélérations je l'ai réalisé en 2012 et  je l'ai vu fonctionner au ralenti pour debugger et en accéléré....
3° J'ai vu les étoiles s'allumer une par une de différentes couleurs en fonction de la valeur des vecteurs accélération.
SI vous saviez programmer, il y a longtemps que vous auriez compris qu'il s'agissait de deux boucles FOR NEXT emboitées.
Tony Nguyen Both are possibilities, that doesn't necessarily make it so.

Yes I do know about loops. I do know that they are automatic. I do know that you wouldn't live long enough to see the day a 200-body simulation complete.
Tony Nguyen Yes, you already told you can you program. So what?
Tony Nguyen Oh I see. You think you managed to complete a 200-body simulation. Remember that there are two calculations between each pair combinations. That equals to 200 factorial. Your equation 2*N*(N-1) isn't as exponential, it's increases very similarly to a parabola opening upwards.
Yanick Toutain
Needs to Be Phrased as a Question: This question needs to be re-worded as a fully formed question

The 1 answer is a good answer. I can see that my oponent don't try to verify the calculus. As Vera Rubin.
THEN MY QUESTION was a good question !
Tony Nguyen Your question and its description are biased and needs to be read by a proof reader.
Yanick Toutain
Neither Google translate nor me know what is a "proof reader".... could you explain with more words what is a "proof reader" THANS
Tony Nguyen A proof reader is someone who goes over a piece of text to spot errors in grammar, style, syntax, etc.
Yanick Toutain
Thanks ! 
A proof reader would have been someone who would read your response on the factorials and you would have said they had nothing to do in this debate? 
Is that right?
Un proof reader aurait été quelqu'un qui aurait lu votre réponse sur les factorielles et vous aurait dit qu'elles n'avaient rien à faire dans ce débat ?
C'est cela ?
http://www.quora.com/Dark-matter...
Tony Nguyen You didn't bother asking what they were for. You just assumed they were irrelevant because of your confirmation bias.

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