# Proof that Gravitational Constant is not constant

Discussion in 'Pseudoscience Archive' started by Robittybob1, Oct 12, 2012.

1. ### Robittybob1BannedBanned

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From the data for the Hulse Taylor binary star system I have shown extensively on Physforum thread "Gravity Waves" thread that the value of G in the region of the star system is less than here in this Solar System. http://www.physforum.com/index.php?showtopic=41123

3. ### AlphaNumericFully ionizedModerator

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I'm not going to wade through 27 pages of crap. Please point to the precise post where you go through the experimental data and show that the observed value of G is outside of experimental error bounds with a high confidence (I'll accept 3 sigma but would prefer 5 sigma).

Actually, sod that. Please link to the paper you are surely in the process of sending to a reputable astrophysics journal where you plan for it to sail through peer review and set the physics world alight. I hope it isn't written in Word....

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wysiwyg

7. ### Robittybob1BannedBanned

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S-M Axis appears to be between 1948672594 and 1948672574 meters with the G being measured at 6.66735E-11 (Earth 6.67384E-11) so it is 6.49E-14 less than here.

Now to see if these figures which are based solely from the Keplerian orbital period formula hold true for the other equations.

dr/dt = 3.56109999 m/y

If you apply the standard S-M Axis and G values you cannot get sensible results that match the measured orbital period.
These are the best fit result so far.
With less than 20 meters of variability in the S-M axis left there is no way the G value will change. It therefore is less that our locality in the Universe.

8. ### rpennerFully WiredStaff Member

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I think Robittybob1 is sourcing http://arxiv.org/abs/astro-ph/0407149
$\begin{array}{ll} \textrm{Parameter} & \textrm{Value} \\ \hline \\ a_p \, \sin \theta_i & 2.3417725 \pm 0.0000008 \, c \cdot \textrm{s} \\ e & 0.6171338 \pm 0.0000004 \\ P_b & 0.322997448930 \pm 0.000000000004 \, \textrm{d} \\ \dot{P}_{b,Obs} & ( -2.4184 \pm 0.0009 ) \times 10^{-12} \\ \hline \\ m_p & 1.4414 \pm 0.0002 M_{\odot} \\ m_c & 1.3867 \pm 0.0002 M_{\odot} \\ \hline \\ c & 299792458 \textrm{m} \cdot \textrm{s}^{\tiny -1} \\ G & ( 6.6738 \pm 0.0008 ) \times 10^{-11} \textrm{m}^{\tiny 3} \cdot \textrm{kg}^{\tiny -1} \cdot \textrm{s}^{\tiny -2} \\ M_{\odot} & ( 1.9885 \pm 0.0002) \times 10^{30} \textrm{kg} \\ \frac{2 G M_{\odot}}{c^2} & 2953.2500770 \pm 0.0000002 \textrm{m} \\ \hline \\ \dot{P}_{b,Gal} & ( -0.0128 \pm 0.0050 ) \times 10^{-12} \end{array}$
where the masses of the stars are derived from the primary observational values, and the last term relates to the geometry of the line-of-sight.

$a_p$ is the semimajor axis of the pulsar's orbit about the center of mass
$\theta_i$ is the angle of inclination relative to the line of sight
$e$ is the eccentricity of the orbit
$P_b$ is the period
$\dot{P}_{b,Obs}$ is the rate of change of the period
$m_p$ is the mass of the pulsar
$m_c$ is the mass of the companion star
$\dot{P}_{b,Gal}$ is the observational effect on the period due to the geometry of the light of sight
$c$ is the speed of light, which is exact in the SI units
$G$ is the gravitational constant
$M_{\odot}$ is the mass of the sun

Note that $2 G M_{\odot} / c^2$ is known to much better precision than either $G$ or $M_{\odot}$.

The contention of the paper is that
$\dot{P}_{b,Obs} - \dot{P}_{b,Gal} = \dot{P}_{b,GR} = - \frac{ 4 G^2 m_p m_c }{c^4} \sqrt[3]{\frac{\pi^8}{P_b^5 c^2 (2 G m_p/c^2 + 2 G m_c/c^2)}} \frac{192 + 584 e^2 + 74 e^4}{ 5 c \sqrt{(1-e^2)^7}} = -( 1.4414 \pm 0.0002 )(1.3867 \pm 0.0002)( 2953.2500770 \pm 0.0000002 \textrm{m} )^2 \quad \quad \times \sqrt[3]{\frac{\pi^8}{(0.322997448930 \pm 0.000000000004 \, \textrm{d} )^5 (299792458^2 \textrm{m}^2 \cdot \textrm{s}^{\tiny -2}) (2953.2500770 \pm 0.0000002 \textrm{m} ) (( 1.4414 \pm 0.0002 )+(1.3867 \pm 0.0002))}} \quad \quad \times \frac{192 + 584 (0.6171338 \pm 0.0000004)^2 + 74 (0.6171338 \pm 0.0000004)^4}{ 5 (299792458 \textrm{m} \cdot \textrm{s}^{\tiny -1}) \sqrt{(1-(0.6171338 \pm 0.0000004)^2)^7}} \quad \quad \times \sqrt[3]{\frac{ (1 \textrm{d})^5 }{ (86400 \textrm{s} )^5}} = ( -2.4020 \pm 0.0004) \times 10^{-12}$

That the authors use a smaller error quite possibly indicates they know that the errors of the estimated masses are anti-correlated in some way. But because the errors in observation are swamped by uncertainty in knowledge of the geometry of the line of sight, any attempt to figure out G from these observations will be affected by these errors which Robittybob1 seems to ignore as well as not communicate his procedures very well.

See http://adsabs.harvard.edu/full/1982ApJ...253..908T for the mass formulas which properly should be inserted into the above relation.

9. ### originTrump is the best argument against a democracy.Valued Senior Member

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That is along this lines of my initial thought at robbitybobbity's post which was; he needs to understand the concept of significan digits!

10. ### Robittybob1BannedBanned

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Thanks R, I'll look at those links. So when I try and use limits in my macros I'll need to run them at the higher and lower values as well to see if that makes any significant differences to my estimates.

11. ### originTrump is the best argument against a democracy.Valued Senior Member

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Out of curiosity what is the exact equation and values you are using?

12. ### Robittybob1BannedBanned

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The standard Keplerian equation for elliptical orbits.

T = 2 * 3.14159265358979 * (a ^ 3 / (GB * (m1 + m2))) ^ (0.5)
Pi = 3.14159265358979
a = the yet to be defined semi-major axis (rough estimates have been made)
GB is the unknown gravitational constant (probably lower than G = 6.67384
M1 and m2 are the masses of the stars

You have to choose both SMA and GB levels so these two following limits apply:
1. T = 27906.97959 period in seconds '(roughly 7.75*60*60)
2. T2 = T - 0.0000765 Which means the yearly change in orbital period is 0.0000765 seconds shorter the next year.

13. ### AlphaNumericFully ionizedModerator

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The problem being GR doesn't give them, it gives something very close to them. This is seen by the precession of Mercury, which Newtonian mechanics (which leads to Keplerian orbits) is unable to explain when you measure precisely enough. Since high precision experiments is precisely what is needed to detect any variations in G you're assuming non-zero non-insignificant effects are zero by working with such equations. When trying to measure some quantity to a particular level of accuracy it is essential you include in your calculations all possible contributions which are large compared to the precision.

For example, suppose you wanted to compute the circumference of a circle and you have a ruler which is accurate to 1mm. You measure the diameter of the circle and measure it to be 100mm. What that means if you know the diameter is between 99mm and 101 mm. So what's the circumference? Well if you ask a computer to give you the value of $100\pi$ it'll return 314.159265359... . However, you cannot quote that value as a justified answer. If you only measured the input to 3 significant figures you can only quote the output to 3 significant figures at best, ie the circumference you can conclude is about 314mm, to within an error of plus/minus $\pi$ mm, as you've multiplied up your error too.

Showing a disregard for how to incorporate errors and precision and significant figures into ones work is a sign of a bad scientist. In your case it undermines your claim completely that you're not accounting for relevant effects.

14. ### brucepValued Senior Member

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He thinks the emission of gravitational radiation, for the PSR1916+13 binary system, is gradually radiating away 'the constant G' rather than orbital energy.

15. ### Robittybob1BannedBanned

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You don't need to put words in my mouth. I will come to some conclusion sooner or later. Currently they say gravitational radiation makes the orbiting bodies lose orbital energy. So if in the same process it loses some G, whatever that is, so what.
But first things first and that means I need to look at the limits of certainty (uncertainty) first.

16. ### OnlyMeValued Senior Member

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That a a very clear and simple explanation, which means its educational value is high.

Good post.

17. ### Robittybob1BannedBanned

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I am using the the formula T = 2 * 3.14159265358979 * (a ^ 3 / (GB * (m1 + m2))) ^ (0.5)
Pi = 3.14159265358979
a = the yet to be defined semi-major axis (rough estimates have been made)
GB is the unknown gravitational constant (probably lower than G = 6.67384
M1 and m2 are the masses of the stars.
OK so what in there needs rounding or whatever due to uncertainty?
T = 2 * 3.14159265358979 * (a ^ 3 / (GB * (m1 + m2))) ^ (0.5)
The period is measured to such a degree they can tell it shortens by 0.0000765 seconds so there must be some uncertainty in that.
2 = 2 , pi to 14 decimal places so that can stay as it is.
we are finding "a" and GB so that can fine tuned to many places
The individual mass of the Stars is not as important as the combined mass. So we will look at the uncertainty here but the mass is so large even if I brought the mass up or down by the uncertainty It won't have much effect on the answer, but we will test it.
^0.5 no uncertainty there either.

You have to choose both a SMA and a G level so these two following limits apply:
1. T = 27906.97959 period in seconds '(roughly 7.75*60*60)
2. T2 = T - 0.0000765 Which means the yearly change in orbital period is 0.0000765 seconds shorter the next year.

18. ### Robittybob1BannedBanned

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I'm able to select the right "G" and semi major axis "a" values that satisfy both that equation and those restraints and there is only 1 value so far. Even though that one value has changed over the weekend as I have corrected errors. So all numbers G and a can be selected to 14 decimal places. Right over the next week I'll work out how to allow for the uncertainty in the masses, the known orbital period and the annual change in the period.
But I will guarantee that the first 3 numbers of G will not be the same 6.66..... for the binary and 6.67.... for the Solar System. But by the end of the week it won't just be my hunch but the figures will be here to show it.

19. ### AlphaNumericFully ionizedModerator

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Did you think about what I said? I specifically addressed your use of Keplerian orbits! Such a formula is not valid at high precision for multiple reasons.

When Kepler and Newton worked out those results they modelled the system as a point object orbiting around a fixed star under Newtonian gravity (Kepler didn't use Newtonian gravity but the results he got follow precisely from it, ie Newton explained Kepler). There are at least 3 flaws with such an assumption when it comes to doing ultra high precision measurements of real world systems.

1. Beyond Newtonian effect. Newtonian gravity is not perfect, it is measurably inaccurate at high precision as illustrated by the precession of Mercury. An orbit which Newton might say is closed Einstein may say is not.
2. Back reactions. In a two body gravitational system the smaller body doesn't orbit the centre of the larger body but rather they both orbit the barycentre, the centre of mass for the system. As such you cannot treat the main star in a system as fixed if you measure to high precision. This is observed in the Solar System, with the Sun 'wobbling' because of the interaction with the planets (mostly Jupiter). In fact it is this wobble which is used to detect large planets in other solar systems.
3. Multi-body effects. Not only is the 2 body problem not as your idealised equation assume but the interaction from multiple planets and other stars causes deviations in orbits. This is why the 3 body problem is so hard, the interactions perturb the system from the nice closed form that Keplerian orbits assume, resulting in wildly different dynamics when measured over long periods of time.

All of these effects are observed in the Solar System and even in the galaxy at large. Assuming Keplerian orbits for the Earth would be wrong because the Earth-Moon system complicates the Earth's orbit about the Sun, which itself is moved about by Jupiter and the other planets, all of whom affect the Earth and Moon directly. It's a very complicated highly coupled chaotic system. Sure, if you want to send a rocket into orbit you don't need to consider this but you do if you want to send a probe to Pluto or do high precision measurements for the GPS network, which also form the basis of experiments to do high precision measurements of G.

Until you're willing (or rather able) to model all of these effects and have accurate measurements of all of their contributions to the dynamics of a gravitational system you cannot measure G in the manner you're trying to with any level of precision. There are better ways to measure G, such as using the GPS network, as G arises in general relativity too and appears in the equations for calculating positions using GPS trackers. But there you still need to do lots of corrections to the naive first approximation. For example, GPS trackers can quickly narrow down your position to a few feet but to get to a precision of centimetres you need to have additional information about the local area, such as gravitational anomalies introduced by denser rock or presence of the ocean or whether you're on a mountain (surrounded by more air than usual), that sort of thing. Keplerian orbits are nice and simple, simple enough to be taught to children, but they only go so far. Knowing the bounds of applicability of a model is something a good scientist should understand because applying a model in a regime where some of the assumptions of the model are violated is going to lead to nonsense results and flawed conclusions.

How rough? If you can only estimate it to within say 10% then you can only measure G to within about 10%, as I just explained. Since a deviation of 10% in the value of G would be blatently obvious and we don't see such a thing even ignoring all the issues with using Keplerian orbits I just went through you're still going to fail to reach any valid conclusion. If you need to measure G to parts per million then you're going to need to measure all of the inputs to your model to at least an order of magnitude higher precision.

Which are only known to a few % accuracy usually, as their masses must be inferred indirectly via their motion and interactions.

Everything Whenever making a prediction about some quantity using an equation and experimental values a good scientific paper will list the uncertainty for every input and then compute the uncertainty in the output resulting from that. It's why you see things like x=1.2345678(32) in papers and on Wikipedia's physics related pages. It means some quantity x is known to 8 significant figures but there's an uncertainty of 32 in the last 2 digits, ie 32 parts out of 12345678.

You clearly have no idea how to deal with uncertainty. If the uncertainty is 10% then regardless of the actual value the true value could still be anywhere from 90% to 110% of that. Have you even looked up how the masses of stars is computed?

It's quite apparent you aren't familiar with the necessary science, neither in terms of uncertainty quantification nor the appropriate models to actually compute the uncertainty from. You give the distinct impression you've have some qualitative arm wavey idea, you've gone onto Google, put in some search words related to gravity, found a page which quotes a formula you don't understand the context or applicability of and are mindlessly trying to apply it to the idea you've had. This is not going to go anywhere but down the drain.

20. ### Robittybob1BannedBanned

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There is no possibility of using GPS on this star system.
Rough estimates have been made previously but with my method the SMA is a chosen value and you test that against the formula, it then tests multiple levels of G to see which one if any will satisfy the Orbital Period formula. Reintroduce the resulting G and "a" values later and the orbital period is as precise as you like. Using the macro now and using the full and best estimates by Weisberg and Taylor I can get a value that will return an answer down to the last fraction of a mm. since I am not using rounded values any longer but ones with the full precision. There is only one solution to the equation and the restraints. Whether the problem is the precession, I can't comment as yet, but your suggestions will certainly be looked at especially now since all the last bugs are being ironed out.
Now I am going back over the other equations to see if the "new" values hold.

There are lot of parameters that have been measured on the binary and I'll probably never get to know how it was done fully , but the results aren't being challenged seriously so we can just take their values and hope for the best. I've got a physics PhD graduate working with me who is looking into the whole thing off the forums.

21. ### AlphaNumericFully ionizedModerator

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Clearly you're utterly failing to grasp my point. GPS networks require extremely accurate models, if you used Newtonian mechanics you'd predict incorrect time effects. They also serve as an experimental test for measuring the value of G to a good precision, provided you use the right models.

How many times do I need to tell you that the orbital formula you're using is wrong? This is the third time now. Either you're extremely bad at reading or you're deliberately ignoring me, neither of which bode well for your plans.

No, they aren't. You can only estimate them and the error in that estimate is non-trivial, even ignoring how the formula itself isn't accurate enough.

You're seriously claiming you're working with an orbital size accurate to a fraction of 1 millimetre? Now you're showing you're just plain daft. No one who understands the experimental measurements involved or even the basic physics of the systems in question will see you cannot get such an accurate measure of the value of a. What's the value for a for the Earth-Sun system? Where do you measure the distance to start and finish?

Then there's the fact the closed orbit associated to the formula you're using doesn't exist due to non-ideal things in reality, like stars and planets not being point particles, not being perfectly spherical, not moving along perfect Keplerian paths, being perturbed by other planets and stars, affecting one another and moving in a non-inertial manner about the centre of the galaxy. There are tons of things you're ignoring, not just in your use of the formula you're talking about but even in how to work out what the values in it are measured as.

Claiming sub millimetre precision is so blatantly ridiculous you're either deliberately trolling or have a terrible grasp of basic physics or engineering principles that a child would be embarrassed to have.

You obviously don't even understand my 'suggestions'.

You are obviously just blindly applying formulae and models you don't understand.

Speaking as a physics PhD graduand I can tell you now (and I have told you!) you're making a slew of mistakes and ignoring a multitude of relevant things which will result in your conclusions being completely worthless.

22. ### Robittybob1BannedBanned

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I am redoing my maths before I say any more sorry. I might have jumped to conclusions a bit quick. Just hold on for a day or so. I do appreciate your input AN. One of the best in my opinion.
So just pause for a while and I'll check if my workings were valid.

23. ### AlphaNumericFully ionizedModerator

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Having just checked with another theoretical physicist I work with to make sure I wasn't being mistaken I stand by what I said. You're using a formula from Newtonian mechanics. Don't get me wrong, it was a triumph of Newton to explain the relationship between orbital time and size observed by Kepler, but it is only approximate. The experiments of that time couldn't detect the time deviations from Newtonian gravity which arise when you measure things carefully enough. In Newtonian gravity you can have a closed orbit, while in GR you cannot because energy is bled from the system via gravitational waves, as well as a modification in precession rates due to relativistic corrections to account for the finite speed of gravity.

Using the Kepler formula you can indeed convert from orbital radius to orbital time or the other way but an error in one means an error in the other, even if the formula were exact, which it isn't.