Proof that Gravitational Constant is not constant

As long as you're willing to be found wrong - that's science.
When you're unwilling, it's dogma.

 

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Plenty of times I'm told that I'm wrong, but very few attempt to show it. In this particular topic there is no easy solution and even Weisberg and Taylor admitted that. As far as I know they never considered a weakening G. Once I did that and lowered G we started to get results that align, like orbit period matches the measured semi major axis and the like. To me that was tantamount to proof once I found the Newtonian formulas started working again!

 

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I've explained why you're wrong many times so quit lying. AN, rpenner and others explained why your wrong in this thread. Your problem is you can't understand any of the arguments or any of the experimental evidence against your point of view. 'All the knowledge in the world is of no use to fools', Don Henley and the Eagles. The knowledge is out there but you're one of the fools Don was talking about. You're lucky that the folks I mentioned take the time to help you so quit lying.

 

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AN thought I hadn't considered the uncertainty adequately in the measurements from memory, but the differences were not expressed as a fraction but whole numbers, and since pointing that out I haven't heard from them again. R Penner has not expressed much opinion on the topic. I acknowledge that you and I tried to sort it out but in the end you just gave up.
I am working my way back over the "Gravity Waves" thread to see if these differences can be settled.
I consciously avoid lying , so I am surprised to be accused of lying. Can you please show me one example of what you consider to be a lie?

 

Succinct.

\(\begin{array}{ll} \textrm{Parameter} & \textrm{Paper Value} & \textrm{Normalized Value} \\ \hline \\ P_b & 0.322997448930 \pm 0.000000000004 \, \textrm{d} & (2.790697958760 \pm 0.000000000035) \times 10^4 \, \textrm{s} \\ a_p \, \sin \theta_i & 2.3417725 \pm 0.0000008 \, \textrm{s} & 2.3417725 \pm 0.0000008 \, \textrm{s} \\ \gamma & 0.0042919 \pm 0.0000008 \, \textrm{s} & (4.2919 \pm 0.0008) \times 10^{-3}\, \textrm{s} \\ e & 0.6171338 \pm 0.0000004 & 0.6171338 \pm 0.0000004 \\ \dot{P}_{b,Obs} & ( -2.4184 \pm 0.0009 ) \times 10^{-12} & ( -2.4184 \pm 0.0009 ) \times 10^{-12} \\ \dot{\omega} & 4.226595 \pm 0.000005 \, \textrm{degree} \cdot \textrm{year}^{\tiny -1} & (2.33761676 \pm 0.00000277) \times 10^{-9} \, \textrm{s}^{\tiny -1} \\ \hline \\ G c^{-3} (m_p + m_c ) & \sqrt{\frac{P_b^5 (1 - e^2)^3 \dot{\omega}^3}{864 \pi^5} } & ( 1.3931567 \pm 0.0000030 ) \times 10^{-5} \, \textrm{s} \\ G c^{-3} m_p & \sqrt{\frac{P_b^5 (1 - e^2)^3 \dot{\omega}^3}{384 \pi^5} } - \sqrt{ \frac{P_b^5 (1 - e^2)^3 \dot{\omega}^3}{3456 \pi^5} + \frac{P_b^3 ( 1 - e^2)^2 \dot{\omega}^2 \gamma}{72 \pi^3 e} } & ( 0.710064 \pm 0.000096 ) \times 10^{-5} \, \textrm{s} \\ G c^{-3} m_c & \sqrt{ \frac{P_b^5 (1 - e^2)^3 \dot{\omega}^3}{3456 \pi^5} + \frac{P_b^3 ( 1 - e^2)^2 \dot{\omega}^2 \gamma}{72 \pi^3 e} } - \sqrt{ \frac{P_b^5 (1 - e^2)^3 \dot{\omega}^3}{3456 \pi^5}} & ( 0.683093 \pm 0.000096 ) \times 10^{-5} \, \textrm{s} \\ G c^{-3} M_{\odot} & & ( 0.49254909491 \pm 0.00000000003 ) \times 10^{-5} \, \textrm{s} \\ \hline \\ m_p & 1.4414 \pm 0.0002 M_{\odot} \\ m_c & 1.3867 \pm 0.0002 M_{\odot} \\ \hline \\ c & 299792458 \textrm{m} \cdot \textrm{s}^{\tiny -1} \\ G & ( 6.6738 \pm 0.0008 ) \times 10^{-11} \textrm{m}^{\tiny 3} \cdot \textrm{kg}^{\tiny -1} \cdot \textrm{s}^{\tiny -2} \\ M_{\odot} & ( 1.9885 \pm 0.0002) \times 10^{30} \textrm{kg} \\ \frac{2 G M_{\odot}}{c^2} & 2953.2500770 \pm 0.0000002 \textrm{m} \\ \hline \\ \dot{P}_{b,Gal} & ( -0.0128 \pm 0.0050 ) \times 10^{-12} \end{array}\)

As you can see, all the observational elements are in units of seconds. All the mass values are always present in the same powers as G so G cannot be extracted from the observational numbers without an independent estimate of mass. So attempting to estimate G from these observations can only be imagined if you don't understand the paper.

Finally, the paper does already test if GR is accurate, because \(\dot{P}_b\) is calculable from the other observations.
\(\dot{P}_b = - \frac{192 \pi}{5} \left( \frac{2 \pi G m_p}{P_b c^3} \right) \left( \frac{2 \pi G m_c}{P_b c^3} \right) \left( \frac{2 \pi G(m_p + m_c)}{P_b c^3} \right)^{-\frac{1}{3}} \left(1 - e^2\right)^{- \frac{7}{2}} \left( 1 + \frac{73}{24} e^2 + \frac{37}{96} e^4 \right) = - \frac{192 \pi}{5} \left( \sqrt{\frac{P_b^3 (1 - e^2)^3 \dot{\omega}^3}{96 \pi^3} } - \sqrt{ \frac{P_b^3 (1 - e^2)^3 \dot{\omega}^3}{864 \pi^3} + \frac{P_b ( 1 - e^2)^2 \dot{\omega}^2 \gamma}{18 \pi e} } \right) \left( \sqrt{ \frac{P_b^3 (1 - e^2)^3 \dot{\omega}^3}{864 \pi^3} + \frac{P_b ( 1 - e^2)^2 \dot{\omega}^2 \gamma}{18 \pi e} } - \sqrt{ \frac{P_b^3 (1 - e^2)^3 \dot{\omega}^3}{864 \pi^3}} \right) \left( \frac{P_b^3 (1 - e^2)^3 \dot{\omega}^3}{216 \pi^3} \right)^{-\frac{1}{6}} \left(1 - e^2\right)^{- \frac{7}{2}} \left( 1 + \frac{73}{24} e^2 + \frac{37}{96} e^4 \right) = - \frac{192 \pi}{5} \left( \sqrt{ \frac{P_b^6 (1 - e^2)^6 \dot{\omega}^6}{186624 \pi^6} + \frac{P_b^4 ( 1 - e^2)^5 \dot{\omega}^5 \gamma}{3888 \pi^4 e} } - \frac{P_b^3 (1 - e^2)^3 \dot{\omega}^3}{216 \pi^3} - \frac{P_b ( 1 - e^2)^2 \dot{\omega}^2 \gamma}{18 \pi e} \right) \sqrt{ \frac{6 \pi}{P_b \dot{\omega}} } \left(1 - e^2\right)^{-4} \left( 1 + \frac{73}{24} e^2 + \frac{37}{96} e^4 \right) \)

 

The values I am most interested in was the period and the projected semi major axis, and the estimates of the masses of the binary stars. OK I have wondered if they needed G to work out the mass of the stars or if the mass was calculated based on inertial mass. True there are things I don't know or understand. But if they have a way of double checking the mass, then when using the period and the projected semi major axis values which are independent of G, and using the Newtonian orbit period formula, the formula only works properly if the value of G (Binary) is lowered.
Are you saying all estimates of the binary masses were dependent on G being 6.6738E-11?
The period change formulas look horrendously complex so I dare not comment, but I tend to agree that the period change could be independent of G.

 

He thinks the pulsar system gravitational radiation is radiating away gravitational attraction rather than orbital energy. Essentially claiming the laws of physics are frame dependent. He told me he was going to 'hold on to that opinion until somebody proves he's wrong'. He's a thread wrecker at the other site.

 

Are you really that dense? Really? The lie is saying nobody has made the effort to show how wrong you are. You haven't made the effort. The difference is I understand the physics and you don't. Compounded by your inability to realize the ignorance you're wallowing in.

 

I'd be surprised if you could tie your shoe laces without getting them mixed up. They're not using Newtonian physics. Since you can't even figure that out you should put a cork in it and learn some physics rather than make a nuisance of yourself.

 

They are not using Newtonian physics because they could not get it to work. It wasn't just a little bit out but miles out which ever way they looked at it, and the difference can't be accounted for by using GR either.
OK the precession rate matched what was accounted for by GR but that is not what we are talking about.

 

Deleted

 

Now where do I say "nobody has made the effort to show how wrong I am"? I acknowledge what AN, R Penner and you Bruce has done, but true it hasn't convinced me yet. R Penner could succeed but just confusing me with some of the most horrendous equations does not do it for me.

 

Normally orbital energy and momentum is transferred by collisions and or via gravity to other bodies. With the HTB there definitely is a loss of orbital energy (well the orbital period has been measured to get shorter, so we assume the orbital energy is lessened too). What happens to that orbital energy is not explained but in my calculations 50% of any orbital energy loss is converted to kinetic energy. So where does the other 50% go?
If I say some part goes into paying the negative gravitational energy back by lowering G, that is a fair statement to be said in a thread discussing alternative theories.

Proving that happens - now that is a tad more difficult.

 

Because I stopped giving a **** by that point since it was clear you don't understand what you're doing and you don't want to understand.

Show your calculations, since I don't believe you are sufficiently competent to be reaching sound conclusions.

Alternative theories isn't a forum where you can just make random claims and not get called on them. It's a place where such claims aren't just deleted but that is quite different from not being criticised.

The devil is in the details and you haven't considered any details.

 

"What happens to that orbital energy is not explained but in my calculations 50% of any orbital energy loss is converted to kinetic energy. So where does the other 50% go?" This is the so called Robittybob Law (named after me of course) by Confused1 in the thread called "Kepler revisited" where we looked at how mass gain speed even after lowering orbital energy. http://www.physforum.com/index.php?act=Post&CODE=06&f=12&t=41174&p=543371

I don't mind the criticism for you and R Penner have been really helpful. I am now going back over the "Gravity Waves" thread and making an effort to double check the calculations.

 

Gravitational radiation caries both energy and momentum away from the binary system. This is covered in all GR textbooks and a good number of GR pop-physics sources and textbooks which gloss over.

It's not a law -- it's a question.

Because classically E = K(v) + U(x) so by looking at just K you aren't looking at the whole picture. However you lose energy, it doesn't mean that both K and U go down in lock-step. See also http://en.wikipedia.org/wiki/Virial_theorem

Which you could have saved us 3 pages of discussion by typesetting so we would know your physics model instead of naked calculations based on no communicated reasoning.

 

I understand "Gravitational radiation caries both energy and momentum away from the binary system." I don't have a problem with that but there is a lock step between orbital energy and kinetic energy and gravitational potential energy. 1 part of GPE = 1/2 part KE + 1/2 Part "other" (which includes Gravitational radiation, collisions, dust, friction, tidal transfer. This is not an exhaustive list of "other"). This is all based on orbital speed and orbital energy calculations, and it is the case for the binary too (I believe, I'm working toward showing it)
So I'll look again at that link and consider your thinking. http://en.wikipedia.org/wiki/Specific_orbital_energy is more along my line of reasoning.

Could you help me with the earliest concept of Gravitational radiation, they say Einstein predicted it, but what exactly did he say or formulate? I tried to find it but came up with nothing other than that "he predicted it".
You make it sound as if all the energy released from the in-fall (GPE change) goes as Gravitational Radiation, but I'm tending to the view it is only a fraction of the energy change. I want to see what Einstein predicted.

 

From 1910-1916 Einstein did the most significant work in his life and rewrote the rules of Newton's Universal Gravitation. The new theory, General Relativity, states that for any smooth, one-to-one space-time coordinate system with dimension 4, the proper time of any interval of any time-like world line is given as \(\tau = \frac{1}{c} \int \sqrt{ \sum_{\mu \in \{ 0,1,2,3 \} } \sum_{\nu \in \{ 0,1,2,3 \} } g_{\mu \nu}(x(\lambda)) \frac{d x^{\mu}(\lambda)}{d \lambda} \, \times \, \frac{d x^{\nu}(\lambda)}{d \lambda} } d\lambda \) where \(x(\lambda)\) is the position in space-time and \(g(x)\) is a geometrical object known as a symmetric tensor. You can think of it as 10 functions of position in space-time where \(g_{\mu\nu}(x) = g_{\nu\mu}(x)\). Einstein went on to say how \(g\) relates to the density of energy, momentum and pressure at every point in space time, \(T\).

Let \(\varepsilon^{\alpha \beta \gamma \delta} \equiv \varepsilon(\alpha, \beta, \gamma, \delta) = \frac{(\alpha - \beta)(\alpha - \gamma)(\alpha - \delta)(\beta - \gamma)(\beta - \delta)(\gamma - \delta)}{12}\).
Let \( \left| g(x) \right| \equiv \frac{1}{24} \sum_{\alpha \in \{ 0,1,2,3 \} } \sum_{\beta \in \{ 0,1,2,3 \} } \sum_{\gamma \in \{ 0,1,2,3 \} } \sum_{\delta \in \{ 0,1,2,3 \} } \sum_{\kappa \in \{ 0,1,2,3 \} } \sum_{\lambda \in \{ 0,1,2,3 \} } \sum_{\mu \in \{ 0,1,2,3 \} } \sum_{\nu \in \{ 0,1,2,3 \} } \varepsilon^{\alpha \beta \gamma \delta} \, \times \, \varepsilon^{\kappa \lambda \mu \nu} \, \times \, g_{\alpha \kappa}(x) \, \times \, g_{\beta \lambda}(x) \, \times \, g_{\gamma \mu}(x) \, \times \, g_{\delta \nu}(x) \)

Then \(\left( g^{\tiny -1} \right)^{\alpha \kappa}(x) = \frac{ \frac{1}{6} \sum_{\beta \in \{ 0,1,2,3 \} } \sum_{\gamma \in \{ 0,1,2,3 \} } \sum_{\delta \in \{ 0,1,2,3 \} } \sum_{\lambda \in \{ 0,1,2,3 \} } \sum_{\mu \in \{ 0,1,2,3 \} } \sum_{\nu \in \{ 0,1,2,3 \} } \varepsilon^{\alpha \beta \gamma \delta} \, \times \, \varepsilon^{\kappa \lambda \mu \nu} \, \times \, g_{\beta \lambda}(x) \, \times \, g_{\gamma \mu}(x) \, \times \, g_{\delta \nu}(x)}{ \left| g(x) \right| }\)
Such that \( \sum_{\alpha \in \{ 0,1,2,3 \} } \left( g^{\tiny -1} \right)^{\alpha \mu}(x) \, \times \, g_{\alpha \nu}(x) = \delta_{\nu}^{\mu}\) (where \(\delta\) is the Kronecker delta on the right side) or \( \sum_{\alpha \in \{ 0,1,2,3 \} } \sum_{\beta \in \{ 0,1,2,3 \} } \left( g^{\tiny -1} \right)^{\alpha \beta}(x) \, \times \, g_{\alpha \beta}(x) = \sum_{\alpha \in \{ 0,1,2,3 \} } \delta_{\alpha}^{\alpha} = \sum_{\alpha \in \{ 0,1,2,3 \} } 1 = 4\)

Let \(\Gamma_{\gamma \delta}^{\beta}(x) \equiv \Gamma(\beta, \gamma, \delta, x) = \frac{1}{2} \sum_{\alpha \in \{ 0,1,2,3 \} } \left( g^{\tiny -1} \right)^{\beta \alpha}(x) \, \times \, \left( \frac{\partial g_{\alpha \gamma} (x) }{ \partial x^{\delta} } + \frac{\partial g_{\alpha \delta} (x) }{ \partial x^{\gamma} } - \frac{\partial g_{\gamma \delta} (x) }{ \partial x^{\alpha} } \right)\)

Let \(R_{\gamma \delta} (x) = \sum_{\alpha \in \{ 0,1,2,3 \} } \left( \frac{ \partial \Gamma_{\delta \gamma}^{\alpha}(x) }{\partial x^{\alpha} } - \frac{ \partial \Gamma_{\alpha \gamma}^{\alpha}(x) }{\partial x^{\delta} } + \sum_{\beta \in \{ 0,1,2,3 \} } \left( \Gamma_{\alpha \beta}^{\alpha}(x) \, \times \, \Gamma_{\delta \gamma}^{\beta}(x) \; - \; \Gamma_{\delta \beta}^{\alpha}(x) \, \times \, \Gamma_{\alpha \gamma}^{\beta}(x) \right) \right) \)
Let \(R (x) = \sum_{\beta \in \{ 0,1,2,3 \} } \left( g^{\tiny -1} \right)^{\alpha \beta}(x) \, \times \, R_{\alpha \beta}(x) \)
Then the "Einstein Field Equations" at the heart of General Relativity are:
\(R_{\alpha \beta} (x) \; - \; \frac{1}{2} R(x) \times g_{\alpha \beta}(x) \; + \; \Lambda \times g_{\alpha \beta}(x) = \frac{8 \pi G}{c^4} T_{\alpha \beta} (x)\)
Which is 10 coupled differential equations in \(g\) and \(T\).

Now I've never taken a course on GR, but I have read more than half of two textbooks on the subject and have assembled the above for your benefit from a variety of sources with some commonly suppressed summation signs and dependency on position spelled out. The whole point of taking a year (or more!) to teach GR is to work out the implications of this simple expression which is wildly non-linear in g.

Here's the simplest application of this:

If \(g_{00} = c^2 , \; g_{11} =g_{22} =g_{33} = -1 \, \; g_{01} = g_{02} = g_{03} = g_{12} = g_{13} = g_{23} = 0, \; \Lambda = 0\)
Then \(g^{\tiny -1} = g, \, \Gamma_{\gamma \delta}^{\beta} = 0, \, R_{\gamma \delta} = 0 \, R = 0 , \, \textrm{and} \, T = 0\) so this is the metric of flat, empty space, expressed in Cartesian coordinates and so \(\tau = \frac{1}{c} \int \sqrt{ c^2 \left( \frac{d x^{0}(\lambda)}{d \lambda} \right)^2 - \left( \frac{d x^{1}(\lambda)}{d \lambda} \right)^2 - \left( \frac{d x^{2}(\lambda)}{d \lambda} \right)^2 - \left( \frac{d x^{3}(\lambda)}{d \lambda} \right)^2 } d\lambda = \frac{1}{c} \int \sqrt{ c^2 - \left( \frac{d X(t)}{d t} \right)^2 - \left( \frac{d Y(t)}{d t} \right)^2 - \left( \frac{d Z(t)}{d t} \right)^2 } dt = \int \sqrt{ 1 - \left( \frac{\vec{v}(t)}{c} \right)^2 } dt\) exactly as in Special Relativity.

 

I see I occasionally use the word "orbital energy" in correctly. Some of the time I'm referring to the gravitational potential energy (GPE) Now it is the loss of distance between the masses that allows for a change in the GPE, the energy released from this change to height "r" is converted to kinetic energy (KE) and energy going to whatever caused it to loose orbital energy. So as it falls it speeds up but the increase in speed always only accounts for 50% of the energy change. Energy going to whatever caused it to loose orbital energy
accounts for the other 50%.

 

You are a bit of a wizz really, for I've never seen anything like this in my life, but looking through all that, which bit specifically makes it certain (prediction) there is gravitational radiation? (Hopefully you don't reply by saying all of it.)