Proof that Gravitational Constant is not constant

No -- you need MORE math to demonstrate that Einstein's theory predicts gravitational waves, because
\(T = 0\) does not require that the special-relativistic solution is the only solution to the Einstein field equations, just like the charge density equaling zero doesn't require that a completely static vacuum is the only solution to Maxwell's equations.

See http://en.wikipedia.org/wiki/Gravitational_wave#Mathematics and http://en.wikipedia.org/wiki/Linearised_Einstein_field_equations for introductory material.

 

Log in or Sign up to hide all adverts.

Looked at http://en.wikipedia.org/wiki/Gravitational_wave#Mathematics and see how right at the end of the article was a simplified formula (can't copy it sorry) but it is the last 3-4 lines of the page after the words "Using mass-centered coordinates, and assuming a circular binary, this is.."
Would you be able to help me fit in some values ( explain what the symbols mean at least) so I can check out what the numerical result would be for the Hulse Taylor Binary - please.)

 

Log in or Sign up to hide all adverts.

I do not understand this post at all. Are you saying you're willing to learn the right way?

Why create a topic in which you argue the wrong way- only to learn the right way? Wouldn't it be much easier to learn about the topic prior to making claims to be 'disproven?'

This action makes absolutely no sense to me.

 

Log in or Sign up to hide all adverts.

I am reluctant to reply further, since I think I would be promoting a delusion of competence.

http://arxiv.org/abs/gr-qc/0501041 The basics of gravitational wave theory -- 45 page review article
(4.43) gives the magnitude for transverse gravitational waves at a distance of r along perpendicular to the circular orbits of mass stars of mass m_1 and m_2 separated by R as with period T as: \( \frac{4}{r} \times \frac{ G m_1 m_2 }{c^2 ( m_1 + m_2 ) } \frac{4 \pi^2 R^2}{c^2 T^2} = \frac{16 \pi^2 G m_1 m_2 R^2}{c^4 T^2 (m_1 + m_2)}\)
Since \(\frac{1}{T^2} = \frac{G (m_1 + m_2)}{4 \pi^2 R^3}\) from Kepler's Law we have the magnitude as: \(\frac{4 G^2 m_1 m_2}{c^4 R r} = \frac{2 G m_1}{c^2} \times \frac{2 G m_2}{c^2} \times \frac{1}{R} \times \frac{1}{r}\) or
The product of the Schwarzschild radii of the stars divided by the distance between them and divided by the distance to the binary system.

See also http://relativity.livingreviews.org/Articles/lrr-2006-3/ where Section 5.1 covers this same binary system again and emphasizes in figure 5 that the masses come from the observations. Or section 4.4 for the discussion about how much energy gravitational waves carry.

 

I believe in the power of simultaneous equations. I was hoping to get a new angle on the problem.

 

I am far from being deluded with my competence. If only you knew how difficult I find the subject. The only hope I have is that with perseverance I might begin to understand it. But I do appreciate your help, but it does often seem a little too far ahead of my understanding.

Thanks for writing this out in words "The product of the Schwarzschild radii of the stars divided by the distance between them and divided by the distance to the binary system. " It helps me know what I need to study...

One question I hope you will answer is what has the "distance to the binary system" got to do with it? Is that the distance between the binary and the Earth? I read that was a hard thing to measure.

 

What would constitute proof to you? The original "G" was more or less discovered the same way as I'm showing that the G in the HT Binary is lower. If you know the force and the masses and their distance between them you get a chance to estimate G. But tell me what would qualify as proof, for I have a feeling any of these equations with "G" in them are not affected (invalidated) by the strength of "G"?
For I am using the Newtonian orbital period formula to find G. I'm not saying the orbital period formula doesn't work across the Universe but I'm saying there are situations where G varies so it appears it doesn't work unless you are willing to vary 'G".

What would be the main reason that G could not vary?

 

No, it wasn't.

It was discovered in the Cavendish experiment as the constant of proportionality that related known laboratory masses to the gravitational dynamics of those masses.
Here, the best you can do is to learn the gravitational dynamics via observation and relate that, with a laboratory-measured value of G, to known units of mass. You don't have any independent estimate of any of the masses or orbital separation.
The researchers went one step further and confirmed that Einstein's model of gravitational dynamics was by far favored over Newtonian dynamics.

Your "finding" that G was lower relates entirely to your inability to figure out what the Kepler parameters of the orbit are.

 

I agree the Kepler parameters have been difficult to determine. OK I used the ones commonly referred to and then later the semi major axis that you helped me calculate. Look I will never know if one is right or wrong but whichever you choose either the period is too short or the gravitation radiation doesn't account for the movement in the orbital period.
Hulse and Taylor received a Nobel prize for their work, so I feel compelled to accept their mass calculations as correct. By "independent estimate of any mass" do you expect me to calculate the masses myself? Obviously I can't do that. Who can? But when we use the formulas on the masses they found, if they don't work we should be able to ask is it "G" that has changed or are their masses or distance calculated incorrectly, and if they got the prize one would assume their work was checked and one would think their maths was correct.
Obviously the Cavendish experiment has been checked against the orbital period formula, and the determined value of G can be used (in most cases).

If we find the value of "G" is needed to calculate the mass of the binary stars in the first place we end up with a circular argument. That is something I need to assure myself of. Are the masses of the stars estimated without resorting to predetermined value of "G".

 

This site suggests it is possible to measure the mass of binary stars without having to refer to "G" "Measuring the Mass of Stars" http://csep10.phys.utk.edu/astr162/lect/binaries/mass.html

"An important applications of binary systems is that under favorable circumstances they provide one of the only ways to determine reliable masses for stars.

Kepler's Laws and Masses

The determination of masses in binary systems generally uses Kepler's 3rd Law,
( m1 + m2 )*P^2 = ( d1 + d2 )^3 = R^3

where P is the orbital period, m1 and m2 are the respective masses, and R = r1 + r2, and the "seesaw equation" for the center of mass:"

I like the R factor here for it may give a good estimate of the R value to get the dr/dt and dE/dt equations to work properly.

Edit: Solving for R
r = ((M1 + M2) * p ^ 2) ^ (1 / 3)
R = 1.63634E+13 meters "R from Kepler's 3rd law"
R = 1950100000 meters "R from Wikipedia"

This new R is only 8391 times too large!

 

Even better equations found on "Using Binary Stars to Determine Stellar Masses"
http://outreach.atnf.csiro.au/education/senior/astrophysics/binary_mass.html

Edit: There seems to be hope in using these equations since one of the stars is a pulsar so it has Doppler shift as it says, "unless v can be measured or inferred directly from Doppler shift in its spectrum it must be calculated from the period" for doppler shift is related to speed and less on period and/or "r" values.
I know it might sound a bit old fashioned but "The forces acting on each star are balanced, that is the gravitational force equals the centripetal force"
That is the bit I was hoping to use to show that the gravitational force which depends on "G" calculates lower than what can be explained if the the "Universal Gravitational Constant" is used. OK it needs mass or velocity to be able to plug them into the Kepler equations, and also trust that the Kepler equations keep on working.

Refer to the article to get the correct versions but here is a taste: 2 stars with mass mA and mB at distances rA and rB from the barycenter.
"FG = FC or
GmAmB/r^2 = mAv^2/rA (5.4)
where v is the orbital speed of A.
Unless v can be measured or inferred directly from Doppler shift in its spectrum it must be calculated from the period, T:
v = 2πrA/T
so substituting this into (5.4) gives:
GmB/r^2 = 4π^2rA/T2

so if we then substitute in (5.3) we get:
GmB/r2 = 4π^2mBr/T^2M or:

M = 4π^2r^3/GT^2 (5.5)
which can be rewritten as:

mA + mB = 4π^2r^3/GT^2 (5.6)

 

So taking that last equation and rearranging it:
from my macro : "find_r"
'mA + mB = 4*pi^2*r^3/G*T^2
'm1 + m2 = 4*pi^2*r^3/(G*T^2)
'(m1 + m2)*(G*T^2) = 4*pi^2*r^3
'(m1 + m2)*(G*T^2)/(4*pi^2) = r^3
'Ratio = r ^ 3 / G
T = 27906.979587552
Ratio = (M1 + M2) * (T ^ 2) / (4 * Pi ^ 2)
Range("A20") = Ratio
Range("B20") = "Ratio r^3/G"

So it appears the same pair of stars have the possibility of having a range of solutions satisfying the ratio r^3/G, where G is a true variable and r is the average separation of the stars.
What does that look like then when for the Hulse Taylor Binary that has a ratio for r^3/G = 1.10985E+38 ?

Using quite big steps in G
G ............. R
6.67E-11 .. 1949304669
6.17E-11 .. 1899355460
5.67E-11 .. 1846631138
5.17E-11 .. 1790710819
4.67E-11 .. 1731061138
4.17E-11 .. 1666990245
3.67E-11 .. 1597574816
3.17E-11 .. 1521537998
2.67E-11 .. 1437031423
2.17E-11 .. 1341210624
1.67E-11 .. 1229303468
1.17E-11 .. 1092171303
6.74E-12 .. 907695234.7
1.74E-12 .. 577835692


But it is only values in and around "6.74E-12 .. 907695234.7 meters" that will satisfy the orbital period formula as I showed earlier in the thread.
One solution to the two equations.
It definitely looks like that a variable G is a strong possibility.

 

What do you want it written in? La Tex equations would be neat wouldn't it?

 

There has been an enormous effort put into measuring the orbital period to the degree that they know it decays by 76.5 microseconds per year. So the period is accurate so how come the orbital radius can't be estimated properly? I know GR will affect the timing as the orbital speeds get closer to light speed but at the moment it is only about 1/1000 c (max 400 km/sec varies quite significantly as they progress around the elliptical orbits.)

 

Going back over the thread, I see it has developed as it went along. There seemed to be a glaring mistake in this post.
"With a lowered "G" factor binary stars are able to orbit at much slower rates, they are able to maintain their separation without falling as much." I think this is correct.
"So there is a greater loss of gravitational potential energy without falling as far." This should have been the other way around "So there is a lesser loss of gravitational potential energy".
"So the rise in kinetic energy does not need to be as great either." This is true . The rest is OK too.

 

Got that a bit wrong too for there are R and r but this time they are talking like the R of the binary not the r of the pulsar, in other words G comes out close to to G = 6.67384 if I use the R calculated by RPenner earlier in the thread 1949138400 +/- 1500. yields a G = 6.67213E-11
How there could be 2 values of G beats me!
I'll look into it tomorrow - tired right now.

 

This is just a monologue by Robbity. There's no discussion, he's just using this as a blog.

 

Well getting a little closer to the truth. The r value i should have been using was the " sum of the semi-major axes" . Not the semi major axis of the pulsar! Not 2 times the semi major axis of the pulsar!
Now fortunately RPenner had helped me calculate that at the beginning of this thread 1949138400 +/- 1500 meters.
Plugged this value into the orbital period formula and see what happens?

2.790698E+04 Measured T (in secs Weisberg and Taylor)
3.570485E+00 T - Tc (calculated T).

2.790341E+04 Tc ... 3.570485E+00 ... 3.221048E-02
2.790344E+04 TcMax ... 3.538274E+00
2.790338E+04 TcMin ... 3.602695E+00 ... -3.221047E-02

The difference in period = 3.5705 seconds +/- 0.0322 seconds
So my calculated "G" for the binary = 6.6721E-11 (calculated GB from orbital measured period and measured A binary)

So it is not as much lower than the Universal Gravitational Constant as I had thought before.
Note I have not attempted to introduce a measurement of the uncertainty yet.

 

You are welcome to comment on the maths like the others are doing. OK there were a few messages of late but that was a result of going back over the thread (correcting the previous errors).

 

So we now have evidence of a very slight change in G for the binary. How will the other formulas pan out when we use this new G in them? So my calculated "G" for the binary = 6.6721E-11 (calculated GB from orbital measured period and measured A binary)
Difference was 1.707624E-14 lower than the usual 6.67384E-11 or 99.97% of the usual.