Proof that Gravitational Constant is not constant

Are you familiar with how gravitational radiation is linked to loss of orbital energy. How does one convert to the other? They are not the same are they?

 

Log in or Sign up to hide all adverts.

Run this Excel macro on a sheet named TMxG Cubed root . You will see the value of lowering the G value as an explanation for the Hulse Taylor Binary anomaly. There are multiple possibilities of semi major axis values, that give a sensible results with a period 76.5 microseconds shorter after a year. (It will set up the sheet as well)

Sub A_Equal_CubedRoot_TMG()
'
'
' Macro recorded 16/10/2012 by Robittybob1
'

Dim Counter As Double
Dim Counter2 As Double
Dim Counter3 As Double
Dim A2 As Double
Dim A3 As Double
Dim G_start As Double 'G_start = original G
Dim T As Double
Dim T2 As Double
Dim PI As Double
'Dim Counter As Double
Dim GB As Double
Dim G2 As Double
Dim GG As Double
Dim dr As Double
Dim G As Double
Dim m1 As Double
Dim m2 As Double
Dim a As Double 'semi major axis 1,950,100,000 m
Dim TM As Double 'a^3/g = (t/(2* PI))^2* (M1 + m2) = TM
Dim mm As Double 'increasing mass by 1%
Dim TM2 As Double
Dim MinA As Double
Dim MaxA As Double 'max Semi-major axis
'Seconds in Julian year 31557600
G = 6.67384E-11
dr = 3.5
'dt = 7.64640648E-05
'Period 'T = 0.322997449 * 86400 = 27906.9795875520
a = 992846625.9 'a = semi major axis = 992846625.9
m1 = 1.4398 * 1.9891 * 10 ^ 30
m2 = 1.3886 * 1.9891 * 10 ^ 30
PI = 3.14159265358979
T = 27906.979587552 '(7.75*60*60)
T2 = T - 0.0000764640648
TM = (T / (2 * PI)) ^ 2 * (m1 + m2)
TM2 = (T2 / (2 * PI)) ^ 2 * (m1 + m2)
GG = G / 100000000 'Keep lowering raising this to get different values G
Range("A3") = GG
G_start = G

Sheets("TMxG Cubed root").Select


Range("G9").Select
ActiveCell.FormulaR1C1 = "dr/dt"
Range("A10").Select
ActiveCell.FormulaR1C1 = "sem-major A"
Range("A10").Select
ActiveCell.FormulaR1C1 = "=sem-major A"
Range("A10").Select
ActiveCell.FormulaR1C1 = "sem-major A"
Range("B10").Select
ActiveCell.FormulaR1C1 = "G "
Range("E10").Select
ActiveCell.FormulaR1C1 = "sem-major A2"
Range("G10").Select
ActiveCell.FormulaR1C1 = "A - A2"
Range("M10").Select
ActiveCell.FormulaR1C1 = "calc. dr/dt"
Range("L11").Select
ActiveCell.FormulaR1C1 = "=RC[-2]>0"
Range("L12").Select

Range("A11:Z4000").Select
'Range(Selection, Selection.End(xlDown)).Select
Selection.Delete Shift:=xlUp

'Range("I11").FormulaR1C1 = "=RC[-7]-RC[-4]"
'Range("J11").FormulaR1C1 = "=RC[-8]-RC[-2]"
'Range("K11").FormulaR1C1 = "=RC[-2]/RC[-1]"
Range("L11").FormulaR1C1 = "=RC[-2]>0"


Range("A1") = G 'Gravitational constant
G_start = G * 1.5 ' (start higher g)
G = G_start
Range("A2") = G_start
MinA = a
MaxA = 2 * a
Range("A4") = a
Range("A5") = MaxA



'XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX


Do Until a < MinA
dr = dr ' - 0.00000001

a = (TM * G) ^ (1 / 3)
A2 = (TM * (G - GG)) ^ (1 / 3)

Range("A11") = a
Range("B11") = G
Range("E11") = A2
Range("G11") = a - A2

If Range("L11") = True Then
G = G - (100000 * GG)
Range("A11

Please Register or Log in to view the hidden image!

11").Select
Selection.Copy
Range("A21").Select
ActiveSheet.Paste
Selection.Insert Shift:=xlDown
GoTo 199
End If

Range("M11") = dr
Range("A11

Please Register or Log in to view the hidden image!

11").Select
Selection.Copy
Range("A21").Select
ActiveSheet.Paste
Selection.Insert Shift:=xlDown
G = G - 100000 * GG
199 'Range("B2") = Counter2
Loop
'

End Sub

 

Log in or Sign up to hide all adverts.

I have a feeling based on work already done that any of the infinite solutions will give off the same amount of gravitational radiation, but it will be worth double checking that. So period change will always be exactly what is predicted by the equations trouble being that the formula Weisberg and Taylor used does not refer to any particular semi major axis,
but if they used the SMA listed in the data table "G" would be in the order of 8.8E-12 for the SMA to be as short as the measured value of 992846625.9 m, but most others have used the SMA value of 1950100,000 m which would be right if G has not decayed. But note 1950100,000 is not a measured value but what is calculated from the Newtonian formula.
If using this formula is good enough to calculate the SMA, then when the SMA is known, how come it is not good enough to calculate G (binary)?
r = 1950100 * 10 ^ 3 'meters in semi major axis.
\(\begin{array}{ll} \textrm{Parameter} & \textrm{Value} \\ \hline \\ a_p \, \sin \theta_i & 2.3417725 \pm 0.0000008 \, c \cdot \textrm{s} \end{array}\)

It is from this value you can calculate Ap (the semi major axis).

 

Log in or Sign up to hide all adverts.

Incorrect. \(a_p\) is the semi-major axis of the pulsar, \(a_c\) is the semi-major axis of its companion, \(a = a_p + a_c\) is the semi-major axis of the pulsar relative to the companion. And \(\frac{a_p}{a_c} = \frac{m_c}{m_p}\).

\(a = c \sqrt{\frac{P_b^3 \dot{\omega} (1 - e^2 )}{24 \pi^3}} = c \sqrt{ \frac{(27906.9795875520 \pm 0.0000003456\, \textrm{s})^3 ( ( 2.3376168123 \pm 0.000002765 ) \times 10^{-9} \, \textrm{radian} \cdot \textrm{s}^{\tiny -1}) (1 - ( 0.6171338 \pm 0.0000004 )^2 )}{24 \pi^3}} = 6.501626 \pm 0.000005 \, c \cdot \textrm{s} = 1949138400 \pm 1500 \textrm{m}\)

\(a_p\) is roughly half of this.

After detailed examination of the older paper, the numerical results only make sense if \(a = c \sqrt{\frac{P_b^3 \dot{\omega} (1 - e^2 )}{24 \pi^3}}\) but the equations in the paper imply \(a = c \sqrt{\frac{P_b^3 \dot{\omega} (1 - e )}{24 \pi^3}}\) so there is a mistake somewhere. Double checking the equation listed for advance of periastron shows that the eccentricity term does in fact have to be squared and its omission is a typo.

 

@rpenner - thanks for that. Was there any possibility in using this expression and showing me how it is used please. I have had 2 different opinions as to how it is used and hence 2 different results.
Sin i was as far as I know sin 45 degrees.

 

This should help me understand what is happening. I am so grateful for your input on this subject. So if I am looking just at the orbital parameters of the pulsar we can ignore the companion for it seems to be logical when the semi major axis of the pulsar =0 the companion is zero too, and therefore a collision has occurred. Is that right? For the orbital period change is measured from the signal coming from the pulsar. So we know how much that orbit decays every year.
Do you have a favorite formula for linking period shortening to shortening of the semi major axis?
I was just running the Newtonian formula for the orbital period twice and solving for a (pulsar) twice, and taking the difference as the change in a (pulsar).

 

That pretty much looks like substitution of a wild guess for an unknown quantity and thus represents the opposite of "knowing." It looks like you should know about as many significant digits for \(\sin \, i\) as you have significant digits for \(a_p\) which is about as many significant digits as you have for \(m_c\).

 

Do the following semi major axis' seem to be about right?
A-comp = 956929497.5
A-pulsar=992208902.5

 

True but so far I haven't seen it mentioned. Just the mention that the orbital inclination was about 45 degrees. It at least gave me something to work with.
But it was knowing what to do with the values when I do find it.

 

You're quoting them to 10 significant figures. In a physics or engineering paper that would mean you have measurements accurately to 10 significant figures, ie parts per ten billion. You do not. This only further illustrates how you lack the capability to properly evaluate the validity of whatever results you'll eventually produce.

If you cannot properly use experimental data and correctly incorporate uncertainties in said data into your work then even if your numbers do imply there's a variation in G such a conclusion will be utterly invalid and dispensable because you will not be able to show the variation is outside of the model uncertainity. As I have explained previously, if you can only measure a metre to an accuracy of 1mm, which is 1 part in 1000, then you cannot then make conclusions about variations in length of size 1mm or smaller, as it's indistinguishable from noise in your ability to measure.

 

I was taking rpenner's figures of a = "1949138400 \pm 1500" and dividing that in the proportion Mass of companion / mass of pulsar. I couldn't see the maths so I did a screen and come up with this solution. Ok it would have the same degree of % error. I use macros a fair bit and I'll have to see if +/- can be incorporated but still looking at the basic math without being too bogged down with detail at this stage.

 

But do you think the differences can account for substantial differences in the "G or a" values. For something seems really at odds here.
I have decided to put the problem aside for the moment as with 2 unknown quantities (even though the semi major axis has been measured and therefore not strictly an unknown, but everyone does not use that value in any of their calculations), it is really difficult to say what is happening.
A reduction in the Gravitational constant in the local region seems like a obvious reason for the anomalous results to the Newtonian orbital period formula.
The measured SMA and the period would suggest the gravity is in the order of 8 times weaker there now as it is on Earth. I thought the other day I was getting close to Earth values but that was using the commonly quoted SMA axis.

In fact I have just run my Orbital Period formulas again using the values Rpenner has helped us generate once we split them in the right proprtion and the calculated period using (G = 6.67384 * 10 ^ -11) and the SMA = a = 992208902 has given me the closest result to date, with measured period being 13.5687 seconds longer.


1.0134397E+04 calculated T (in secs) 27906.9795875520 Measured T (in secs) -17772.5822134628 Difference in seconds


And that is what I would expect if G has decayed somewhat.
Solving for G
8.8012829E-12 calculated GB from known orbital period and SMA.

If G was normal the orbital period would be 10,134 seconds not 27,906

So that is the proof so far!

 

Those that are familiar with macros can check the math by running it.

Sub Orbital_Period()
'
' Orbital_Period Macro
' Macro recorded 27/09/2012 by Robittybob1
'
'Julian year of 31,557,600 seconds

Dim thrs As Double
Dim T As Double
Dim Tc As Double
Dim r As Double
Dim e As Double '(eccentricity -0.617131)
Dim K2 As Double
Dim dr As Double
Dim dr_dt As Double
Dim GB As Double
Dim c As Double
Dim M1 As Double
Dim M2 As Double
Dim mu As Double '(std gravitational parameter (mu = G*(M1+M2))
Dim a As Double 'semi major axis 1,950,100,000 m

Sheets("Orbital Period").Select
Range("A1").Select

G = 6.67384 * 10 ^ -11
c = 299792458
M1 = 1.4398 * 1.9891 * 10 ^ 30
M2 = 1.3886 * 1.9891 * 10 ^ 30
ap = 992208902.5
PI = 3.14159265358979

Tc = 2 * 3.14159265358979 * (ap ^ 3 / (G * (M1 + M2))) ^ (0.5)
thrs = Tc / (60 * 60) 'secs to hours
Range("A10") = K2
Range("A11") = Tc
Range("B11") = " calculated T (in secs)"
Range("A12") = thrs
Range("B12") = "t(hrs))"
Range("A13") = a
Range("B13") = "Semi Major Axis in meters"

T = 27906.979587552
Range("D11") = T
Range("E11") = "Measured T (in secs)"
T2 = 27906.9795110896

T = 27906.97959

a = (((T / (2 * PI)) ^ 2) * G * (M1 + M2)) ^ (1 / 3)
Range("A14") = a
Range("B14") = " calculated Semi Major Axis in meters"
A2 = (((T2 / (2 * PI)) ^ 2) * G * (M1 + M2)) ^ (1 / 3)
Range("A15") = A2
Range("B15") = "Calculated semi Major Axis in meters A2"
da = A2 - a
Range("A16") = da 'a2 - a
Range("B16") = "da in meters based on orbital period change"

'solve for Gb

GB = ap ^ 3 * (2 * PI / T) ^ 2 / (M1 + M2)

Range("A17") = GB
Range("B17") = " calculated GB from orbital period and SMA"

Range("A9:A24").Select
Selection.NumberFormat = "0.0000000E+00"

Range("A9:E24").Select
Selection.Insert Shift:=xlDown

End Sub

 

It's like you're not even listening. Blindly applying formulae and assuming that because a number with 10 decimal places comes out of your computer then you can quote it to 10 decimal places is a pit fall I'd expect 1st year engineers to be embarrassed to do (no offence to engineers).

The fact the formula you're using isn't valid at high precision, a point I've explained 2 or 3 times now, just shows you aren't interested in doing this properly, you're just wanting to put on a white coat and play scientist.

 

The difference in the two G values is not in any way dependent on the finer decimal places.
The Gravitational constant in the region of the binary is 8.8012E-12 compared to 6.67384E-11 in our local region. So there is not arguing over decimal points; we are talking of a factor of 8 times the difference, not piddly fractions.
The decimal points in the cut and paste are simply the raw data from the Excel macro and will have to be adjusted for uncertainty in the measured quantities but it would be unlikely to change the answer to any significant amount.
I will attempt to introduce the uncertainty values into the equations on the macro, now that I understand what needs to be done.

But can you believe that gravity can be 8 times less potent due to gravitational radiation?

 

Fixed up the text and a couple of formulas forthe Orbital Period macro.

This is the results it gave:
2.79070E+04 Measured T (in secs Weisberg and Taylor)
1.01344E+04 calculated T (in secs when using G= 6.67384E-11)
2.81511E+00 t(hrs))calculated T (in hours when using G= 6.67384E-11)"
9.92209E+08 Measured Semi Major Axis in meters Weisberg and Taylor
1.94930E+09 calculated Semi Major Axis in meters using G= 6.67384E-11
1.94930E+09 Calculated semi Major Axis in meters A2 using G= 6.67384E-11
-3.67460E+00 da in meters based on orbital period change using G= 6.67384E-11
8.80128E-12 calculated GB from orbital period and SMA
9.92209E+08 calculated Semi Major Axis in meters using G Binary
9.92209E+08 Calculated semi Major Axis in meters A2 using G Binary
-1.87039E+00 da in meters based on orbital period change using G Binary

Trying running this macro.

Sub Orbital_Period()
'
' Orbital_Period Macro
' Macro recorded 27/09/2012 by Robittybob1
'
'Julian year of 31,557,600 seconds

Dim thrs As Double
Dim T As Double
Dim Tc As Double
Dim r As Double
Dim e As Double '(eccentricity -0.617131)
Dim K2 As Double
Dim dr As Double
Dim dr_dt As Double
Dim GB As Double
Dim c As Double
Dim M1 As Double
Dim M2 As Double
Dim mu As Double '(std gravitational parameter (mu = G*(M1+M2))
Dim a As Double 'semi major axis 1,950,100,000 m


Sheets("Orbital Period").Select
Range("A1").Select

G = 6.67384 * 10 ^ -11

c = 299792458
M1 = 1.4398 * 1.9891 * 10 ^ 30
M2 = 1.3886 * 1.9891 * 10 ^ 30
ap = 992208902.5

PI = 3.14159265358979
e = 0.617131
mu = G * (M1 + M2)

Tc = 2 * 3.14159265358979 * (ap ^ 3 / (G * (M1 + M2))) ^ (0.5)
thrs = Tc / (60 * 60) 'secs to hours
Range("A11") = Tc
Range("B11") = " calculated T (in secs when using G= 6.67384E-11)"
Range("A12") = thrs
Range("B12") = "t(hrs))calculated T (in hours when using G= 6.67384E-11)"""
Range("A13") = ap
Range("B13") = "Measured Semi Major Axis in meters Weisberg and Taylor"

T = 27906.979587552
Range("A10") = T
Range("B10") = "Measured T (in secs Weisberg and Taylor)"
T2 = 27906.9795110896 'Measured shorter period after 1 year, T (in secs Weisberg and Taylor)"

T = 27906.97959

a = (((T / (2 * PI)) ^ 2) * G * (M1 + M2)) ^ (1 / 3)
Range("A14") = a
Range("B14") = " calculated Semi Major Axis in meters using G= 6.67384E-11"
A2 = (((T2 / (2 * PI)) ^ 2) * G * (M1 + M2)) ^ (1 / 3)
Range("A15") = A2
Range("B15") = "Calculated semi Major Axis in meters A2 using G= 6.67384E-11"
da = A2 - a
Range("A16") = da 'a2 - a
Range("B16") = "da in meters based on orbital period change using G= 6.67384E-11"

'solve for Gb

GB = ap ^ 3 * (2 * PI / T) ^ 2 / (M1 + M2)

Range("A17") = GB

Range("B17") = " calculated GB from orbital period and SMA"

a = (((T / (2 * PI)) ^ 2) * GB * (M1 + M2)) ^ (1 / 3)
Range("A18") = a
Range("B18") = " calculated Semi Major Axis in meters using G Binary"
A2 = (((T2 / (2 * PI)) ^ 2) * GB * (M1 + M2)) ^ (1 / 3)
Range("A19") = A2
Range("B19") = "Calculated semi Major Axis in meters A2 using G Binary"
da = A2 - a
Range("A20") = da 'a2 - a
Range("B20") = "da in meters based on orbital period change using G Binary"



Range("A9:E24").Select
Selection.NumberFormat = "0.00000E+00"

99
Range("A9:E24").Select
Selection.Insert Shift:=xlDown

End Sub

 

Do you all agree that Gravitational constant may not be constant after all?
Is everyone happy now with the concept that gravitational radiation is like radiating away the system's gravitational attraction?
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. So there is a greater loss of gravitational potential energy without falling as far. So the rise in kinetic energy does not need to be as great either. (there could be a relativity aspect here for the energy associated with a moving mass goes up higher (E = MC^2/SQRT(1-v^2/c^2)) and plays a bigger part as the orbital velocity becomes more relativistic.

 

No one agrees with you on this, because that which is asserted without evidence may be dismissed without evidence.

That isn't what is happening. It is energy and momentum which is being radiated away.

You haven't understood the content of the 2004 paper or the content of the (slightly buggy) 1982 paper which requires that you understand undergraduate physics, astronomical conventions, how to read a science paper, and elements of General Relativity (to correct the formula). But both the 1982 and 2004 papers double check the predictions of general relativity against observation, and one of the postulates of general relativity is that G is constant throughout all space and time. Therefore your calculations which are based on incorrect, even reckless, guesses about the meaning of various terms in the papers doesn't begin to challenge physical theory.

 

Thanks. I hadn't heard from anyone so I didn't know if that meant they all agree, for usually if they disagree I hear the news rapidly. The only way the pulsar can orbit the companion at the SMA that you helped me calculate is for the "G" in the region of the binary to be much less than the value of "G" measured elsewhere.
It seems to me to be the most likely explanation and regardless of whether anyone else has proclaimed " G is constant throughout all space and time" I am standing by the analysis of the physical facts.

How do you explain the fact that the pulsar would have to do its orbit 2.75 times faster (at SMA of 9.92209E+08 meters) than it does if G is what it assumed to be 6.67384E-11?

That bit about the energy and "gravitational radiation is like radiating away the system's gravitational attraction" etc probably didn't come across the best for it was the first time I have tried to say this. True "energy and momentum .... is being radiated away" but that energy and momentum was "stored" in the system's gravitational potential energy (GPE) and that is dependent on "G". If "G" were to fade, GPE is released as "energy and momentum".
It is hard concept to say how it works precisely but that is my opinion at the current time.

 

"(they).....check(ed) the predictions of general relativity against observation, and one of the postulates of general relativity is that G is constant throughout all space and time."
One case where it can be shown that "G" is reduced seems enough to break that postulate.
What is a postulate anyway? What does Google offer?
"pos·tu·late/ˈpäsCHəˌlāt/
Verb:
Suggest or assume the existence, fact, or truth of (something) as a basis for reasoning, discussion, or belief.
Noun:
A thing suggested or assumed as true as the basis for reasoning, discussion, or belief."

So it is a type of assumption assumed true but not necessarily based on fact, a belief.

So there is definitely possible to have the a general postulate with certain exceptions.

The energy calculations they did were based on "G" so it doesn't matter what level G was the answer is still the same. If they were using an equation that had the semi major axis value in it it wouldn't have come out so well.