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?
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 $$\frac{a_p}{a_c} = \frac{m_c}{m_p}$$.It is from this value you can calculate Ap (the semi major axis).
See http://adsabs.harvard.edu/full/1982ApJ...253..908T for the mass formulas which properly should be inserted into the above relation.
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).
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.
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 $$m_c$$.Sin i was as far as I know sin 45 degrees.
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.
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.Do the following semi major axis' seem to be about right?
A-comp = 956929497.5
A-pulsar=992208902.5
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.
But do you think the differences can account for substantial differences in the "G or a" values. For something seems really at odds here.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.
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.
No one agrees with you on this, because that which is asserted without evidence may be dismissed without evidence.Do you all agree that Gravitational constant may not be constant after all?
That isn't what is happening. It is energy and momentum which is being radiated away.Is everyone happy now with the concept that gravitational radiation is like radiating away the system's gravitational attraction?
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.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.