The speed of gravitational field propagation

[Jonathan Scott]

The effect that "predicts" the location of the source is very simple, and a close non-relativistic equivalent can be seen in the way ripples spread from an object moving with a slow constant velocity on the surface of water. Circular waves spread out from each instantaneous location, but the source location moves. If an observer some distance away runs a straight line through a series of points on nearby wave fronts where the front is aligned in the same direction, the direction of the line points to where the object is expected to be at the current time.

To put it another way, if you plot those circular waves horizontally with the time at which they were emitted time as a vertical axis, the result for a static source is a circular cone, but for a moving source it is a skewed cone, where the center line is sloped according to the velocity. The straight lines up the sides of the cone for the static case remain straight lines but now point to the shifted apex for the moving case. The projections of those lines on to the base correspond to the projected direction of the source in the plane of the water.

A trivial calculation of the "static field" without taking motion into account is like assuming that the field is perpendicular to the local wave front. However, when motion is taken into account the effective field is like the line that joins parallel parts of each wave front, pointing to the extrapolated future position.

[Disclaimer: Please check this for yourself as this is from my head rather than any textbook]

 

[Estif]

You are getting dangerously close to crackpottery here - claiming a theory is wrong when you yourself don't understand it.

There are no instantaneous forces in either GR or in EM.

If you have questions, please ask them. If you want to push an agenda that a particular theory is wrong, I would suggest that you devote the effort into understanding the theory (at the level of being able to do the calculations yourself, and not relying one what someone else - even me - tells you it says) before advancing that claim.

I'm not pushing agenda, I'm simply interpreting the article that was given here by someone else and is actually supposed to represent mainstream GR opinion, as far as I can figure.

http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

I want to know if that is the "official" explanation of how GR works, and if not I want to know what is. I'm doing exactly what you want me to do, I am devoting the effort into understanding the theory by coming here and asking questions, here they are again:


1.) Article says the delay gets canceled, is that true?

2.) Does Earth orbit Sun, or the point where Sun was 8 minutes ago?

3.) Why can we measure this 8 minutes delay with light, but not with gravity?

4.) Do gravity waves propagate as transverse or longitudinal waves and how do you know?

 

[Estif]

To start with, Newton's law of gravitation for a system of point masses is

\ddot {\mathbf r}_i = \sum_{j\ne i}\frac{\mu_j (\mathbf r_j-\mathbf r_i)}{r_{ij}^3}

The majority of the other terms in that equation are first-order post-Newtonian corrections to Newton's law.

Can you explain that equation?
Why inverse-CUBED, where did inverse-SQUARE law go? Where are the masses, m(i) * m(j)?


Newton's law of universal gravitation
http://en.wikipedia.org/wiki/Newton's_law_of_gravitation

Vector form:

 

[D H]

Can you explain that equation?
Why inverse-CUBED, where did inverse-SQUARE law go? Where are the masses, m(i) * m(j)?

Start with the vector form you quoted from wikipedia:

\mathbf F_{12} = -G\frac{m_1m_2}{|\mathbf r_{12}|^2}\hat{\mathbf r}_{12}

This is "the force applied on object 2 due to object 1". Dividing this by the mass of object 2 yields the acceleration of object 2 induced by the gravitational interaction between the two objects.

\mathbf a_{12} = -\,\frac{Gm_1}{|\mathbf r_{12}|^2}\hat{\mathbf r}_{12}

The unit vector \hat{\mathbf r}_{12} is the vector from object 1 to object 2 scaled by the distance between the objects. Thus

\mathbf a_{12} = -\,\frac{Gm_1}{|r_{12}|^3}\mathbf r_{12}

Now denote |\mathbf r_{12}|\equiv r_{12} = r_{21} and using \mathbf r_{12} = - \mathbf r_{21} yields

\mathbf a_{12} = \frac{Gm_1}{r_{21}^3}\mathbf r_{21}

Compare this to the form in the link,

\mathbf \ddot{\mathbf r}_i =<br /> \sum_{j\ne i} \frac{\mu_j}{r_{ji}^3}\mathbf r_{ji}(1 + \text{relativistic terms})

The only differences in these two forms are

1. The former represents the acceleration due to the one interaction while the latter represents the acceleration due to the totality of mass in the solar system,
2. The latter expression incorporates "relativistic terms", and
3. The former expression uses Gm_1 while the latter uses \mu_j.

Astronomers can assess the http://en.wikipedia.org/wiki/Standard_gravitational_parameter" \mu_j to a high degree of accuracy (more than 10 decimal places for the Sun, almost 9 for the Earth). In comparison, G is know to a paltry 4 decimal places.

 

[Estif]

Start with the vector form you quoted from wikipedia:

\mathbf F_{12} = -G\frac{m_1m_2}{|\mathbf r_{12}|^2}\hat{\mathbf r}_{12}

Let m(i) = 100 kg
Let m(j) = 999,999 kg

Let m(i) have its center at coordinates [0, 0, 0] meters
Let m(j) have its center at coordinates [500, 0, 0] meters

m(i)= mass 1, m(j)= mass 2, G= constant, R= unit vector
r= distance from the center of mass1 to the center of mass 2
=================================================

r= sqrt(dx*dx + dy*dy + dz*dz)
r= sqrt(500*500 + 0*0 + 0*0)
r= sqrt(250,000)
r= 500 meters


Solve for mass 1:
-----------------
F(i,j)= -G * m(i)*m(j) * R /r^2

F(i,j)= -G * 100 * 999,999 * [(0 - 500)/500, 0/500, 0/500] / 500^2
F(i,j)= -G * 99,999,900 * [-1, 0, 0] / 250,000
F(i,j)= -G * [-400, 0, 0]
F(i,j)= [G * 400, 0, 0]

F = m * a
a(i)= F(i,j) / m(i)
a(i)= [G * 400, 0, 0] / 100

vector_a(i)= [G * 4, 0, 0]



Solve for mass 2:
-----------------
F(j,i)= -G * m(j)*m(i) * R /r^2

F(j,i)= -G * 999,99 * 100 * [(500 - 0)/500, 0/500, 0/500] / 500^2
F(j,i)= -G * 99,999,900 * [1, 0, 0] / 250,000
F(j,i)= [-G * 400, 0, 0]

F = m * a
a(j)= F(j,i) / m(j)
a(j)= [-G * 400, 0, 0] / 999,999

vector_a(j)= [-G * 0.0004, 0, 0]


If you do this for all the planetary pairs in the solar system, all the planets, moons and the sun, it works very well. Depending on the precision and time you want to invest you can get extremely precise results (past or future predictions), not only for days, months and years but the system stays stable for millions of years and errors you get are largely due to floating point precision, size of delta-time and the measurement data you set as initial conditions.

\mathbf F_{12} = -G\frac{m_1m_2}{|\mathbf r_{12}|^2}\hat{\mathbf r}_{12}

This is "the force applied on object 2 due to object 1".

This is probably just a matter of phrasing in the context of what you going to say next, but let me say that the force is not relative in that way, it's the same from object 2 to object 1 and from object 1 to object 2, only direction is opposite and that is why we have unit vector in that equation, so to give us the direction vector and the same equation works for both objects equally well.

Dividing this by the mass of object 2 yields the acceleration of object 2 induced by the gravitational interaction between the two objects.

\mathbf a_{12} = -\,\frac{Gm_1}{|\mathbf r_{12}|^2}\hat{\mathbf r}_{12}

The unit vector \hat{\mathbf r}_{12} is the vector from object 1 to object 2 scaled by the distance between the objects. Thus

\mathbf a_{12} = -\,\frac{Gm_1}{|r_{12}|^3}\mathbf r_{12}

Unit vector has scalar value of 1, it is only there so to gives us correct sign, correct direction, but I'm afraid that equation will not come up right with three dimensional vectors, as you just lost two dimensions with that conversion. Anyway, for the example given above, with both masses on the same axis, that equation does work, but it produces scalar not vector value.

a(j)= -G * m(i) * r / r^3
a(j)= -G * 100 * 500 / 125,000,000

X_axis_a(j)= -G * 0.0004


To make this correct it needs to be noted that it works with individual vector components or just with two objects on the same axis, so if you want to make it work in 3-D, then "r" is not the real distance "r", but it's either r[X], r[Y] or r[Z] component, and "r^2" is then really equal to (dx*dx + dy*dy + dz*dz).

Now denote |\mathbf r_{12}|\equiv r_{12} = r_{21} and using \mathbf r_{12} = - \mathbf r_{21} yields

\mathbf a_{12} = \frac{Gm_1}{r_{21}^3}\mathbf r_{21}

Compare this to the form in the link,

\mathbf \ddot{\mathbf r}_i =<br /> \sum_{j\ne i} \frac{\mu_j}{r_{ji}^3}\mathbf r_{ji}(1 + \text{relativistic terms})

The only differences in these two forms are

1. The former represents the acceleration due to the one interaction while the latter represents the acceleration due to the totality of mass in the solar system,
2. The latter expression incorporates "relativistic terms", and
3. The former expression uses Gm_1 while the latter uses \mu_j.

Astronomers can assess the http://en.wikipedia.org/wiki/Standard_gravitational_parameter" \mu_j to a high degree of accuracy (more than 10 decimal places for the Sun, almost 9 for the Earth). In comparison, G is know to a paltry 4 decimal places.

1.) I take it that "r(i)" stands for acceleration then, but why not "a(i)"? Anyway, if you put the "sum sign" in front of the first equation and instead of the index (1,2) place (i,j), then you would also get the "totality of the mass". It's just a way to denote an algorithm, it's nothing more but vectors addition where you cycle through all the pairs of the system.

It means that if you have Sun, Venus, Mars, Earth and Moon, you need to calculate the force between all the pairs, like this: F(Sun-Venus), F(Sun-Mars), F(Sun-Earth), F(Sun-Moon), F(Venus-Mars), F(Venus-Earth), F(Venus-Moon), F(Mars-Earth), F(Mars-Moon), F(Earth-Moon)... then calculate acceleration for each side and then sum all those acceleration vectors for each mass.

2.) I wonder how much different result you get with all the relativistic terms. Could you solve that equation for the example given above so we can compare?

3.) You have to have both masses and so I suspect that variable 'mu' contains the mass of the second body m(j). Therefore we can use it just the same in classical equations without any relativistic terms. In other words, I'm pretty sure that if we divide 'mu' by m(j) we get constant G with whatever precision 'mu' has.


I'm not sure what to conclude... equations do seem to be very similar, if not identical, but your equation is not 3-D friendly, it throws away that vector and if not careful that can lead to wrong implementation in three dimensional space.

 

[Jonathan Scott]

I'm not pushing agenda, I'm simply interpreting the article that was given here by someone else and is actually supposed to represent mainstream GR opinion, as far as I can figure.

http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

I want to know if that is the "official" explanation of how GR works, and if not I want to know what is. I'm doing exactly what you want me to do, I am devoting the effort into understanding the theory by coming here and asking questions, here they are again:


1.) Article says the delay gets canceled, is that true?

2.) Does Earth orbit Sun, or the point where Sun was 8 minutes ago?

3.) Why can we measure this 8 minutes delay with light, but not with gravity?

4.) Do gravity waves propagate as transverse or longitudinal waves and how do you know?

You asked that before, and were given lots of answers, but if you want the points answered specifically:

According to standard GR (and SR in the weak field limit):

1) The effect is as if the delay is cancelled. However, that is not how it actually works.

2) Earth orbits the point where the Sun is expected to be now given what it was doing 8 minutes ago.

3) The 8 minutes delay is the same for light and gravity.

4) Gravitational waves only occur when there are changes in a gravitational field. They propagate at c. They are transverse quadrupole. See the Wikipedia article on http://en.wikipedia.org/wiki/Gravitational_wave" for a picture.

 

[D H]

[Example of computing gravitational acceleration via force]
If you do this for all the planetary pairs in the solar system, all the planets, moons and the sun, it works very well.

No, it doesn't work very well. The reason is that the gravitational constant G is one of the, if not the, least well known of the fundamental physical constants. Moving from force to acceleration eliminates one of the masses in Newton's law. Replacing G*M_j with μ_j eliminates the other mass and makes the resulting equation much more accurate. The only way scientists know to estimate the masses of a planet is to divide the gravitational parameter for the planet by G. A planet's gravitational parameter is observable; it's mass is not. Why bring G into the equation when the product G*M is directly observable, and in many cases, accurately observable. For example, the Sun's gravitational parameter is known to 10+ decimal places, Earth's to almost 9 places, Jupiter's to 6 places, Saturn's to 5 places.

Depending on the precision and time you want to invest you can get extremely precise results (past or future predictions), not only for days, months and years but the system stays stable for millions of years and errors you get are largely due to floating point precision, size of delta-time and the measurement data you set as initial conditions.

The uncertainties in the gravitational parameters alone limits the time span over which predictions remain close to accurate. Add in the uncertainties in planetary states (positions+velocities at some epoch) and its game over in terms of accurately predicting the state of the solar system for hundreds of thousands of years or more.

Unit vector has scalar value of 1, it is only there so to gives us correct sign, correct direction, but I'm afraid that equation will not come up right with three dimensional vectors, as you just lost two dimensions with that conversion.

The way to compute a unit vector in the direction of some known vector \mathbf v in any cartesian space is \mathbf v = \hat{\mathbf v}/||\mathbf v||. Whether the space is a one dimensional cartesian space or a 10,000 dimensional cartesian space doesn't matter.

Anyway, for the example given above, with both masses on the same axis, that equation does work, but it produces scalar not vector value.

It is a vector and it is pointing in the right direction. Look at the equation.

1.) I take it that "r(i)" stands for acceleration then, but why not "a(i)"? Anyway, if you put the "sum sign" in front of the first equation and instead of the index (1,2) place (i,j), then you would also get the "totality of the mass". It's just a way to denote an algorithm, it's nothing more but vectors addition where you cycle through all the pairs of the system.

That is exactly what that sum is doing. It computes the second time derivative of position for each body in the solar system. Only it does so very accurately.

It means that if you have Sun, Venus, Mars, Earth and Moon, you need to calculate the force between all the pairs, like this: F(Sun-Venus), F(Sun-Mars), F(Sun-Earth), F(Sun-Moon), F(Venus-Mars), F(Venus-Earth), F(Venus-Moon), F(Mars-Earth), F(Mars-Moon), F(Earth-Moon)... then calculate acceleration for each side and then sum all those acceleration vectors for each mass.

There is no need to compute forces. The end goal is to compute accelerations, and this can be done without computing forces. All that computing forces does is to add a source of error.

2.) I wonder how much different result you get with all the relativistic terms. Could you solve that equation for the example given above so we can compare?

That general relativity solved a well-known problem, the anomalistic precession of Mercury, is one of the reasons general relativity gain fairly quick acceptance in the scientific community. Even after accounting for the interactions of the planets, 19^th astronomers could not fully explain Mercury's anomalistic precession. Theory and observation differed by 43 arcseconds per century. General relativity fully explains that difference.

3.) You have to have both masses and so I suspect that variable 'mu' contains the mass of the second body m(j). Therefore we can use it just the same in classical equations without any relativistic terms. In other words, I'm pretty sure that if we divide 'mu' by m(j) we get constant G with whatever precision 'mu' has.

You are assuming we know the masses of the planets. We don't. We do however know the product μ=G*M for many of the planets by observations of the planets' moons.

I'm not sure what to conclude... equations do seem to be very similar, if not identical, but your equation is not 3-D friendly, it throws away that vector and if not careful that can lead to wrong implementation in three dimensional space.

Look at it this way. There are two primary sources of planetary ephemerides: The Jet Propulsion Laboratory in Pasadena, California and the Institute of Applied Astronomy in St. Petersburg, Russia. Both use equations of the cited form. You are in essence arguing with the world renowned experts in this domain.

 

[Estif]

Look at it this way. There are two primary sources of planetary ephemerides: The Jet Propulsion Laboratory in Pasadena, California and the Institute of Applied Astronomy in St. Petersburg, Russia. Both use equations of the cited form. You are in essence arguing with the world renowned experts in this domain.

Look at it this way. NASA is using the same equations I do, and they landed on the Moon, believe it or not. You are in fact arguing with the world renowned expert in this domain, me.


It's really easy to resolve every single point of this argument, probably in less than 5min.

Do you think you're up to it?


Let m1 = 100 kg
Let m2 = 999 kg

Let m1 have its center at coordinates [100, 700, 400] meters
Let m2 have its center at coordinates [500, 200, 300] meters
=============================================================

acceleration1= ?
acceleration2= ?

 

[A.T.]

Look at it this way. NASA is using the same equations I do, and they landed on the Moon, believe it or not.

What does landing on the Moon has to do with exact predictions of the planet's movement over long periods of time? The Moon-Earth distance is tiny compared to the solar system, the involved masses are tiny compared to Sun's mass, the flight time is only a few days and most important: you have humans on board to correct the course.

You are in fact arguing with the world renowned expert in this domain, me.

An world renowned expert in the domain of silly arguments like: You don't need GR to fly to the Moon so GR must be wrong.

 

[D H]

Look at it this way. NASA is using the same equations I do, and they landed on the Moon, believe it or not. You are in fact arguing with the world renowned expert in this domain, me.

Oh, please! That you are arguing this is prima facia evidence that you are not. And believe it or not, NASA used (and still uses)

\frac{d^2\mathbf r}{dt^2} =<br /> \frac{\mathbf F_{\text{ext}}}{m} +<br /> \sum_j \frac{\mu_j}{||\mathbf r_j - \mathbf r||^3}(\mathbf r_j - \mathbf r)

to propagate the state of an interplanetary spacecraft , where F_ext is the total of the non-gravitational forces acting on the vehicle and where the sum is over the gravitational bodies. On the time scale that it takes a vehicle to go from the Earth to Mars, for example, relativistic effects are very small, and are much smaller than the uncertainties in those external forces.

It's really easy to resolve every single point of this argument, probably in less than 5min.

Let m1 = 100 kg
Let m2 = 999 kg

Let m1 have its center at coordinates [100, 700, 400] meters
Let m2 have its center at coordinates [500, 200, 300] meters
=============================================================

acceleration1= ?
acceleration2= ?

\begin{aligned}<br /> \mathbf r_{12} &amp;\equiv \mathbf r_2 - \mathbf r_1 =<br /> \left[ 500 , 200, 300 \right]\,\text{m} -<br /> \left[ 100, 700, 400 \right] \,\text{m}=<br /> \left[ 400, -500, -100 \right]\,\text{m} \\<br /> \mathbf r_{21} &amp;\equiv \mathbf r_1 - \mathbf r_2 =<br /> -\mathbf r_{12} = \left[-400, 500, 100\right]\,\text{m} \\<br /> r_{12} &amp;= r_{21} = ||\mathbf r_{12}|| = 100\sqrt{42}\,\text{m} \\<br /> \mathbf a_1 &amp;= \frac{Gm_2}{r_{12}^3}\mathf r_{12} =<br /> \left[97.96, -122.45, -24.49\right]\times 10^{-15}\,\text{m}/\text{s}^2 \\<br /> \mathbf a_2 &amp;= \frac{Gm_1}{r_{21}^3}\mathf r_{21} =<br /> \left[-9.806, 12.26, 2.452\right]\times 10^{-15}\,\text{m}/\text{s}^2<br /> \end{aligned}

Compare using forces,

\begin{aligned}<br /> F &amp;= \frac{Gm_1m_2}{r^2} = 15.872\times 10^{-12}\,\text{kg}\,\text{m}/\text{s}^2 \\<br /> a_1 &amp;= \frac{F}{m_1} = 158.72 \times 10^{-15}\,\text{m}/\text{s}^2 \\<br /> \hat{\mathbf r}_{12} &amp;= \frac{\mathbf r_{12}}{||\mathbf r_{12}||}<br /> = \left[0.617213, -0.771517, -0.154303 \right] \\<br /> \mathbf a_1 &amp;= a_1 \hat{\mathbf r}_{12}<br /> = \left[97.96, -122.45, -24.49\right]\times 10^{-15}\,\text{m}/\text{s}^2 \\<br /> a_2 &amp;= \frac{F}{m_2} = 15.888 \times 10^{-15}\,\text{m}/\text{s}^2 \\<br /> \hat{\mathbf r}_{21} &amp;= -\hat{\mathbf r}_{12}<br /> = \left[-0.617213, 0.771517, 0.154303 \right] \\<br /> \mathbf a_2 &amp;= a_2 \hat{\mathbf r}_{21}<br /> = \left[-9.806, 12.26, 2.452\right]\times 10^{-15}\,\text{m}/\text{s}^2<br /> \end{aligned}

The same result as before.

 

[S.Vasojevic]

Things to do for today:

1. switch on GPS, see if it works.

2. Look at the picture below, and describe in terms of mass and force what is happening.

 

[George Jones]

Locked for moderation.