The speed of gravitational field propagation

[Estif]

You mean like

https://www.physicsforums.com/showthread.php?p=1091901#post1091901?

Thanks, but I don't see any equations there. Can you explain how do you calculate trajectories with GR? What are the equations necessary to simulate solar system?

Note that Physics Forums rules to which you agreed when you registered,

https://www.physicsforums.com/showthread.php?t=5374,

in part, state Discussion of non-published non-mainstream personal theory is not allowed.

I want to discuss mainstream theory.

So, what does mainstream theory say - does the delay get canceled or not?

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

Why can we measure this delay with light, but not with gravity?


Thank you.

 

[Jonathan Scott]

Thanks, but I don't see any equations there. Can you explain how do you calculate trajectories with GR? What are the equations necessary to simulate solar system?

I want to discuss mainstream theory.

So, what does mainstream theory say - does the delay get canceled or not?

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

Why can we measure this delay with light, but not with gravity?Thank you.

I think you've missed the fact that the force on a test object due to the overall field of a moving source, both in electromagnetism and in gravity, includes components due to the motion of the source as well as static components. In electromagnetism, this is of course magnetism. For gravity, there are multiple components due to motion, but the important one here is closely analogous to magnetism and is often called "gravitomagnetism".

The static fields (electric or conventional gravitational) point to where you see the source to be, at the time that light left it. However, the dynamic fields mean that the overall force points to where the source will be right now. [Edit: That's not really technically accurate - that would be the "static fields as calculated without taking the motion into account"; it depends how you define it, but that's roughly how it works.]

In both cases, for a slow-moving source, you can look at it from the reference frame of the source instead, and see that the overall force on the test object (in terms of rate of change of momentum) points to the source in that frame, and hence that it must in other inertial frames too. This is in fact what we find; when you transform both the combined fields and the force law to find out how a test particle will move as seen from another inertial frame, you find that the force is directed towards the extrapolated location of the source at the current time.

Obviously, if the motion of the source was disturbed, the extrapolated location may no longer match the actual location. However, for gravity the conservation laws of energy and momentum mean that it's very difficult to create a "gravitational surprise", because the center of mass of any complete system always moves with constant velocity. Even if for example the sun could explode into two equal parts which moved apart at relativistic speed, the total energy and center of mass of the combined system would still be unaffected, and it would take some time for the parts to separate far enough for the overall gravitational field to change.

 

[A.T.]

You told me to Google it?
Two things are wrong with your response:

1.) None of the links actually answer the question.

2.) The question was rhetorical.

1) So you read all the Google results? Wow!

2) Why complain about the insufficient answer then?

If you tried to answer the question you would understand.

Understand what? You have shown nothing to understand so far.

They base their conclusion on some equations full of assumptions, it is not an experiment. It is not experimental data, you know?

The propagation speed of changes in the electric field is experimentally well determined to be c. There are many exmaples of that. Take the the x-ray tube:
http://en.wikipedia.org/wiki/X-ray_tube
Fast electrons are suddenly stopped at the anode. The information about that sudden change of the E-field propagates as x-rays at a finite speed. If the propagation of the E-field was instantaneous, you could not create EM-waves propagating at finite speed by accelerating charges. There would be actually no EM-waves at all.

 

[Estif]

The propagation speed of changes in the electric field is experimentally well determined to be c. There are many exmaples of that. Take the the x-ray tube:
http://en.wikipedia.org/wiki/X-ray_tube

Fast electrons are suddenly stopped at the anode. The information about that sudden change of the E-field propagates as x-rays at a finite speed. If the propagation of the E-field was instantaneous, you could not create EM-waves propagating at finite speed by accelerating charges. There would be actually no EM-waves at all.

That link does not support what you said. Any other links?

 

[A.T.]

That link does not support what you said. Any other links?

http://en.wikipedia.org/wiki/Bremsstrahlung
http://en.wikipedia.org/wiki/Electromagnetic_radiation#Speed_of_propagation

Do you at least understand that if the propagation so of the E-field was instantaneous, you could not have EM-waves propagating at a finite speed?

 

[Estif]

http://en.wikipedia.org/wiki/Bremsstrahlung
http://en.wikipedia.org/wiki/Electromagnetic_radiation#Speed_of_propagation

Do you at least understand that if the propagation so of the E-field was instantaneous, you could not have EM-waves propagating at a finite speed?

No, I do not understand that, and your links do not talk about anything like it at all. You seem to be confusing EM fields and EM waves. Perhaps I'm confusing something, but you are not explaining it. I also don't understand why is it hard to point some paper or article that makes these claims you do, and explains it. I want to understand how GR explains its results closely match Newtonian mechanics, which makes calculations based on instantaneous gravity. There was only one link presented here that says this delay gets canceled, but it turned out majority does not agree with it.

So, what is the explanation then that majority do agree with?I also want to understand how do you simulate solar system with GR. What equations do we use, and how. Without being able to that I do not see how GR can make better predictions about planetary trajectories than classical physics.

How did they calculate Mercury precession, what equations did they use?

 

[Estif]

I think you've missed the fact that the force on a test object due to the overall field of a moving source, both in electromagnetism and in gravity, includes components due to the motion of the source as well as static components. In electromagnetism, this is of course magnetism. For gravity, there are multiple components due to motion, but the important one here is closely analogous to magnetism and is often called "gravitomagnetism".

I like the sound of that, thanks. But, that does not seem to be a standard part of GR. It actually seem to be completely independent calculation maybe even able to produce results all on its own, without GR, so to say.

Can you show some equation where we can see how parameters vary due to velocity? Shouldn't it be the same if some system, say a binary star, is moving 9000m/s or 90m/s as a whole? It seems that theory implies different relative orbits depending on some absolute speed through the metric of spacetime or something. Did I get that right?

The static fields (electric or conventional gravitational) point to where you see the source to be, at the time that light left it. However, the dynamic fields mean that the overall force points to where the source will be right now. [Edit: That's not really technically accurate - that would be the "static fields as calculated without taking the motion into account"; it depends how you define it, but that's roughly how it works.]

In both cases, for a slow-moving source, you can look at it from the reference frame of the source instead, and see that the overall force on the test object (in terms of rate of change of momentum) points to the source in that frame, and hence that it must in other inertial frames too. This is in fact what we find; when you transform both the combined fields and the force law to find out how a test particle will move as seen from another inertial frame, you find that the force is directed towards the extrapolated location of the source at the current time.

Obviously, if the motion of the source was disturbed, the extrapolated location may no longer match the actual location. However, for gravity the conservation laws of energy and momentum mean that it's very difficult to create a "gravitational surprise", because the center of mass of any complete system always moves with constant velocity. Even if for example the sun could explode into two equal parts which moved apart at relativistic speed, the total energy and center of mass of the combined system would still be unaffected, and it would take some time for the parts to separate far enough for the overall gravitational field to change.

Is this based on some experiments? Have you got some links that explain the math and show equations for that? Are you talking about the delay cancellation? If yes, then let me repeat that is a paradox, it breaks continuity. You can not cancel time, you can not cancel distance. You can only traverse them both in a continuous manner, and the time it takes will tell you what was the speed; velocity= distance/delay. Equations can cancel time and space, sure, but I'm skeptic those equations actually correspond to reality.

 

[D H]

If so, can you pick an "inertial" coordinate system where the Sun has some constant velocity and figure out whether the Earth is being accelerated towards its current position at each moment?

The Sun does not have a constant velocity in an "inertial" frame; Jupiter's mass is about 1/1000 that of the Sun. The International Celestial Reference Frame with origin at the solar system barycenter is a better bet. See http://www.usno.navy.mil/USNO/astronomical-applications/astronomical-information-center/icrs-narrative for a description.

As far as equations of motion, see section 8.3 (page 3) of chapter 8 of The Explanatory Supplement to the Astronomical Almanac (to be published), http://iau-comm4.jpl.nasa.gov/XSChap8.pdf . These are what JPL uses as the basis for its Development Ephemerides.

 

[pervect]

To calculate the radiation from an electric charge, the standard thing to do is to use Larmor's formula:

http://en.wikipedia.org/wiki/Larmor_formula

The gravitational case is discussed, using approximations (weak field, nearly Newtonian, velocities << c ) in MTW, pg 989. MTW = "Gravitation" by Misner, Thorne, and Wheeler.

Various people have already posted links to past discussion on the topic, so I won't repeat it.

There was no formalism at the time MTW was written to describe the general case of high velocities/strong fields. Perhaps things have changed since then, though I doubt it. However, for the situation of the Earth orbiting around the sun, the approximations in MTW should be just fine. They wouldn't be so good for more extreme situations, where more detailed calculations specific to the problem at hand would be required. Some such calculations have been done, for instance for the pulsar inspiral that's been experimentally measured.

In any event, the useful way to approach the problem has already been described. To figure out the amount that the force appears as "non-central" for the electromagnetic case, you calculate the momentum that's radiated away by electromagnetic and gravitational radiation.

For the GR case, I'll direct you to the discussion in MTW.

You won't find anything that's detailed that's a "simple web page" on the topic. Not all technical subjects are reducible to "simple web pages". Sorry.

As far as interpreting "forces" in GR, I would assume that Carlip is using Newton-Cartan theory to perform the conversion. See http://en.wikipedia.org/wiki/Newton–Cartan_theory, or MTW chapter 12 pg 289.

What Newton-Cartan theory does is to translate Newtonian gravity into the geometric language of GR. By performing the translation "backwards", you can sometimes turn GR concepts into Newtonian ones. The answers you get will be highly coordinate dependent, though, they won't be nicely coordinate independent as they are in GR.

To get any meaningful results, the contribution of the space curvature components (by curvature components I mean the Christoffel symbols) to the motion of the body in the coordinate system you choose have to be negligible. This turns out to be the case for low velocities. As objects approach the speed of light, the contribution of the space curvature components can't be ignored, and the conversion process of viewing things as a "force" fails.

But you should really ask him (Carlip) - politely - if you need more details. Cartan theory is the approach I'd take to the problem, though.

 

[Estif]

As far as equations of motion, see section 8.3 (page 3) of chapter 8 of The Explanatory Supplement to the Astronomical Almanac (to be published), http://iau-comm4.jpl.nasa.gov/XSChap8.pdf . These are what JPL uses as the basis for its Development Ephemerides.

Can someone explain how does "r^3" come about in those formulas?

How does 'distance squared' get replaced by 'distance cubed' and why?--------------------------------
pervect,

I'm not sure if you were answering my question as to how GR can simulate solar system or something else, so can you confirm if that is the "easiest" way to actually practically implement GR? And if so, why isn't that conversion included in GR equations to start with?

 

[A.T.]

Do you at least understand that if the propagation so of the E-field was instantaneous, you could not have EM-waves propagating at a finite speed?

No, I do not understand that, and your links do not talk about anything like it at all. You seem to be confusing EM fields and EM waves.

An EM wave is a change of the EM-fields propagating at finite speed. If changes of the E-field would be instantaneous the waves would propagate at infinite speed. Well, they wouldn't be waves anymore, just instantaneous changes of the E-field, felt immediately in the entire universe.

 

[Estif]

An EM wave is a change of the EM-fields propagating at finite speed.

Electromagnetic waves CONSIST of electric and magnetic fields which oscillate. Wave-particles do propagate at or below their terminal velocity of lightspeed, but this has nothing to with field propagation. You should imagine a moving particle describing a helicoid path and you get your wave and your change of EM fields regardless of their propagation speed.

If changes of the E-field would be instantaneous the waves would propagate at infinite speed. Well, they wouldn't be waves anymore, just instantaneous changes of the E-field, felt immediately in the entire universe.

I think you are picturing it the wrong way. For example, we can have two kinds of waves, say in water.

Here a light would correspond to slow transverse waves on water surface, depending on viscosity, while field propagation would correspond to longitudinal waves, depending on compressibility, and these longitudinal waves would be spherical, much faster and pretty much how you should imagine fields, as opposed to transverse propagation of EM waves.To take this illustration further, you can imagine these waves would have their different terminal velocity, depending on some "drag coefficient" which would further be influenced by density, that represents gravitational potentials in this example. Interesting enough, this illustration can even explain some peculiar properties of magnetic fields in analogy to vorticity in fluid-dynamics.

You can see a right-hand rule just like with Lorentz force, that we were talking about earlier, and funny enough Biot-Savart law ('magnetic field' from your previous example) can be equally used in electromagnetics and aerodynamics. -- Hope this helps with visualization of these otherwise very abstract concepts.- "Imagination is more important than knowledge." (Albert Einstein)

 

[D H]

Can someone explain how does "r^3" come about in those formulas?

How does 'distance squared' get replaced by 'distance cubed' and why?

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.

 

[A.T.]

while field propagation would correspond to longitudinal waves,

See, it's you who is confusing waves and fields here, not me. The analogy to the change of the E-field is not some longitudinal waves, but simply the change of amplitude of the transverse wave.

 

[Vanadium 50]

The problem is that GR produces very similar results as Newtonian gravity with much more dubious interpretation visa vi 'instantaneous action', and I don't even know what is that interpretation if it's not the one where delay gets canceled. What are the other interpretations for this "instant" action beside "delay-cancel" one?

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.

 

[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,

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

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,

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

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)

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

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= ?

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

Compare using forces,

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

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.