# Why two mass attracts each other?

Discussion in 'Physics & Math' started by hansda, Mar 19, 2013.

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1. ### brucepValued Senior Member

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As you know the Gravity Probe B measured the geodetic effect. Apparently Wellwisher doesn't know about this test for GR and spacetime geometry?

3. ### Markus HankeRegistered Senior Member

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Well, I suppose the main difference would be that the geometric model gives the correct amount for the perihelion precession, whereas the force-based method gives wrong predictions

The crucial difference is that the Newtonian force field is linear, whereas the Einstein field equations are not; in other words, the gravitational field in GR is self-interacting. Hence the difference in predictions. GR does not use any forces - see also my GR primer thread for details.

5. ### hansdaValued Senior Member

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I asked a general question. Instead of giving general answer, you are explaining a very particular case of perihelion precession.

What do you mean by "self-interacting gravitational field of GR"?

Can you explain a "contact force(F=ma)" in terms of GR(geometry of spacetime)?

7. ### Markus HankeRegistered Senior Member

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I gave the general answer underneath. Newtonian force fields are linear, whereas the geometry of GR space-time is highly non-linear. That is the reason they don't give the same results, and that is where they differ.

In GR, all forms of energy are a source of gravity. This obviously holds for the gravitational field itself as well, since it contains energy. Thus, gravity is self-interacting - the gravitational field is in itself a source of gravity. Mathematically, this is why the Einstein field equations are highly non-linear, in contrast to Newtonian gravity. This is also ultimately why the two models can never be equivalent for strong fields - the stronger the field, the more deviation from Newtonian results you will get.

No, because GR is a model for gravity, not contact forces.
The concept of "forces" is not inherent in GR, but only a secondary derivation from it. Gravity can be modelled perfectly in terms of geometry alone, without any mention of forces. In fact that is ultimately the whole point of GR.

If you are asking whether F=ma can be generalised to curved space-times, then the answer is of course yes - the acceleration then becomes a 4-vector which is the covariant derivative ( defined in terms of space-time geometry ) of 4-velocity with respect to proper time, yielding a 4-force as result. But as I said already, this is "artificial" in that GR needs no forces to work.

I really do urge you to take a look at my GR primer thread, it is all explained there.

8. ### hansdaValued Senior Member

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So in Newtonian case a mass will move linearly whereas in GR the mass will move non-linearly?

Energy content of a mass is same as energy content of its gravitational field?

Let us consider a practical example. Consider "precession of a gyro". If a 'contact force' is applied to a "gyro", it will precess. Can this "precession of a gyro" be explained by GR?

9. ### TachBannedBanned

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The correct statement is that:

1. In Newtonian physics, the equation of motion is:

$\frac{d^2u}{d \phi^2}+u=\frac{m}{h^2}$

where

$\frac{1}{r}=u=\frac{m}{h^2}(1+e cos (\phi))$ is the solution of the above ODE

whereas

2. In GR, the equation of motion is:

$\frac{d^2u}{d \phi^2}+u=\frac{m}{h^2}+3mu^2$

The term $3mu^2$ is responsible for the existence of the non-Newtonian orbits.

10. ### Markus HankeRegistered Senior Member

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No, this is not what I said. I stated that the field equations in GR are non-linear, leading to a geometry of the gravitational field which is different from the Newtonian ( linear ) description. Hence the differences between Newton and GR. In other words - if formulated in terms of forces, the gravitational field in GR is not a simple inverse square law, but something much more complicated.

I didn't say that ( though it's an interesting point ). All I was saying is that the gravitational field itself is also a form of energy, and thus a source of gravity. This is the physical result of the non-linearity of the field equations.

Of course. You will indeed find that, just like in the case of Mercury's orbit, the Newtonian prediction for gyroscopic precession becomes increasingly inaccurate the stronger the gravitational influence in question is. Once again Einstein gives the correct numbers here - in GR this effect is called Lense-Thirring precession :

http://en.wikipedia.org/wiki/Lense–Thirring_precession

It is one of the classic tests of General Relativity, and ultimately is the result of frame dragging which is a purely geometric effect.

11. ### rr6BannedBanned

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Quasi-physical "force"---"purely geometri"c vs spin

"Purely geometric", here again is reminiscent of shape/form/pattern of vehicle affecting air-flow around the vehicle i.e. the position a collective set of atoms/molecules that create a 2ndary ~~wave/shape/pattern~~~ ergo there can be no puerly geometric affect, without the existence of a medium that occupies a space and a set of positions in space.

Ultra-micro ergo quasi-physical yet affects the more macro-physical, tho we do not see the connection, except indirectly between loss of energy via two observations far away of very massive set of two orbiing each other celestial objects-- binary pulsars or something like the -- and one lab experiement that shows neutrons falling but hesitating at discrete levels--- apparrently similar or reminiscent of to early discrete levels of electron shell identificaltion via photons coming in or going out.

And similar to our even more indirect assessment of virtual particles popping into and out of spactial existence, to rapidly for us to directly observe. This reminds me of my belief that gravity may operate just a fraction greater than the speeds-of-radiation. There is a least one experiement to verify the speed-of-gravity using Jupitor--- as best as I recall ---and if we take the allowed tolerance of errors given, we see that on the far side of error, gravity could be operating at .2% faster than speed-of-radiation.

So, if a virtual particle pops in and out of existence to quickly to be observed, why not have a gravitational force( gravitons ) that is too fast to observe?

So, in summary;

Newton = linear static background space

Einstien = complex( 4D ) dynamic background graviational spacetime

Precession = affects of bodies upon other bodies--- Fuller approximation "usually at 90 degrees see moon moving nearly 90 degrees to radial line to Earth"

Frame dragging = spin( vector? ) affecting/interaction-with gravitational spacetime and that may or may not translate to "purely geometric".

Trying to bring GR down to the common human comprehension abilities. The sports fans stand and sit in sequence to create a geometric pattern/shape/form called a ~~~wave~~~ or more specific as a people wave.

Air/wind blows the water to create water ~~~waves~~~~. No wind leaves no water waves on semi-static pond surface.

r6

12. ### hansdaValued Senior Member

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You are just explaining the difference between 'GR field equations' and 'Newtonian equations'. This is not the exact answer to my question. If you refer my original question in the post #759, it is basically to find the difference between 'effect of GR field equations on a mass' and 'effect of Newtonian equations on a mass'.

EFEs may be non-linear but is there effect on a mass is also non-linear?

Newtonian prediction of precession is inaccurate because at that time 'frame-dragging effect' was not known. Otherwise his calculation was correct.

So, from this a conclusion can be made that "frame dragging" has a effect of "anti-gravity" or "repulsive gravity" which causes 'anomaly in perihelion precession of Mercury'.

13. ### Q-reeusValued Senior Member

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How do you reconcile your sofar unchallenged claim above, repeated also in #767, with the oft quoted passage from http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
That last bit is best understood as an "it-sure-looks-like-it-here-but-looks-are-deceiving" style of statement - i.e. the real position is that stress-energy tensor has no gravitational energy density term - period. I happen to believe gravitational field should be a source term (as is the case in e.g. much maligned Yilmaz Gravity), but in GR it is not. As another German GR buff put it elsewhere: "gravity does not somehow walk around from the LHS to the RHS of the EFE's and become it's own source." But you obviously disagree with that individual and above reference too. Why?

Last edited: Jun 6, 2013
14. ### TachBannedBanned

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You are mixing up two very different phenomena: orbital precession and the gyroscopic precession.

No, it wasn't, the Newtonian equation I gave you above shows a (closed) ellipse with no precession while GR equation of motion shows an open trajectory with precession.

15. ### hansdaValued Senior Member

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What is the difference in "basic principle(not effect)" for "orbital precession" and "gyroscopic precession"?

If your above statement is right; how Newton could calculate "perihelion precession" for all the planets(except Mercury) in the Solar system almost correctly.

16. ### TachBannedBanned

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He didn't. That is the whole point.

17. ### hansdaValued Senior Member

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You have reference?

18. ### TachBannedBanned

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Any book on GR. have you ever opened one? Maybe now is the time for you to get started. I can recommend a few, are you interested?

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this? yes.

21. ### hansdaValued Senior Member

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YES. See chapter 12.12 and 12.13. Newtonian calculations for precession of all the planets in the solar system is mentioned there.

22. ### TachBannedBanned

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Yes, did you understand this:

"If the calculation described in the previous section is carried out more accurately, taking into account the slight eccentricities of the planetary orbits, as well as their small mutual inclinations, and retaining many more terms in the expansions (1015) and (1017), then the perihelion precession rate of the planet Mercury is found to be $5.32$ arc seconds per year. However, the observed precession rate is $5.75$ arc seconds per year. It turns out that the cause of this discrepancy is the general relativistic correction to Newtonian gravity. "

This is precisely what I have already shown you a few posts ago.

23. ### hansdaValued Senior Member

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Did you mean the above by your statement in post #773? This fact I already mentioned in my post #772.