# Orbit precessions - General Relativity vs Newton

1. Feb 4, 2016

### Cobalt101

What is the underlying feature of general relativity that, unlike newtonian mechanics, results in the correct calculation of orbits i.e. including precession (e.g. Mercury). I not asking for the mathematics (i.e. the additional term in the equation) but rather what underlying "physical" feature causes it to not just be a perfect ellipse ?

2. Feb 5, 2016

3. Feb 5, 2016

4. Feb 5, 2016

### sandy stone

Those slides certainly help to visualize what is going on, but I have a question. The two slides treat curvature in the time dimension and in spacial dimensions separately. Is this an accurate description of the situation, or is it done to simplify the explanation? For instance, in the depiction of curved space surrounding the sun, is there also curvature in the time dimension as well, that is just too hard to illustrate?

5. Feb 5, 2016

### A.T.

Yes. You can only show 2 of the 4 space-time dimensions in such diagrams. What you see in part 1 (space-time) happens locally everywhere along the orbit shown in part 2 (space-space).

6. Feb 5, 2016

### pervect

Staff Emeritus
I don't think that precession can be attributed to a single underlying physical feature of general relativity. If onr consider pertubation of a circular orbit - i.e. one takes a particle in a circular orbit and gives it a small perturbing "nudge", the perturbed particle will appear to oscillate relative to the non-preturbed particle. Precession exists when the period of this oscillation is different from an orbital period.

7. Feb 5, 2016

### Staff: Mentor

I would say it's the fact that in GR, gravity is not a simple inverse square central force the way it is in Newtonian physics. In general it isn't a force at all; but even in the weak field, slow motion limit, where it can be viewed as a force, it isn't a simple inverse square central force. There is also an extra piece of the interaction which is velocity-dependent (some sources refer to this as the "gravitomagnetic" interaction, and the inverse square Newtonian force as the "gravitoelectric" interaction). But only a pure inverse square force can produce closed elliptical orbits.

8. Feb 6, 2016

### haushofer

Looking at that visualization, you can wonder how scalar theories of gravity account for precession (they do, in fact, but with the wrong amount). So I'm a bit sceptic about such a visualization.

9. Feb 6, 2016

### m4r35n357

I would say that the precession really is caused by the additional term in the gravitational potential (derived from the Schwarzschild solution). There is simply "more gravity" very close to a black hole than to a sphere (or point) of the same mass in Newtonian gravity. See for example equation 7.48 in Carroll.

As I understand it, it is entirely possible (but masochistic) to use curved spacetime to model Newtonian gravity, so that can't be the explanation.

10. Feb 6, 2016

### PAllen

Note, this formulation does not involve a spacetime metric at all, nor spacetime curvature. Only spatial components of the curvature tensor are non-vanishing, same for the connection.

11. Feb 6, 2016

### m4r35n357

OK, thanks for the clarification!

12. Feb 6, 2016

### haushofer

It isn't. In Newton-Cartan theory there is no precession. I agree with your statement that precession is caused by such an additional term, and I'm not sure if this visualization is really an explanation. As I understand, already for scalar theories in flat space(time) one can find that an orbit does not close after a full rotation, hence giving precession.

13. Feb 6, 2016

### haushofer

Newton-Cartan can be casted in a metrical formulation, but this metric structure is degenerate because there is no spacetime interval being invariant under the local isometry group (rotations+boosts). But this is no problem; the Riemann tensor is then completely given by the degenerate metrics and an additional closed two-form.

14. Feb 6, 2016

### Hornbein

Aha. So precession is greatest with Mercury because it has the greatest angular velocity amongst the planets?

15. Feb 6, 2016

### Staff: Mentor

It's a combination of angular velocity and closeness to the Sun, because the Sun's "gravitomagnetic field" gets weaker further away, just like the "gravitoelectric" field does.

16. Feb 6, 2016

### Hornbein

Right. But these quantities are directly related for an orbiting body.

17. Feb 6, 2016

### Staff: Mentor

Note, however, that this "additional gravity" depends on the angular momentum of the test object, so it isn't that the ordinary Newtonian force gets larger; it's that there is a different kind of force present, which is velocity-dependent, and therefore works more like a "magnetic" force than an "electric" force.

18. Feb 6, 2016

### PAllen

What I said is consistent with what you say. There is no spacetime metric. That is equivalent to saying there is no spacetime invariant interval. The artificial combination (which is wholly unnececssary to the formulation of the theory, anyway) is not a spacetime metric because it lacks essential properties. In my view, the logical formulation has connection (and curvature tensor constructed from it), but no metric. Connection is all you need for the geodesic equation and curvature. Trying to make the connection metric compatible adds nothing because the metric is not a true metric.

Last edited: Feb 6, 2016
19. Feb 6, 2016

### Staff: Mentor

Yes, they are, but the relationship depends on the orbital energy and angular momentum, which means it depends on the mean distance from the Sun, and also on the orbital eccentricity. I actually should have mentioned eccentricity as another factor that increases Mercury's precession.

20. Feb 7, 2016

### haushofer

If you think in terms of equations of motion. If you e.g. want to derive the geodesic equation from an action principle, you need the metric formulation. The metric formulation also emerges naturally if you consider Newton-Cartan theory as a gauge theory of the underlying Galilean algebra. So it depends on what you call "the formulation of the theory".

21. Feb 7, 2016

### PAllen

Ok, I can see that point of view, up to a point - the degeneracy of the metric flows from the degeneracy of the Galilean algebra (over 4 dimensions). However, I have never seen the value of the action principle derivation of geodesic equation in relativity (as opposed to all other uses of action principles), because making it work for null geodesics is highly artificial. Conceptually (and pedagogically), for relativity, I strongly prefer the 'straightest possible path' formulation of geodesic, made precise via parallel transport. Introduction of metric comes (IMO) logically after discussing geodesics and curvature (and it is important to realize that a metric is wholly unnecessary to define geodesics and curvature).

22. Feb 7, 2016

### bcrowell

Staff Emeritus
Well, $F\propto r^2$ also works, but that's obviously not applicable here.

23. Feb 7, 2016

### bcrowell

Staff Emeritus
The standard way to approach this is to use conserved quantities to reduce it to a one-dimensional problem with an effective potential. That works great, but it's also nice to have a description that works for a less mathematically adept audience.

As Peter Donis notes, any full explanation would need to explain why orbits *are* closed ellipses in the Newtonian limit. However, we can always take that as an established fact and then argue about perturbations on top of Newtonian gravity.

One such approach is to consider the spatial part of the curvature of the spacetime surrounding the sun. This spatial curvature is positive, so a circle's circumference is less than $2\pi$ times its radius. This causes Mercury to get back to a previously visited angular position before it has had time to complete its Newtonian cycle of radial motion. (I've forgotten where I first saw this argument, but I didn't originate it. This web page http://www.relativity.li/en/epstein2/read/i0_en/i1_en/ attributes it to Epstein, Relativity Visualized.)

One way to check that this argument makes some sense is to consider the case where the eccentricity approaches zero. The argument implies that the perihelion advance should approach a nonzero limit in this case, and that is in fact the correct result. (I used to have a different favorite explanation for the effect, but people here on PF convinced me that it was wrong for this reason.)

24. Feb 7, 2016

### Jonathan Scott

Why not be specific? See MTW "Gravitation" section 40.5, which discusses perihelion shift in the PPN approximation, especially equation (40.18):
$$\delta \phi_0 = \frac{2-\beta+2\gamma}{3} \frac{6 \pi M_{\odot}}{a(1-e^2)}$$
In this expression, $\beta$ is the non-linearity correction to the time component and $\gamma$ is the space curvature factor, both equal to 1 in General Relativity. As usual $e$ is the eccentricity and $a$ is approximately the semimajor axis (which is like the average radius of the orbit).
I see that setting both $\beta$ and $\gamma$ to zero does not give the Newtonian result of zero precession; I think that the Newtonian result actually corresponds to $\beta=2$ and $\gamma=0$, although I'm not sure about that.

25. Feb 7, 2016

### A.T.

I think here is the old thread on this:
https://www.physicsforums.com/threa...-newtons-gravity-and-g-r.626729/#post-4028708