What we recognize as non-relativistic (Newtonian) gravity comes from the timelike part (i.e., the 4th dimension) of the Einstein field equations. If you take the nonrelativistic limit, this part of the field equations will decouple from the rest and will appear as a scalar, not tensor, field equation; indeed, it is none other than the Newton-Poisson equation for gravity, ∇2ϕ = 4πρ. The magnitude of the contribution of spatial curvature (which would be in the form of a three-dimensional tensor equation) would be (ϕ / c2) times the Newtonian potential, thus negligible (unless you want to calculate small corrections, like those needed to explain Mercury's perihelion advance or the bending of light by the Sun.)

OK. But that is a posteriori. I meant to ask if, for example for a small test mass in the presence of a central large point mass, starting with a non-relativistic Lagragian (that is, the metric tensor would be the unit matrix) without forces, one can introduce changes in the metric tensor and, applying the Euler-Lagrange equations, get motion equations that in the limit of large radius.

A nonrelativistic Lagrangian means that time is absolute. Any metric, therefore, would be 3-dimensional. The Euler-Lagrange equations you get out of that three-dimensional part will correspond to velocity-dependent post-Newtonian corrections to the Newton-Poisson equation, not to Newtonian gravity. So I believe that the answer to your question is a no.

Another way of thinking about it is that the source of gravity in the rest frame of a gravitating object is that object's rest mass. There is no term representing the rest mass in a non-relativistic Lagrangian; for a free particle, all you have is the kinetic energy term (1/2)mv2. However you combine this with a spatial metric, you will get a velocity-dependent term in the Lagrangian, not the static Newtonian potential −Gm/r.

Viktor,

I believe that if we add to the classical metric (dx2+dy2+dz2) a term (2GM/r dt2), the geodesic satisfies the classical Newton equation.

I attached a file where I wrote it for a free fall problem (null angular momentum).

Could you take a look at it and tell me if that makes sense to you?

Thanks,

Alejandro

Alejandro: Unless I am mistaken, I do not think that you can recover the Newtonian acceleration law from your metric. Please see attached.

An even bigger, conceptual problem with this approach is that by giving up relativity, you are explicitly eschewing general covariance. Under Galilean transformations, the geodesic structure will be observer dependent. So while an observer at rest may recover something like Newton's acceleration law, a moving observer will see something different.

Thank you very much Viktor.

I had done all wrong. My curiosity stems from the fact that generalizing the classical Lagrangian with time as a coordinate it is possible to obtain Newton's equation (see attached). I was trying to look at the generalized Lagrangian as a metric. After all, the only difference between the classic action with time as coordinate and the line integral is a factor dt/ds. But as you showed me, that leads to a different motion equation.

Thanks again.

Alejandro

Q: Is it necessary to work in the Minkowski space-time to introduce the gravity as a perturbation of the metric tensor? A: No , but it does not do any harm.

A geometric formulation of Newtonian mechanics was actually developed by Elie Cartan in 1923-1925 and by Kurt Otto Friedrichs in 1927. This formulation can be found in standard textbook, e.g. Gravitation from Misner, Thorne and Wheeler.

Ten years ago, with Brandon Carter we showed that Newtonian mechanics can be formulated in a 4D fully covariant framework based on a variational (action) principle, and we derived the expression of the Lagrangian in the general case of self-gravitating (super)fluid mixtures. This formalism, which was later extended to elasticity, relies on exterior calculus. Please find attached the first paper of the series.

Article Covariant Analysis of Newtonian Multi-Fluid Models for Neutr...

Nicolas:

Wonderful! This is exactly what I was looking for. Misner et al Gravitation Chapter 12.

Thanks a lot.

A couple of comments on Cartan's theory: 1) there is no spacetime metric. The fundamental objects of the theory are the connection coefficients. 2) A spatial metric exists, but it does not determine the theory of gravity; that comes from the only nontrivial connection coefficient, which is Γj00.

I also wonder how Cartan's approach can be combined with a Lagrangian formulation (the topic of the original question). In general relativity, the "universal and minimal" coupling to the metric comes as the metric a) defines inner products, and b) defines the volume element. In Cartan's theory, there is no metric, no volume element I guess, so the geometric effect would need to come solely from the definition of the covariant derivative.

Looking at it more closely, I note that given a free particle Lagrangian L = (1/2)mv2, the Euler-Lagrange equations of motion will be given by (d/dt)(mv) = 0. In order to use Cartan's connection coefficient in the first place, I think it is necessary to treat v as a four-velocity. If I do that, the equation translates into ma0 = 0, and mak + Γk00v0 = 0 (aμ = (d/dt)vμ). This replicates the Newtonian acceleration law only if v0 = m. Hmmm... interesting.

Effectively, I was reading Cartan's approach and, as Viktor says, there is no metric but just Γj00.

Now, assuming a diagonal metric and only 1 spatial coordinate, following Viktor’s previous derivation the motion equation would be:

x’’ – 2 Γttx (x’)2 + Γxtt =0

where the prime indicates temporal derivative and:

 Γttx = 1/2 gtt dgtt/dx

 Γxtt = -1/2 gxx dgtt/dx

I believe that Newton’s law cannot be derived from here, for it would require gtt = 0.

Now, one can think in approaching Newton’s law for small velocities, which would require:

x'2

Alejandro: In your notation, the geodesic equation is x'' + [1 - 2x'2/(v02 + 2U)]dU/dr = 0. Indeed, if you make v0 large enough such that x'2/v02

First of all, note that tensors are intrincally covariant or contravariant in Newtonian spacetime because of the absence of any metric (tensor indices can no longer be raised or lowered).

Even though the Newtonian spacetime does not have any metric, we can still define a 2-rank spacetime "pseudo" metric tensor by pushing forward and pulling back the spatial Euclidian metric gij and gij respectively. Strictly speaking these "pseudo" metric tensors are not true metric tensors, as remarked by Alejandro, because they are degenerate: they possess a null eigenvector. The null eigenvector of gab can be interpreted as the gradient of the absolute Newtonian time t. The degeneracy of gab implies that there exists another null eigenvector ea, which can be interpreted as the observer's 4-velocity with respect to the absolute "ether" frame. We can now define a coordinate system such that e0=1 and ei=0: this corresponds to an observer at rest in the ether frame. Of course, we know from Galilean invariance that the laws of Newtonian mechanics remains unchanged for observers in relative uniform motion with respect to the ether frame. Therefore, any physical tensor should be invariant under the shift ea+ba. It turns out that when gravitation is included, these transformations can be generalised to time-dependent velocity (boost) ba(t). Regarding the spacetime connection, we can no longer use the Christoffel expression since there is no metric. Still one can define a spacetime connection, as discussed in the paper attached in my previous message.

P.S. The correspondence between Newtonian and relativistic spacetimes is discussed in Section 3 from the attached paper.

Article Covariant Newtonian and Relativistic dynamics of (magneto)-e...

It is not necessary to introduce gravity as a perturbation of the metric tensor. Gravity can be introduced as a deviation to the flat Minkowski metric in the form of a ratio of Newtonian gravitational acceleration and the measured acceleration. This ratio happens to be 1 (exact 1) for the flat Minkowski metric. In case of rotation curves and perihelic precession of planets, it has values other than 1. Following is the link to a complete theory based on this criteria. This theory satisfies Einstein's field equations and does not use the weak field approximation.

Article Periodic relativity: Basic framework of the theory

Article Supplement to periodic relativity

If you think in terms of fundamentals instead of mathematical constructions, the following conclusion is easily reached... without constraining the speed of light to be a constant locally, which amounts to a Minkowskian metric, it is hard to imagine deriving features of curvature such as Shapiro delay and so-called double bending.

However, you can get a sort of flat gravity if you want, because you can get time dilation without using SR.  Einstein showed this in 1911.  This is good enough to get the precession of Mercury, and misled me for some years.  It cannot get the double bending (at least not without invalid sleight of hand) or the Shapiro delay.

Please read the attached file . A topically irrelevant but sill important correction I have to add as to the picture reproduced therein, painted in the 'Chamber of signatures' in the Vatican: it is not meant to show Heraklit ,  but rather  represents Michelangelo in its middle and is painted 1509-1511 by Raphael. I thank Professor Doplicher  for pointing these errors out to me recently in Corfu.

Continuing from my previous post on page 2.

The metric of general relativity (Schwarzschild metric) cannot explain Hydrogen spectra. The metric of periodic relativity can explain Hydrogen spectra.

Article Hydrogen spectra using Einstein's field equations