# Why isn't general relativity equivalent to Newtonian gravity?

I know this question may seem a bit laughable, but the way the equations for general relativity are formed is through Poisons equation: $$\nabla^2\phi=4 \pi G \rho$$ Which are formed using Newton's law of gravity? You may say 'General relativity describes curved space-time through the metric and Newtonian gravity describes a vector field". However, wouldn't that be equivalent? If we draw an axis of space and time and draw an accelerated world-line and made it straight, it would look like space-time is curved. So why is Newton's law different to Einstein's laws, if we can describe the phenomena through 2 different ways? Is it the energy-momentum tensor? When we use the geodesic equation I interpret that as changing curved space-time into a vector field to describe the motion relative to flat space-time, why wouldn't that vector field be equivalent to newtons laws? Sorry if these question seems trivial, I'm just confused that's all.

• Related post by OP: physics.stackexchange.com/q/541697/2451 Apr 8, 2020 at 17:26
• @JoshuaPasa That is why QMechanic said "related" and not "duplicate" Apr 8, 2020 at 17:31
• @JoshuaPasa No, posting a link definitely does not cause a question to get closed. I suggest you read about the process of casting close votes (and specifically identifying duplicates) in our help center to understand how it works. Apr 8, 2020 at 17:39
• @Edouard: You are misunderstanding Guth's argument. It is not that Newtonian space (infinite with constant avg. density) can't exist, it's just it cannot be static. It is either must be expanding (just like our universe) or collapsing. Apr 10, 2020 at 18:49
• @A.V.S. How does space expand in Newtonian spacetime, because I thought that was a feature of general relativity? More specifically the cosmological constant in Einstein's field equation. Apr 10, 2020 at 18:56

Why isn’t General Relativity equivalent to Newtonian gravity?

Poisson’s equation is linear in the potential. Einstein’s equations are nonlinear in the metric. There is no mapping that could make them equivalent, because they don’t even have the same number of degrees of freedom. (The potential is one number at each point; the metric is ten numbers at each point.) However, Einstein’s equations do reduce to the Poisson equation in the weak-gravity limit.

• So, can you interpret the metric as a "tensor potential"? Apr 8, 2020 at 17:28
• I don’t recommend it. To me, that just confuses the two theories. There is, however, a weak-gravity correspondence, which is $g_{00}\approx -1-2\phi$. Apr 8, 2020 at 17:35

Why isn't general relativity equivalent to Newtonian gravity?

Poisson equation alone does not allow for relativistic causality.

An important ingredient of relativistic theory is the existence of finite speed of signal propagation. Poisson equation contains only spatial derivatives, so any changes in the sources are instantaneously reflected in the potential, and if the potential alone determines some observables of the theory we would have instantaneous propagation.

Note, that it is still possible to have Poisson equation and relativistic causality if the theory also has additional degrees of freedom that propagate relativistically and that observables of the theory are gauge invariant functions of the potentials. This is the case with electrodynamics in Coulomb gauge. The equation for scalar electrostatic potential is precisely the Poisson equation but there is no causality violation, since contributions from the vector potential would compensate for the effects of instantaneous propagation of electrostatic potential.

Similar situation could occur in (linearized) theory of gravitational field: by imposing appropriate gauge condition one could have Poisson equation for a specific component of the gravitational field (which we could identify with Newtonian gravitational potential), but the theory must also have additional degrees of freedom independent on this one potential to retain relativistic causality.

So just knowing from astronomical observation that Newtonian gravitational theory describes the Solar system really well and postulating principles of relativity we would inevitably come to the conclusion that the theory must be relativistic theory of spin–2 field (spin–0 is eliminated by absence of gravitational aberration established by Laplace with high degree of precision, while for spin–1 field like charges would experience repulsion rather than attraction), i.e. general relativity.

• Can we define a vector potential for gravity to make it causal and Lorentz invariant? The reason I'm asking for this is because the metric doesnt, directly, define the EOM of general relativity, however, we can use it in the geodesic equation to get a vector field. I don't know how to exactly do that but it may be possible. Would the degrees of freedom match up then? Apr 10, 2020 at 19:52
• There is an approximation to GR called gravitoelectromagnetism, where one can define a vector potential analogously to EM field. It would correspond to metric components $h_{0i}$. But for a fully relativistic theory one would also need $h_{ij}$ components. because the metric doesnt, directly, define the EOM of general relativity what do you mean? Einstein field equations (EFE)? Metric does not define EFE it satisfies EFE. Apr 10, 2020 at 20:15
• What I mean by that is the metric had to be used again to define the EOM, but doesn't describe the EOM by itself. The EOM would describe a vector field analogous to Newton's laws. Apr 11, 2020 at 17:19
• Also why would gravitoelectromagnrtism not be Lorentz invariant if it's in the same form as Maxwell's equations, which are Lorentz invariant Apr 11, 2020 at 17:20
• @Edouard: … and your point? Not every family of Lorentzian spacetimes has a consistent Newtonian limit (or when they do, interesting features may disappear), it is quite possible that your “multiverse” model does not have such limit. If you do have questions about Newtonian cosmology, make a dedicated Q&A thread for them. Apr 19, 2020 at 6:56