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.
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Related post by OP: https://physics.stackexchange.com/q/541697/2451 – Qmechanic Apr 08 '20 at 17:26
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@JoshuaPasa That is why QMechanic said "related" and not "duplicate" – BioPhysicist Apr 08 '20 at 17:31
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@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. – David Z Apr 08 '20 at 17:39
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Although you didn't include a cosmology tag, no physics would work without a cosmos to work in, so let me point out that, on p.296-297 in the 1997 ed. of Guth's pop. sci. book titled "The Inflationary Universe" (available on inter-library loan in most parts of the U.S.), he uses a very simple algebraic proof to show that a universe formed in Newtonian gravity would collapse in the instant of its formation. – Edouard Apr 08 '20 at 18:10
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@Edouard: that a universe formed in Newtonian gravity would collapse in the instant of its formation this is wrong statement and Guth never proves such thing. Newtonian homogeneous and isotropic cosmology evolves under the same Friedmann equations as FRW solution in GR (with irrelevant in Newtonian gravity parameters set to zero). There is no instantaneous collapse. – A.V.S. Apr 09 '20 at 18:44
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@A.V.S. -My point isn't that space instantaneously collapses: It's that Newtonian space can't exist. Please look at the algebraic solution on p.296-297 in the 1997 ed. of the book I've cited, than tell me what's wrong with it. (I won't be detailing it here, as the book will remain under copyright until 2027.) Thanks. – Edouard Apr 09 '20 at 21:42
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Re my last comment, Guth cites no source for his printed proof, so the solution may've been either original, or given to him verbally. His proof won't fit into a PSE "comment", but whatever errors it might contain (-I sure don't see any) might. – Edouard Apr 10 '20 at 16:43
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1@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. – A.V.S. Apr 10 '20 at 18:49
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1@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. – Joshua Pasa Apr 10 '20 at 18:56
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@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. – Joshua Pasa Apr 10 '20 at 18:56
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@JoshuaPasa: First of all expanding universe in GR could exist without cosmological constant, it is accelerating expansion that is being explained by c.c. And Newtonian cosmology is a perfectly consistent class of solutions, see for example https://doi.org/10.1119/1.18398 or works by Ehlers (one, two, three ). – A.V.S. Apr 10 '20 at 19:52
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@A.V.S. So could a Newtonian universe have a "cosmological constant" – Joshua Pasa Apr 11 '20 at 17:14
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@JoshuaPasa: Yes, you could instroduce c.c. into Newtonian gravity. For example the equation for gravitational potential would be $\Delta \Phi = 4 \pi G \rho - \Lambda$ (Note, that in the context of Newtonian cosmology potential is somewhat ambiguously defined, so one should treat it only as locally defined function). – A.V.S. Apr 11 '20 at 18:49
2 Answers
I’ll answer your first question. You are supposed to ask one, not six.
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.
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1I don’t recommend it. To me, that just confuses the two theories. There is, however, a weak-gravity correspondence, which is $g_{00}\approx -1-2\phi$. – G. Smith Apr 08 '20 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.
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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? – Joshua Pasa Apr 10 '20 at 19:52
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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. – A.V.S. Apr 10 '20 at 20:15
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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. – Joshua Pasa Apr 11 '20 at 17:19
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Also why would gravitoelectromagnrtism not be Lorentz invariant if it's in the same form as Maxwell's equations, which are Lorentz invariant – Joshua Pasa Apr 11 '20 at 17:20
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why would gravitoelectromagnrtism not be Lorentz invariant … Because it is not the full theory, but a truncation of GR. There are basically two widely used approximations to GR: weak field (when metric differs from Minkowski only slightly) and post-Newtonian (when characteristic velocities are assumed to be small compared to $c$). GEM is intersections of both, it works when velocities are small and metric is almost flat. There are other terms (e.g. proportional to $1/c^4$, so of a higher order in post-Newtonian expansion) that have no direct analogs in Maxwell's equations. – A.V.S. Apr 11 '20 at 18:44
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In the last (2003's) revision of the BGV Theorem, its authors accepted (in a footnote) the Aguirre-Gratton cosmology "steady-state eternal inflation" (which balances expansion against contraction) as plausible within their theorem's parameters, so it appears to me that the possibility that contraction in either of two causally-separated regions might be balanced by expansion in the other to arrive at a net expansion or contraction equaling zero in a multiverse divided between causally-separated local universes or temporal iterations. This is compatible with GR, but not in Newton's universe. – Edouard Apr 19 '20 at 03:59
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1@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. – A.V.S. Apr 19 '20 at 06:56
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My point had just been that Newtonian physics was based only on a uni-verse, with the possibility of local causal separations (mainly, "black holes") having been theorized only some years after his death. Given the resemblances between the cosmological and event horizons, black hole cosmology has (as recognized by Smolin, Poplawski, and others) some explanatory potential that Newtonian physics lacks, especially in the vicinity of non-negligible curvature. It also has some observational basis, including the elliptical orbits of formerly binary stars whose partner's collapsed. – Edouard Apr 19 '20 at 13:46
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It should be clearer if "possibility exists" is substituted for "possibility" alone, in my comment before my last one. Sorry for that goof. – Edouard Apr 19 '20 at 13:57