# Why general relativity over other similar theories?

Since all theories where gravity is seen as a result of space curvature automatically satisfies the equivalence principle, why should one prefer GR over all other alternatives? Surely you can make the argument that GR fits with experimental data but Einstein when formulating his theory had no experimental data.

Why is GR a "simpler" theory to endorse over the alternatives? What made Einstein assume what he did? Are there any good reasons why space as described by GR and the Einstein field equations should behave in a such way? What makes other descriptions of space in alternative theories to GR "less likely?"

• What alternative theories are you considering? – Ernie Jul 23 '15 at 0:41
• There is a great deal of research that goes into looking at non-GR theories (see this book springer.com/us/book/9789400701649). GR has a lot of features that make it attractive but some which make it not so much (dark matter, dark energy, UV-IR incompleteness). – Arthur Suvorov Jul 23 '15 at 1:13

Let me try to break down the question into several parts, in the context of seeking a gravity theory that satisfies an action principle. That is, we are looking for a Lagrangian density that describes the theory.

First, the equivalence principle tells us that the gravitational field must couple universally to matter.

Second, the theory has to be (at least) a tensor theory. A vector theory would yield a repulsive force between like charges. A scalar theory would produce the wrong sign for potential energy, and this would violate observations concerning the equivalence principle. A theory with half-integral spin is even worse, as the exchange of such a particle cannot result in a static inverse-$r$ force. (This is explained very nicely in Feynman's Lectures on Gravitation -- Addison-Wesley, 1995.)

So we are looking for a tensor theory that couples universally to matter.

The simplest field that couples universally to matter would be the metric: it couples universally and minimally, meaning that the metric is used to form inner products and provides the volume element for the action integral, and nothing else.

So then, this suggests that we are looking for a metric theory. Variations of the Lagrangians of matter fields will yield their corresponding stress-energy tensors... so far so good. On the other side is the gravitational field. We wish to make sure that whatever field equations we obtain, in the weak field limit we get back Poisson's equation for gravity. So... what kind of an invariant scalar quantity, formed from the metric tensor and its derivatives, can represent the gravitational field?

For starters, it has to contain at least first derivatives of the metric tensor, otherwise we will not get the second derivative in the field equations that is required to reproduce Poisson's equation. It may contain second derivatives of the metric, but we have to be careful: if the resulting field equations contain higher derivatives, we may run afoul of the Ostrogradsky instability.

One of the simplest scalar quantities that we can form from the metric and its derivatives is the curvature scalar. If we make the curvature scalar the gravitational Lagrangian, the resulting field equations are Einstein's: this is general relativity.

A trivial and permissible modification is to add a constant to the curvature scalar. (This changes the field equations because even a constant is multipled by the volume element, formed from the metric, in the action integral.) The resulting field equations are again Einstein's, but with a cosmological constant.

Other scalar quantities formed from the metric do not reproduce Poisson's equation. For instance, there is conformal gravity, in which the gravitational Lagrangian is formed from the conformal (Weyl) tensor. In the weak field approximation, this results in a quartic equation of motion, and it is incompatible with the notion of a perfect fluid approximation of nonrelativistic matter. Its proponents address this criticism by arguing that the underlying fermion fields are not perfect fluids, but I find that argument unpersuasive.

What if we replace the curvature scalar with a scalar function of it in the Lagrangian? These would be the so-called $f(R)$ theories. They turn out to be more-or-less equivalent to another family, the family of scalar-tensor theories: in this case, an additional scalar field is introduced that also couples minimally, in essence replacing the gravitational constant with a field.

These and many other metric theories of gravity (metric meaning no direct coupling between additional fields and matter, so matter particles move along geodesics) can be studied using the so-called parameterized post-Newtonian (PPN) formalism. Described in detail in Clifford Will's book (Theory and experiment in gravitational physics, Cambridge, 2000) this formalism has as many as 11 parameters that can be subjected to constraints derived from observation. To date, general relativity wins: in all precision gravitational experiments, constraints are zeroing in on GR, and other theories are either excluded or require ridiculously fine-tuned parameters.

These considerations, of course, were not available to Einstein back in 1915. What he was seeking was a theory in the form of $G_{\mu\nu}=8\pi GT_{\mu\nu}$ where $G_{\mu\nu}$ is some tensor that is a) formed from the metric, and b) satisfies a conservation law (i.e., it is divergence-free). The first condition was a reflection on the fact that the equivalence principle implies that the theory has to be geometrical: i.e., given that material particles respond to gravity the same way, independent of their composition, the effects of gravity can be transformed away, at least locally, by a geometric transformation. As for the second condition, it was a reflection of the fact that the matter stress-energy tensor, $T_{\mu\nu}$, is itself divergence-free.

He had many false starts: he tried $G_{\mu\nu}=R_{\mu\nu}$ (works fine in the vacuum but $R_{\mu\nu}$ is otherwise not divergence-free) and he was not aware of the Bianchi identities that, when contracted, yield $(R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R)_{;\mu}=0$. Nonetheless, in late 1915, he stumbled upon $G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R$ and the theory was born. However, it was not Einstein but Hilbert who derived this form of the left-hand side of the Einstein equations from what is today called the Einstein-Hilbert Lagrangian, ${\cal L}_G=R$.

And lest we forget, Einstein's original goal was not to seek a new theory of gravity. What he was seeking was a generalization of the theory of relativity (later to be known as the special theory) to accelerating frames. It was the realization that the equivalence principle means that acceleration and gravity cannot be distinguished that led him to the conclusion that the theory necessarily must account for gravity. Even the name reflects this: Einstein's theory is the general theory of relativity.

• "he was seeking was a generalization of the theory of relativity (later to be known as the special theory) to accelerating frames." -- Just wrong. STR has no absolutely no problems with accelerated frames, and Einstein was actually seeking a relativistic theory of gravity. The oft-repeated claim that one needs GTR to deal with accelerated frames is a very silly myth that has no basis in reality. – Stan Liou Jul 23 '15 at 23:43
• From what I know, SR handles accelerated frames quite well. – Madde Anerson Jul 24 '15 at 0:13
• SR can handle accelerated frames, but they are not treated as equivalent to inertial frames. The sought-after theory was a theory that generalizes the concept of equivalence of reference frames from inertial frames to all frames. This is why the new theory was called the general theory. And while Einstein recognized fairly early on that gravity must be part of the picture, my impression (from Pais's scientific biography; obviously, I have no direct knowledge of Einstein's thoughts) is that the initial motivation was this generalization to noninertial frames, not gravity. – Viktor Toth Jul 24 '15 at 0:39
• @ViktorToth it's rather trivial to treat all frames equally in STR (in the same sense that GTR does): just express your equations of motion in tensorial form. Then no coordinate system, inertial or otherwise, is intrinsically distinguished. The difference in GTR lies in the fact that metric is dynamical and not necessarily flat. ... Saying that Einstein's primary motivation was treatment of accelerated frames is a really bizarre claim. Rather, it was treatment of gravity that obeyed the equivalence principle, a goal presented from the earliest (1907 and 1911) papers on the subject. – Stan Liou Jul 24 '15 at 1:08
• @StanLiou, in SR, noninertial frames are second-class citizens. That's why Einstein himself asks the question (Jahrbuch der Radioaktivität und Elektronik, 1907, quoted by Pais), "Is it possible that the principle of relativity also holds for systems which are accelerated relative to each other?" Einstein also states, "[the equivalence principle] assumption extends the principle of relativity to the case of uniformly accelerated motion of the reference frame" [ibid]. (He considered nonuniform acceleration in 1912; that's when he first used "equivalence principle" to describe this assumption.) – Viktor Toth Jul 24 '15 at 1:22

It was the first to fit the observations (e.g. the anomalous precession of Mercury), while having no free parameters to "adjust". The only parameter is the gravitational constant, which was already known with high precision in 1916. Its every prediction since either has been confirmed or is consistent with observations without any massaging. It has attractive theoretical structure induced by general covariance, that fits intuitions about role of symmetry in physics currently prevailing among physicists. And of course it has authority of Einstein behind it. It is not without flaws, but there is no clearly better alternative. Until then "if it is not broken don't fix it" is at work.

• If an observation is not consistent with GR, it is then called dark matter or dark energy. And all is well. – Andrey Sokolov Jul 24 '15 at 2:47
• Dark matter has now been photographed, so it was a correct prediction theday.co.uk/science/… Dark energy is essentially glorified cosmological constant, which is an exception to no massaging, but one that confirms the rule. Einstein originally proposed it not to fit observations, and then observations confirmed the expanding universe prediction made without it. On dark energy the jury is out, but it may yet turn out to be a correct prediction. If not, GR will be modified or replaced. But accelerated expansion is an issue in alternatives too. – Conifold Aug 2 '15 at 0:57
• Dark matter was never a prediction. It just refers to the discrepancy between the matter that we see and what needs to be there to account for the motion of stars and galaxies. To my knowledge no one knows what dark matter is. So, for the time being it is a discrepancy between GR and observations. – Andrey Sokolov Aug 2 '15 at 5:46

As you correctly say, any theory grounded on a $C^2$ manifold kitted with a metric encodes the equivalence principle. The curvature tensor then encodes all the physics. This is the first intellectual step. All that is left is to work out what the evolution equation for the curvature tensor / metric will be that determines the metric for our manifold.

So it seems reasonable to postulate that the geometry, or the curvature tensor, is to be set somehow by the matter / energy distribution. This is a theory of gravity, so we have a vague idea that $\mathbf{G}\propto \mathbf{T}$ is going to be the "look" of our theory, where $\mathbf{G}$ is some geometrical object that will fix the Riemann curvature in some way and $\mathbf{T}$ models the energy distribution in the World. We want our theory to be co-ordinate free, so a good choice for the geometrical beast $\mathbf{G}$ is some tensor related to the Riemann tensor, and $\mathbf{T}$ is some other tensor field that models the distribution "stuff" in the World. This is our second intellectual step.

There is other physics, though, aside from the equivalence principle, that we wish to encode. We wish to encode the local conservation of energy-momentum in our theory. The stress energy tensor $\mathbf{T}$ models the "flow" of the momentum four vector in spacetime and we express the local conservation of energy by the vanishing of the divergence $\nabla\cdot \mathbf{T} = T^{\mu\,\nu}{}_{;\nu}=0$

So if we are going to describe the distribution of "stuff" by the stress energy tensor $\mathbf{T}$, then we should like the left hand side $G$ to be (i) a purely geometric object, (ii) whose divergence automatically vanishes as a general property and is (iii) a second rank tensor like $\mathbf{T}$. What object does this? It turns out that the only second order tensor with these properties is the Einstein tensor, and its divergence vanishes by dint of the Bianchi identity (let's see whether I can get the square brackets in the right spots, I hate (anti)symmetrization brackets) $R_{\mu\nu[\rho\sigma;\tau]} = \, 0$ where $\mathbf{R}$ is the Riemann tensor. It does this because the Bianchi identity is the expression of the general behavior that the boundary of a boundary of a volume is always empty.

So we postulate $\mathbf{G}=\kappa\,\mathbf{T}$ where $\mathbf{T}$ is the stress energy tensor and $\mathbf{G}$ the Einstein tensor as the simplest possible rank 2 tensor equation with the "look" of "geometry proportional to stuff". It now remains to "calibrate" our scaling constant $\kappa$, which one does by looking at the nonrelativistic limit and observing that the field equations asymptote in this limit to Newton's law (or, more correctly, the gravitational Poisson equation) as long as we take $\kappa = 8\pi\,G/c^4$.

So $\mathbf{T}$ fixes the rank two Ricci tensor. This is not the same as fixing the curvature tensor or the metric. There are still unfixed degrees of freedom in the Riemann tensor, namely the part called the Weyl Tensor. However, it turns out the derivatives of the Weyl tensor are fixed by the derivatives of the Ricci, so that a specification of the Ricci, through the Einstein tensor and, by the Einstein equations, through the mass/energy distribution, together with suitable boundary conditions will indeed fix the Weyl. See Cesaruliana's wonderful answer to the Physics SE question "Why do the Einstein field equations (EFE) involve the Ricci curvature tensor instead of Riemann curvature tensor?".

## protected by Qmechanic♦Oct 9 '15 at 5:57

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).