Einstein described his discovery of the equivalence principle as the "happiest thought of my life". Why? What, in broad conceptual terms, is the logical chain of reasoning that leads from the equivalence principle to general relativity?


4 Answers 4


In special relativity, unaccelerated observers cannot do any experiment to determine their state of motion in any absolute sense--all meaningful motion is relative to something else.

  • We want to generalize the principle of relativity, that laws of physics are independent of inertial motion, as far as possible.

Mathematically, this suggests using tensors, to make equations covariant with nigh-arbitrary coordinate transformations. But let's begin with the Eötvös experiment:

  • If under the influence of only gravity, every test particle behaves in an identical manner, no matter its mass or composition.

This is suggestive of making gravity a property of spacetime, rather than an interaction within spacetime. But even more important is that:

  • Objects in freefall don't feel their own weight.

In Newtonian mechanics, if you have two test masses near each other in a gravitational potential $\Phi$, they will accelerate in the same way, to first order. They will be at rest relative to each other. Of course, they are still at different positions, so experience different gravitational force, but this shows up as the tidal force proportional to $\partial^2\Phi/\partial x^i\partial x^j$, a second-order effect. Intuitively, gravity "doesn't exist" for infinitesimal separations $dx$ (and times $dt$): you need to look at larger scale, in space or time or both, to detect its presence.

In other words, over small enough regions of space and time i.e., experiments of short duration), physics is just as if you're an in an inertial frame. But since special relativity, not Galilean relativity of Newtonian mechanics is right,

  • Spacetime is locally Minkowski, the flat spacetime of STR.

Geometrically, things fall into place: in mathematics, a manifold is a structure that locally "looks like" flat Euclidean space. So both criteria--gravitational effects independent of mass or composition and physics behaving like special relativity over small enough regions can be done simply by having spacetime be a manifold that's locally Minkowski.

Geodesic deviation is also a second-order effect: that's the geometrical meaning of the Riemann curvature tensor, which is determined only by the second derivatives of the metric. (But Einstein didn't think of curvature as the actual gravitational field; more on that later.)

... my little box is pushed over a cliff and is in freefall. Presumably my box has now become a locally inertial reference frame, its spacetime is now flat and therefore described by the Minkowski metric.

That's right, to first order. To see what's going on more precisely, it may help to explicitly build coordinates out of this. Say you have a local frame of one timelike direction and three spacelike directions $\{\hat{e}_\alpha\}$, orthonormal (that's actually not necessary, but it's more intuitive if you start with a Cartesian-like situation), i.e., four vectors spanning the tangent space at your location in spacetime.

Pick any direction in spacetime... any vector can be written in terms of components of your frame vectors. Consider the geodesic in that direction; it will have some affine parameter along it (say, proper time/length along it if it wasn't a lightlike direction). You can therefore label each point on that geodesic with coordinates consisting of that affine parameter and the vector components of the direction you picked.

We've constructed Riemann normal coordinates corresponding to your local frame. The metric evaluated at your location (!!) is just Minkowski $\eta_{\mu\nu} = \mathrm{diag}(-1,+1,+1,+1)$, as it corresponds to the dot products of the frame vectors. And since the coordinates were defined in terms of geodesics and geodesic deviation is second-order, the metric in a neighborhood of your location is Minkowski to first order.

I'm still on planet Earth, but the spacetime has changed.

No, we've basically just decided to view the metric in coordinates $x^\mu$ such that at a particular event $p$, $g_{\alpha\beta}|_p = \eta_{\alpha\beta}$ and $\partial g_{\alpha\beta}/\partial x^\gamma = 0$. The metric is Minkowski at $p$ and its derivatives vanish.

It doesn't mean it's exactly Minkowski on a region around $p$--there will be a difference in second order. We can't make the second derivatives vanish in general because there's not enough degrees of freedom in a smooth coordinate transformation. But since the second-order derivatives are what determines curvature = geodesic deviation = gravitational tidal forces, we don't want those to go away.

Einstein came to the conclusion that gravity is given by a geometry of spacetime and free-falling objects are locally inertial. But from a modern perspective, we can use the latter to motivate GTR as well:

  • If objects in freefall have no proper acceleration (what an accelerometer measures), could they be actually be inertial?

On a manifold, the geodesics are curves that have the following property: transporting a tangent vector along the geodesic itself gives a parallel vector. In more physical terms, the velocity of a particle traveling along it is unchanged as it follows the geodesic. Geodesics naturally generalize the notion of "inertial motion = no acceleration". Of course, that's exactly what Einstein concluded, but the mathematics explicating these various relationships between the metric and the connection were published a few years after GTR.

But it's interesting that Einstein did not think of the spacetime curvature as corresponding to the presence of a gravitational field. Rather, he thought of the connection coefficients the geodesic equation $$\frac{d^2x^\alpha}{d\lambda^2} + \Gamma^\alpha_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda} = 0$$ as the components of the gravitational field. Some modern authors don't like this because $\Gamma^\alpha{}_{\mu\nu}$ are not tensorial, but conceptually Einstein's view makes more sense: GTR marries gravitation and inertia, and with the above notion of "inertial", the connection is exactly what determines parallel transport and hence inertial motion.

  • $\begingroup$ How does the observation that "objects in freefall don't feel their own weight" lead to the conclusion that gravity is local? Also, from the assumption that gravity is local, how do you conclude that "over small enough regions of space and time... physics is just as if you're an in an inertial frame"? I found useful the notion of gravity as a property of spacetime, but I still can't quite see the chain of logic that leads from free falling in a lift to GR. Thank you $\endgroup$
    – Peter4075
    Aug 12, 2011 at 19:58
  • $\begingroup$ Since freefalling objects have no proper acceleration (feel no weight), local physics is as if gravity didn't exist. If you're in a little box in space, you would not be able to tell whether you're far from any gravitating bodies or in fact falling towards one. The restiction of locality is simply because this can only work locally--by doing measurements over large enough regions, you can detect gravity by things like tidal forces, etc. (not to mention simply looking outside your box). Fortunately, "locally STR" corresponds in a precise mathematical way to how geometrical manifolds work. $\endgroup$
    – Stan Liou
    Aug 13, 2011 at 2:45
  • $\begingroup$ I'm in my little box on Earth where the spacetime in my box is slightly curved and described by the static weak field metric. Next, my little box is pushed over a cliff and is in freefall. Presumably my box has now become a locally inertial reference frame, its spacetime is now flat and therefore described by the Minkowski metric. I'm still on planet Earth, but the spacetime has changed. Does the action of being in freefall change the manifold of spacetime in my box so it's now described by a different metric? How? What type of spacetime would an external observer see in my freefalling box? $\endgroup$
    – Peter4075
    Aug 13, 2011 at 12:23
  • $\begingroup$ Peter, now that I've a clearer picture of your what you want to know, it's more convenient to simply rewrite the post focusing on that rather than answering in the comments. I hope the result is an improvement. $\endgroup$
    – Stan Liou
    Aug 13, 2011 at 19:58
  • $\begingroup$ So, in my little box I drop 2 stones and they experience the same gravitational acceleration (to 1st order). To 2nd order, they drift ever so slightly together (tidal forces). My little box is shoved over a cliff. I then again drop' 2 stones. In my box's new coordinate system, to 1st order, they remain in the same position. To 2nd order (tidal forces) they move slightly together. Freefall therefore automatically/'magically' creates a new coordinate system which (to 1st order) is flat spacetime. The spacetime hasn't changed but the coordinates have. That's clearer. Thanks for persevering. $\endgroup$
    – Peter4075
    Aug 14, 2011 at 10:56

perhaps Einstein was happy due to the fact that his general relativity resolved an age old dilemma in Newtonian Mechanics: why is inertial mass = gravitational mass?

here are some ways that the weak equivalence principle may be stated:

  1. The trajectory of a point mass in a gravitational field depends only on its initial position and velocity, and is independent of its composition.

  2. All test particles at the alike spacetime point in a given gravitational field will undergo the same acceleration, independent of their properties, including their rest mass.

  3. All local centers of mass vacuum free fall along identical (parallel-displaced, same speed) minimum action trajectories independent of all observable properties.

  4. The vacuum world line of a body immersed in a gravitational field is independent of all observable properties.

  5. The local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception.

  6. Mass (measured with a balance) and weight (measured with a scale) are locally in identical ratio for all bodies (the opening page to Newton's "Principia").

  • $\begingroup$ These are lots of different formulations of the equivalence principle, but you haven't addressed the OP's question of how this leads to GR. $\endgroup$
    – user4552
    Aug 10, 2011 at 23:06
  • $\begingroup$ Ben - what's an OP? $\endgroup$
    – Peter4075
    Aug 11, 2011 at 15:33
  • 1
    $\begingroup$ OP is used for either "Original Post" or "Original Poster", depending on what makes more sense in context. $\endgroup$
    – Stan Liou
    Aug 12, 2011 at 2:29

The basic idea is that since small, electrically neutral bodies with given initial conditions follow universally defined trajectories, you can define those trajectories to be straight lines. This leads to a theory of gravity that is geometrical in character. By taking the trajectories, which we'd normally think of as being curves, and calling them straight, one is forced to take the background of space and time, which we normally think of as being flat, and call it curved.


Short answer: the famous 'rocket elevator' thought experiment.

Long answer: It turns out that the motion due to gravitational force acting on some mass is exactly the same as the motion due to a mass in an accelerated reference frame. Einstein famously imagined the experience of standing inside a windowless box, with no way to contact the outside, accelerated by a rocket at $9.8 m/ sec^2$ which, in fact and in theory, is identical to the experience of standing inside a box that is sitting still on the surface of the earth. An observer in the box cannot ever tell the difference between these experiences.

This is, of course, only an abstract way of thinking about $(\text{gravitational mass}) = (\text{inertial mass})$. But, proceeding from this point of view and incorporating Special Relativity, it is possible to construct the entire theory of General Relativity.


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