6
$\begingroup$

Context: One of the key reasons Einstein first suspected space is curved by mass is due to the falling elevator thought experiment. This experiment implies that mass must curve the path of a light ray. From this, he inferred that space must be curved. But why come to that conclusion? (It could rather be for instance that space is flat and the light simply "falls" through flat space like any other object). Apparently the reason is that it was believed that light must follow a geodesic. Thus if space was flat, light would have to travel in a straight line. The thought experiment suggested light actually follows a curved path, and hence by implication, space must curved.

What first lead physicists to believe that light must follow a geodesic, and why?

I suspect the reason may be just a logical extension of Fermat's theorem for light, but this may be entirely wrong.

$\endgroup$
1
  • $\begingroup$ I don't know the answer to your question Rory. But I do know that Einstein didn't think space was curved by mass. In his 1920 Leyden address he described a gravitational field as a place where space was "neither homogeneous nor isotropic". He also said "As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable". Space is altered by a massive body, but it isn't curved. $\endgroup$ Dec 31, 2019 at 15:29

1 Answer 1

4
$\begingroup$

It is the so-called Fermat Principle in geometric optics: light chooses the path which takes it least time to travel. Since the speed of light (in the given homogeneous medium) is constant, this must be the shortest path. Later refinements tell us that it is not always shortest but anyway its length must be stationary. Which exactly means a geodesic for some metric. For empty space this is Minkowski's metric, otherwise, some other metric.

This was the simple answer. But one can give a deeper answer, if one looks at the notions of "geodesic" and "straight line" in their historic development. Long before the invention of mathematics, people came with the notion of a straight line, the line on which light travels. (Just think of practical method of checking that a line is straight. The most basic of them just compare it with a light ray.) Mathematics is a later invention. The notion of the straight line (distilled from from observations of nature!) was axiomatized and became a part of mathematics. The ancients also knew that a straight line minimizes distance. Calculus of variations is a still later invention, as well as the general idea of idea of geodesic. So it should not come as a surprise if in the end we obtain that light travels on geodesics:-) All these relevant notions of mathematics developed from the original observations of light rays. This explanation depends on the point of view that our mathematics comes from the observation of nature as its final source. This point of view is not shared by everyone.

$\endgroup$
2
  • $\begingroup$ Yes. I reckon you could make an inner product using the refractive index and the regular inner product space. I think this might turn our flat spaced into a curved space (embedded in a higher dimensional flat space) . From this viewpoint Fermat's theorem could indeed be viewed as a curved path that takes the shortest distance between points. Here space would be curved by refractive index rather than mass. (This all assuming the space I propose actually can exist. It might not do) $\endgroup$ Dec 23, 2019 at 22:59
  • $\begingroup$ Yes indeed regarding your more detailed analysis of Fermat 's theorem. I never say minimised or maximised myself. I alway say "stationary" point as this alows for the other possibility, a point of inflection. Regarding metric. Yes indeed. I suspect if you use the inner product I suggested above this will give.you a Rieimann Metric on the "optical" space. This space then I suspect has all you need for it to be a Reimann manifold. By the Witney embedding theorems this optical space will exist as a curved surface in a higher dimensional flat space. $\endgroup$ Dec 24, 2019 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.