Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Calculating using the equivalence principle only accounts for half the deflection of light, whereas the other half is from curvature of space-time.

But isn't the equivalence principle the same thing as curvature of spacetime?

Deflection of light has two parts: The first part is falling people measure the deflection. Why is it significant for falling people to do this and how does this follow principle of equivalence? Falling people means gravity and curvature of spacetime doesn’t play a role. This means we’ll know how much deflection is caused by the motion of the photon and general relativity? The second part is the curvature of space. For example, if you have a rod, the rod itself would bend around the Sun. These two parts are like two halves, each of it is one half the deflection.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

There's a very intuitive, visual explanation of how the equivalence principles relates to the bending of light here, which includes an explanation as to how the geometry of space enters the picture in terms of link up up infinitesimal patches of space together.

Note that it the freely-falling frame is important here because according to the principle, physics should follow special relativity (over a small enough patch). That's what allows one to conclude that the light beam should be straight in that frame... and therefore must bent in the other.

But answer your question directly, no, it is simply not the case that the equivalence principle is "the same thing as" the curvature of spacetime. Rather, the equivalence principle establishes that gravity and the geometry of spacetime are inextricably linked, but it does not by itself prescribe a particular geometry that spacetime must follow.

Einstein's version of the equivalence principle, from which the deflection of light follows, is:

The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

In any spacetime geometry, a small enough patch looks flat, so in a free-falling frame spacetime looks just like special relativity (to first order). But this is true in any theory of gravity in which gravity manifests as spacetime geometry with gravitational free-fall following a geodesic. It does not have to be the geometry predicted by general relativity.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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