# Observing light from a falling elevator

Today in my physics class, my teacher said that if one would point a laser into a falling elevator (acceleration because of a gravitational field), an observer inside the elevator would see the light form like a parabola, like i have tried to illustrate here: He told us this is because the elevator accelerates downwards, and therefore the light would "bend" to an inside observer. But i'm wondering how that is, since the light also falls inside gravitational fields. If the light falls at the same rate as the elevator, an inside observer would just see a straight line. Like the image below: My question is now: who is right? Will the light accelerate at the same rate as the elevator, or what will happen?

By the way, this is my first question on the physics stackexchange, so please tell me if i could improve my question, thanks!

• The laser is fixed, outside the elevator, pointing into the elevator. Is that correct? Oct 14, 2021 at 21:15
• Yes, the laser is fixed @PM2Ring Oct 14, 2021 at 21:50

You are right; teacher was a bit misleading but not totally wrong.

Proof: motion under gravity satisfies a principle called the Strong Equivalence Principle. This says that in a reference frame in free-fall physics is locally just the same as if there were no gravity at all and you had a reference frame in constant velocity motion. In such a frame light travels in a straight line.

Your elevator falling under gravity is an example of a reference frame in free-fall. Therefore relative to the elevator light travels in a straight line.

Relative to the Earth the light beam bends down owing to gravity, and the bend is exactly enough to make it be a straight line relative to the elevator. You can use this to deduce the bend of a light beam owing to gravity! Congratulations: you will then have done an exact calculation in general relativity. (If you do the calculation you will find the bend is very small; the radius of curvature of the light beam track relative to Earth is something like a light-year).

All the above describes any particular light wave emitted by the laser. Now let's think about the fact that the laser source is being accelerated upwards relative to the elevator. This means that each successive moment it emits at a different height relative to the elevator, so you have a succession of light waves each of which travels straight (relative to elevator), but the later ones started out from a higher-up location relative to the elevator. I think the teacher was trying to convey this aspect of the situation but maybe was struggling to do so clearly.

Added remark on direction of the light

If you shine a light beam into something that is moving relative to you (e.g. shine a torch into the window of a passing train) then there is a slight change of direction, just as you would expect from Newtonian physics. e.g. If the light beam is at right angles to the moving window surface in your rest frame, then in the train frame the light beam is not at right angles to the window surface. This affects the elevator experiment too, only now the relative velocity between laser and elevator is increasing so this effect gets larger as the elevator falls. The light still propagates in a straight line relative to the elevator, but its direction of travel will be changing (assuming the laser is fixed relative to Earth). (Thanks to PM2Ring for pointing this out.)

• You don't need the Strong Equivalence Principle, only the Einstein Equivalence Principle, because the propagation of light is a non-gravitational experiment. Oct 15, 2021 at 6:12
• The light beam is tilted up very slightly by aberration. Oct 15, 2021 at 6:26
• @PM2Ring Thanks for this comment; I see what you mean and therefore added a remark. Oct 15, 2021 at 8:13

If the elevator is in outer space without gravity or any acceleration, the light goes straight.

If it is in outer space, without gravity, but accelerated by some engine, the light bends to the opposite direction of the acceleration. By the equivalent principle, the same happens to an elevator at rest on the surface of Earth. The light bends downwards.

For a free fall elevator, under Earth gravity, due to the same equivalent principle, the light goes straight as in the first case.

• But isn't the gravitational field an inertial frame of refference, and therefore the line would be straight when viewing from inside the elevator? Oct 14, 2021 at 21:33
• In general relativity, the concept of inertial movement is extended by the notion of geodesics. In the absense of gravity both coincide: objects move in straight lines without net forces applied. On a gravitational field, the geodesics don't mean constant velocity trajectories. Oct 14, 2021 at 21:40

There is an ambiguity in what is meant by the path of the light.

If you mean the path of any particular photon (i.e., the light emitted at a given time), then your picture is right: This path is straight in the elevator frame by the Einstein Equivalence Principle. However, it is horizontal only if it was emitted at the moment the elevator was at rest relative to the laser. Otherwise, it is tilted because the velocity of the photon has a vertical component in the elevator frame.

If you mean the instantaneous locus of all photons in flight (i.e., the light existing at a given time in the elevator frame), then this path bends downward in the elevator frame -- opposite to the picture you attribute to your teacher. This may be counterintuitive, but consider the elevator after it has been falling at acceleration $$-g$$ for a time $$T$$ since being at rest relative the laser. Assume the vertical velocities involved are small compared to $$c$$, so the horizontal velocity of light is essentially $$c$$ (up to higher-order terms). In the elevator frame, the photon at horizontal coordinate $$x$$ (the laser is at $$x = 0$$) was emitted at time $$\tau = T - x/c$$, when the laser was at vertical position $$+g\tau^2/2$$ and had vertical velocity $$+g\tau$$. Thus the photon, after traveling in a straight line for time $$x/c$$, has current vertical position $$g\tau^2/2 + g\tau(x/c) = g(T^2 - x^2/c^2)/2$$. This is a downward-bending parabola. The same effect would be seen in Newtonian physics with, say, a hose emitting water or a machine gun emitting bullets.

Note that the above "paths" are defined in terms of the reference frame of the elevator (notional grid of clocks and rulers), and for the case of light, neither of these paths would be straightforward to observe by what someone naively (optically) sees in the elevator. For example, if there is smoke that scatters the light so it can be seen from elsewhere in the elevator, the instantaneous appearance of the light beam would not correspond to the second interpretation (locus of all photons in flight), because of the differing time for propagation to the observer's eye from different parts of the beam.