Does a light beam bend in an accelerating elevator?

[inertiaforce]

My understanding is that a light beam will bend in an upward accelerating elevator. I told my uncle this. He responded that the velocity of the elevator cannot be added to the velocity of light since light's velocity always remains the same no matter the motion of the source or the observer. Therefore, he says, that an upward accelerating elevator would not bend the light beam since you cannot add the velocity of the elevator to the velocity of the light beam.

He is essentially arguing that bending the path of the light beam by adding the upward velocity of the elevator is attempting to add to the velocity of the light beam. Since light's velocity remains constant, the path of the light beam should therefore remain straight he says. Is he correct? Or am I correct?

 

[A.T.]

My understanding is that a light beam will bend in an upward accelerating elevator.

Yes, if "accelerating" means proper acceleration that an accelerometer measures:
http://en.wikipedia.org/wiki/Proper_acceleration

He responded that the velocity of the elevator cannot be added to the velocity of light since light's velocity always remains the same no matter the motion of the source or the observer.

That's only true in inertial frames of reference, where proper acceleration is zero.

 

[HallsofIvy]

Your uncle is mistaken. The speed of light does not change. Velocity is a vector quantity and it direction can change.

 

[WannabeNewton]

My understanding is that a light beam will bend in an upward accelerating elevator.

The spatial trajectory of the light beam will curve yes. Any freely falling object inside of the elevator will appear to have a downward acceleration relative to the rest frame of the elevator due to the proper upward accelerator of the elevator. This applies just as well to light as it does to massive objects since light is freely falling (in the geometric optics approximation). If it helps you can think of it through the equivalence principle. Imagine you're standing on top of a cliff in the Earth. If you throw a ball directly across then it follows a parabolic trajectory through space; the ball is of course freely falling. Ignoring inhomogeneities in the Earth's gravitational field, if we transformed to a frame freely falling with the ball but not necessarily with the same velocity as the ball, the ball will in this freely falling frame appear to travel in a straight line.

The same exact thing happens with a light beam as well, except the trajectory won't be a parabola in the gravitational field (the exact trajectory can be worked out easily by writing down the null geodesic equation in Rindler coordinates). But it will still be a curved trajectory. Again this is entirely due to the proper acceleration of the elevator or equivalently the uniform gravitational field in which the elevator is at rest because the light beam itself is freely falling and thus experiences a downward coordinate acceleration in the elevator rest frame. In a freely falling frame the light beam will travel in a straight line.

It is true that locally the speed of light is always unity and when I say locally I mean a measurement of the light speed using local rods and clocks right when the light beam passes by these measuring instruments. But this has no bearing on the coordinate speed of light, which is the speed of its spatial trajectory through a coordinate system comoving with the elevator with clocks that are at rest in it and synchronized with one another and rigid rods comoving with the clocks. The coordinate speed can certainly act in a way such that it leads to a coordinate acceleration of the light beam in the elevator frame. Unlike the locally measured speed, which is physical since it is performed in a momentarily comoving local inertial frame, the coordinate speed has in general no physical meaning. In the case of the accelerating elevator its upwards proper acceleration just manifests itself as a downwards coordinate acceleration of the light beam in the elevator rest frame and hence yields a curved spatial trajectory of the light beam in this frame.

 

[pervect]

My understanding is that a light beam will bend in an upward accelerating elevator. I told my uncle this. He responded that the velocity of the elevator cannot be added to the velocity of light since light's velocity always remains the same no matter the motion of the source or the observer. Therefore, he says, that an upward accelerating elevator would not bend the light beam since you cannot add the velocity of the elevator to the velocity of the light beam.

He is essentially arguing that bending the path of the light beam by adding the upward velocity of the elevator is attempting to add to the velocity of the light beam. Since light's velocity remains constant, the path of the light beam should therefore remain straight he says. Is he correct? Or am I correct?

Consider an outside observer in an inertial frame who is not accelerating. He will see the light traveling in a straight line and the elevator accelerating upwards. If you ignore relativistic corrections, the vertical position of the elevator will be z=1/2 a t^2 , the vertical position of the light beam will be z=0.

Now convert to the elevator frame. If you are concerned about tiny relativistic effects, you'll need to use the Lorentz transform. But for short time intervals (much less than a year) they won't matter. You can then note that the elevators height in the elevator frame is zero, and the light beam's height in the elevators frame is now z' = -1/2 a t^2 which means the light accelerates downwards.

 

[georgir]

Interesting thing... if the elevator is free-falling, it is in fact a "truer" inertial frame than our fixed-to-the-surface frames. So a light beam in such elevator would be completely straight, but must appear somewhat bent upwards to us outside the elevator.

 

[WannabeNewton]

Interesting thing... if the elevator is free-falling, it is in fact a "truer" inertial frame than our fixed-to-the-surface frames. So a light beam in such elevator would be completely straight, but must appear somewhat bent upwards to us outside the elevator.

Yes but it shouldn't come as too much of a surprise. Like I said earlier, the same thing happens when we throw a ball over a cliff. To the person on the cliff the ball takes a parabolic path down to the base of the cliff but in a freely falling elevator it moves in a straight line (with the usual caveats of curvature scales and acceleration scales of course).

So the bending of the spatial trajectory of a light beam due to acceleration of a reference frame is in actuality quite mundane. We've been dealing with its basic premise ever since we started doing Newton's 2nd law problems for free fall back in high school physics.

 

[rlshuler]

This is a good question which took nearly 100 years to answer. First let me clarify what everyone agrees on (including the posters above, I believe). The light and a freely falling reference mass stay at the same "height" in all reference frames. If you look at it from a frame moving relatively to the reference mass, the light trajectory is curved.

Many papers appeared, mostly in the American Journal of Physics (which has educational interests) saying that indeed light bends by exactly the amount expected from Newtonian gravity. This is called the principle of equivalence, which the question didn't mention but I assume you are familiar with it since the setup is the same.

The problem is that light bending near the sun or other astronomical object is twice the Newtonian amount. All of the early papers said the extra bending was not covered, explained or included in the elevator experiment (equivalence). Some people, even myself for a while, supposed equivalence to be broken, and GR experts talked of accumulated bending over distances as if that answered the question. Unfortunately, all bending is accumulated over distances, including the Newtonian kind, so that is not really an answer.

In 2002 a paper appeared in AJP by Ferraro deriving all of the bending from equivalence. And it suggests your uncle was on to something, even if he didn't have a complete explanation. The light cannot move straight across the elevator in the same amount of time that it moves through a curved arc across the elevator. this causes some "turning" of the wavefronts which are in addition to the relative "falling" (because actually the elevator accelerated away).

Viewed from a later time in the frame of the elevator, when it is moving relative to the reference mass and (formerly horizontal) light path, time is dilated in the frame of the reference mass. That means all velocities are slower in that frame. So for the moving observer, who sees a curved light path, it takes longer for the light to cross the elevator. So what, you may be asking. I will tell.

Suppose you are at a fixed point in the elevator, and you release a reference mass and a horizontal pulse of light at the same time. You observe objects below you, including the reference mass and light ray. The farther they are below you, the more you have accelerated since you were in the same reference frame (at the same height). Therefore, the more you see them as time dilated, and the horizontal progress of the light as slow (though it its curved track it still goes at the speed of light).

So, you see the horizontal component of the velocity of light, the part subject to time dilation, as slower and slower at greater distances below you. This is like light passing through any substance which reduces its speed, and leads to Huygens refraction bending.

While that is not the official explanation given by Ferraro, I think having read that, you can read the conclusion of Ferraro's paper and get something out of it, even if you don't follow all the math. Ferraro's paper is also on arXiv and you can download it for free: http://arxiv.org/pdf/gr-qc/0209028.pdf