Really parabolic light ray in an accelerated elevator?

[Thankall]

In an elevator "vertically" accelerated at g in outer space, the equivalence principle says a "horizontal" light ray in the elevator looks like a parabola. I completely understand that the light ray is curved but don't understand why the deflected light ray is an exactly parabola.

Almost all articles use the analogy of throwing a stone horizontally on Earth to draw a parabolic trajectory. However, light is different from a stone since light has constant speed. Going back to the elevator, I understand the vertical shift of the light ray is 0.5*g*t^2 downward which gives the light ray a vertical component of speed. If that is true, the horizontal speed of light cannot be C anymore in order to maintain the total light speed of the curved ray still at constant C. Thus, the horizontal speed component of light must be decreasing from C while but the vertical speed component increases as g*t.

Therefore, the shape of the light ray must be different from a parabola; otherwise, the light speed will increase. Is there anything wrong?

 

[jartsa]

Light is not actually different from a stone. Vertical motion causes time dilation of horizontal motion of both objects. And that causes the trajectories of both objects to deviate from a parabola.

(There's also a similar "problem" of the vertical speed of the stone approaching the speed of light.)

 

[PeterDonis]

I understand the vertical shift of the light ray is 0.5*g*t^2 downward which gives the light ray a vertical component of speed. If that is true, the horizontal speed of light cannot be C anymore in order to maintain the total light speed of the curved ray still at constant C.

This argument would be correct if you were working in an inertial frame. But you aren't; you are working in a frame in which the elevator is at rest, which is not inertial. In a non-inertial frame, the speed of light does not have to be C.

 

[PeterDonis]

Light is not actually different from a stone. Vertical motion causes time dilation of horizontal motion of both objects. And that causes the trajectories of both objects to deviate from a parabola.

There's also the problem of the vertical speed approaching speed of light.

Not within a small enough patch of spacetime for the equivalence principle to apply. In a small enough patch of spacetime, there is not enough time for vertical motion to get large enough for the issues you describe to apply.

 

[Dale]

light has constant speed

That is only true in an inertial frame. In a non inertial frame the speed can be different. The elevator is non inertial

Edit: I see @PeterDonis already made this same point

 

[jartsa]

Not within a small enough patch of spacetime for the equivalence principle to apply. In a small enough patch of spacetime, there is not enough time for vertical motion to get large enough for the issues you describe to apply.

I don't mean there's a real problem when the speed of a 'falling' stone approaches c. The trajectory just deviates from the Newtonian parabola trajectory.

 

[PeterDonis]

I don't mean there's a real problem when the speed of a 'falling' stone approaches c. The trajectory just deviates from the Newtonian parabola trajectory.

Yes, but this is outside the purview of the EP, because you have to look at a large enough patch of spacetime to see it. It's not that it doesn't happen; it's just that it happens outside the range of applicability of the EP.

 

[sweet springs]

Hi.

In an elevator "vertically" accelerated at g in outer space, the equivalence principle says a "horizontal" light ray in the elevator looks like a parabola. I completely understand that the light ray is curved but don't understand why the deflected light ray is an exactly parabola.

You refer to Rindler coordinate system. I am also interested in whether null geodesic, i.e. a light course, is parabola. Is it mathematically shown easily?

 

[jartsa]

Yes, but this is outside the purview of the EP, because you have to look at a large enough patch of spacetime to see it. It's not that it doesn't happen; it's just that it happens outside the range of applicability of the EP.

That's too easy. Anytime an experimenter tells us that she has found a deviation from EP, we just say "if you found a deviation then your laboratory is too big".

"If you can measure tidal forces, then your laboratory is too large" That's okay, because ... I don't know why.

 

[PeterDonis]

Anytime an experimenter tells us that she has found a deviation from EP, we just say "if you found a deviation then your laboratory is too big".

No, we ask if tidal forces are measurable.

"If you can measure tidal forces, then your laboratory is too large" That's okay, because ... I don't know why.

Because the lack of measurable tidal forces is the definition of "small enough patch of spacetime for the EP to apply".

Obviously this depends on the magnitude of the tidal forces and the accuracy of your measuring instruments. In other words, the EP is really an approximation, and how good an approximation depends on the situation. Welcome to physics.

 

[Ibix]

That's too easy. Anytime an experimenter tells us that she has found a deviation from EP, we just say "if you found a deviation then your laboratory is too big".

"If you can measure tidal forces, then your laboratory is too large" That's okay, because ... I don't know why.

The equivalence principle arises in general relativity because you can make the first and second derivatives of the metric vanish by a clever choice of coordinates. But you can only make the second derivatives vanish at a point by that choice, at least in general. So the equivalence principle only applies at a mathematical point. So it is inevitably violated to some extent by any actual experiment. All you can say is that if you neglect curvature you should get less wrong as you experiment in smaller and smaller patches of spacetime.

So if you ever found a circumstance where the errors grew, not shrank, as you worked on a smaller scale you would have foubd a violation of the equivalence principle. An analogy: the equivalence principle for the surface of the Earth says that you should be able to use a Cartesian coordinate system on a small area of the surface of the Earth. But at some point the discrete nature of matter means that the surface is not well modeled by a smooth manifold - and here be quantum dragons.

 

[A.T.]

You refer to Rindler coordinate system. I am also interested in whether null geodesic, i.e. a light course, is parabola. Is it mathematically shown easily?

Yes, that seems the core of the question, not the EP.

 

[pervect]

In an elevator "vertically" accelerated at g in outer space, the equivalence principle says a "horizontal" light ray in the elevator looks like a parabola. I completely understand that the light ray is curved but don't understand why the deflected light ray is an exactly parabola.

I would not say that the path is exactly a parabola - just that that's a good approximation for short distances. More below.

Almost all articles use the analogy of throwing a stone horizontally on Earth to draw a parabolic trajectory. However, light is different from a stone since light has constant speed. Going back to the elevator, I understand the vertical shift of the light ray is 0.5*g*t^2 downward which gives the light ray a vertical component of speed. If that is true, the horizontal speed of light cannot be C anymore in order to maintain the total light speed of the curved ray still at constant C. Thus, the horizontal speed component of light must be decreasing from C while but the vertical speed component increases as g*t.

Therefore, the shape of the light ray must be different from a parabola; otherwise, the light speed will increase. Is there anything wrong?

The constancy of the speed of light in this (and other similar problems) should be interpreted as follows. If two sufficiently close observers on the elevator exchange light signals, the distance between the two objects is equal to c times half the two-way propagation delay of the light signals.

There's several key points here: The first point is that this approach gives the expected answer for distance even though light travels in a parabola as long as the distances are short. Let d be the distance between two points defined as c times the round-trip time T divided by 2, so that ##d = \frac{T}{2} \,c##. Then the amount of parabolic deflection is ##\frac{1}{2} a \left( \frac{T}{2} \right) ^2 = \frac{1}{2} a \left( \frac{d}{c} \right) ^2##

The deflection is second order in d, thus by taking T sufficiently small, we can make the deflection negligible in comparison to the distance. We can do a further analysis to show how the deflection affects the distance, but hopefully it's obvious that given the deflection is negligible in comparison to the distance, it won't matter.

I noted that the exact path of the light wasn't parabolic. To get the exact path that light travels when T is not negligible can be most easily solved by noting that light travles in a straight line in an inertial frame of reference. One then needs to know how to mathematically translate between the inertial frame of reference and the accelerated frame of reference, by having a specific map that gives the inertial coordinates in terms of the accelerated frame coordinates (or the reverse, the mapping needs to be invertible).

One can find this problem worked out in, for instance, MTW's text "Gravitation", but the text uses advanced tensor methods to do so. It is perhaps sufficient to note that the equation of motion in an inertial frame for a rocket with a constant (proper) acceleration is not ##d = \frac{1}{2} a t^2##, but is rather given by the "relativistic rocket equation" <<link>>. The fact that if a rocket accelerates long enough it doesn't follow a parabolic path should demonstrate that the parabolic path idea works for Newtonian physics, but is not the exact solution for relativistic physics.

 

[DrGreg]

I am also interested in whether null geodesic, i.e. a light course, is parabola. Is it mathematically shown easily?

(Uniformly accelerated) Rindler coordinates ##(T,X,Y,Z)## are related to (inertial) Minkowski coordinates ##(t,x,y,z)## by $$\begin{align*}
ct &= Z \sinh \frac{gT}{c} \\
x &= X \\
y &= Y \\
z &= Z \cosh \frac{gT}{c} .
\end{align*}$$A particle moving parallel to the ##x## axis at constant velocity ##v## (##|v| \leq c##) in Minkowski coordinates is given by$$\begin{align*}
t &= \frac{x}{v} \\
y &= 0 \\
z &= \frac{c^2}{g}. \\
\end{align*}$$Substituting,$$
\begin{align*}
Z \sinh \frac{gT}{c} &= \frac{cX}{v} \\
Z \cosh \frac{gT}{c} &= \frac{c^2}{g}. \\
\end{align*}$$Squaring and subtracting,$$
Z^2 = \frac{c^4}{g^2} - \frac{c^2 X^2}{v^2},
$$i.e.$$
\frac{X^2}{(v/c)^2} + Z^2 =\frac{c^4}{g^2},
$$the equation for the arc of an ellipse. In the case where ##v=c##, it's the arc of a circle.

So, in Rindler coordinates, the spacelike part of a timelike geodesic is an elliptical arc, and the spacelike part of a null geodesic is a circular arc.

(Note: Rindler coordinates are defined only for ##Z>0##. ##Z=0## is the Rindler horizon, whose properties are similar to the event horizon of a black hole. In particular, it takes infinite Rindler time ##T## for the particle described above to reach the horizon.)

 

[sweet springs]

Oh, thanks DrGreg.

If that is true, the horizontal speed of light cannot be C anymore in order to maintain the total light speed of the curved ray still at constant C.

You do not have to worry about it. Approaching the event horizon light speed by coordinate time goes down slower than c but speed by proper time is kept c.

 

[pervect]

lipse. In the case where ##v=c##, it's the arc of a circle.

So, in Rindler coordinates, the spacelike part of a timelike geodesic is an elliptical arc, and the spacelike part of a null geodesic is a circular arc.

(Note: Rindler coordinates are defined only for ##Z>0##. ##Z=0## is the Rindler horizon, whose properties are similar to the event horizon of a black hole. In particular, it takes infinite Rindler time ##T## for the particle described above to reach the horizon.)

Very nice, I may have seen this mentioned before but if I have, I'd totally forgotten it. I did have to think a bit why we set ##Z = c^2/g## though. I'm not sure I could explain it really well, my attempt would be to say that the product of the proper acceleration of the particle and the distance from the Rindler horizon always being equal to ##c^2##. So if you know the particle's proper acceleration is g, then it must be at the specified location. But this seems rather informal.

 

[DrGreg]

I did have to think a bit why we set ##Z = c^2/g## though.

You don't have to set it to this value; any positive constant will do, but for this value the proper acceleration is ##g##. If you put ##Z = c^2/g## and then differentiate ##(ct,x,y,z)## twice with respect to proper time ##\tau##, you get the 4-acceleration, then measure its length to get the proper acceleration. The intermediate expression for 4-velocity shows you that ##T=\tau## for this constant value of ##Z##.

 

[pervect]

I should probably just work this out, but perhaps you know the answer already. If, for the Rindler metric, (the spatial projection of) null geodesics are circles, and timelike geodesics are ellipses, what are (said projections) for space-like geodesics? I'm guessing it's the other conic section, the hyperbola.

 

[DrGreg]

I should probably just work this out, but perhaps you know the answer already. If, for the Rindler metric, (the spatial projection of) null geodesics are circles, and timelike geodesics are ellipses, what are (said projections) for space-like geodesics? I'm guessing it's the other conic section, the hyperbola.

Well, I suppose there's nothing wrong in putting ##|v| > c## in the equations above. They no longer represent a physical particle but just a hypothetical point moving faster than light. So you still get an ellipse except in the limit ##v \rightarrow \infty##, in which case you get the straight line ##Z = c^2/g##.

All of this analysis is for lines that are "horizontal", i.e. spatially orthogonal (not 4-orthogonal) to the direction of acceleration in Minkowski space. I haven't thought through how other lines transform.

 

[stevendaryl]

I just wanted to add, in case anyone is confused about this point: A lot of people seem to think that the Equivalence Principle is required to be able to do physics in an accelerated coordinate system, to figure out how light beams and clocks and so forth behave in an accelerated rocket. That's not true. If you know how clocks and light beams behave in good old inertial Cartesian coordinates, then you can figure out how they behave in any other coordinate system by just doing a coordinate transformation. All you need is calculus, not any theory of gravity.

It's the other way around: If you want to know how clocks and light beams behave in a gravitational field, then you need the Equivalence Principle, to get an approximate answer, or a full theory of gravity, to get an exact answer.