View Full Version : Light-Reflection Clocks and Gravity

xylophobe

2012-Feb-09, 12:27 AM

I remember reading somewhere that if a transverse mounted light-reflection clock was positioned in a free-falling elevator then the light would bounce back-and-forth between the mirrors (in red below) for the entire free-fall journey. This would look similar to the crude text example below.

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It was stated that this happens because the light "falls" at the same rate as the elevator ...

...but wouldn't the light trace out a parabola-shaped trajectory which would result in the light hitting the second mirror at an incident angle less than 90-degrees which would screw up the 2nd and latter reflections?

The illustration that I remember has the light going in straight-line, hypotenuse-like, angles between the mirrors to create triangle shapes.

a1call

2012-Feb-09, 01:45 AM

Interesting concept. Never heard of it before but makes sense.

Reminds me of a book I read that stated that a bullet shot horizontally will reach the earth at the same instance that a dropped bullet would from the same height.

As for slight angular deviation, you would need it to direct towards the falling/lower position of the mirrors. If you'd bounce back perpendicularly you could miss the other mirror if it was small enough to fall out of sight before reaching it.

larrew

2012-Feb-09, 01:56 AM

According to general relativity, the free-falling elevator should be indistinguishable from an elevator in space, in which a light clock should work, so a light clock would work in a free-falling elevator.

Now for your post.

Yes, the diagram is wrong; the diagram is depicting a spaceship moving at a constant velocity. Yes, the incidence angle is less than 90-degrees. But the clock still works. Bouncing off the mirror doesn't change the y-velocity of the light, only reverses the x-velocity, and and so the only thing affecting the y-velocity of the light is the falling. This falling is also affecting the mirrors in exactly the same way, hence their y-velocities are always the same, and so their y-positions are always the same, the latter reflections aren't messed up, and the clock works.

larrew

2012-Feb-09, 02:22 PM

As for slight angular deviation, you would need it to direct towards the falling/lower position of the mirrors. If you'd bounce back perpendicularly you could miss the other mirror if it was small enough to fall out of sight before reaching it.

I believe you are assuming that the light would travel in a straight line. The gravity of the earth will (I believe) correct for this, bending the light to hit the mirror.

profloater

2012-Feb-09, 06:31 PM

I think this like Einstein's original thought experiment about what it would be like to travel with the light. Suppose the light beam is visible due to smoke. Each mirror is like an observer. An observer in the box would see nothing strange as you would expect, but an observer outside accelerating relative to the box, and assuming he can see inside, can only view the light beam using light which reaches him at different times due to his changing position. This is very different from observing a projectile. So it would not look like the diagram.

xylophobe

2012-Feb-09, 06:31 PM

According to general relativity, the free-falling elevator should be indistinguishable from an elevator in space, in which a light clock should work, so a light clock would work in a free-falling elevator.

Now for your post.

Yes, the diagram is wrong; the diagram is depicting a spaceship moving at a constant velocity. Yes, the incidence angle is less than 90-degrees. But the clock still works. Bouncing off the mirror doesn't change the y-velocity of the light, only reverses the x-velocity, and and so the only thing affecting the y-velocity of the light is the falling. This falling is also affecting the mirrors in exactly the same way, hence their y-velocities are always the same, and so their y-positions are always the same, the latter reflections aren't messed up, and the clock works.

Thanks for the reply but according to my calculations/drawing the light reflected off the second mirror misses the next reflection.

If the left mirror is mirror-1 and the right mirror is mirror-2 and the reflector vertical positions are labeled 1, then 2, then 3, etc. then after striking mirror-2 at position 2 the light misses mirror-1 at position 3.

The incident angle: if the next trajectory is drawn as a straight line then it results in the light striking mirror-1's position 4, a point one position down from where it should strike. If a true parabolic trajectory is drawn/calculated then the light strikes the mirror-1 side even farther down!

This is, by the way, a different scenario than a light-reflection clock traveling at a constant velocity as discussed in special relativity.

korjik

2012-Feb-09, 07:14 PM

I remember reading somewhere that if a transverse mounted light-reflection clock was positioned in a free-falling elevator then the light would bounce back-and-forth between the mirrors (in red below) for the entire free-fall journey. This would look similar to the crude text example below.

|\

...\

....|

.../

|/

|\

...\

....|

It was stated that this happens because the light "falls" at the same rate as the elevator ...

...but wouldn't the light trace out a parabola-shaped trajectory which would result in the light hitting the second mirror at an incident angle less than 90-degrees which would screw up the 2nd and latter reflections?

The illustration that I remember has the light going in straight-line, hypotenuse-like, angles between the mirrors to create triangle shapes.

It would cause the later images to form in a lower position than the stationary case. This would mess up the later reflections, except that the mirrors move downward also. Oddly enough, the displacements cancel.

Jeff Root

2012-Feb-09, 07:53 PM

xylophobe,

My understanding of your scenario:

A pulse of light is emitted horizontally from one vertical mirror

toward a second mirror, so that the pulse bounces back and

forth between the two mirrors forever. The mirrors are on

opposite walls of an elevator. The elevator is in free fall.

As the elevator accelerates downward due to gravity, the light

pulse also accelerates downward at the same rate. Anyone

inside the elevator will see the light bouncing back and forth

in a straight line and constant speed. The reflections occur

at constant intervals.

An observer outside the elevator sees the light pulse following

a zigzag path in which the reflection points get farther and

farther apart as the speed of the elevator increases. The light

pulse has to travel an increasing distance between reflections,

so the time between reflections increases.

-- Jeff, in Minneapolis

larrew

2012-Feb-09, 08:03 PM

Thanks for the reply but according to my calculations/drawing the light reflected off the second mirror misses the next reflection.

If the left mirror is mirror-1 and the right mirror is mirror-2 and the reflector vertical positions are labeled 1, then 2, then 3, etc. then after striking mirror-2 at position 2 the light misses mirror-1 at position 3.

The incident angle: if the next trajectory is drawn as a straight line then it results in the light striking mirror-1's position 4, a point one position down from where it should strike. If a true parabolic trajectory is drawn/calculated then the light strikes the mirror-1 side even farther down!

This is, by the way, a different scenario than a light-reflection clock traveling at a constant velocity as discussed in special relativity.

Don't forget that the vertical displacement between the mirrors is increasing. In the drawing this also is not shown. Jeff Root explained this nicely.

xylophobe

2012-Feb-10, 12:29 AM

Don't forget that the vertical displacement between the mirrors is increasing. In the drawing this also is not shown. Jeff Root explained this nicely.

I think I understand what is happening - as the elevator accelerates its speed increases and time slows down more and more and more and ...

So there is an element of GR because the light travels a parabolic path but there is also an element of SR because of the velocity.

The GR effect would be the only evident time dilation for a couple of mirrors located on the surface of a planet since the acceleration is constant and the speed is constant at zero. The mirrors would have to be angled upward like two ping pong paddles need to be angled upward in order to keep the ball bouncing between the paddles. On a body with an intense gravitational field the light would be like mortar shots between the two mirrors (see attached picture).

Jeff Root

2012-Feb-10, 02:39 AM

The path of light in an ordinary gravitational field is always

an hyperbola, not a parabola. Even massive objects do not

usually have parabolic trajectories. If not for atmospheric

drag and other extraneous effects, a cannonball's trajectory

from cannon to target would be a portion of an ellipse, with

the center of mass of the planet at one focus of the ellipse.

Give the cannonball a speed greater than escape speed, and

it will follow a hyperbolic trajectory. Give it exactly escape

speed -- no more -- and it will follow a parabolic trajectory.

Except in the case of being very close to a black hole, light

always has escape speed. At the photon sphere of a black

hole, light could theoretically orbit in a circular trajectory.

Light which crosses the photon sphere downward follows a

spiral trajectory to the black hole's center. I haven't yet

found out exactly which spiral that is, but I suspect that it

becomes less sharply curved as it approaches the center.

That's the opposite of, for example, an Archimedean spiral,

which becores *more* sharply curved toward the center.

When the light reaches the center, I believe it will finally

be going straight down. If the black hole weren't rotating,

anyway...

-- Jeff, in Minneapolis

larrew

2012-Feb-10, 03:44 PM

the time between reflections increases.

Wait a second. I'm not a physicist, but as far as I know you shouldn't be able to tell the difference between an elevator in free fall and one just floating in space. This is known as the weak equivalence principle (http://en.wikipedia.org/wiki/Equivalence_principle#The_weak_equivalence_princip le). In space, the time between reflections wouldn't increase (why would it?), so doesn't this mean that the time between reflections wouldn't increase in a free falling elevator?

Strange

2012-Feb-10, 04:28 PM

Wait a second. I'm not a physicist, but as far as I know you shouldn't be able to tell the difference between an elevator in free fall and one just floating in space. This is known as the weak equivalence principle (http://en.wikipedia.org/wiki/Equivalence_principle#The_weak_equivalence_princip le). In space, the time between reflections wouldn't increase (why would it?), so doesn't this mean that the time between reflections wouldn't increase in a free falling elevator?

The observer Jeff referred to is stationary (not in free fall) and so sees the period increase as the free-faller speeds up. The free-faller, of course, doesn't notice anything. (Until he suddenly stops free falling with a thump!)

larrew

2012-Feb-10, 08:06 PM

The observer Jeff referred to is stationary (not in free fall) and so sees the period increase as the free-faller speeds up. The free-faller, of course, doesn't notice anything. (Until he suddenly stops free falling with a thump!)

Oh, right. I understand now.

xylophobe

2012-Feb-11, 04:55 PM

The path of light in an ordinary gravitational field is always

an hyperbola, not a parabola.

-- Jeff, in Minneapolis

The trajectory (http://en.wikipedia.org/wiki/Trajectory) equation seems to be more similar to a parabola (http://en.wikipedia.org/wiki/Parabola) than a hyperbole (http://en.wikipedia.org/wiki/Hyperbola).

A question that occurs to me is that as the light traveling between two mirrors of a light-reflection clock in a strong g-field traces out a parabola then at the vertex the velocity has lost all of its y-component velocity, there is only x-component so doesn't that mean that the light is traveling slower than "c" at the vertex? Is this the cause of light not following Newtonian calculations since light falls about twice the rate as calculated by Newton's law of gravitation formula? It also means that time moves more slowly for GR than the simple:

v0*cos(theta) with v0 = initial velocity; theta = launch angle (angle off of x-axis)

16282

Shaula

2012-Feb-11, 08:48 PM

The problem is that you are thinking of photons being accelerated by gravity. They are not. They always follow geodesics. You do not need GR and SR individually to do this - GR incorporates SR (SR is a simplified picture of GR). The photons always travel at c and always follow the geodesics at this speed. What is happening is that spacetime is warped by gravity and the acceleration of the photons you apparently see are simple because you are assuming a flat spacetime where there isn't one. You cannot go back to Newtonian models and patch together SR conceptual pictures when you are in a domain that requires GR.

xylophobe

2012-Feb-13, 04:52 PM

The problem is that you are thinking of photons being accelerated by gravity.

If the photons are not accelerated then how do they fall in sync with the elevator mirrors - the elevator is accelerating. Many examples shown on the internet show parabolic trajectories for the photons - if the photon is changing direction then it is undergoing an acceleration. This was also confirmed using a solar eclipse that measured the change in position, due to the sun's gravity, of a background star.

Kuroneko

2012-Feb-13, 10:17 PM

The trajectories of test particles are conic sections in the approximation of weak fields and low speeds. Light is never slow in that sense. But a conceptually important piece you're missing is that aberration distorts the angles in the mirror frame as compared to the stationary frame. Though there will be some deviation due to gravitational tidal forces.

If the photons are not accelerated then how do they fall in sync with the elevator mirrors - the elevator is accelerating.

Actually, neither of them are accelerating. It is the stationary observers in a gravitational field that are accelerating by the effect of whatever force is keeping them stationary. But if we're talking about coordinate acceleration, yes, light and elevator do that--but this is a coordinate/frame artifact, not what is measured (by, say, an accelerometer on the elevator).

Shaula

2012-Feb-14, 06:22 AM

if the photon is changing direction then it is undergoing an acceleration

But it is not changing direction as it moves - it is being emitted from one mirror, following the equivalent of a straight line to the other mirror where it is absorbed and re-emitted.

This was the thing I was trying to get across. You are mixing SR type concepts into a GR question. In GR the photons follow geodesics, which are curved in the presence of gravity. They are not accelerated sideways by a force acting on them - they follow the shortest path through curved space-time.If we assume spacetime is flat (i.e. watch from a frame where it is) then we see an apparent acceleration purely because of our choice of co-ordinates.

xylophobe

2012-Feb-15, 11:17 PM

But it is not changing direction as it moves - it is being emitted from one mirror, following the equivalent of a straight line to the other mirror where it is absorbed and re-emitted.

This was the thing I was trying to get across. You are mixing SR type concepts into a GR question. In GR the photons follow geodesics, which are curved in the presence of gravity. They are not accelerated sideways by a force acting on them - they follow the shortest path through curved space-time.If we assume spacetime is flat (i.e. watch from a frame where it is) then we see an apparent acceleration purely because of our choice of co-ordinates.

I guess I would just like to know a simple way to equate the time dilation from gravity to a distant observer who is not affected by SR or GR effects. In SR it is simple to use geometry to calculate the time dilation due to velocity. If the parabolic curve is the path that the light travels in a GR setting (stationary on surface of gravitating body (uniform gravity field)) then is the length of the parabolic path compared to the straight-line path the difference in the rate of time for both observers? In other words if I know the length of the parabolic path and the length of the straight-line path then it should be a simple ratio of the lengths to determine the time dilation.

Just using formulas does not lead people (not all people, of course) to develop an intuitive understanding of a subject - I would like to be able to explain it and show how to calculate it via geometry. Everyone uses different tools to understanding subjects.

Kuroneko

2012-Feb-16, 03:02 AM

I guess I would just like to know a simple way to equate the time dilation from gravity to a distant observer who is not affected by SR or GR effects. In SR it is simple to use geometry to calculate the time dilation due to velocity.

In relativity, both GR and SR, time dilation is given by the metric. For example, for the usual flat, Minkowski metric ds^2 = -dt^2 + d\sigma^2, where d\sigma^2 = dx^2 + dy^2 + dz^2 is the Euclidean metric, the proper time along any path is given by dτ˛ = -ds˛, and therefore:

\frac{d\tau^2}{dt^2} = 1 - \frac{d\sigma^2}{dt^2} = 1-v^2

in other words, dt = γdτ, the standard relativistic time dilation.

We could have the metric be asymptotically flat, though the choice of frame is still not unique. In that case, we have an isolated gravitational system, and the total energy and momentum are definable, so we can further restrict our frame to have the zero total momentum. But there's no particular conceptual point here; GTR is founded on the basic fact that physics is completely independent of what frame we use or coordinates we use to map spacetime. The proper time along some curve in spacetime is given by the metric in all cases.

If the parabolic curve is the path that the light travels in a GR setting (stationary on surface of gravitating body (uniform gravity field)) then is the length of the parabolic path compared to the straight-line path the difference in the rate of time for both observers?

A uniform gravitational field has vanishing tidal forces, thus the spacetime it describes is flat. Therefore, we have ordinary Minkowski spacetime, in a uniformly accelerated frame: ds^2 = -g^2x^2dt^2 + d\sigma^2, where g is the acceleration due to gravity. This is the Rindler coordinate chart of Minkowski spacetime.

In other words if I know the length of the parabolic path and the length of the straight-line path then it should be a simple ratio of the lengths to determine the time dilation.

No. If the observers are at the same location, the time dilation between them is given by the metric, the inner (dot) product of their four-velocities being the Lorentz gamma. If the observers are not at the same location, the time dilation between them simply does not and cannot have a unique answer.

Shaula

2012-Feb-16, 06:11 AM

I guess I would just like to know a simple way to equate the time dilation from gravity to a distant observer who is not affected by SR or GR effects.

There isn't one. In order to do it you need to get into the more complicated maths of GR. Sorry. It is still largely geometric in nature but way more complicated than path lengths being different due to motion.

Kuroneko

2012-Feb-16, 07:04 AM

It's isn't a matter of involving complicated maths; the problem is more fundamental than that. The frame of an observer is local. To extend over spacetime, you need a frame field, for example one given by a multitude of observers throughout spacetime. You have a lot of freedom to do that, but the cost you can have a multitude of frame fields that include a particular observer's local frame but differ widely elsewhere. The frame field cannot be said to belong to any one of those observers.

Sometimes, the spacetime has a symmetry that makes a particular choice more interesting. This happens in the case of Schwarzschild coordinates, for example--we say that it represents the frame of a stationary observer at infinity, but what is really going on is that there is frame field of infinitely many static observers surrounding the black hole. It's still a choice of convention; the only difference is that Schwarzschild spacetime has a particular symmetry that relates those observers in a particularly nice way.

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