# Thought experiment revisited

1. Jan 22, 2008

### ch@rlatan

You know the 'thought-experiment'....the spacehip travelling at a speed close to the speed of light....where a photon is emitted upwards to a mirror, bounces off and returns to its source. The space traveller (observer 1) standing by the emitter observes a simple up and down motion. A second observer some distance away observes the photon moving diagonally upwards and downwards to the right (assuming the craft is moving left to right AND the part of the craft between him and the emitter is open for him to witness the event).

Well, if we place the observer in the craft (observer 1) between the emitter and the second observer and remove the back of the craft (or better still make the craft glass) will he not see the same as the second observer?

2. Jan 22, 2008

### JesseM

What do you mean "see the same"? Observer 2 won't see the photon moving sideways relative to the craft, but he does see the craft itself moving sideways, so the photon still moves diagonally relative to himself.

3. Jan 22, 2008

### yuiop

You seem to be getting relative velocity and distance mixed up here. If the second observer (some distance away) has the same velocity as the glass rocket he too will see the bouncing photon going straight up and down. If the distant observer is moving relative to the rocket then he will see the photon moving diagonally. If observer 1 inside the rocket moves relative to the emitter he will see it moving diagonally (imagine a very long rocket so he can get some speed up). The distance away is not very important here.

4. Jan 22, 2008

### ch@rlatan

What I mean is....when the craft is made of glass (let's say invisible) and there is a reference frame behind the craft both observers perceive the sideways motion of the photon.

Let me take it further. An invisible craft moving from left to right has on board an invisible emitter (remember this is a thought-experiment). Both observers are situated as before. This emitter differs from the first as it emits a pulse of photons moving out from the source as an expanding light speed sphere. This event will have happened in a particular space-time. The craft and the mirror have moved on. The photons are not aware of them. The source of the photons is now not in the craft but in time-space that the craft has left behind.

5. Jan 22, 2008

### JesseM

Why would an observer on the craft perceive any sideways motion? Just to be clear, are you assuming that the axis between the two mirrors is at right angles to the direction the craft is moving (as seen in the frame where it's moving)? If so, wouldn't you agree that an observer who sees the craft at rest would just see the light bouncing up and down on that axis?
Do you understand that there is no "objective" notion of movement in relativity? The craft may be moving from left to right in the frame of one observer A, but in the frame of another observer B on board the craft, the craft is totally at rest while it is A who is moving from right to left.
What does it matter how the source is moving? All photons move at c in a given frame, regardless of the motion of the source--if you see an emitter at rest relative to you and another emitter moving relative to you, and they both emit photons at the moment their positions coincide, the photons from the "moving" emitter won't behave any differently from the photons from the "resting" emitter which were emitted at the same position and time.

Anyway, in the frame where the craft is moving from left to right, some photons on the expanding light speed sphere which started at the position of the bottom mirror will happen to be moving on a diagonal that leads them to hit the top mirror, which will have moved to the right some amount by the time the photons reach it. In the frame where the craft is at rest, these same photons move straight up, and hit the top mirror a little later, since the top mirror is not moving to the right or left in this frame. It's the same photons being emitted from the position of the bottom mirror and hitting the top mirror in both cases, just viewed from different frames.

6. Jan 23, 2008

### ch@rlatan

Dear JesseM

Because the observer 'on board' has no visible reference frame and perceives the photon as moving left to right. He appears to himself and other observers to be travelling at light speed alone and unaided. The craft, the emitter and everything on board is invisible (though the mirror will detect and reflect light in the normal manner).

Light is not aware of the craft nor are any observers.

Light is not aware of the emitter nor are any observers. Does light behave according to the things that surround it?

That's ballistics!. Relative time is based on a single photon being emitted directly upwards and reflected directly downwards according to the observer on the craft. According to you, if we were to isolate that photon from the sphere we would find that it missed the mirror which has now moved on. The photon you propose as hitting the mirror has actually travelled angularly to the mirror and been reflected angularly. Both observers could then agree on the distance covered by that photon and the time took....there would be no relative time. Though, the observer on board would perceive the photon as travelling (slower than normal) directly up and down but would be aware that it was merely a trick of the light.

7. Jan 23, 2008

### JesseM

I don't know what you mean by "visible reference frame", a reference frame is just a coordinate system that an observer uses...the observer on the ship could just define his reference frame such that the origin of his spatial axes was always located wherever he was, in which case the two mirrors on the ship would be at rest in this coordinate system as well (since the mirrors are not moving relative to the observer). If the left-right axis is the x-axis, and the up-down axis is the y-axis, then perhaps he would see the bottom mirror 2 meters to his right on the floor x=2, y=0 and the top mirror 2 meters to his right on the ceiling (let's say this is 4 meters above the floor) at x=2, y=4. So, he will see the light moving only vertically on the y-axis, with the light moving from y=0 to y=4 and back, but always having the same x-coordinate x=2. On the other hand, if the other observer outside the ship sees both him and the mirrors moving to the right on the second observer's x'-axis at 200,000,000 meters/sec (a significant fraction of c, which is 299,792,458 m/s), then in this observer's coordinate system, if the bottom mirror starts at the origin at x'=0 and y'=0 when the light leaves it, it works out that the light reaches the top mirror in a time of 1.79108535257604 *10^-8 seconds (using http://www.math.sc.edu/cgi-bin/sumcgi/calculator.pl [Broken] to get nice precise numbers), so the top mirror will have moved to the right by 200,000,000 * 1.79108535257604 *10^-8 = 3.58217070515209 meters in this time, meaning the light had to go diagonally to hit the top mirror at x'=3.58217070515209, y'=4 (so using the pythagorean theorem, the total distance it covered was sqrt[3.58217070515209^2 + 4^2] = 5.36953880336569 meters, and dividing by the time of 1.79108535257604 *10^-8 s gives a speed of 299792458.000001 m/s, with that 0.000001 just being a matter of roundoff error.

If you don't like these numbers, pick some of your own and we can repeat the calculation...just give me the x,y coordinates where the light leaves the bottom mirror in the frame of the observer on the ship, the x,y coordinates where it hits the top mirror in the same frame (assuming the mirrors are oriented vertically, both events should have the same x-coordinate, so the light moves vertically only), and then the x',y' coordinates where it leaves the bottom mirror in the frame of the observer who sees the ship moving left to right on the x' axis at some significant fraction of c, then we can figure out the corresponding x',y', in this frame where it hits the top mirror.
He can't be moving at light speed, only some significant fraction of it.
What does it mean to say light is "aware" or "not aware" of anything? What does this have to do with my question above?
No, the point of that paragraph was that it wasn't affected by the speed of the emitters! That's why I said the photons from the "moving" emitter won't behave any differently from the photons from the "resting" emitter.
Um, what? I specifically said that if we isolated the photon from the sphere which went directly up to hit the top mirror in the frame of the observer on the craft, and we isolated the same photon from the sphere in the frame of the observer who sees the craft moving, that same photon would now move diagonally in the second frame so it would still hit the mirror. It's certainly true that, just as with projectiles in ballistics, a photon which is emitted directly upwards in one frame will be emitted along a diagonal in another frame in motion relative to the first.
In the frame where the two mirrors are moving left to right, yes, it moved on a diagonal. In the frame where the two mirrors are at rest (the ship-observer's frame), no, it moved purely vertical. There is no "objective" truth about this, every frame's perspective is equally valid in relativity.
Nonsense, this would imply the observer on board knows that he is "really" moving. But the whole point of the term relativity is that things like movement are all relative, it is equally valid to consider the ship-observer at rest and the second observer in motion as it is to consider the second observer at rest and the ship-observer in motion. Relativity rejects the notion that there is any experiment that can determine an "absolute truth" about such notions, the laws of physics will all work exactly the same way in each frame (meaning that if two observers in windowless labs that are in relative motion through empty space perform the same experiment inside their respective labs, they'll get the same results).

Last edited by a moderator: May 3, 2017
8. Jan 23, 2008

### Staff: Mentor

Just in case it isn't clear, just by removing the back of the spacecraft, you don't make the external and internal observers equivalent. One is still moving and the other still isn't.

9. Jan 23, 2008

### JesseM

One is still moving relative to the mirrors and one isn't...I'm sure you meant something like this, but from some of charl@tan's comments I think one of the problems here may be that he/she is stuck on the notion of absolute motion, the idea that there is some objective truth that the ship is "really" moving and that the path of the light beam is "really" diagonal. So to be clear, I want to emphasize that relativity denies there can be any physical meaning to the notion of absolute motion or absolute space.

10. Jan 25, 2008

### ch@rlatan

Hi JesseM

You're right! I am confused. This is what you said you said...

...and this is what you said...

????

The first part of your detailed response seemed to me to be Relativity 101 wrapped in smoke...and a couple of mirrors. I think we are moving away from the crux. Allow me to propose another experiment.

A craft leaves the Earth from the Northpole at light speed (I know that's not possible but it doesn't detract from the point and helps with the numbers). It's flight path is directly towards the Sun. Imagine an x-axis that joins the two. Along this axis, at the halfway point is a y-axis at which we can place an observer (though it's not so important as the event will take place without him anyway).

11. Jan 25, 2008

### Staff: Mentor

Can you specify where you are confused? What JesseM is saying seems perfectly clear. The same photon when viewed (pretend you can view it) within the moving frame will have a vertical path, but when viewed from the other frame will have a diagonal path.

This shouldn't be all that surprising. Consider a train moving with constant velocity. If you drop a ball, it falls straight down. But view that same ball from the viewpoint of the ground, and it appears to take a curved path. Same idea.

So? What's the experiment?

12. Jan 25, 2008

### ch@rlatan

Hi JesseM

You're right! I am confused. This is what you said you said...

...and this is what you said...

????

The first part of your detailed response seemed to me to be Relativity 101 wrapped in smoke...and a couple of mirrors. I think we are moving away from the crux. Allow me to propose another experiment.

A craft leaves the Earth from the Northpole at light speed (I know that's not possible but it doesn't detract from the point and helps with the numbers). It's flight path is directly towards the Sun. Imagine an x-axis that joins the two at that moment - an axis created at the point of departure. Along this axis, at the halfway point is a y-axis at which we can place an observer (though it's not so important as the event will take place without him anyway).

On board the craft (which again is invisible) is an emitter. This emitter will, at the halfway point emit 3 photons. One towards the Sun, one back towards the Earth (both these along the x-axis)and one directly up (along the y-axis).

The observer on board sees all these photons moving away from him at light speed. For this to happen the photon moving back to Earth along the x-axis will be frozen at y=0, x=0. The photon emitted toward the Sun will have to move along the x-axis at twice the speed of light. The photon that went up along the y-axis is happily bouncing between the mirrors. This is light behaving ballistically and is necessary for the light to hit the mirror.

Or we can say an event occurred at y=0, x=0 where light is doing as it should....travelling in a straight line with speed c. But now the observer on the craft sees the photon travelling back to Earth at twice the speed of light and the photon travelling towards the Sun is frozen at the emitter. The photon travelling upwards will continue up along the y-axis having missed the mirror that was in constant motion along the x-axis toward the Sun. Here the frame of reference is in local space instead of belonging to an observer. Can't we?

13. Jan 25, 2008

### Mentz114

No, no, no. It makes a nonsense of the whole thing.

No again. You've missed a fundamental point about light - it does not behave ballistically. All observers will see light travelling at c. This is compensated by time dilations and length contractions.

Inertial observers have to be associated with matter - so I assume you are placing another observer somewhere.

14. Jan 25, 2008

### JesseM

You didn't quote the full paragraph in the second case. What I said was:
So what I'm saying here is that if you select/isolate the photons from the expanding sphere that hit the mirror in one frame, and then you select/isolate the same photons in the other frame, those same photons will also hit the mirror in the second frame. If you use the frame where the mirrors are moving first, the photons that hit the mirror were moving diagonally in this case, and then when you switch to the other frame, you find that these same photons were moving vertically. On the other hand, if you use the frame where the mirrors are at rest first, the photons that hit the mirror were moving vertically in this case, so then when you switch to the other frame, you find that these same photons were moving diagonally. This is exactly what I meant when I said "I specifically said that if we isolated the photon from the sphere which went directly up to hit the top mirror in the frame of the observer on the craft, and we isolated the same photon from the sphere in the frame of the observer who sees the craft moving, that same photon would now move diagonally in the second frame so it would still hit the mirror."
Actually it does detract from the point, because relativity simply doesn't allow us to talk about the "reference frame" of something that's moving at light speed. Also, if the mirrors were moving at light speed in the Earth's frame, then a photon starting from the position of the bottom mirror could never catch up with the top one without moving faster than light. Pick some sublight velocity instead, like 0.6c.
Joins the north pole and the sun, you mean?
So the observer is at the origin of this x,y coordinate system, correct? Coordinates x=0, y=0?
All right...
Again, it's invalid to consider things from the perspective of one moving at light speed. Also, you seem to be implying that the observer on board is using the same x-y coordinates as the observer at rest between the Earth and the Sun, but the whole point of different "reference frames" in relativity is that each observer uses their own coordinate system, one which they are at rest in. The observer on board will not be at rest in the x-y coordinate system you described above, he should use his own set of x'-y' coordinate axes which "move along with him" as seen from the Earth's point of view. The statement that each observer measures photons to move at c is based on the assumption that each observer is using a coordinate system in which they are at rest; if the observer on the ship uses the x-y coordinate system of the other observer who is sitting between the Earth and the sun, the ship-observer's speed relative to the photons will not be c in this case.
What mirrors? You never specified any details of a set of mirrors in your setup. Which observer are the mirrors at rest relative to? Is the axis joining them parallel to the y-axis? If the mirrors are at rest relative to the observer on the ship, and the photon went up along the y-axis of the observer sitting between the Earth and the Sun, then if the photon starts out at the same position as the bottom mirror, the top mirror will have long moved on in the x-direction by the time the photon reaches the same y-height as the top mirror, so it'll miss the top mirror completely.
No, again, you seem very confused about coordinate systems here, the claim that the observer on the craft sees all photons moving at c relative to himself is based on the assumption that the observer on the craft is using a coordinate system that moves along with him, not the x-y coordinate system of the observer at a fixed distance between the Earth and the Sun.

Last edited: Jan 25, 2008
15. Jan 26, 2008

### ch@rlatan

Dear JesseM,

Firstly, Yes - both those paragraphs do say the same thing about the photons moving up. I misread the reference frame that you were talking about (not once but three times...I think I read (red) what I wanted to read (reed)). Apologies.

Secondly,
and
Mmmmm.....

And thirdly.....I still haven't made my point (points actually) possibly due to poor thought-experiments.

So I'll concentrate on one at a time.

There is a piece of very long string in each of my hands. Attached to each of those strings is a craft. These craft(s) will depart in opposite directions at 3/4c each for 1 second when they will stop. There are no observers of the craft or in the craft. I am in the middle of them with my strings that are suitable marked for measuring. At what speed will these craft be moving apart relative to one another?

16. Jan 26, 2008

### JesseM

No problem.
These quotes aren't really contradictory. I didn't say photons behaved like ballistic projectiles in all respects, only in the sense that a photon emitted up in one frame will move diagonally in another. But a ballistic projectile would have different speeds in the two frames, while a photon will have the same speed in both--that was Mentz114's point, and I agree.
You mean, at what rate is the distance between them increasing as measured in your frame (using the string, say)? The distance between them is increasing at a rate of 1.5 light years per year in your frame. However, in either craft's own rest frame, as measured by rulers and clocks at rest in that frame, the other craft is only moving away at 0.96c. Here I have used the formula for addition or relativistic velocities given on this page, which says that if craft #2 is moving at 0.75c to the right in your frame, and you are moving at 0.75c to the right in the frame of craft #1, then the velocity of craft #2 in the frame of craft #1 is (0.75c + 0.75c)/(1 + 0.75*0.75) = 1.5c/1.5625 = 0.96c. The claim that nothing can move faster than c only applies to individual objects as measured in any given frame, not the distance between objects as measured in a frame where both are moving.

Last edited: Jan 26, 2008
17. Jan 31, 2008

### ch@rlatan

Dear JesseM,

OK last one.....I promise

So between you, you are saying that, yes...the photon in the experiment does move further in the frame of the observer at distance from the spaceship, but because time moves faster for this observer relative to the observer on the spaceship the observed speed of light is still c.

So in my frame I have let out 2.25*10^8 metres (0.75*c) of string on each side after 1 second - at which point the craft have stopped (assume c = 3*10^8 m/s). I radio them both to tell them they are 1.5 light seconds apart according to my measurements. And they say "No, no, no. We are only 0.96 light seconds apart according to (y)our measurements!". Can we both be right?

Sure this equation will always return a value of less than one...it is designed to. It is a transformation (think about...'of what?') formulated to prove Einstein's theory of the speed of light in an at-rest frame. Why did Einstein postulate this theory? Because he imagined himself travelling at light speed with a photon that would appear to him to be at rest...and he just didn't like the idea!.

Anyhooo, on the back of our discussions I have hit (what I think is) a paradox in the observation of Einstein's bouncing photon, which I will shortly post as a new thread.

18. Jan 31, 2008

### JesseM

Well, I'm happy to continue to answer your questions if you have any more of them.
We both agree that in the frame of the observer who sees the ship in motion, the photon travels diagonally a greater distance than the vertical distance between the mirrors, and the time it takes to travel this distance in the frame of the observer outside the ship is greater than the time it takes to travel the vertical path in the frame of the observer on the ship. Note that time dilation is relative though...the observer who sees the ship in motion does say that the clock of the observer moving along with the ship is running slow, but the observer moving along with the ship says that his own clock is running normally and the other observer's clock is running slow (just imagine the other observer outside the ship had his own light clock--in the frame of the observer moving along with the ship, the photon in this second light clock would be moving diagonally, so it'd take longer for the light to go from one mirror to another in this second light clock as seen in the frame of the ship-observer).
No, they won't say that, because the rules of SR that I mentioned only apply to inertial frames--if the two craft stop, then they have to accelerate, so they aren't sticking to a single inertial rest frame throughout the problem. If you analyzed things from the point of view of their non-inertial frame, you might find that in this frame the distance expands suddenly from 0.96 ls to 1.5 ls during the acceleration that brings them to a stop in your frame (note that in non-inertial frames there is no restriction on objects having a coordinate velocity larger than c, the light-speed restriction is only meant to apply to inertial frames).
It can be derived from the Lorentz transformation, which transforms the coordinates assigned to a given event in one frame to the coordinates assigned to the same event in another frame, under the assumption that each observer defines their coordinates using readings on a network of rulers and clocks at rest relative to themselves, with the different clocks in a given observer's network "synchronized" with one another using the Einstein synchronization convention, which assumes the speed of light must be the same in all directions in that observer's frame (so an observer can 'synchronize' two clocks by setting off a flash at their midpoint, then setting each to read the same time at the moment the light from the flash reaches them). Assuming the two postulates of relativity hold, it's easy to show that if one observer uses coordinates (x,y,z,t) and a second observer uses coordinates (x',y',z',t'), and the spatial origins of the two coordinate systems coincide at t=0 and t'=0, and the spatial origin of the second coordinate system is moving along the x-axis of the first coordinate system in the +x direction at speed v, then the transformation will be:

x' = gamma*(x - vt)
y' = y
z' = z
t' = gamma*(t - vx/c^2)

where gamma = 1/squareroot(1 - v^2/c^2)

From this coordinate transformation it is easy to derive the formula for addition of velocities I linked to above, as well as the standard time dilation and length contraction equations.
Nonsense, Einstein's formulation of SR had little to do with his childhood thought-experiment about riding along with a light beam (which is impossible in relativity anyway). Rather, it had to do with the fact that physicists had previously assumed Maxwell's laws of electromagnetism (which say that the speed of light is the same regardless of the motion of the emitter) would only hold in a particular reference frame, the rest frame of the luminiferous aether (which light waves were imagined to be vibrations in, like sound waves are vibrations in air), but this hypothesis suggested that observers in motion relative to the aether would measure light to travel at different speeds in different directions, yet experiments such as the Michelson-Morley experiment which tried to verify this prediction had been unable to detect any differences in the speed of light in different directions. Einstein wanted to see if he could construct a theory where Maxwell's laws would work for every inertial observer, and the result was SR. One of the postulates of SR was that all fundamental laws of physics would have the property of being invariant under the Lorentz transformation (meaning the equations would be the same in all the different coordinate systems given by this transformation), a mathematical property known as "Lorentz-invariance", and all the most fundamental laws which have been discovered since 1905 (like the laws of quantum field theory) have indeed had this property.

Last edited: Jan 31, 2008