# SR and Light Clocks - Confused!

1. Aug 4, 2008

### aeromedic

Hello all. I wonder if you could help explain something to me. No matter how much I read I am not sure if my understanding of special relativity is correct. My confusion is with respect to a light clock constructed as one mirror above the other and a photon reflecting between the two mirrors.

Scenario 1: There are two spacemen, each with an (initially) synchronised light clock and no relative velocity between the spacemen. If each spaceman looks at the other's clock, they would both agree that the cycle times of the photons in either light clock are identical and that the photons are “hitting” the mirrors in absolute synchrony. For the purpose of simplicity, let’s assume 1 full cycle takes 1 “second” in either clock.

Scenario 2: Now assume one spaceman has constant velocity relative to the other. Again each has a light clock as mentioned. If each spaceman looks at his own clock he will continue to assert that a full cycle takes 1 second as he will be unaware of his own velocity.

However, when spaceman 1 looks at spaceman 2’s clock he will disagree with spaceman 2. He will say that spaceman 2 has forward velocity and therefore, as light has constant velocity, the angulation of the photon’s path means that each cycle now takes longer than 1 second in spaceman 2’s clock. All well and fine.

If this is true and the velocity of light is truly identical for both parties then by all accounts, the light should strike the mirror of spaceman 1 earlier than spaceman 2. But here is the problem I have… if the velocity is relative then the reciprocal should apply to spaceman 2: he will believe that he is stationary and spaceman 1 has velocity in the opposite direction. Therefore spaceman 2’s photon will hit his mirror earlier than spaceman 1. But if both can see the others photon they cannot both believe that their own photon hits their own mirror first!

NoEinstein!

2. Aug 4, 2008

### JesseM

Time dilation is in fact symmetrical--if you and I are in relative motion and both moving inertially, then I will observe your clock to be running slower than mine (whether it's a light clock or any other type), while you will observe my clock to be running slower than yours.

3. Aug 4, 2008

### aeromedic

So, to clarify, if each spaceman had an event started simultaneously in their worlds and if they could immediately communicate with the other upon the events conclusion (disregarding every law of conventional phsyics) they would both believe that they were the first to communicate with the other and that they were, effectively, both awaiting a response from the other?

One other question: If I departed from earth at a high velocity, then there is a common perception that I would be able to return at a relatively younger age than my earthly colleagues. But surely when I changed direction to come back to earth I would age aggresively, such that I would return back at my future earthly age anyway. I would only remain relatively younger as long as I maintained my distance from earth?

4. Aug 4, 2008

### MeJennifer

There really is no such thing as an order for spatially separated events in spacetime. Even if one were to confuse the issue by insisting on a synchronization convention then still different observers would not necessarily agree on the order of a given set of events.

This is one of the key differences between Galilean spacetime and Minkowski spacetime as is used by special relativity.

5. Aug 4, 2008

### yuiop

No, you would return younger than your colleagues. You age slower whether you are going away from the Earth or coming towards it. In the time dilation equation t'=t*sqrt(1-v^2/c^2) the velocity term is squared and the time dilation is the same whether v is positive or negative.

6. Aug 4, 2008

### aeromedic

Thank you for your reply... it makes complete sense. Thinking about my spaceman I can see that the direction of his relative velocity is irrelevant.

MeJennifer: Thank you for your response. If I understand, then what you are saying is that if two objects are spatially isolated and there is no way to compare their velocities then their actions are chronologically independent?

Last question: When photons exit from an energy source (eg a light bulb) does one photon speed away from the subsequent photon at the speed of light or do they both travel with a constant distance separating them?

Thank you all. This site has been very helpful and has answered some questions I have had for ages!

7. Aug 4, 2008

### JesseM

There is always an objective truth about the order of events on a given observer's worldline, like sending a message and receiving a message. The problem with your thought-experiment is that different reference frames define simultaneity differently (events at different locations which have the same t-coordinate in one frame can have two different t-coordinates in another), and so the idea of "instantaneous communication" is problematic, a signal which is received at the same time it was sent in one frame might actually be received at an earlier time than when it was sent in another (this is one of the main reasons that tachyons, hypothetical particles which would move faster than light, are considered very unlikely). But if we simply assume that each observer has a method to communicate "instantaneously" relative to their own frame's definition of simultaneity, there still must be an objective truth about the order of sending and receiving for each observer. If the situation is perfectly symmetrical, so the two observers start their clocks at a common point in space and then move apart inertially, and both send their signals after a prechosen amount of time has passed on their own clocks, then each will receive the signal from the other one at the same time on their own clocks too (the exact time would depend on the velocity they move apart).
No, in any inertial frame a moving clock is always running slower, not faster (this is demonstrated by the light-clock thought experiment), and the rate a clock is ticking in a given frame depends only on its speed at that moment in that frame. So, if we treat the Earth as being at rest in an inertial frame (perhaps it would be better to use a space station moving inertially in deep space rather than an orbiting planet, but the speed of the orbit and the gravity at the surface aren't large enough to make too much difference), and in that frame your ship has the same speed during its outbound trip away from Earth as it does during the inbound trip, in this frame your clock must be slowed down equally on both legs of the trip. And since both legs last an equal amount of time in this frame (since the ship has the same distance to cover to return as it did to travel out, and is moving at the same speed), this frame predicts that you age the same amount during the outbound leg as during the inbound leg. And there always has to be an objective truth about local events in relativity, like what time your clock reads when you turn around and what time it reads when you arrive back at Earth, so that means all frames agree that your clock elapses the same amount of time on the outbound leg as on the inbound leg.

Last edited: Aug 4, 2008
8. Aug 4, 2008

### JesseM

In relativity you must distinguish to types of "relative speed" between objects--the first is the speed of the second object in the first object's rest frame (assuming the first object is moving inertially), the second is the rate that the distance between them changes in some other frame where neither is at rest, which is sometimes called the "closing speed" (the speed one is object is 'closing in on' the other one in this frame). In Newtonian physics there would be no difference between these two notions, but because of the way rulers and clocks distort in relativity, the value of speed when using the first notion can be different than the value of the speed when using the second. When people say that light always moves at speed c, they are talking about the first notion of speed. And photons cannot have their own rest frame in SR, so you can't talk about the speed of the first photon relative to the second in terms of this first notion. On the other hand, with the second notion of speed, the "closing speed" between a photon and another object doesn't have to be c, and the closing speed between two photons would be zero.

9. Aug 5, 2008

### aeromedic

Hello JesseM

Thank you so much for answering these questions… I hope I am not wasting too much of your time.

Following your replies I spent most of last night thinking about the implications and at 1am, having thought I understood it all, I got a little confused! My girlfriend is bewildered by the fact that I even want to know the answer to these questions but I just do.

I think I understand the concept of the frames (I have read about this before.) Essentially if I took a series of 2 dimensional images of “now” and stacked them in chronological order, I create a “time loaf”. (I am aware that it is in fact not 2D but it certainly makes it easier to mentally process) From the observer’s point of view (looking through the time loaf,) the greater the distance between 2 objects then the more distorted his perception of the events in the slice. Thus objects at a distance appear to be in the present whereas they are actually in the past. In my mind, this is simply analogous to me looking at the sun. What I am seeing is the sun as it was 8 minutes ago and somebody on the sun would see me as I was 8 minutes ago. But “God” (for want of a better word) would see us both in realtime as his time slice would be perfectly perpendicular to the flow of time (mine being distorted by the speed of light).

The problem I have with this is that it is merely an optical illusion, a bit like looking at a fish in a pond. Nothing is actually happening with time itself. By contrast the light clock thought experiment tells me that time truly is distorted. Is this distortion merely an illusion from the point of view of the observer???

If in the time loaf, a space-object has a closing velocity of, say, -0.5c in the 2 dimensional plane, then from the observers point of view, the space-object has slowed (and I guess turned slightly red and dim) – but this is just a Doppler effect. It is simply an illusion of time warping that occurs as the object gets further away in my slice of the time loaf. When the object turns around, although I will continue to calculate that the light clock on the object is slower, I will see it accelerate (as the light will take a shorter and shorter time to get to me(???)) turn blue and get brighter. The space-object, were it wearing a watch, would return to my “world” and our watches would still be synchronised even though I would believe that its watch has been persistently slow.

That said, if the space-object returned and our watches were not synchronised, then things get even messier. The other postings suggest that the space-object would return to me and its watch would truly be in the past. However, if we assume that the distortion of time has perfect symmetry, then one could argue that I sped away from the space-object and therefore I should return with my watch in the past. Well we can’t both have watches in the past when we are standing next to each other. Things appear to fall apart once the distance between 2 objects is reduced / reducing. What is going on?

10. Aug 5, 2008

### yuiop

Your assumption of perfect symmetry is incorrect and is the cause of your confusion. One of the most misleading statements in pop science treatments of relativity is that "everything is relative". This is simply not true!

The "twins paradox" experiment is superficially symmetrical. One twin goes away in a rocket from her twin sister that she leaves behind on a space station for a given time, turns around and comes back. For the purpose of this experiment the spacestation is far away form any large gravitational bodies. The travelling twin on the rocket might consider themselves to stationary and from her point of view the "stationary" twin left behind on the spacestation appears to go away and return. There are a number of things we can do to show that there is no real symmetry in the situation. The first thing we can do is that we can attach one accelerometer to the space station and another to the rocket. The accelerometer on the rocket shows that the "travelling" twin reverses the direction of her motion halfway through her journey while the same is not true for the twin left onboard the space station. We could also get the twins to send signals to each other of a given wavelength at one second intervals (as timed by their own clocks). When the travelling twin departs they both recieve signals that are redshifted, but when the travelling twin turns around she immediately sees the signals from the space station as blue shifted, while the twin on the spacestation does not see blue shifted signal from the rocket until much later in the journey when the travelling twin is well on her way home. For a quite a while the twin on the space station is not even aware that the travelling twin has turned around and is heading back. So now we have two things that are provably not symmetrical about the two twins during the experiment. What they measure is not the same.

There is another way to prove the asymmetry. We introduce a third inertial observer. The third observer also has an accelerometer and to qualify as an inertial observer their accelerometer must read zero during the entire experiment. Not let's say the third observer has a velocity equal to the velocity of the outward leg of the travelling observer. For the initial part of the journey the third observer wuld say the travelling twin is stationary and the space station is moving away with constant velocity. When the travelling twin turns around, the third observer sees the travelling twin as changing velocity while the space station continues with constant velocity. So the third observer would say the space station had constant velocity during the entire journey while the travelling twin did not. The same would be true from the point of view of any inertial observer whatever their velocity relative to the space station. the situation of the two twins is not symmmetrical from the point of view of any inertial observer. All inertial observers agree that the travelling twin changed her velocity at some point during the experiment and that she has aged less than the twin on the space station when the twins get back together. The twins do not see their measurements as symmetrical. The independent inertial observers do not see the motion of the twins as symmetrica. Since there is no observer that sees the situation as symmetrical there is no observer that expects the twins to have aged the same amount and there is no paradox.

Last edited: Aug 5, 2008
11. Aug 5, 2008

### DrGreg

As you raised the idea of a "time loaf", the thing you need to know is this: each observer slices their bread in a different direction!

See diagram at bottom of this post.

Two events are simultaneous, according to an observer, if they are in the same slice. But as their slices are at different angles, two observers disagree what "simultaneous" means.

Each observer calculates the other observer's clock rate as the rate of "blobs per slice" (in the diagram), using the other observer's blobs but their own slices. (Let's say each blob represents a one-second tick of a clock, and each slice is one second thick.)

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12. Aug 5, 2008

### JesseM

No problem, you're not wasting my time at all.
I think you're misunderstanding the "time loaf" concept, i.e. the idea of slicing up 4D spacetime into a series of 3D slices. The slicing is not based on when you see images of events--i.e. if your clock reads 12 noon, August 5 2008 at the same instant you see a distant star going nova, the event of your clock striking that time and the event of the nova would not be part of the same "slice". Instead, slices are collections of events that are assigned the same t-coordinate in your rest frame. One way of thinking about this is in terms of your compensating for the delays in images reaching you, under the assumption that light always travels at exactly c in your frame; for example, if that distant nova was exactly 100 light years away in your frame, then since you saw it at a time-coordinate of 12 noon, August 5 2008 according to Earth clocks, the event itself would be assigned a time coordinate of 12 noon, August 5 1908, and thus be simultaneous with events on Earth at that date--part of the same "slice".

As DrGreg said, a key idea here is that different frames slice the same time loaf at different angles, which is another way of saying they disagree about the simultaneity of different events--if two events at different locations are assigned the same t-coordinate in my frame, they may be assigned two distinct t-coordinates in the frame of an observer in motion relative to me. This has to do with the fact that each observer assumes light moves at c in all directions relative to themselves. So for example, if I am on a spaceship which I treat as being at rest, and I set off a flash at the midpoint of the ship, and light-detectors at either end of the ship will click when they detect the light from the flash, then in my frame the light moves at the same speed in both directors so the two clicks are simultaneous. But if in your frame the ship is moving forward, that means in your frame the detector at the front of the ship is moving away from the point where the flash went off and the detector at the back is moving towards it, so if you assume the light moved at the same speed in both directions in your frame, naturally you must say that the click from the light detector at the back of the ship went off before the click from the light detector at the front.
Moving clocks still dilate in each observer's frame when they assign t-coordinates to events (like successive ticks of a moving clock) using the technique I described above where you compensate for the time the light from each event must have taken to cross the distance between the event and yourself, in your own frame.
No, the relativistic Doppler effect takes into account both "genuine" time dilation and also the fact that successive peaks emitted by the source have a different distance to travel to reach you. In fact, if an object is moving towards you, you'll actually see its clock sped up, not slowed down; but if you compensate for the fact that the light from each successive tick was emitted at a shorter distance from you, you'll find that the ticks were "really" happening at the slowed-down rate predicted by the time dilation equation in your frame.
This is the classic "twin paradox" of special relativity--see this page for an excellent discussion--but the flaw is assuming symmetry between an inertial observer and an observer who has to accelerate to turn around, in fact the time dilation equation is only supposed to work in inertial reference frames.

13. Aug 5, 2008

### cryptic

There is also a symmetric set-up of twin paradox, where both observer move in opposite direction with respect to a third observer at rest. In this arrangement SRT does not work.

14. Aug 5, 2008

### JesseM

Of course SR works here--do you understand that simultaneity is relative in SR, so that different observers disagree about whether events at different locations happened at the same time-coordinate (simultaneously in their frame) or different time-coordinates? As long as the three observers don't turn around to reunite, they each have their own inertial rest frame with its own judgments about simultaneity, so if the third observer C saw the other two A & B depart in opposite directions at 0.8c when they were all aged 20, C would say that the event of his turning 40 was simultaneous with the events of A & B turning 32, but A would say the event of his turning 32 was simultaneous with the event of C turning 27.2, and with the event of B turning 22.634.

15. Aug 5, 2008

### cryptic

Your conception of SRT is not precise enough. According to SRT moving clocks run slower, thus clock A must be slower then clock B because A is moving with respect to B. On the other side, clock B must run slower then clock A because B is moving with respect to A.

Logically expressed:

t(A) < t(B) & t(B) < t(A) ==> t(A) ≡ t(B).

16. Aug 5, 2008

### JesseM

Not in any objective sense, no, only from the perspective of the frame where it's moving (you understand there is no absolute notion of motion or rest in relativity, right?) If B is moving relative to A, then in A's rest frame B's clock runs slower than A's, while in B's rest frame A's clock runs slower than B's. There is no physical sense in which either of these clocks is "really" running slower or faster than the other.

17. Aug 5, 2008

### cryptic

But Hafele-Keating experiment shows that moving clocks, with an absolute speed around the center of the earth, actually run slower.

18. Aug 5, 2008

### MeJennifer

There is no such thing as an absolute speed.

While I do not question the validity of general relativity the Hafele–Keating experiment failed to demonstrate it and the researcher's conslusions where shoddy to say the least. Something similar happened to the solar eclipse experiment performed by Eddington on Recife. In both cases the "post processing" of the data was clearly biased.

19. Aug 5, 2008

### JesseM

Those clocks were not moving inertially, the time dilation equation of SR only applies to inertial frames. Of course, as in the twin paradox, there is an objective truth about which of two clocks has elapsed more time in total if they move apart and then reunite later, although there is no object truth about which was running slower at any given instant during the trip, and no truth about which of two clocks moving apart inertially has elapsed more time after one of them has ticked forward by a given amount.

20. Aug 5, 2008

### cryptic

Rotation is always an absolute speed.