# Stationary and moving clocks

1. Mar 26, 2008

### arbol

1. Let S be an x-coordinate system. The origin of S is located at x = 0m. Let S' be an x'-coordinate system. The origin of S' is located at x' = 0m. This information is not sufficient in order to determine where the origin of one system is located with respect to the other.
2. Let system S' move along the x-axis of system S with a constant velocity v in the direction of increasing x.
3. Let observer A (who is able to count the time t like a clock in such a way that each of his count is exactly 1s) be stationed at the origin of S, and let observer B (who is able to count the time t' in all respects like observer A) be stationed at the origin of system S'.
4. Let the origin of the moving system S' be located, with respect to the stationary system S, at x = a at the time tA = t'A = 0s, so that the origin of the stationary system S, with respect to the moving system S', is located at x' = -a at the time tA = t'A = 0s.
5. If a ray of light departs from the origin x = 0m of the stationary system S at the time tA = 0s, and arrives at the origin of the moving system S' at the time t'B, then
t'B = a/(c - v).
6. If the ray of light is reflected at the time t'B back to the origin x = 0m of the stationary system S, arriving there at the time TA, then
TA - t'B = a/(c - v).
7. If the ray of light departs from the origin x' = 0m of the moving system S' at the time t'A = 0s, and arrives at the origin of the stationary sytem S at the time tB, then
tB = a/c.
8. If the ray of light is reflected at the time tB back to the origin of x' = 0m of the moving system S', arriving there at the time T'A, then
T'A - tB = (a + v*tB)/(c - v).

Last edited: Mar 26, 2008
2. Mar 26, 2008

### belliott4488

I think I followed all that ... did you have a question?

3. Mar 26, 2008

### RandallB

????
In 3. & 4. You have established events separated by the same distance as measured in both frames.
Also there is no known t’A at x=a there would be a t’ observable in the moving frame from x=a it could not be observed by A and certainly would not be 0.

These distances and times will not be the same as implied here, not if you use SR anyway.

4. Mar 26, 2008

### JesseM

I agree with RandallB, if at time tA=0 in the stationary system, the origin of S' is located at position x=a and the clock moving along with the origin of S' reads t'A=0, then it will not be true that at time t'A=0 in the moving system, the clock at the origin of S reads tA=0, not unless you set a=0 so that the origins of the two systems coincide at tA = t'A = 0. This is a consequence of the relativity of simultaneity, which says that if two events at different spatial locations (like the clock at the origin of S reading tA=0 and the clock at the origin of S' reading t'A=0) happen simultaneously in one frame, they happen non-simultaneously in other frames.

5. Mar 26, 2008

### Staff: Mentor

I have a more basic question. What are tA and t'A? If A is an observer then it doesn't make sense to multiply an observer by a time. If you mean tA is the time on observer A's clock then what is t'A, perhaps the time on observer A's clock as somehow measured by observer B?

For simplicity why don't you just use the standard configuration where the clocks are both set to 0 when the two observers pass each other? (same origin)

6. Mar 26, 2008

### RandallB

Yes starting from t’=t=x’=x=0 only makes sense.
For t’A since A is stationary in the x frame t’A would need to be the time A would directly observe in x’ provided the moving frame was equipped appropriately with x’ synchronized clocks positioned for viewing by “A” presumable each clock marked with its x’ location in that frame. Both passing t’ & x’ observables are available to A and worth collecting.

It will show that time appears to pass FAST in x’ when “A” collects the information directly that way. But when collecting information from other observes “A” also knows that time on any one x’ clock runs slow; thus A knows that from the x view x’ has a serious problem with getting clocks correctly synchronized. Of course from the x’ view of “B” it is “A” and the x clocks that seem to not be correctly synchronized clocks within their x frame.

The lesson is that as an inertial observer you and others in your frame of reference can synchronize clocks; but you cannot automatically assume that the same times at separate points in your frame are really simultaneous no matter how good the light signal synchronization may appear to be. That is the point of simultaneity. Not that anyone can assume their own clocks a synchronized correctly, but that no frame can assume their version of what is simultaneous is correct based on SR alone.

7. Mar 26, 2008

### belliott4488

EDIT: Whoops - scratch this. I'll leave this post in case some of it is useful, but on rethinking the problem, I see that I was too quick to apply the "by symmetry" operator and conclude that observer A's world line passes through the point t'=0, x'=-a in the S' frame; rather, it should pass through t'=0, x'=-a*gamma, I believe. So now I have to agree that I can't makes sense of the original post, or rather, I think I can follow the reasoning, but I believe that reasoning is flawed.
-----------------------------------------------------------------
While I agree that the OP's use of notation is confusing, I think there is a way to read it that is consistent with SR, although I'm still missing the point of the post.

Initially, he says that at time t=0 in the unprimed frame, the world line of the observer in the primed frame crosses the point x=a, so that's one event. Now, he also says that the primed observer's clock reads zero for that event, i.e. the primed observer gives that event the coordinates (0,0), i.e. t'=0 and x'=0 at this event. No problem so far, other than the absence of a clear reason to set things up this way.

Now he states that the "origin of the stationary system S, with respect to the moving system S', is located at x' = -a at the time tA = t'A = 0s." This confused me at first, but I don't think he means to say that the event with coordinates in the unprimed system x=0, t=0 has the coordinates (-a,0) in the primed system. The "origin" of the unprimed system, ie. (0,0) clearly does not occur at t'=0 in the primed system. Nonetheless, if you construct the world line of the unprimed observer (assumed to be at x=0) in the primed frame, the point where that world line crosses the x' axis will indeed have coordinates (-a,0). <edit:not so> I believe this is the event the OP was referring to, although I don't believe it's right to refer to this as the "origin" of the S system, since it occurs at some delta-t before zero in that frame.

In any case, if you take this to be the meaning you can now construct two space-time diagrams, one from the point of view of each observer, with these three events marked:

1. $$(x=a,t=0), (x'=0,t'=0)$$
2. $$(x=0,t=0), (x'=a-v \delta t', t'=\delta t')$$
3. $$(x=0,t=-\delta t), (x'=-a,t'=0)$$
where $$\delta t$$ and $$\delta t'$$ are the time differences between the two events on the unprimed t axis, which can be expressed in terms of a and v, but which I haven't bothered to do.

If you then consider the two reflected beams of light, I think the algebra works out for the rest of it, although you have to figure out the notation along the way (note that t'A and T'A aren't the same, for e.g.).

What this all shows still escapes me, however.

Last edited: Mar 27, 2008
8. Mar 28, 2008

### arbol

Thank you for your feedback. In my post, either one event or (exclusively) the other takes place (the ray of light departs from the origin of the stationary system S to the origin of the moving system S', or, exclusively, the ray of light departs from the origin of the moving system S' to the origin of the stationary system S).

Nevertheless, from your feedback I thought of the following new version of my post (where I only cover the set up of system S and system S'):

1. Let S be a stationary x-coordinate system. Let S' be a stationary x'-coordinate system. Let the x'-axis of system S' coincide with the x-axis of system S, and let system S' move along the x-axis of system S with constant velocity v in the direction of increasing x. Let observers (each of them with the ability to count the time t in such a way that each count is exactly 1s, being synchronous to each other) be stationed along the x-axis of S, and let observers (each of them with the ability to count the time t' in all respects like the observers stationed along the x-axis of S) be stationed along the x'-axis of the moving system S'.

2. Let the origin of the moving system S' coincide with the point x = a on the x-axis of the stationary system S at the time ta = t'A = 0s. (The observer that counts the time ta = 0s is stationed at the point x = a on the x-axis of the stationary system S, and the observer that counts the time t'A = 0s is stationed at the origin of the moving system S', but two distinct observers cannot occupy the same space-time point where the two space-time points (x, ta) = (a, 0s) and (x', t'A) = (0m, 0s) coincide. Therefore, it is necessary that the observer that counts the time ta = 0s be the same observer that counts the time t'A = 0s.)

3. Since the origin of the moving system S' is located, with respect to the stationary system S, at x = a on the x-axis of system S at the time ta = t'A = 0s, and since every observer that is stationed along the x'-axis of the moving system S' has the ability to count the time t' in all respects like the observers that are stationed along the x-axis of the stationary system S, the origin of the stationary system S, with respect to the moving system S', is located at the point x' = -a on the x'-axis of system S' at the time t'a = tA = 0s. (The observer that counts the time t'a = 0s is stationed at the point x' = -a on the x'-axis of the moving system S', and the observer that counts the time tA = 0s is stationed at the origin of the stationary system S, but two distinct observers cannot occupy the same space-time point where the two space-time points (x', t'a) = (-a, 0s) and (x, tA) = (0m, 0s) coincide. Therefore, it is again necessary that the observer that counts the time t'a = 0s be the same observer that counts the time tA = 0s.)

Last edited: Mar 28, 2008
9. Mar 28, 2008

### belliott4488

Why do you say that "two distinct observers cannot occupy the same space-time point"? Is this a literal statement about two physical observers with some mass, physical extent, etc.? If so, then I think this is an unnecessary restriction. "Observers" in SR are usually somewhat abstract concepts, generally taken to be point-like, so that they exist at one point without bringing up concerns about their right hands being at different space-time points than their lefts hands and so forth.

If you really insist on seeing your observers as physical beings, then you'll run into trouble since the set of moving observers are moving on the same axis as the stationary ones. Won't they run into each other? It's kind of like a train of cars attempting to travel down a track while another train of stationary cars is resting there.

In any case, you can get around all this by simply stating that your observers have clocks that are synchronized in their respective rest frames, aligned along their x-axes in such a way that they can unambiguously determine the time and location of any space-time event that occurs. The means for doing this is well-established, and no one should object to your invoking this mechanism here.

This is where you're getting into trouble. First, by "origin", I take it you mean simply the point x=0 in the given reference system, not the space-time point (0,0), which also has a time coordinate equal to zero. Is that correct?

Now, you state that at time t=0 in S, the origin (x'=0) of the S' system is at the location x=a in S. Also, you state that this event, i.e. the coinciding of the point x=a on the S x-axis with the point x=0 on S' x-axis, also occurs at time t'=0 in S'. This is fine: you have one distinct event and have defined the space-time origins (i.e. the x=0 and t=0) of your two reference frames such that this event has coordinates (x,t)=(a,0) in S and (x',t')=(0,0) in S'. No problem.

Here is the problem. Because these two frames are in motion relative to each other, they do not agree on what space-time events are simultaneous. In particular, observers in S can look at all the markings on the S' axis (noting that it is Lorentz-contracted), and they can make a note of where those markings fall along their own x-axis at time t=0. For example, they can take note of which position along the x' axis happens to coincide with the point x=0 on the x=axis at time t=0. The observers in S', however, will not agree that those locations on their x'-axis coincided with the stated locations on the x-axis at time t'=0; rather, they will say that the given points all coincided at other times.

In particular, when you say that the origin of S will cross the point x'=-a at time t'=0, this is not true - it cannot be.

Observers in S will say that the distance between the two origins was equal to a according to their axis markings at time t=0, but they will also note that since the x'-axis is Lorentz-contracted, that it is not the point x'=-a that coincides with the point x=0 on their axis at time t=0, but rather the point x'=-b=-gamma*a
(gamma=Lorentz factor: $$\gamma=[1-v^2]^{-1/2}$$).

Moreover, if these observers in S ring a bell at this point (x=0) when t=0, the observers in S' will not agree that the bell was rung at t'=0; rather it will be rung slightly later.

In other words, the observers in S' will not see the the point x'=-b in their frame cross the point x=0 in S at time t'=0. Rather, they will see it cross x=0 at an earlier time. Thus, we really have three separate events:

1. The point (x,t)=(a,0) <=> (x',t')=(0,0)
2. The point (x,t)=(0,0) <=> (x',t')=(-gamma*a,gamma*va)
3. The point (x,t)=(0,-av) <=> (x',t')=(-a/gamma,0)

This is all due to the relativity of simultaneity, for which you have not properly accounted in your description.

Last edited: Mar 28, 2008
10. Mar 28, 2008

### RandallB

I agree with belliott This is not true at all – especially in a though experiment.

You are making the set up much too hard to follow.
Therefore I’m going to assume the purpose of your posting is to get help in how to describe SR type problems.

I like to use is a system of trains and stations along the x axis.
Train cars and stations are all numbered out from zero both negative and positive.
Stations all have clocks displaying (x,t) just as all train cars have clocks displaying (x’, t’).
The x & x’ on each clock of course never changes as it is the station or train car number for that clock. Everyone use the same units for time and distance with (unit x) = c(unit t); thus you can leave units off and assume all use any scale you like eg (foot, nanosecond) or (Light-year, Year) it all the same.
The only synchronization between the frames in that x=x’=0 when t=t’=0 meaning when station 0 and train car 0 will be together or were together both their clocks show t=t’=0.

You must pick convection for numbering the train cars. I number them as negative to the train reference frame distance. Stations have the same sign as the station reference frame numbering. That way any stationary observer on train or in a station will see increasing numbers co by on the moving clocks. But always seeing decreasing numbers works OK too, just pick a standard.

That’s it you can start from any position, each car while passing a station allows both the station attendant and the train conductor to record four bits of info (x,t)(x’,t’). The same info is know and shared with all observers within a frame but the two frames will disagree as to which frame has properly synchronized if the trains one x is longer of shorter the one x’.
Any problem you can make should fit into that system without much trouble, just be sure to define and collect all four pieces of information for every event, along with what each frame thinks is happening at the x=0 and x’=0 from their point of view for that event time. That is where the two frames will never agree.

Note: passengers are allowed to move between station and train i.e. twins; but due the speed they must take place instantly, no time allowed for any gravitational affects, this set up is only good for SR work.