# The Switch Paradox

1. Jul 28, 2010

### calebhoilday

This is a paradox that I have derived based on my understanding of relativity of simultaneity. Upon someone resolving it, I should be better able to understand the concept, which seems rather contradictory of the greater theory at this point in time.
If events are considered to be simultaneous to a stationary observer, then the lack of simultaneity for a moving observer can be calculated based on the velocity of the moving observer and the distance between events in the moving observers frame of reference.
T= DV/C^2
Or
Duration between events in moving observers frame = (Distance between events in moving observers frame * Velocity of moving observer)/ The speed of light^2
In this paradox, a prisoner has been locked a really long opaque room, which he knows is being transported through space. In this tube he is left with two cuckoo clocks at either end, each one a light second away from him. He notices that one is ahead of the other, as he can see the hour indicated by the actions of the cuckoo clock, that one of them starts 1s before the other.
He considers if they are simultaneous to a stationary observer, then he must have a velocity of 0.5C, but due to not knowing if this discrepancy is just simply due to not setting the clocks correctly, he cannot determine his velocity.
After some time he considers that if he switches the clocks, placing one in the place the other was, he would be able to determine the speed at which he travels
When he does this he notices that the clocks maintain the 1s discrepancy with the one that is forward, only being due to which end of the room it hangs from. He is then able to rule out a fault in synchronisation and can safely assume that his speed is 0.5C
The principle of relativity forbids such a method from determining an absolute velocity.
I am very interested in how this paradox is resolved.

2. Jul 28, 2010

Staff Emeritus
I don't understand how this is supposed to work. He sees two clocks that run at the same rate - clocks that may or may not be set to different times. What can he possibly conclude from that?

3. Jul 28, 2010

### calebhoilday

There is three possible causes for a difference between the two clocks:
(a) The frame has velocity
(b) Human error
(C) A mixture of both.

The error that results from the velocity of the frame is dependent on the position of the clock. The human error is dependent on the clock itself. Swapping clocks revealed in the hypothetical that it is not the clocks themselves that are causing the difference but rather the position the assume.

Clock X was one second in front of clock Y. Swapping positions resulted in clock Y being one second in front of clock X. This reveals that the discrepancy was totally position based, due to the room having velocity and had nothing to do with human error.

4. Jul 28, 2010

Staff Emeritus
But all that tells you is the relative velocity of the person who synced the clocks.

5. Jul 28, 2010

### Fredrik

Staff Emeritus
Uhh Caleb, are you actually saying that if you swap the positions of the clocks, the clock that was ahead is now behind?

6. Jul 28, 2010

### Austin0

Hi
Clock desynchronization is not observable within a frame. It is only as observed from other frames and of course will vary according to the relative velocity of those fraMES.

7. Jul 28, 2010

### Aaron_Shaw

What the guy above said...
If the front clock was 1 ls ahead then when he swaps the same clock (now at the rear) will stay 1 ls ahead. The clocks will only swap the 1 ls difference, eventually, if they were moving ahead away from the bloke who swapped them.

8. Jul 28, 2010

### calebhoilday

According to my understanding yea, hence the problems.

Why wouldn't that happen?

"But all that tells you is the relative velocity of the person who synced the clocks." Vanadium 50

This is what I thought also at first and the fact that vanadium 50 pointed this out undermines the other posts.

Consider if you got two clocks out of a box at each end. You set them at each so that they are synchronised. Based on previous parameters an observer considering the clocks to have a velocity of 0.5C would consider the trailing clock to be 0.5 seconds behind clock the leading clock (in terms of the event being the animation of the clock each hour). If this observer happened to be the absolute reference frame in the universe, then this discrepancy will be carried across. When the switch takes place on board, the clocks would go from being synchronised to being 2 seconds out. You could use this logic to determine absolute speed, without having to synchronise the clocks in another reference frame.

9. Jul 28, 2010

### starthaus

There is no such thing as "an absolute reference frame of the universe".
Besides, all observers in motion wrt the two clocks will see them "desynchronized" by a factor of $\gamma(v)\frac{vL}{c^2}$ where L is the distance between the two clocks.

This is false. Are you making up all this stuff or are you reading it from a sci-fi book?

Err, no.

10. Jul 28, 2010

### JesseM

It seems you are assuming that clocks will "naturally" adjust themselves to keep simultaneous in some preferred "stationary" frame, but this isn't so. The only reason to expect two clocks to be simultaneous in any frame is if someone set them that way, moving clocks around without resetting them can cause them to go out-of-sync in whatever frame they were initially synchronized in.

11. Jul 28, 2010

### DrGreg

calebhoilday, you are forgetting that the act of moving the clocks within the room causes time dilation. If you move both clocks at the same speed relative to room, then both clocks will gain the same amount relative to the room and they will still be one second apart -- the one that was ahead before will still be ahead now, regardless of whether it's at the front or back.

From the point of view of someone outside the room relative to whom both clocks were originally in sync, the clocks have been moved at unequal speeds so one has dilated more than the other and they are now out of sync.

There is no such thing as "the absolute reference frame in the universe" or "absolute speed".

12. Jul 28, 2010

### Fredrik

Staff Emeritus
I hope you understand that the only thing that can answer a question like that is a theory, so I'm going to have to use SR to answer it. SR says that the time displayed by a clock is the proper time of the curve in spacetime that represents its motion. It's very easy in this case to ensure that the curves that describe the motion of the clocks while you're doing the switch have the same proper time. That ensures that the one that was ahead is still ahead. Nothing fancy or high-tech is needed to make sure that this happens. It's sufficient that the switching process looks symmetrical to the guy in the middle (and I was always assuming that that's what you had in mind anyway).

His comment applies to the situation before you switch the clocks (which is actually going to be identical to the situation after you have switched the clocks). It doesn't apply to the full SR-contradicting scenario you described.

I assume that I'm comoving with the ship when I'm doing that.

That's not what I get. If x and y are simultaneous (x0=y0) and two light-seconds apart (y1-x1=2), the difference between their time coordinates in the other frame is

$$(\Lambda x)_0-(\Lambda y)_0=\gamma(x_0-vx_1)-\gamma(y_0-vy_1)=\gamma v(-x_1+y_1)=\frac{1}{\sqrt{1-0.5^2}}\times 0.5\times 2\approx 1.15$$

13. Jul 28, 2010

### Janus

Staff Emeritus
Switching the clocks will not change the synchronization of the clocks according to the prisoner. Clock x will still be one second ahead of clock y as far as he is concerned. It will change their synchronization according to the observer for which they were initially in sync. According to him, the Clocks will no longer be in sync and clock y will be ahead of clock x.

You mistakenly assumed that the clocks remain in sync according to this "stationary" observer regardless of how the prisoner moves them. This is not the case. Consider what happens as the prisoner moves the clock from the back of the room to the front. For our observer, during this period, the clock is moving, relative to him, more than 0.5c. Conversely, the clock moving from front to back is moving slightly less than 0.5c. the clock moving forward will experience more time dilation than the one moving backwards, and the clocks will no longer be in sync after the moves.

On the other hand, as long as the prisoner moves both clocks at the same speed(relative to the room) from one wall to the other, he will see no difference in time dilation between the two and they will keep their respective time difference.

14. Jul 28, 2010

### calebhoilday

I also considered this, but one will find that the amount of time lost is not the same over a particular distance regardless of the velocity of the object.

this can be show mathematically using these parameters:
lenght to cover: 1 light year
velocity of craft A : 1/3 C
velocity of craft B : 1/2 C

The assumption you have made is: Time taken for craft A to cover the distance to stationary frame - Time taken to cover the distance according to craft A frame = Time taken for craft A to cover the distance to stationary frame - Time taken to cover the distance according to craft A frame

or

3 - (3*(8/9)^0.5) = 2 - (2*(3/4)^0.5)

but
3 - (3*(8/9)^0.5) = 0.1715 years
and
2 - (2*(3/4)^0.5) = 0.2679 years

they are not equal.

15. Jul 28, 2010

### Fredrik

Staff Emeritus
Why is there no gamma in your calculations? (Compare yours to mine above).

The phrase "distance to stationary frame" doesn't make sense. You can talk about the distance between two objects, but not about the distance between an object and a frame. (Also, there's no such thing as a "stationary frame").

You also need to be more careful with the details. I have read your description of this new scenario, but I have no idea what you're describing.

16. Jul 28, 2010

### calebhoilday

Disregard my last post it it isnt what janus is infering. I'll be back in 2 hours or so to crunch some numbers, concerning janus's post, but still may not work out.

Sorry fredrik for the lack of clarity; will give better detail soon.

17. Jul 29, 2010

### calebhoilday

If you consider clock X to be the leading and clock Y to be trailing with clock X being the one that is ahead one second.

When considering the room frame, if the clocks are switched, at the same velocity, then if special relativity (or my understanding of it) will infer that clock X is always ahead of clock Y regardless of the position by the same magnitude.

Special relativity will also require that when this switch occurs, that what is predicted by an outside observer, as to what the prisoner observes, must also arrive at the no change due to position. This is rather involved and I will come up with a less involved situation to highlight issue quantitatively, when time permits (in the next 24 hours hopefully).

The reason that Lorentz transformations have not been factored into the paradox is because, i think that if the switch happens at very slow rate that Lorentz transformation effects can be ignored. meaning that i think that the Lorentz transformations will result in differing outcomes dependent on the velocity; not resulting in the required 1 second in the observing frame difference as a result of the switch with every velocity ( I haven't crunched the numbers yet, so i'm not 100% on this being the case)

I will admit that this paradox is poorly presented, hopefully the next hypothetical will be found to be less opaque.

Last edited: Jul 29, 2010
18. Jul 31, 2010

### calebhoilday

Sorry for the delay on writing an explanation, here is the reason.

(((1-VS)/(S-V))*(((1-((S-V)/(1-VS))^2)^0.5)-1))-(((1-V^2)/(V-S))*(((1-S^2)^0.5)-((1-V^2)^0.5))=V

To explain what this is step by step, is a very involved process and will take time (to do it well). Number crunching worked out not so well (for me or special relativity) as it confirmed the qualitative assumption that I made. That being if one were to reposition the clocks at slow speeds then the effects of time dilation would not correct repositioning enough to maintain congruency. It also confirmed that if S=-V a result would be congruent (will be explained more), meaning if repositioning happens at a fast enough velocities, then you can get a congruent result.

tomorrow is sunday, no work no plans, should have it posted tomorrow.

19. Jul 31, 2010

### Mentz114

It's been pointed out to you that this is nonsense. He doesn't have a speed, except that which can be measured relative to another observer. It is a tenet of SR that 'it is not possible to distinguish a state of rest from a state of uniform motion.'

(my bold ) in what frame ? You're talking about absolute velocity here and there's no such thing.

No, don't waste your time on this. There's nothing to resolve.

20. Jul 31, 2010

### calebhoilday

what i was trying to say with the switch paradox, was that if you switched clocks at a slow enough rate and you found that the clock discrepancy was entirely position based, then you could safely assume that you had a velocity of 0.5C compared to some kind of absolute velocity.

The argument against this is time dilation will modify each clock by a particular amount regardless of the velocity, so that the discrepancy moves with the clocks and the the depicted situation would never happen.

What iv said is that you do not loose a fixed amount of time off a clock if it moves a particular distance. the amount of time lost due to time dilation, varies with velocity over a particular distance and it can be shown that this can become negligible at very low velocities.