# Considering Relative Simultaneity - Relying on clocks instead of observers

1. Jul 23, 2007

### ryancomplete

Einstein’s train-carriage example:

A train-carriage is moving at a constant velocity v relative to the ground. An observer is stationed on the ground perpendicular to the midpoint of the train at time x relative to the ground’s frame of reference. At this time, lightning strikes the front and back of the train-carriage which reflects and reaches the ground observer at the same time. The ground observer therefore concludes that the lightning struck the front and back of the train at the same time. But, for an observer who is stationed on the train at the midpoint of the train-carriage, the light from the front lightning strike will reach him/her before the back lightning strike and thus he/she will conclude that the front of the train carriage must have been struck first. And therefore, based on their respective observations, simultaneity is relative.

But what if clocks were used instead of observers in order to detect simultaneity? For example:

Two clocks that are stationary with respect to the ground’s frame of reference, clock gr1 and gr2, are synchronous, and two clocks that are stationary with respect to the train-carriage’s frame of reference, clock tc1 and tc2, are synchronous. Clock tc1 is at the very back of the train-carriage and clock tc2 is at the very front of the train carriage. Each clock is light sensitive and will immediately stop ticking the moment light reaches it. At a specific time with regard to the ground’s frame of reference, clock tc1 is at the exact position as clock gr1 and clock tc2 is at the same position as clock gr2. At this specific time with regard to the ground’s frame of reference, lightning strikes the back clocks, gr1 and tc1, and the front clocks, gr2 and tc2. As a result, clocks gr1 and gr2 stop ticking and each record the time to be x. The clocks tc1 and tc2 also stop ticking and both these clocks record the time to be y. Now the fact that the back and front clocks, tc1 and tc2, have recorded the same time would surely demonstrate simultaneity regardless of the differences in the times that the light from the front and back strikes took to reach the observer on the train-carriage.

2. Jul 23, 2007

### JesseM

And what procedure do you use to synchronize them? Einstein began his 1905 paper by discussing the difficulty of defining what it means for clocks at different locations to be synchronized, and looked at what kind of coordinate systems would result if you synchronized them in a given frame using the assumption that light moves at c in both directions in that frame. So, using the Einstein synchronization procedure, an observer on the train could synchronize clocks at the front and back of the train by setting off a flash at the midpoint of the train, and setting them to read the same time when the light from the flash reached each one. Likewise, an observer on the tracks could set off a flash at the midpoint of his own two clocks, and set them to read the same time when the light reached them. But this procedure necessarily means they'll disagree about simultaneity--in the track frame, the front end of the train is moving away from the point where the flash was set off while the back end is moving towards that point, so naturally if the observer on the track assumes the light moves at c in his frame, according to his own clocks the light reaches the clock at the back end of the train before the one at the front end, meaning the train's clocks are out-of-sync in his frame.

Last edited: Jul 23, 2007
3. Jul 23, 2007

### SpitfireAce

wait, im lost, why are you assuming that tc1 and tc2 will record the same time... the principle of simultaneity says that tc2 should stop before tc1

4. Jul 23, 2007

### Staff: Mentor

Tomato, tomato.

5. Jul 23, 2007

### ryancomplete

Hi JesseM,

Many thanks.

6. Nov 30, 2007

### wsellers

Is it necessary to use light flashes?

What if the observer uses a method that doesn't involve light flashes? To use a fanciful example, suppose the observer has two identical robot mice and sets them going in opposite directions from the midpoint. Regardless of the direction of the train, each mouse would reach the end of the car after having traveled for the same amount of time (at which point each clock would start).

The same method could be used when the lightning flashes, if the robots were set at each wall facing each other and programmed to start crawling upon "seeing" a lightning flash. Wouldn't each mouse reach the midpoint after having traveled for the same amount of time?

7. Nov 30, 2007

### RandallB

Sure that would all work fine in that train reference frame:
But from the POV of the Train Stations how fast is each ‘mouse’ moving?
As we classically expect as measured in the station frame because of the speed of the train one is moving faster than ‘mouse’ speed the other slower.
The whole reason Relativity was even developed; is every measurement ever made said that effect does not happen when you use light speed photons instead of the mice.
Light speed c still measured the same c in the other frame even if it required E & Hz of the photons to change. Therefore, what is simultaneous in one frame is not simultaneous in the other.

8. Nov 30, 2007

### Janus

Staff Emeritus

You're forgetting about length contraction. If in the ground frame gr1 aligns with tc1 at the same instant that gr2 aligns with tc2, then in the ground frame, the distance between gr1 and gr2 is the same as the distance between tc1 and tc2.

BUT, in the ground frame, the distance between tc1 and tc2 is due to length contraction. IOW, if the relative velocity between train and ground is 0.866c and the distance between gr1 and gr2 as measured in the ground frame is 1km, that means that the distance between tc1 and tc2, in the train's frame will be 2km.

Also, the in the train's frame, it is the distance between gr1 and gr2 which is contracted, and from this frame this distance is 1/2 Km.

Thus, tc1 cannot be aligned wth gr1 at the same time as tc2 aligns with gr2 in the train's frame if tc1 aligns with gr1 at the same time as tc2 aligns with gr2 in the ground frame. You still end up with a difference in simultaneity.

9. Nov 30, 2007

### pervect

Staff Emeritus
This is a variation of what's known in the literature as "slow clock transport", if you do a google you'll find many peer-reviewed papers (and a plethora of websites, probably including a bunch of cranky ones) that talk about it.

However, slow clock transport still requires one to specify a frame of reference, the one in which the robot mice move slowly.

Relativity predicts that in any given frame, synchronization by light signals in that frame and synchronization by slow clock transport in that frame will be equivalent.

10. Nov 30, 2007

### JesseM

I gave a derivation of why the slow clock method gives the same definition of simultaneity as the light-signal method in the second half of this post.

11. Nov 30, 2007

### wsellers

Just so I understand, does this mean then, in the case where (1) light rays go from each end of the car to the center, (2) robot mice go from each end to the center, and (3) each light ray and mouse movement is initiated by a lightning flash at either end of the car, THAT: the light rays do not reach the center at the same time and (much later) the mice also don't reach the center at the same time (from the point of view of the observer in the car)?

12. Nov 30, 2007

### JesseM

All observers agree on local events, so if both light beams are predicted to hit the center of the train at the same place and time in the train-observer's frame, and the same is true of the robot mice, then other frames will predict this too. Of course, in other frames the center of the train is moving, so this doesn't mean they agree that the meeting-place of the two light beams or the two mice is at the midpoint of the positions in space where each one started.

13. Nov 30, 2007

### RandallB

NO Your problem comes from not giving simultaneity its due.
Two flashes of light separated by any distance will always meet at the half way point at the same time – only makes sense.
Doesn’t matter if your on the train or in train stations.

BUT only if the two flashes start out at the same time! – again only makes sense right.

The issue is: IF lighting strike #1 hits at time = 0.0 for both the train and the station the frame that sees the flashes meet at the halfway point in their frame is the one where strike #2 hits at that same time = 0.0 in their frame.
That is only possible in one frame because the relative speed between the two frames, only one frame can have the #2 strike happen at the same time as #1 strike.
Again, this makes sense since the two locations (the middle point for the train) and (the middle point for the stations) will be a different places by the time the light reaches either middle point.
It was seeing that this only made sense is how SR was found in 1905.

You won’t ‘get’ SR till you ‘get’ simultaneity. Focus on that.

edit: Then by understanding simultaneity, knowing the speed between them and observing just one clock directly in the other frame an observer in either frame can define all their observations and predict those in the other frame. And ALL will agree with the same locations and times as observed in each frame. Neither observer will be “incorrect”

Last edited: Nov 30, 2007
14. Nov 30, 2007

### wsellers

Hope I am not beating a dead horse, but something is still not clear to me. The observer standing on the embankment will (I think) predict that the light beams hit the center at different times while the mice will hit the center (which by now has moved quite a bit relative to that observer) at the same place and time. Presumably this is what the observer on the train would indeed experience (assuming she had superpowers to detect the differences). However, I must be mistaken because then she would be able to infer that the train was moving. And this is what I just don't get: if the observer on the train sees the mice arriving at the center at different times (in conformance with the light beams having earlier arrived at different times), then the predictions of the observer on the embankment would be incorrect.

15. Nov 30, 2007

### wsellers

I think I understand the relativity of simultaneity when the only "tool" one uses is light beams. What I can't tell is whether the logic holds up when other supplementary tools are available.

16. Nov 30, 2007

### JesseM

Relativity postulates that all the laws of physics are the same in all inertial (non-accelerating) reference frames, and so far all the evidence supports this. What this means is that if you are in a closed windowless box moving inertially through flat spacetime and you perform an experiment, you'll get the same results as another experimenter in another closed windowless box who performs the same experiment (aside from any inherent randomness in the laws of physics), regardless of your velocities relative to one another or any external landmarks. So, it's not possible that the light-signal method could agree with the robot-mouse method in one frame but differ from it in another frame; if there's one frame where the speed of light is the same in both directions, so that sending light-signals out from the center of the train would give the same results as sending slower objects moving at the same speed in opposite directions from the center, this must be true in all frames. And since we know that the light-signal method means that clocks which are synchronized in their own frame must appear out-of-sync in other frames, the same would have to be true of other methods as well.

17. Dec 1, 2007

### Staff: Mentor

Agreed, but the pronunciation doesn't come through so well :)

It is generally understood that "observers" use clocks to measure time and rods to measure distance, both of which are at rest wrt the observer.