# Simultaneity in Relativity

1. Jul 30, 2009

### stovepipe

Hi again,

So I'm still relatively new to working with Relativity (no pun intended) and to these forums and I have a question about simultaneity. I have read that simultaneity is broken when events are viewed from different frames. I wasn't quite sure what this meant until I worked out an example for myself today, and it puzzles me.

I set up a theoretical experiment in which two clocks were synchronized and placed equidistant (L/2) from a flash bulb. Obviously, to anyone in the same frame as the apparatus the time shown on both the clocks would be (L/2)c + tflash , if the flash went off.

An observer now travels at .6c parallel to the apparatus. At the instant this observer passes the bulb, it triggers the flash. Observer A then records the time shown on each clock when the reflected light reaches him. He records the clocks to display (L/5)c and (4L/5)c!!

How can this be?!? Of course the answer is that the simultaneity of events was broken. But it doesn't make sense that the same sets of photons could hit the same sets of clocks at 2 different times, or in fact at an infinite number of times! (One could imagine an infinite number of reference frames all with different speeds less than c all passing the origin at the same time.)

I took the time to fully work through the example with the 2 frames I set up, and found that both frames can be used to predict what was seen by the other frame. Just that the two frames disagree on what actually happened. No wonder refs make bad calls!!

While mathematically this seems feasible, it didn't really make physical sense. I mean what if instead of a flash bulb it was a lethal dose of gamma radiation, and my cats were standing in front of the clocks. The cats wander off of the target platforms just before the clocks strike (L/2)c. But according to other frames the radiation has already hit. It's like Schrodinger's cat all over again! But instead of whether or not it was observed, it's who observed it. So do my cats live?

Thanks,
-Steve

2. Jul 30, 2009

### diazona

Nice thought experiment Steve, but there's something I'm not understanding... how are you coming up with those numbers Lc/5 and 4Lc/5? Where is the light reflected from - one of the clocks? And when A records the time on the clocks, does he correct for the time light took to travel from where the clocks are to where he is?

Regardless, the answer to your question is time dilation. The fact is, in the reference frame of the apparatus, the time that each clock reads when the flash of light hits it is Lc/2 (say tflash=0 for simplicity). Those are two simultaneous, colocated events (same time AND same place) - "the clock reads Lc/2" and "the flash of light hits the clock" - and so they will be simultaneous and colocated for all observers. That is, all observers, no matter what their velocities, will see each clock reading Lc/2 at the same time the observer sees the flash of light hitting that clock. BUT (and this is key) a moving observer like observer A, if he is keeping his own time and not going by what the clock reads, will notice that it takes longer than Lc/2 for the flash light to hit the clock. The only way to reconcile these two observations is for observer A to see the clock running slow.

3. Jul 30, 2009

### vin300

Have a look at this http://galileo.phys.virginia.edu/classes/252/synchronizing.html" [Broken]

Last edited by a moderator: May 4, 2017
4. Jul 31, 2009

### stovepipe

The site vin300 posted seems to agree with me.

Last edited: Jul 31, 2009
5. Jul 31, 2009

### stovepipe

Here are my graphs of the events in Minkowski space. The flash bulb is represented by world line B, the two clocks are represented by the world lines C and D. The observer that sees the events as not simultaneous is represented by the world line A. The first diagram is for a frame that is co-moving with B, C, and D. The second diagram is for a frame co-moving with A. In both diagrams, green lines are lines of constant proper time, cyan lines are for simultaneous events wrt A, and magenta is for simultaneous events wrt B. The dotted blue lines are the paths that the photons travel.

The green lines are to account for time dilation. And you will notice that in the frame co-moving with A, the distance between C and D appears smaller due to the apparent length contraction of the apparatus.

Both frame agree that B sees image on the clock at Lc/2, while A sees the image on the clock at 4Lc/5 and Lc/5.

If the diagrams are fundamentally wrong I would like to know.

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6. Jul 31, 2009

### vin300

One sec, the time of flash is L/2c not Lc/2

7. Jul 31, 2009

### stovepipe

Well that may be. I did the problem with L = 1 ly and was quickly trying to generalize. Either way, I'm still confused by the concept of how things can be simultaneous in one frame, but not in another.

Last edited: Jul 31, 2009
8. Jul 31, 2009

### Saw

Hi, stovepipe. Nice example, I agree.

And in SR both frames should agree that that happens. It’s “only” that they disagree on how to label (which time and space coordinates they assign to) those events. The frame of the apparatus will say that both cats have wandered off at the same time, after the radiation has travelled equal distances to reach them. The frame of the other observer will agree that both cats have wandered off and saved their lives, it’s “only” that she labels those events with different coordinates: for her, one cat has escaped earlier, after the radiation has travelled a shorter distance, while the other cat has jumped off later, after the radiation has traversed a longer distance. Another way to put it: they reach the same solution, though they follow different paths.

You may have this objection: Well, if it is “only” a question of labels or paths, it’s not dramatic. But we’re talking about “time” and that’s a serious thing. For example, imagine that the cats do not escape and do die. Someone at rest with the observer’s frame witnesses the death of the cat that, in his frame, dies earlier. He decides to save at least the other cat. In his frame that looks, in principle, possible, since the lethal signal will only reach the other cat quite later. However, in the cat’s frame, it is already dead… Solution? Technical rule: if the two events are simultaneous in one frame, they are spacelike separated and not causally connected. More clearly: to do so, the observer should send a faster-than-light (FTL) signal, which is not possible in SR.

You may have a second objection: add to that GR and space-time curvature. In the light of this, the generous observer could go through a wormhole to do his deed. That is a shortcut in spacetime, no violation of the rule that bans FTL. So we may find such observer sitting by the cat and trying to convince it to run away… Solution? For popular science books: either the observer will not manage to persuade the cat (because a mysterious invisible hand will prevent him from doing so) or the observer has jumped into a parallel universe where the cat survives (although in the original universe it dies). For common sense: well, whoever is saying so is either trying to sell more books or has honestly got lost somewhere in his reasoning… ( = Just a little provocation, with best wishes, the forum was lately getting a little boring without this sort of discussions... )

9. Jul 31, 2009

### Ich

I think you diagrams are ok, except that you label the clocks in the second one with your wrong conclusion t=0,2/0,8.
The point is that these cyan lines in no way justify that a clock in B's frame shows 0,8 when B shows 0,8.
They say that a clock synchronized and at rest in A's frame at the respective position shows 0,8.
Clock 2, at the same time at the same position, of course reads 0,5.

10. Jul 31, 2009

### Staff: Mentor

The time recorded on each clock when the flash strikes it is the same for all observers--it is not affected by the observer's motion.

Of course, it can't be. What is different for each observer is the time that the photons arrive at the original clocks according to the moving observer's own clocks. That's where the relativity of simultaneity comes in.

11. Jul 31, 2009

### sylas

Lots of people beat me to it; but I am posting anyway since I went to the trouble of making some diagrams. (I got delayed by a poker game.) Here's my take...

If the observer is observing the clocks with the light reflected from the flash, the observer will see (L/c)c + tflash from both clocks. This doesn't change. However, the moving observer will consider that the two clocks are not in synch, and they are both running slow.

Let me try to nail it down. Let me assume that the speed of light is 3*108 m/s (it is actually a fraction less; sue me, or imagine I'm using "short" meters) and that the length L is 6 kilometers.

1. The clocks' frame

In the frame of the flash gun and the clocks, the flash goes off at time t=0. Both clocks read zero at this instant (simultaneous in the clocks' own frame). The moving observer is passing the flash gun at this instant.

10 microseconds later (10-5 seconds) the light from the flash reaches the clocks. At this time, the clocks read 10. (They count microseconds.) At this point, the observer moving at 0.6c is now a distance 1.8 kilometers from the flashgun.

Another 2.5 micro seconds later, the moving observer is 2.25 kilometers from the flash gun, and meets the light reflected from the nearest clock coming back the other way.

37.5 microseconds later (50 microseconds after the flash) the moving observer is 9 kilometers from the flash gun, and 12 kilometers from the further clock. The light reflected in the flash from the further clock just now catches up with the observer.

In both cases, what the moving observer (or any observer, in fact) will see in the light reflected off the two clocks is the reading 10 microseconds.

Here is a space time diagram, marked off in kilometers, and microseconds, in the frame of the flashgun and the clocks. The magenta lines are light from the flashgun, reflected off the clocks, and then intersecting the path of the moving observer, in blue.

2. In the moving observer's frame

Now, let's put the same thing into the frame of the observer moving past the flash gun when it goes off. I am using c = 0.3 km/ microsecond, and the velocity of the observer relative to the flash gun is 0.18 km / microsecond.

Using Lorenz transformations, we map points (time, dist) as follows:
$$\begin{array}{l|rl|rl} & \multicolumn{2}{c|}{\text{flash frame}} & \multicolumn{2}{c}{\text{obs frame}} \\ \hline \text{general} & t & x & t' = \gamma ( t - v x / c^2 ) & x' = \gamma ( x - v t ) \\ & & & = 1.25 (t - 2 x) & = 1.25 ( x - 0.18 t ) \\ & & & = 1.25 t - 2.5 x & = 1.25 x - 0.225 t \\ \hline \text{flash} & 0 & 0 & 0 & 0 \\ \text{right clock reads 0} & 0 & 3 & -7.5 & 3.75 \\ \text{light hits right clock} & 10 & 3 & 5 & 1.5 \\ \text{obs sees right clock} & 12.5 & 2.25 & 10 & 0 \\ \text{left clock reads 0} & 0 & -3 & 7.5 & -3.75 \\ \text{light hits left clock} & 10 & -3 & 20 & -6 \\ \text{obs sees left clock} & 50 & 9 & 40 & 0 \end{array}$$​

Note that the clocks are not synchronized in this frame. The left clock reads 0 at time 7.5 microseconds, and the right clock reads 0 at time -7.5 microseconds. They are also running slow, in this frame, and moving left at 0.6c.

Here is the space time diagram transformed to the frame of the passing observer.

Cheers -- sylas]

12. Jul 31, 2009

### stovepipe

Ok, I see now. In A's frame the clocks didn't all read 0 at the bottom of the diagram. That's where I went wrong. According to A's frame in my diagram, while 0.2 years elapsed on clock C since the flash went off, it "read" 0.3 years when the flash went off. Similarly for clock D, 0.8 years elapsed on the clock, but it "read" -0.3 years when the flash went off. I needed to use the line of simultaneity for B's frame to determine the where the clocks read 0.

Thanks,
-Steve

Last edited: Jul 31, 2009
13. Oct 17, 2009

### Hacky

My understanding is that this experiment would not yield the same result with say two baseballs launched from the center of the train because they would follow classical Galilean addition of velocities (the one launched forward would be moving at the launch velocity plus the velocity of the train, the one launched rearward is launch velocity minus velocity of train, so each observer on train and outside train would still see balls arrive at either end of train at same time (due to back and front of train moving forward). However, the velocity of the light is c in both directions regardless of how fast the train is moving, so the light gets to rear in less time and front in longer time?

One other question, with light why does the observer on the train not detect what the outside observer sees (light getting to rear of train earlier than front)? Is this due to the opposite effect of a signal that would need to be relayed back to observer in center of train to let them know when original signals hit front and rear of train (would take less time from back of train than from front to reach the observer in center)?

Thanks, Howard