# Measurement in special relativity

1. Jul 27, 2013

hi
I read this statement in site www.askamathematician.com and I couldn't understand it.
is it possible to explain this?

"if measurements A and B are taken enough apart, they will be “space-like separated” according to SR, meaning that neither event precedes the other. Some observers will correctly believe that A happened first, others will know that B came first, and SR says that nobody is wrong. Time doesn’t work the way we usually think it does, so watches won’t agree for observers moving relative to each other. "

2. Jul 27, 2013

### Staff: Mentor

That is all essentially correct. What is your specific question?

3. Jul 27, 2013

if two happen was spacelike, they don't have any causal relationship, so, state which one occur first isn't true, I think

4. Jul 27, 2013

### HallsofIvy

Staff Emeritus
Yes, that's exactly what your quote says. So again, do you have a question about it?

5. Jul 27, 2013

yes, can you please explain for me how it possible?

6. Jul 27, 2013

### WannabeNewton

There is no frame independent time ordering of events in special relativity for space-like separated events. If I fix an event $p$ and we consider an event $q$ that is not in the light cone of $p$ then $q$ can occur before, at the same time as, or after $p$ in different inertial frames.

7. Jul 27, 2013

how another frame can say it occur before or after? i can't understand!

8. Jul 27, 2013

### Staff: Mentor

Try googling around for "relativity of simultaneity train" - that will find you some good pointers to Einstein's classic thought experiment describing the how simultaneity can be relative. Read aboutthat, see if it helps you understand, and if not you can come back with some more focused questions.

9. Jul 28, 2013

### Staff: Mentor

This is a necessary consequence of the two postulates. If the laws of physics are the same in all inertial reference frames (principle of relativity) and if the speed of light is c in all inertial reference frames (invariance of c) then it logically follows that different frames disagree about the order of events.

Consider a train car with a flashbulb in the middle of the car and two detectors, one at the front of the car and one at the rear. In what order does the light from the flash reach the detectors? In the reference frame of the train the light travels at c the same distance in both directions and therefore the light reaches the detectors at the same time. In the reference frame of the ground the light travels at c but the light heading towards the back of the train travels a shorter distance than the light heading towards the front and therefore it reaches the rear detector first.

10. Jul 28, 2013

### ghwellsjr

Your article does an excellent job of explaining so I'm not sure I can offer anything more but I'll try. I'm going to draw some spacetime diagrams for you. The article explains what a spacetime diagram is so I won't go into details about that.

In order to make things a little simpler, I'm going to show three observers, all traveling at different speeds and we will consider the first event A to occur when they happen to coincide. The second event is labeled B in the diagrams. Each observer measures the time of the event by sending a radar signal at an appropriate time which hits the second event B and bounces off it and returns to the observer. Each observer considers the time at which the radar signal hits event B to be half way between when he sent the radar signal and when he received its echo.

I'm using the speed of light (and the radar signals) to be 1 foot per nanosecond and because of the way that I have draw the diagrams, the radar signals will travel along 45-degree diagonals (as the article pointed out).

In the first diagram, focus your attention on the blue observer. The dots mark off 1-nanosecond increments of time. At the dot labeled with the blue 0, he sends the black radar signal which propagates upward and to the right, reflecting off of event B and arriving back at the blue observer at his time of 12 nanoseconds. (Don't be confused by the fact that the red observer has previously sent his radar signal which then propagates "in parallel" with blue's and finally with green's.) Therefore, he concludes that event B occurs at his time of 6 nanoseconds which is the same time as event A.

Now do the same thing for the red observer who is traveling to the right at 0.6c. He emits his black radar signal at his dot marked with the red 0 and receives its echo at his time of 15 and so he concludes that event B occurred at his time 7.5 nanoseconds which is before his time for event A.

Finally do the same thing for the green observer who is traveling to the left at 0.8c. He emits his black radar signal at his dot marked with the green 0 and receives its echo at his time of 20 and so he concludes that event B occurred at his time 10 nanosecond which is after his time for event A.

Note that all of these measurements were made in the same Inertial Reference Frame (IRF), the rest frame of the blue observer. Key to this working is the fact that the dots for the other two observers are Time Dilated according to their speeds and spaced out accordingly.

But we can use the Lorentz Transformation process to redraw the diagram in the IRF in which the other two observers are at rest. First the rest frame for the red observer:

Please note that all the information that was contained in the first diagram is also contained here, the only difference being in the coordinates. The diagram for the rest frame of the red observer clearly shows that event B occurs before event A.

Finally, the rest frame for the green observer:

Note that in the diagram for the rest frame of the green observer, event B occurs after event A.

Also, please note that the reason why each observer comes to a different conclusion about the ordering of the two events is because they each are assuming that their own radar signal takes the same amount of time to reach event B as it takes for the echo to get back to them.

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Last edited: Jul 28, 2013