# Relativity of simultaneity

Tags:
1. Jul 1, 2015

### Noctisdark

Good morning,
Yesterday I was reading a book about special relativity, It focused a lot about consequences of the theory, but there's only one thing I couldn't understand, which is as the title suggests, Relativity of simultaneity, that two different observers can't agree if two event are simultanious, I understand Einstein's thought expirement, but I get often confused about it, for example Imagine an observer at rest, let's name it A, whose origin is O, now draw the (x,t) space-time along with A's world line and O, let's place a point G in Elsewhere (Not in the absolute future or past of O), how can I draw the wordline of a moving observer B passing through O that record O and G to be simultanious,

Last edited: Jul 1, 2015
2. Jul 1, 2015

### Mentz114

I think what is meant can be seen more clearly on a space-time diagram. In this plot, time is the vertical axis and x is a spatial dimension on the horizontal axis.

The first image shows the events and worldlines in the coordinates of the blue worldline and second is in the coordinates of the green WL. The events A and B are simultaneous in 'blue' coordinates but not in 'green' coordinates. The two diagrams are connected by a Lorentz transformation.

Hence 'relativity of simultaneity'.

File size:
1.4 KB
Views:
77
File size:
1.5 KB
Views:
84
3. Jul 1, 2015

### phinds

Mentz, I guess I'm missing something here. Don't you have the relative vertical positions of A and B reversed in the second drawing?. In other words, is it not appropriate to simply rotate the entire 1st picture so that the green goes vertical?

4. Jul 1, 2015

### Staff: Mentor

In the first picture, the green observer's x and t axes don't intersect at a 90-degree angle, and that property will be preserved if you just rotate the picture to get the green axis vertical. That's a valid picture, but it's not the one that's usually drawn, and it's not the one that Mentz drew. Try drawing lines of constant green t (hey - that sounds funny when you say it ) through the points A and B, see where they intersect the green t axis, and that will give you the right vertical relationship.

5. Jul 1, 2015

### phinds

I guess I'm particularly dense this morning. What I'm interpreting from the first picture is that blue sees A and B happening at the same time whereas green in that pic sees A first and then B. The second pic shows green seeing B happening first, which seems wrong to me.

6. Jul 1, 2015

### Mentz114

The diagrams are correct and can be switched by a Lorentz transformation.

My interpretation of 'simultaneous' is $t_A=t_B$. Maybe not standard.

7. Jul 1, 2015

### Staff: Mentor

In the first picture, the lines of constant green t slant from bottom left to top right, so the line of constant t (all events on such a line are simultaneous in the green frame) that passes through B is lower on the page and intersects the green t axis below (earlier) than the line that passes through A.
If you draw both t axes in both pictures, it will be clear.

8. Jul 1, 2015

### phinds

OK, thanks guys. I was being dense this morning.

9. Jul 1, 2015

### Noctisdark

Guys, thanks for replies, can I ask another question, suppose an event happened at (0,0) in A's frame of reference [(t,x) space-time], by lorentz transformation, I find that for any observe x' = y*(x-vt) = 0, t' = y*(t-x*v/c^2) = 0, I know what I am wrong, can someone explain why ?

10. Jul 1, 2015

### Mentz114

Under the the transformation $t'=\gamma t + \beta \gamma x,\ x'=\gamma x + \beta \gamma t$, (0,0) goes to (0,0). That makes no difference. You can put the origin ( t=0,x=0) anywhere you like. What do think is the problem ? On my diagram I did not specify an origin because it is not needed.

11. Jul 1, 2015

### Staff: Mentor

No, you're right. The form of the Lorentz transformation you're using assumes, for simplicity, that the two coordinate systems are chosen such that they use the same event as the origin. That's physically equivalent to saying that we set the two clocks, one at rest in each frame, to zero at the exact moment that they were colocated.

12. Jul 1, 2015

### ghwellsjr

Yes. And all you really need are the two events A and B. The two worldlines are irrelevant except to let us know that the second frame is moving at 0.3c with respect to the first one.

You could assign the coordinates xA=0, tA=0 and xB=5, tB=0 then transform to a speed of 0.3c and get the new coordinates of x'A=0, t'A=0 and x'B=5.24, t'B=-1.57. Since tA=tB, the two events are simultaneous in the first frame and since t'A≠t'B, the two events are not simultaneous in the second frame.

Looks standard to me. Why are you thinking your interpretation might not be standard?

13. Jul 1, 2015

### phinds

I think my question confused things.

14. Jul 1, 2015

### Mentz114

More generally we could say that transforming two points $(t,x_1), (t,x_2)$ then $\gamma t + \gamma \beta x_1 <> \gamma t + \gamma \beta x_2$ unless $x_1=x_2$ (colocation).

Panic.