# Question on simultaneous events.

1. Feb 5, 2004

### lky

Suppose if I am in a ship travelling from points A to B (10 light years apart) at a relavistic speed of say 0.8c.

Then suppose if there is a very strong light bulb at both points A and B, and assuming that the light rays do not get weakened along the way.

Now if I am in the ship moving from A to B at 0.8c and I am in the midpoint of my journey, when I observe the 2 light bulbs turn on simultaneously.

Am I correct to conclude that the 2 light bulbs are indeed turned on simultaneously, as if viewed by a stationary observer, since that the speed of light is constant to all observers, irregardless of their motion?

Or would the motion of the ship have any effect on this simultaneity?

Or I am incorrect to assume that the midpoint of my journey means light has to travel the same distance for both the cases of points A and B?

I would appreciate any help to clarify my doubts.

2. Feb 5, 2004

### Staff: Mentor

Everyone agrees that the light from each bulb arrived at the ship at the same time. But they disagree on whether the lights were switched on at the same time.

The fact that the two flashes reach the midpoint at the same time is evidence that they were turned on simultaneously according to observers in the rest-frame of the light bulbs. Observers in the ship will disagree that the lights were turned on at the same time. (Observers in the ship will conclude that light B must have turned on first, since it is moving towards the ship.)
Simultaneity is relative to the observer's frame. Observers at rest with the bulbs and those in the ship will disagree on what is simultaneous.
The light from each bulb is only seen to have traveled the same distance according to the observers at rest with the bulbs. Folks in the ship disagree.

3. Feb 5, 2004

### lky

Re: Re: Question on simultaneous events.

Just to be sure if I got you correct,

Because if both bulbs were turned on simultaneously, I (on the moving ship) would see the bulb at B turn on first.
Therefore if I am on the moving ship and I saw the 2 bulbs turn on simultaneously, it would mean that the bulb at A was turned on earlier than the bulb at B because I am moving from A to B.

And if C was instead a stationary observer at the mid point, he would see the bulb at A turn on earlier than B.

4. Feb 5, 2004

### gnome

This is very confusing to me.

Here's the way I see it:

Let's change the scenario, and say the lights were turned on simultaneously as viewed by an observer O who is at rest wrt points A and B and who is located at the midpoint between A and B, and at that same instant (before either observer sees the lights) observer O' in the spaceship passes directly by the same midpoint.

Edit: Just to clarify: here the lights are turned on at the instant that the spaceship passes the midpoint, so both observers will see the lights at some (different) later time.

In this situation, O' will first see the light from B at time T1', and sometime later, at time T2' will see the light from A. He measures the distance to B to be, say, d1' and concludes that it was turned on at time T1' - d1'/c. Similarly, when O' sees the light arriving from A, he measures the distance to A to be d2' and concludes that that light was turned on at time T2' - d2'/c. He then compares those two times he calculated in order to determine which light was switched on first.

But in the situation iky described, O' sees light coming at him from both points A and B, and each point is the same distance (call it d') away, and in each case the speed of the light is the same, c, so wouldn't he conclude that the time it took for the light to arrive at his location from each point was &Delta;t' = d'/c and therefore the lights were turned on simultaneously? In this situation, stationary observer O, measuring the distances to A and B to be d, would be concluding that the time for the light to arrive at the same central location was &Delta;t = d/c so he also concludes that the lights were turned on simultaneously. However, I think that because d' is smaller than d, observer O' concludes that the time that has elapsed since the lights were turned on is less than the elapsed time that would be determined by observer O.

If this is not correct, I'm in trouble. Please straighten me out.

Last edited: Feb 6, 2004
5. Feb 5, 2004

### LURCH

I'm going to disagree with you DocAl, which probably means I'm wrong, but maybe you could show me where I'm wrong. I would have said that if light from A and B meet at Mid-point M when M is occupied by the ship, then the observer on the ship will see the two come on simultaneously. And that an observer stationary relative to A and B (and M) would also agree. He would see that both lights were turned on simultaneously, but the ship was nowhere near M when this occured. In fact this observer would say that the ship was one year out from A when both lights came on, and the light from A took four years to catch up to the ship, arriving at the same moment as light from B reached the ship head-on.

This would be consistant with time-dilation, since to an observer inside the ship, the light from A closed the gap in one year, because the ship was one lightyear out from A at the time the bulb came on. Is this not so?

6. Feb 5, 2004

### gnome

Yes, I think the stationary observer O finds that it took the spaceship (5ly)/(.8c)=6.25 years to travel from A to his position at the midpoint, whereas it took 5 years for the light from A to cover that distance, so as LURCH says, the ship was 1.25*.8 = 1ly from A when the light turned on.

But for observer O' in the spaceship, the trip from A to the midpoint took 6.25*&radic;(1-.82)=3.75 years, and he measures the distance from A as v&Delta;t'=(.8c)*3.75 years = 3 ly. So he concludes that only 3 years have elapsed since the lights were turned on. Agree?

7. Feb 6, 2004

### lky

LURCH, your explanation sounds plausible, but I'm assuming that the ship is indeed physically at the midpoint between points A and B.

Drawing a rough space-time diagram (see attached file), with really bad symmetry ( the gradient for both light from A and from B should be perpendicular instead).

From my not-so-good-understanding and space-time-diagram, it seems that the 2 light turns on at time t together. And both the ship and the stationary observer would see the light bulbs turn on together.

Or is it that the axis for the space time-daigrams are different for the moving ship and the stationary observer and the axis are instead skewed to each other?

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8. Feb 6, 2004

### Staff: Mentor

All observers will agree that the light arrives simultaneously at the ship. So, the observer on the ship would see the the two light beams arrive simultaneously. This does not mean that observers would agree that the lights were turned on simultaneously; that is a deduction, not a direct observation.
See my comments above. No one "sees" the lights turned on simultaneously. But observers at rest with A-B will insist that they were turned on simultaneously. Observers in the ship will not.

9. Feb 6, 2004

### gnome

Please tell me where I am wrong. If the observer on the ship doesn't think both lights were turned on 3.75 years ago, exactly when does he think each one was turned on?

10. Feb 6, 2004

### Staff: Mentor

Much of this discussion hinges on how this statement by lky is interpreted:
I've been interpreting this to mean: The light from both bulbs arrives at the ship exactly as the ship is at the mid-point between the bulbs. (That's what "I observe the 2 light bulbs turn on simultaneously" means to me.)

Am I wrong about what you meant, iky?

11. Feb 6, 2004

### Staff: Mentor

No. The light bulbs are moving. So, when the light that reaches the ship first started out, the light bulbs were not equidistant from the ship.
Right! Observer O will agree that each light beam traveled the same distance in getting to midpoint. So observer O thinks they were emitted simultaneously.
Observer O' (ship) disagrees that the light beams were emitted when the bulbs were equidistant from the ship; so he thinks each beam traveled a different distance. Also, observer O' sees all the usual relativistic effects:
1) Moving clocks slow down
2) Moving clocks are no longer synchronized (if they are synched in their own frame)
3) Moving lengths shrink
I think we'll both survive!

12. Feb 6, 2004

### Staff: Mentor

When did the lights go on? Mysteries revealed?

Here's how to think of it.
Pretend that observers in O have three clocks: one at A, one at M (midpoint), and one at B.

According to O, these clocks are synchronized. When they all read T = 0, that's when lights at A & B were turned on. When the light reaches M, O thinks that his clocks all read T = 5 years.

What does O' think? First, he knows that the clocks in O are not in synch: they are way off. O' thinks that clock A is 4 years behind clock M! (Use Lorentz transformations to check this. $\Delta T = \frac{vL}{c^2}$) Thus, when the light reaches M, O' says that clock M reads T = 5, but clock A only reads T = 1. So O' thinks only 1 year has passed (on the clocks in O) since the light left A. But since the moving clocks are slow, that means that O' says that 1/.6 = 1.66 years have passed.

Similary, O' says that the clock at B is 4 years ahead of the clock at M. So, when the ship is at the midpoint, O' says that clock B reads T = 9. So O' thinks 9 years have passed (on the clocks in O) since the light left B. So O' says that 15 years (of his time) have passed.

So, according to the observers in the ship, the light:
- left A 1.66 years ago
- left B 15 years ago

(Assuming I didn't mess up my transformations; it's been a while )

13. Feb 6, 2004

### Eyesaw

If my head was not simultaneous in time with my body,
I would expect it to fall off when I walk up a
flight of stairs. Relativity of simultaneity...time
dilation...lol.

Last edited: Feb 6, 2004
14. Feb 6, 2004

### lky

Re: Re: Question on simultaneous events.

Yup, that's what I meant.

15. Feb 6, 2004

### gnome

Hurrah! I don't want to get too excited about this, but at least I did finally get the same result as you. Thanks Doc.

But I still don't see exactly what you are doing. Here's what I did.

First, I let O' claim that he is stationary & A and B are moving toward the left at .8c. Since the distance from A to B in the A-B frame is 10ly, it is 6ly in the O' frame, and when O' is at the midpoint, it is 3ly to each of A and B in the O' frame. We all seem to agree on that.

Next I let x' = the distance that A and B travelled since the lights went on, and t' = the time elapsed since that time. So, as to the light from B:
ctB' = 3 + xB'
And as to B itself:
.8ctB' = xB'
Solving those equations gives tB' = 15 y

Similarly, as to the light from A:
ctA' = xA'
And as to A itself:
.8ctA' = 3 - xA'
Solving those equations gives tA' = 1.67y

---------------------------------------------------

Alternatively, using the Lorentz transformations I find in my textbook:
t' = &gamma;(t - vx/c2)
In frame O, xA = xB = 5y, and tlightson = -5 (i.e. 5 yrs ago).
&gamma; = 1/&radic;(1-.64) = 1.67, so as to A:
tA' = 1.67[-5 - (.8)(-5)] = -1.67y

and as to B:
tB' = 1.67[-5 - (.8*5)] = -15y

----------------------------------------------------

Now, Doc, does my second calculation express EXACTLY the same approach as yours ('cause even though we got the same answers, it doesn't look the same to me [b(] )
You seem to be using a very abbreviated form of the transformation.
Am I still missing something that would make it easier for me to see?

16. Feb 6, 2004

### Janus

Staff Emeritus
I did a couple of animations some time back to illustrate this very thing. Now my illustration deals with a railway car and lightning strikes rather than spaceships and light bulbs, the the principle it illustrates is exactly the same.

The first one shows the sequence of events according to the observer stationary at the mid point.

http://home.teleport.com/~parvey/train1.gif [Broken]

The exanding circles represent the leading edges of the light spheres.

Note that both lights turn on at the same time and that the light spheres reach both observers at the same time, so each of them "sees" the lights turn on at the same time.

The second one shows the sequence of events according to the observer on the Ship/train. Remember, from his position, he can consider himself as stationary and the lights as moving.

http://home.teleport.com/~parvey/train2.gif [Broken]

First note that as each light turns on, the leading edge of the light still expands as a sphere from the initial point of emission according the viewpoint of the spaceship/train observer. But, the light sources move away from these points as time goes on. By the time the midpoint is reached, the lightsources and their intial points of emission are far removed from each other. Not also that the intial emission points always maintain the same distance form the train observer.

Now, since the train observer is not at the midpoint when either of the lights is initially emitted, this means he is closer to one of the emission points than the other at that time.

Thus in order for him to "see" the lights turn on at the same time, the light he is heading for has to start emitting first, followed later by the light he is departing from.

this means that as the midpoint passes by, the light spheres both arrive, and again both observers "see" the light come on at the same time.

Again, in order for order for both observers to "see" the lights turn on at the same time, the lights actually turn on at the same time for the observer with no relative velocity with respect to the lights, but for the observer with a relative velocity, they will have to turn on at different times.

IOW, events that are simultaneous in one frame are not simultaneous in another.

Last edited by a moderator: May 1, 2017
17. Feb 6, 2004

### Staff: Mentor

Yes! It's exactly the same.
For a complicated problem, I would just go right to the lorentz transformations and turn the crank. But for simple problems like this, I like to apply the "rules" for how clocks and metersticks behave. Of course, these rules are exactly equivalent to the lorentz transformations, but it makes me think I understand what's going on.

1) moving metersticks shrink (by a factor of &gamma;)
2) moving clocks slow down (by a factor of &gamma;)
3) moving clocks are out of synch (by a factor of &Delta;X V/c2)

Make sense? Get these rules in your bones and you can solve some problems quickly. (Derive them for yourself.)

18. Feb 6, 2004

### @Gents

Lets assume the spaceman dont noe anything abt physic, and he is moving from A to B with a man stationed at M. If he reach M and saw both the light from A and B reached him at the same time, doesnt he think both are lighted simultaneously?

19. Feb 7, 2004

### lky

If the spaceman don't know anything about physics, not only would he think the explosions are simultaneous, he would also think that the Earth is flat.

20. Feb 7, 2004

### lky

A big thanks to you guys for clarifying my doubts, especially Doc Al and his detailed explanation, and also Janus' animations.