Question on simultaneous events.

  • Thread starter lky
  • Start date
  • #1
lky
6
0

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
Doc Al
Mentor
44,940
1,201
Originally posted by lky
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?
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.)
Or would the motion of the ship have any effect on this simultaneity?
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.
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?
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
lky
6
0


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
1,036
1
(Observers in the ship will conclude that light B must have turned on first, since it is moving towards the ship.)
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 Δt' = 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 Δt = 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:
  • #5
LURCH
Science Advisor
2,549
118
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
1,036
1
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*√(1-.82)=3.75 years, and he measures the distance from A as vΔt'=(.8c)*3.75 years = 3 ly. So he concludes that only 3 years have elapsed since the lights were turned on. Agree?
 
  • #7
lky
6
0
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?

Please help.
 

Attachments

  • #8
Doc Al
Mentor
44,940
1,201
Originally posted by LURCH
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.
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.
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.
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
1,036
1
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
Doc Al
Mentor
44,940
1,201
Much of this discussion hinges on how this statement by lky is interpreted:
Originally posted by lky
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.
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
Doc Al
Mentor
44,940
1,201
Originally posted by gnome

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 Δt' = d'/c and therefore the lights were turned on simultaneously?
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.
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 Δt = d/c so he also concludes that the lights were turned on simultaneously.
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.
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.
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
If this is not correct, I'm in trouble. Please straighten me out.
I think we'll both survive! :smile:
 
  • #12
Doc Al
Mentor
44,940
1,201
When did the lights go on? Mysteries revealed?

Originally posted by 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?
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. [itex]\Delta T = \frac{vL}{c^2}[/itex]) 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 :wink:)
 
  • #13
129
0
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:
  • #14
lky
6
0


Originally posted by Doc Al
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?
Yup, that's what I meant.
 
  • #15
1,036
1
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' = γ(t - vx/c2)
In frame O, xA = xB = 5y, and tlightson = -5 (i.e. 5 yrs ago).
γ = 1/√(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
Janus
Staff Emeritus
Science Advisor
Insights Author
Gold Member
3,526
1,269
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:
  • #17
Doc Al
Mentor
44,940
1,201
Originally posted by gnome

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(] )
Yes! It's exactly the same.
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?
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 γ)
2) moving clocks slow down (by a factor of γ)
3) moving clocks are out of synch (by a factor of ΔX V/c2)

Make sense? Get these rules in your bones and you can solve some problems quickly. (Derive them for yourself.)
 
  • #18
1
0
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
lky
6
0
Originally posted by @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?
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
lky
6
0
A big thanks to you guys for clarifying my doubts, especially Doc Al and his detailed explanation, and also Janus' animations.
 
  • #21
1,036
1
Make sense? Get these rules in your bones and you can solve some problems quickly.
Solve problems: yes.
Make sense? NO!

[Janus: I'd like to learn how to write animations like those you posted. Would you mind showing me the code? Can you tell me what specifically to search for to find tutorials, etc.?]

I guess I'm able to use these transformations well enough, but play around with the numbers a little & it immediately becomes clear that even weirder things happen as soon as acceleration is involved. For example, in iky's problem O' travels from A to the midpoint in 3.75 years by his clock, or 6.25 years by the O clock. But suppose, instead, that he flew at .825c for [6 yrs O-time = 3.39 yrs O'-time] and .2c for [0.25 yrs O-time = 0.245 yrs O'-time], so now he arrives at the midpoint in the same 6.25 yrs O-time, but it's only been 3.64 yrs O'-time ("his" time). Now, he arrives at the midpoint, sees the light arriving from A and calculates that the light left A (1/√(1-.2^2) * (-5 + 5*.2) = 4.08 yrs ago, O'-time. But wait! The spaceship left A only 3.64 yrs ago (O'-time). The spaceship passed the light? Obviously not. So the acceleration must screw things up (unless I made a mistake in my conversion).

Anyway, if the space-travelers left A to fly to B, why would they claim that they (the space-travelers) are stationary and that A and B are moving? Isn't that just being ornery? After all, they know that they are the one's that are going to be accelerating and decelerating. Why not just keep O-time and make appropriate adjustments to the readings of their instruments?

EDIT: Spaceman says, "I'll fire my rockets & make that planet come to me."
HUH?
Yeah, yeah -- I'm just ranting.
 
Last edited:
  • #22
Doc Al
Mentor
44,940
1,201
Originally posted by lky
A big thanks to you guys for clarifying my doubts, especially Doc Al and his detailed explanation, and also Janus' animations.
I am glad that you found the discussion helpful. And, yes, Janus... cool animations!
 
  • #23
dlgoff
Science Advisor
Gold Member
3,824
1,756
And, yes, Janus... cool animations!
ditto. Thanks. I would like to use them to explain to my young daughter if that's okay.
 
  • #24
1,036
1
Thanks, Doc. I very much appreciate your help, but I'm not quite "there" yet. I'm still wrestling with the question of why the space traveler moving from A to B, having begun in the AOB frame, and planning to end up in that frame, wouldn't choose to continue using AOB as his frame of reference, making appropriate conversions or even calibrating his instruments to measure time and distance as they would be measured in that frame. (As opposed to taking the - to me, at least - very strange position that he is the stationary "center of the universe" and everything else is moving relative to him.)
After all, if I can use the Lorentz transformations to determine how things look to him, he can also.

So, is he compelled to use his local frame of reference just to avoid making these adjustments, or is there something more fundamental that I'm missing?
 
  • #25
Doc Al
Mentor
44,940
1,201
Originally posted by gnome
Solve problems: yes.
Make sense? NO!
You're killin' me.
I guess I'm able to use these transformations well enough, but play around with the numbers a little & it immediately becomes clear that even weirder things happen as soon as acceleration is involved.
The lorentz transformations relate measurements between inertial frames. When things accelerate, you have to be careful in applying special relativity.
For example, in iky's problem O' travels from A to the midpoint in 3.75 years by his clock, or 6.25 years by the O clock. But suppose, instead, that he flew at .825c for [6 yrs O-time = 3.39 yrs O'-time] and .2c for [0.25 yrs O-time = 0.245 yrs O'-time], so now he arrives at the midpoint in the same 6.25 yrs O-time, but it's only been 3.64 yrs O'-time ("his" time). Now, he arrives at the midpoint, sees the light arriving from A and calculates that the light left A (1/√(1-.2^2) * (-5 + 5*.2) = 4.08 yrs ago, O'-time. But wait! The spaceship left A only 3.64 yrs ago (O'-time). The spaceship passed the light? Obviously not.
Bah! Careful. In lky's problem, we can't have the spaceship changing speeds. But we can fudge it OK by considering 3 frames of reference: O, O'(spaceship 1, v=.825c), O''(spaceship 2, v=.2c). After our astronaut travels for 6yrs (O time) he jumps to the new spaceship. (We assume that he and his pocketwatch makes the transistion in zero time with no problems.) Now he travels in the new ship (O'' frame) until he makes it to the midpoint. Yes, he claims that only 3.64 yrs have passed for him; and his new O'' friends say that the light left A 4.08 yrs ago. But they laugh and say, "Dude, your watch was wacky and your time means nothing to us! What you thought was ΔT (when you were on the other, faster ship) was really longer since you were moving compared to us."
 

Related Threads on Question on simultaneous events.

  • Last Post
Replies
18
Views
1K
  • Last Post
Replies
18
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
12
Views
3K
Replies
4
Views
2K
Replies
280
Views
41K
Replies
14
Views
2K
Replies
9
Views
701
Replies
1
Views
1K
Replies
9
Views
863
Top