Events can be simultaneous in one time frame but not so in others

[stevmg]

A recent forum thread went into a long disussion of simultaneity. The term "ansible line" was used (later referred to as a "simultaneity line.")

Einstein demonstrated with a thought experiment with regards to lightning and a moving train that events can be simultaneous in one time frame but not so in others.

1) Can two events which are simultaneous in one frame of reference (FOR) be simultaneous in another - that is - THE SAME TWO EVENTS, or is the simultaneity limited to only ONE FOR.

2) Can one always find an FOR in which any two event pairs are simultaneous?

3) It would seem that an event A can be simultaneous with an event B in one FOR and also be simultaneous with event C in a second FOR. This does not mean that B and C are simultaneous but it should be possible to find an FOR in which they are. Again, this would NOT mean that \exists an FOR in which A, B and C are simultaneous mutually, or does it?

 

[espen180]

1) If you take a look at a Minowski diagram, you will notice that the slope of the simultaneity line is linearly proportional to the velocity. Therefore, the answer is no.

2) No. Mark the two events in a spacetime diagram and draw a straight line from one to the other. If the slope of this line is greater than or equal to c, there is no intertial frame in which they are simultaneous.

 

[yossell]

I *think* the answer to (a) is yes. Imagine a frame F traveling in the x direction compared with frame G, and think about events that lie on the y-axis - that is, events that lie on a line at right angles to the direction of motion. I think both frames will agree on the times of such events, and two events simultaneous in G will be simultaneous in F.

 

[stevmg]

Thanks - have to get ahold of a Minkowsky diagram somewhere. Don't have one on hand.

Still need the answer to part 3...

stevmg

 

[stevmg]

I *think* the answer to (a) is yes. Imagine a frame F traveling in the x direction compared with frame G, and think about events that lie on the y-axis - that is, events that lie on a line at right angles to the direction of motion. I think both frames will agree on the times of such events, and two events simultaneous in G will be simultaneous in F.

Thanks!

A special case - OK

Still need an answer to part 3

stevmg

 

[yossell]

Ah 3 - I'm even less sure of my answer here - but

I think if A and B are spacelike related and A and C are space-like related, then B and C are space-like related. If there's a frame where A and B are simultaneous, then they're space-like related; ditto for A and C. So B and C are space-like related, so there's a frame where they're simultaneous.

But this certainly doesn't mean you can find a frame where all events are simultaneous.

But I would wait until a professional confirms or disproves what I say here!

 

[starthaus]

1) If you take a look at a Minowski diagram, you will notice that the slope of the simultaneity line is linearly proportional to the velocity. Therefore, the answer is no.

The correct answer is "yes".

x'=\gamma(x-vt)
t'=\gamma(t-\frac{vx}{c^2})

so:

dx'=\gamma(dx-vdt)
dt'=\gamma(dt-\frac{vdx}{c^2})

If there is a frame where dx=0 and dt=0 then, in all frames dt'=0 (and dx'=0)

 

[starthaus]

A recent forum thread went into a long disussion of simultaneity. The term "ansible line" was used (later referred to as a "simultaneity line.")

Einstein demonstrated with a thought experiment with regards to lightning and a moving train that events can be simultaneous in one time frame but not so in others.

1) Can two events which are simultaneous in one frame of reference (FOR) be simultaneous in another - that is - THE SAME TWO EVENTS, or is the simultaneity limited to only ONE FOR.

2) Can one always find an FOR in which any two event pairs are simultaneous?

3) It would seem that an event A can be simultaneous with an event B in one FOR and also be simultaneous with event C in a second FOR. This does not mean that B and C are simultaneous but it should be possible to find an FOR in which they are. Again, this would NOT mean that \exists an FOR in which A, B and C are simultaneous mutually, or does it?

You can start answering your own questions using only the prototype at post 7. All you need is the differential form of the Lorentz transforms.

 

[espen180]

think about events that lie on the y-axis - that is, events that lie on a line at right angles to the direction of motion.

I am unable to make sense of this statement. How can an event (a point) lie at right angles to anything?

Here's a small Minowski diagram applet i threw together. (http://sites.google.com/site/espen180files/Minowski.html?attredirects=0&d=1"

You can vary the speed of frame S'. The movable line parallel to the x' axis is the S'-frame's simultaneity planes. Please convince yourself that two events can only be simultaneous in one frame.

 

[starthaus]

I *think* the answer to (a) is yes. Imagine a frame F traveling in the x direction compared with frame G, and think about events that lie on the y-axis - that is, events that lie on a line at right angles to the direction of motion. I think both frames will agree on the times of such events, and two events simultaneous in G will be simultaneous in F.

Yes, you are correct, no doubt about it.

 

[starthaus]

I am unable to make sense of this statement. How can an event (a point) lie at right angles to anything?

Look at post 7, it gives a rigorous explanation.

 

[espen180]

The correct answer is "yes".

x'=\gamma(x-vt)
t'=\gamma(t-\frac{vx}{c^2})

so:

dx'=\gamma(dx-vdt)
dt'=\gamma(dt-\frac{vdx}{c^2})

If there is a frame where dx=0 and dt=0 then, in all frames dt'=0 (and dx'=0)

I think you're missing the point.

Assume two events which are simultaneous in S. Their spacetime coordinates are
E₁=(x₁,t₁)
and
E₂=(x₂,t₂)
with t₁=t₂

Frame S' is traveling relative to S with velocity v. So we have
x'₁=γ(x₁-vt₁)
x'₂=γ(x₂-vt₂)=γ(x₂-vt₁)

t'₁=γ(t₁-vx₁/c²)
t'₂=γ(t₂-vx₂/c²)=γ(t₁-vx₂/c²)

There is only two ways the events E₁ and E₂ can be simultaneous in S':
1) x₁=x₂
2) v=0

In the case of 1), there is only one event, and simultaneity is not a useful concept here.
In the case of 2), S=S', there is only one frame.

This proves the statement that a pair of events can only be simultaneous in one frame.

 

[starthaus]

I think you're missing the point.

I don't think so.

There is only two ways the events E1 and E2 can be simultaneous in S':
1) x1=x2
2) v=0

In the case of 1), there is only one event, and simultaneity is not a useful concept here.

Err, no. You forgot the fact that you have two other coordinates , y and z , that can separate the events. Event means E(x,y,z,t).yossell already explained that to you in post 6. You need to pay attention.

 

[DrGreg]

think about events that lie on the y-axis - that is, events that lie on a line at right angles to the direction of motion.

I am unable to make sense of this statement. How can an event (a point) lie at right angles to anything?

The events lie on a line that is at right angles to the direction of motion

Here's a small Minowski diagram applet i threw together. (http://sites.google.com/site/espen180files/Minowski.html?attredirects=0&d=1"

You can vary the speed of frame S'. The movable line parallel to the x' axis is the S'-frame's simultaneity planes. Please convince yourself that two events can only be simultaneous in one frame.

You are thinking in two dimensions only (1 space + 1 time). You need to think in three dimensions (2 space + 1 time). That's why yossell referred to the y axis, at right angle to both the t and x axes.

 

[my_wan]

Without me providing the math, I suggest you remain skeptical of my answer. I haven't actually done the math myself to prove it.

What I see here in 1) is that if the two events have a spacelike separation, then it's possible to define their positions and timing in such a way that more than one frame will define them as simultaneous. This wouldn't work if the two events are in the same place. Also note that the two observers may note agree on where the events occurred.

2) Again, my guess is yes.

3) This is much more difficult, but I can imagine special cases might exist where it should be possible in principle.

The way I picture this, to justify my answers, is to consider two observers with some arbitrarily defined time horizon encircling each of them, such that any event on their respective time horizons are simultaneous. If those two horizons overlap, like overlapping circles (note that what is round for one observer is not for the other), then the two points where the time horizons overlap meets is the event time and locations that make it possible. Such time and locations would change for each moment in time, if we are talking solely in terms of Special Relativity.

 

[yossell]

2) Again, my guess is yes.

I'm pretty sure that espen180's `no' answer to this question is correct. There's no frame where two time like events are simultaneous.

 

[espen180]

The events lie on a line that is at right angles to the direction of motion

You are thinking in two dimensions only (1 space + 1 time). You need to think in three dimensions (2 space + 1 time). That's why yossell referred to the y axis, at right angle to both the t and x axes.

Ah, I see! My bad.

 

[my_wan]

It looks to me like espen180's numbers are almost valid in post #12, but the OP conditions were misstated.

There is only two ways the events E₁ and E₂ can be simultaneous in S':
1) x₁=x₂
2) v=0

x₁ does not have to equal x₂, but t₁ must equal t₂. This does not mean t₁ will equal t'₂. Simultaineiety only depends on the time difference between events, not the proper time at which the events occurred.

In the case of 1), there is only one event, and simultaneity is not a useful concept here.

No, there are two event and two observers in 1). Note the original wording:

1) Can two events which are simultaneous in one frame of reference (FOR) be simultaneous in another - that is - THE SAME TWO EVENTS, or is the simultaneity limited to only ONE FOR.

In the case of 2), S=S', there is only one frame.

Again, it's the exact same physical conditions as provided in 1), only with the question generalized for certain specific two observers defined in relation to ANY two events. If there's only one event in the first question then there's ony one in the second, same for frames. Thus the conclussion is invalid.

 

[espen180]

x₁ does not have to equal x₂, but t₁ must equal t₂. This does not mean t₁ will equal t'₂. Simultaineiety only depends on the time difference between events, not the proper time at which the events occurred.

For simultaneity in S', you need t'₁=t'₂. t₁=t₂ is assumed. In 2D spacetime, 1) and 2) are the only way to achieve this in two frames simultaneously.

No, there are two event and two observers in 1). Note the original wording:

Even if two events occur at the same time in the same location, you consider it two separate events?

Again, it's the exact same physical conditions as provided in 1), only with the question generalized for certain specific two observers defined in relation to ANY two events. If there's only one event in the first question then there's ony one in the second, same for frames. Thus the conclussion is invalid.

If v=0 between two frames with a common origin and parallel axes, you still consider them two separate frames?

 

[stevmg]

Getting sleepy...

Later...

stevmg

 

[my_wan]

Let's explicitly define two separate frames AND two separate events in which both events are simultaneous in both reference frames.

Two observers are in a head on collision course at 86% c (γ=0.5). The event locations are stationary relative to each other, the line between them as defined from the event frames intersects the point where the observers are about to collide, and are equidistant from that collision point. This is a special case such that if the events are simultaneous from the frame of an observer at the collision point, and both observers are approaching this collision point at the same speed, then those two events will be simultaneous at any time prior to the collision.

Note how, in this situation, the two observers will not agree on how far they are from the line between the event locations, or where the events originated, yet they will both agree the events are simultaneous regardless of when during the approach to that collision the events occurs.

Do you disagree with this, or do I need to post a schematic with numbers.

 

[my_wan]

Also, given the above special case, an arbitrary number of frames, approach or receding from the impact point in opposite directions, can be paired such that they agree the events are simultaneous. Thus, wrt question 3), the answer is yes. At first thought it appears that, based on this special case, it would be contingent on the source of the two events sharing a common frame. Yet in the absolute sense this is a meaningless condition when specifying a pointlike location in spacetime.

This would not mean that two events could be defined such that they were simultaneous for any arbitrary set of three frames. Only that three frames can be defined such that any two arbitrary events will be mutually simultaneous. Thus 3) would seem to be valid in special cases where the triplets of frames are chosen for consistency wrt the events, but not necessarily when the events are attempted to be chosen for consistency wrt three frames.

 

[djy]

A recent forum thread went into a long disussion of simultaneity.
3) It would seem that an event A can be simultaneous with an event B in one FOR and also be simultaneous with event C in a second FOR. This does not mean that B and C are simultaneous but it should be possible to find an FOR in which they are. Again, this would NOT mean that \exists an FOR in which A, B and C are simultaneous mutually, or does it?

No. Your premises are that A and B are spacelike separated, and A and C are spacelike separated. It is easy to come up with an A, B, and C such that B and C are timelike separated. This means there's no FOR in which they're simultaneous.

For example, consider one FOR with (x, t) coordinates (assume c = 1)

A = (0.0, 0.0)
B = (1.0, 0.0)
C = (1.0, 0.1)

A is clearly simultaneous with B. A is also spacelike separated from C, so it is simultaneous with it in another FOR. But B and C are clearly timelike separated. They are not simultaneous in any FOR.

Another way of stating this is that it's possible for a massive object to go from B to C without reaching the speed of light. This means that B precedes C causally, and this must be true in every FOR, thus they are simultaneous in none.

 

[Rasalhague]

Assuming, due to my own ignorance (rather than lack of curiosity!), that the discussion is restricted to inertial frames of reference in flat spacetime (a.k.a. Minkowski space)...

1) Can two events which are simultaneous in one frame of reference (FOR) be simultaneous in another - that is - THE SAME TWO EVENTS, or is the simultaneity limited to only ONE FOR.

Yes. For example, if the FORs are at rest with respect to each other, differing only by a translation and/or rotation. A second example, which others have mentioned: if the FORs are in relative motion but the events lie on a line orthogonal to the direction of motion. Also in general if the two events (in the colloquial sense) happen at the same point in space; but the word event can also be used in a special sense, "a point in spacetime", in which case this third example is the trivial statement that every event is simultaneous with itself. Otherwise no.

2) Can one always find an FOR in which any two event pairs are simultaneous?

No. (Assuming you mean "any two events" or "any pair of events".) One can always find an FOR for which any two events with a spacelike separation are simultaneous. But if there's a causal (timelike or lightlike) separation between two events, there are no FORs for which they're simultaneous.

3) It would seem that an event A can be simultaneous with an event B in one FOR and also be simultaneous with event C in a second FOR. This does not mean that B and C are simultaneous but it should be possible to find an FOR in which they are. Again, this would NOT mean that \exists an FOR in which A, B and C are simultaneous mutually, or does it?

I disagree with the bit I've emboldened. It might be possible to find such an FOR, but it won't necessarily be possible. Whether such an FOR exists will depend on the absolute position of C relative to B in spacetime. Among the infinitely many events that A is simultaneous with in the second FOR, there will always be some at causal separations from B (timelike and lightlike), and there are no FORs in which such events are simultaneous with B.

(I assume, in 3, you're saying "this does not meant that B and C are simultaneous" in the FOR where A and C are simultaneous (which is true: B and C are not necessarily simultaneous unless they're identical)).

Maybe someone more knowledgeable than me can answer the questions for general coordinates in flat spacetime and for curved spacetime.

 

[starthaus]

3) It would seem that an event A can be simultaneous with an event B in one FOR and also be simultaneous with event C in a second FOR. This does not mean that B and C are simultaneous but it should be possible to find an FOR in which they are.

Assume event A(x_A,t_A) . Then, event B(t_B=t_A,x_B). So, A and B are simultaneous in frame F.
Assume that there is a frame F' moving at speed v wrt F. In frame F', A has the coordinates A'(t'_A,x'_A) and it is simultaneous with C(t'_A,x'_C)

where:

t'_A=\gamma(v)(t_A-\frac{vx_A}{c^2})

If you want for B and C to be simultaneous, this simply means t_A=t'_A

i.e.

t_A=\gamma(v)(t_A-\frac{vx_A}{c^2})

The above is a simple equation in v with the solution:

v=\frac{2x_At_A}{t_A^2+(x_A/c)^2}<c

So, you can always find a frame F' that satisfies your condition 3. It is interesting to note that the speed v between F and F' depends only on the spacetime coordinates of A.

 

[djy]

Assume event A(x_A,t_A) . Then, event B(t_B=t_A,x_B). So, A and B are simultaneous in frame F.
Assume that there is a frame F' moving at speed v wrt F. In frame F', A has the coordinates A'(t'_A,x'_A) and it is simultaneous with C(t'_A,x'_C)

where:

t'_A=\gamma(v)(t_A-\frac{vx_A}{c^2})

If you want for B and C to be simultaneous, this simply means t_A=t'_A

i.e.

t_A=\gamma(v)(t_A-\frac{vx_A}{c^2})

The above is a simple equation in v with the solution:

v=\frac{2x_At_A}{t_A^2+(x_A/c)^2}<c

So, you can always find a frame F' that satisfies your condition 3. It is interesting to note that the speed v between F and F' depends only on the spacetime coordinates of A.

I'm pretty sure this isn't right. For example, it's possible that A = (0, 0), and then v = 0/0, undefined.

The velocity of the frame in which B and C are simultaneous is, I believe,

v = c^2 \frac{t_C - t_B}{x_C - x_B}.

Of course, if v \ge c, then the frame is not physically realistic.

 

[starthaus]

I'm pretty sure this isn't right. For example, it's possible that A = (0, 0), and then v = 0/0, undefined.

No, you don't:t_A=\gamma(v)(t_A-\frac{vx_A}{c^2})

For t_A=x_A=0 the above reduces to 0=0, satisfied for any v.

The velocity of the frame in which B and C are simultaneous is, I believe,

v = c^2 \frac{t_C - t_B}{x_C - x_B}.

Of course, if v \ge c, then the frame is not physically realistic.

Welll, I showed my derivation, let's see yours and we'll figure out which one is wrong.

 

[yossell]

hmph.

My guess is djy is using the idea that an object's worldline is orthogonal to that object's simultaneity line to find the pertinent velocity. But Starthaus' method seemed plausible too - the only part of Starthaus' derivation I couldn't follow was the final line. Starthaus, care to give us a hint how you got your expression for v from your expression from tA?

 

[starthaus]

hmph.

My guess is djy is using the idea that an object's worldline is orthogonal to that object's simultaneity line to find the pertinent velocity. But Starthaus' method seemed plausible too - the only part of Starthaus' derivation I couldn't follow was the final line. Starthaus, care to give us a hint how you got your expression for v from your expression from tA?

Sure, solve for v:t_A=\gamma(v)(t_A-\frac{vx_A}{c^2})

BTW, if you take t_A=x_A=0 as djy wants to, you get 0=0, satified for any v.

 

[yossell]

Sure, solve for v:

lol - well of course, but I couldn't as I was getting fourth powers of v - but I see now I had gamma all wrong, and it all works out as you said. So now both of you have convinced me and I've got to think again...