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 occured.

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 occured.

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...

 

[Rasalhague]

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

I don't follow your reasoning here. I don't know why you only discuss two frames. Could it be that we're interpreting the question differently? I've attached a Minkowski diagram, drawn to scale with geometric units, c = 1, of a case where B and C are simultaneous in one frame (yellow) but t_A (red time coordinate of A) does not equal t'_A (orange time coordinate of A). The three lines are lines of simultaneity in the three different frames.

So, you can always find a frame F' that satisfies your condition 3.

Not for any choice of C. In the attachment, I've shown an example where it is possible because there's a spacelike separation between C and B. But if we chose a C at a timelike separation from B (that is, if the line in the diagram connecting C and B was steeper that 45 degrees from horizontal), then we couldn't find an inertial frame of reference according to which B and C are simultaneous. djy gives an example of this in #23; in my diagram, djy's case would be represented by a C directly above B (C and B lying on a vertical red line).

 

[Rasalhague]

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

I think this is what djy had in mind. Let F'' be the frame where B and C are simultaneous. Let v(F'' in F) be the velocity of frame F'' with respect to F. I'll abbreviate it to v in this post, as no other velocities are mentioned.

t''_C - t''_B=0=\frac{(t_C - t_B)-(x_C-x_B) \cdot vc^{-2}}{\sqrt{(1-(v/c)^2}}

(t_C - t_B)=(x_C-x_B) \cdot vc^{-2}

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

djy's equation looks okay to me.

starthaus, I don't understand how your derivation relates to the question. I don't understand your assumption that "If you want for B and C to be simultaneous, this simply means t_A = t'_A". I don't know what relevant information we get from an equation that doesn't depend on the choice of event C. There are infinitely many events simultaneous to A in F'. The velocity of F'' depends on which of these we call C.

 

[starthaus]

I think this is what djy had in mind. Let F'' be the frame where B and C are simultaneous. Let v(F'' in F) be the velocity of frame F'' with respect to F. I'll abbreviate it to v in this post, as no other velocities are mentioned.

t''_C - t''_B=0=\frac{(t_C - t_B)-(x_C-x_B) \cdot vc^{-2}}{\sqrt{(1-(v/c)^2}}

(t_C - t_B)=(x_C-x_B) \cdot vc^{-2}

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

djy's equation looks okay to me.

starthaus, I don't understand how your derivation relates to the question. I don't understand your assumption that "If you want for B and C to be simultaneous, this simply means t_A = t'_A". I don't know what relevant information we get from an equation that doesn't depend on the choice of event C. There are infinitely many events simultaneous to A in F'. The velocity of F'' depends on which of these we call C.

I simply put the OP in mathematical form. There are only two frames in the OP, F and F'. There is no F". I think that we are interpreting the OP differently.

 

[espen180]

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

How does that work? If B and C are to be simultaneous in some frame F'', you need t^{\prime \prime}_B=t^{\prime \prime}_C

t_A=t'_A means either x_A=t_A=0 or v=0.

 

[starthaus]

How does that work? If B and C are to be simultaneous in some frame F'', you need t^{\prime \prime}_B=t^{\prime \prime}_C

Once again, my reading of the OP is that there are only two frames, F and F'. There is no F".

 

[Rasalhague]

I simply put the OP in mathematical form. There are only two frames in the OP, F and F'. There is no F". I think that we are interpreting the OP differently.

Yes, I think we must be.

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?

The three frames, F, F', F'', I mentioned are: (1) a frame in which A and B are simultaneous; (2) a frame in which A and C are simultaneous; (3) a frame in which B and C are simultaneous. If we take the existence of F and F' as givens, then the existence of F'' depends on the nature of the separation between B and C.

 

[starthaus]

Yes, I think we must be.



The three frames, F, F', F'', I mentioned are: (1) a frame in which A and B are simultaneous; (2) a frame in which A and C are simultaneous;


(3) a frame in which B and C are simultaneous. If we take the existence of F and F' as givens, then the existence of F'' depends on the nature of the separation between B and C.

Sure, so depending how you interpret the OP, you get different results. I understand your answer (it is elementary) and , in the context of your interpretation of the OP, it is correct.

 

[Rasalhague]

Sure, so depending how you interpret the OP, you get different results. I understand your answer (it is elementary) and , in the context of your interpretation of the OP, it is correct.

Could you explain how you interpreted the question?

 

[starthaus]

Could you explain how you interpreted the question?

See post 25.

 

[stevmg]

This is the post that started this from another ungodly long thread - hence we started anew.

Gentlemen, Please!

Humour me, have a go at this restatement of the paradox in a different form.

********************************
Observer Alice, says to Bob ,who is just passing by at nearly the speed of light 'My Granny on Proxima Centauri is just sitting down to kippers for her tea'

Alice knows that because she has an Ansible (which allows her to see what Granny is doing right now without having to wait for the light to arrive)

Bob, who also has an Ansible, takes a quick look and says 'No she isn't, your Granny had her kippers for tea three days ago'

Explain.

You might like to show how Bob's Ansible allows him to travel back in time and re-experience events that have already happened.
What would Alice need to do to 'freeze' her Granny in time so that she is always having tea?
How long can she hold Granny frozen?

This led to my original three questions:
My original 3 questions (post 1):

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?

There were a lot of answers posted in that thread to AJ Bentley's question. My answer is noted below:

By Einstein et al there can be no such thing as an Ansible. Basically, an Ansible literally establishes simultaneity between two events (the event itself and its observation in the Ansible.) This is not possible wrt to all frames of reference. Simultaneity "changes" (or can change) when looking at two distict events from the two different FORs.

The scenario descibed above is literally the Einstein train in which the lightning flashes are perceived as simultaneous in one FOR while in the other FOR, the lead flash occurs before the back flash. [Section IX, The Relativity of Simultaneity, Relativity, Albert Einstein.]

In this case, Bob is traveling towards Granny and sees her event to occur before Alice.

Lorentz time transformation and SR:

t' = \gamma[t - xv/c²]

Set t = 0 (for Alice,) x = whatever (call it x) for the distance between Alice and Granny in a common FOR. v is Bob's speed wrt Alice and going towards Granny.

Question 1)

WE ARE LIMITED TO ONE DIMENSION (two, if you include ct, and we translate the origin of Bob (who is traveling fast) to overlay the origin of Alice

Below is a schematic depiction of my answer`(the left members of the pairs are the distances, the right elements are the times). v ("approximately the speed of light") and x (the distance from Alice to Proxima Centauri) are arbitrary so are merely noted by v and x. Note that both v and x are positive. x', or the distance in Bob's frame, from him as crosses Alice to Granny is noted but not necessary for the discussion, while t' is calculated by the Lorentz time transformation. v for the Alice-Granny frame of reference is zero (= 0) while we just denote the velocity of the Bob-Granny frame of reference as v. is a positive number but not actually calculated here. t' calculates, by using the above information, to a negative number. This goes along with your supposition as presented.

Bob(0,0)......moving frame(v)......Granny(x',t' - a negative number)
Alice(0,0)......reference frame(v=0)...Granny(x,0)

In this depiction Bob is moving towards Granny. The Lorentz transformation changes to
A) t' = \gamma[t + xv/c²] if Bob is moving away from Granny.
B) t' = \gamma[t - xv/c²] is the original equation with Bob mving towards Granny.
We know that t = 0 at the origin and at the end of x (because Alice and Granny are stated to be simultaneous.) Since we are overlaying the origin of Bob's FOR on Alices and the story starts at the same instant for both, t' = 0. Assuming x and v are non-trivial, at the "end" of x' (Granny again) t' is NOT zero (t' \neq 0) and thus the "beginning of x' and the end of x' are not simultaneous. This goes along with the story that Bob "sees" Granny eating her kippers before Alcice (both are looking through their simultaneity telescopes at t' = 0 and t = 0 respectively and t' and t are defined as being simultaneous - same point point, same time.)

By this scenario, the only way there can be simultaneity in Bob's FOR is for x or v to be trivial.

It is most important here to understand that I have limited this to a 2-dimensional (including the dimension ct) world. From now on I will refer to this as a one-dimensional world.

Starthaus interjected this beautiful calculus restatement of the same thing:

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)

Keep in mind that in one dimension (say, the x-axis) the only way to differ in FOR's is by velocity or v. Using one FOR as inertial, another FOR is different only if v . In this one-dimensional world, t \neq 0 for any x (at x'.) If v = 0, then we have simultanaeity for any x and its hidden twin x'.

By Yossels post, if we have a two or more dimensional world in which the axes are all orthogonal to the x-axis (say, the standard three dimensional world), we have as stated:

I *think* the answer to (a) [There are more than one FOR's that have points that are simultaneous with the original points] 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.

There are thus an infinite number of simultaneity lines as the number of points on the y or z axes is unlimited.

I think starthaus's final statement, which I quote:

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

I can't think of anything that satisfies dt = 0 and dx = 0 other than an FOR which is orthogonal to the x-axis which agrees with yossell. If there is another I would like to know what it is.

Question 2) Does there \exists an FOR for any two non-simultanous events that makes them simultaneous?

I haven't got a clue. I don't even know where to start.

Question 3) Answered by starthaus in post 25. I have another algebraic answer which is very tedious and I'm not sure is correct, so we'll leave it out.

 

[Ich]

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

No, only if the events are spacelike separated.

 

[Rasalhague]

See post 25.

I'm saying "if A is simultaneous with B in one FOR, and A is simultaneous with an arbitrary C in a second FOR, then it's not always possible to find a third FOR in which B is simultaneous with the given C; whether it's possible depends on the separation vector between B and C." (Did I misunderstand the question, Steve?)

Starthaus, are you saying "if A is simultaneous with B in one FOR, it's always possible to find a second FOR and an event C such that C is simultaneous in that second FOR with both A and B"? If so, then the condition is

t'_C=t'_A=t'_B

which leads to djy's formula again, by eliminating t'_C, except that now it's v(F' in F) instead of v(F'' in F), and A and B instead of C and B. If x_A \neq x_B, then

v=c^2 \frac{t_B-t_A}{x_B-x_A}

Since t_B=t_A[/tex], v = 0. Alternatively, if x_A = x_B[/tex], then A = B = C, so they'll be simultaneous in all frames, albeit trivially.<br /> <br /> I still don't understand the statement in #25: "If you want for B and C to be simultaneous, this simply means t_A=t'_A". The condition t_A=t'_A makes x'_A=0, right? How does this ensure that t'_C=t'_A=t'_B?

 

[Rasalhague]

Keep in mind that in one dimension (say, the x-axis) the only way to differ in FOR's is by velocity or v.

Or a translation in time or space, or a reflection in time or space.

Question 2) Does there exist an FOR for any two non-simultanous events that makes them simultaneous?

I haven't got a clue. I don't even know where to start.

The answer is the same as for your original question 2: no. Such a frame only exists if there's a spacelike separation between the two events. If it's possible for information sent from one event to be received at the other event, you can't find a frame in which they're simultaneous. And if you could, it would lead to paradoxes.

Question 3) Answered by starthaus in post 25. I have another algebraic answer which is very tedious and I'm not sure is correct, so we'll leave it out.

Ah, so did I mistake your meaning in #31, etc? Were you comparing three frames or two?

 

[stevmg]

I simply put the OP in mathematical form. There are only two frames in the OP, F and F'. There is no F". I think that we are interpreting the OP differently.

What's "OP?"

 

[stevmg]

Ah, so did I mistake your meaning in #31, etc? Were you comparing three frames or two?

In this starthaus did the work for three frames. I'll use that.

Starthaus used the term "OP." What does that mean?

 

[Altabeh]

In this starthaus did the work for three frames. I'll use that.

Starthaus used the term "OP." What does that mean?

"OP" is widely used among many posters throughout different forums on the Internet as a shorthand for "Original Poster" or "Original Post".

AB

 

[Rasalhague]

In this starthaus did the work for three frames. I'll use that.

Did you spot starthaus's post #35?

Once again, my reading of the OP is that there are only two frames, F and F'. There is no F".

 

[stevmg]

"OP" is widely used among many posters throughout different forums on the Internet as a shorthand for "Original Poster" or "Original Post".

AB

Got it...