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

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  • #26
djy
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Assume event [tex]A(x_A,t_A)[/tex] . Then, event [tex]B(t_B=t_A,x_B)[/tex]. 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 [tex]A'(t'_A,x'_A)[/tex] and it is simultaneous with [tex]C(t'_A,x'_C)[/tex]

where:

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

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

i.e.

[tex]t_A=\gamma(v)(t_A-\frac{vx_A}{c^2})[/tex]

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

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

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 [itex]A = (0, 0)[/itex], and then [itex]v = 0/0[/itex], undefined.

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

[itex]v = c^2 \frac{t_C - t_B}{x_C - x_B}[/itex].

Of course, if [itex]v \ge c[/itex], then the frame is not physically realistic.
 
  • #27
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I'm pretty sure this isn't right. For example, it's possible that [itex]A = (0, 0)[/itex], and then [itex]v = 0/0[/itex], undefined.
No, you don't:


[tex]t_A=\gamma(v)(t_A-\frac{vx_A}{c^2})[/tex]

For [tex]t_A=x_A=0[/tex] the above reduces to 0=0, satisfied for any v.


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

[itex]v = c^2 \frac{t_C - t_B}{x_C - x_B}[/itex].

Of course, if [itex]v \ge c[/itex], 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.
 
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  • #28
yossell
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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?
 
  • #29
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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:


[tex]t_A=\gamma(v)(t_A-\frac{vx_A}{c^2})[/tex]

BTW, if you take [tex]t_A=x_A=0[/tex] as djy wants to, you get 0=0, satified for any v.
 
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  • #30
yossell
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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...
 
  • #31
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If you want for B and C to be simultaneous, this simply means tA = 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 tA (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).
 

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  • #32
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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.

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

[tex](t_C - t_B)=(x_C-x_B) \cdot vc^{-2}[/tex]

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

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 tA = 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.
 
  • #33
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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.

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

[tex](t_C - t_B)=(x_C-x_B) \cdot vc^{-2}[/tex]

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

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 tA = 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.
 
  • #34
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If you want for B and C to be simultaneous, this simply means [tex]t_A=t'_A[/tex]
How does that work? If B and C are to be simultaneous in some frame F'', you need [tex]t^{\prime \prime}_B=t^{\prime \prime}_C[/tex]

[tex]t_A=t'_A[/tex] means either [tex]x_A=t_A=0[/tex] or [tex]v=0[/tex].
 
  • #35
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How does that work? If B and C are to be simultaneous in some frame F'', you need [tex]t^{\prime \prime}_B=t^{\prime \prime}_C[/tex]
Once again, my reading of the OP is that there are only two frames, F and F'. There is no F".
 
  • #36
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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 [tex]\exists[/tex] 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.
 
  • #37
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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.
 
  • #38
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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?
 
  • #40
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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.:tongue:

********************************
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 [tex]\exists[/tex] 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' = [tex]\gamma[/tex][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' = [tex]\gamma[/tex][t + xv/c2] if Bob is moving away from Granny.
B) t' = [tex]\gamma[/tex][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' [tex]\neq[/tex] 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".

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

so:

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

If there is a frame where [tex]dx=0[/tex] and [tex]dt=0[/tex] then, in all frames [tex]dt'=0[/tex] (and [tex]dx'=0[/tex])
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 [tex]\neq[/tex] 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 travelling 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 [tex]dx=0[/tex] and [tex]dt=0[/tex] then, in all frames [tex]dt'=0[/tex] (and [tex]dx'=0[/tex])
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 [tex]\exists[/tex] 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.
 
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  • #41
Ich
Science Advisor
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2) Can one always find an FOR in which any two event pairs are simultaneous?
No, only if the events are spacelike separated.
 
  • #42
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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

[tex]t'_C=t'_A=t'_B[/tex]

which leads to djy's formula again, by eliminating [itex]t'_C[/itex], 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 [itex]x_A \neq x_B[/itex], then

[tex]v=c^2 \frac{t_B-t_A}{x_B-x_A}[/tex]

Since [itex]t_B=t_A[/tex], v = 0. Alternatively, if [itex]x_A = x_B[/tex], then A = B = C, so they'll be simultaneous in all frames, albeit trivially.

I still don't understand the statement in #25: "If you want for B and C to be simultaneous, this simply means [itex]t_A=t'_A[/itex]". The condition [itex]t_A=t'_A[/itex] makes [itex]x'_A=0[/itex], right? How does this ensure that [itex]t'_C=t'_A=t'_B[/itex]?
 
  • #43
1,384
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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?
 
  • #44
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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?"
 
  • #45
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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?
 
  • #46
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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
 
  • #47
1,384
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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".
 
  • #48
696
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"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...
 

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