Required speed for two events to be simultaneous

[Celso]

I'm doing some exercises about special relativity and one of them asks to find the speed in an arbitrary frame of reference (1) in such a way that it perceives two events at the same time that didn't happen simultaneously in other frame of reference(2).

Is it correct to state that if the distance between the two events in the frame (2) is $ct_{2}$, and the first event happened when $t_{2} = t_{1} = 0$, an observer in the frame of reference (1) could only perceive these two events as simultaneous if it were traveling at the speed of light?
What I tought was
$t' = \gamma (t - \frac{ux}{c^2})$
Since $x = ct_{2}$
$t' = \gamma (\frac{x}{c} - \frac {ux}{c^2})$
For this to be $0$: $\frac{x}{c} = \frac{ux}{c^2} \therefore u = c$
Is there anything incorrect here?

 

[Ibix]

Is it correct to state that if the distance between the two events in the frame (2) is $ct_{2}$, and the first event happened when $t_{2} = t_{1} = 0$, an observer in the frame of reference (1) could only perceive these two events as simultaneous if it were traveling at the speed of light?

What does relativity say about frames with a relative speed of $c$? (Hint: look at the second postulate)

 

[kent davidge]

You could consider a standard Lorentz boost in the $x$-direction of frame $1$ with a speed $v$. Then setting the coordinate time of the simultaneous events in frame $1$ to zero to make things easy, you get that $v$ is given by either of the ratios $$|v| = \Bigg|c^2 \frac{t_2^1}{x_2^1} \Bigg| = \Bigg|c^2 \frac{t_2^2}{x_2^2} \Bigg|$$ where the sub-indices labes the frames and the superindices indicates the events, so that e.g. $t_2^1$ is the coordinate time of event $1$ in frame $2$.

find the speed in an arbitrary frame of reference

the frames cannot be too arbritary, they should be connected by a Lorentz transformation, if they are not, then you are not doing Special Relativity.

perceive these two events as simultaneous if it were traveling at the speed of light?

and the Lorentz transformations are between inertial frames. A frame traveling at speed of light is not inertial, thus cannot be related to a inertial frame through a Lorentz transformation.

 

[sweet springs]

Is it correct to state that if the distance between the two events in the frame (2) is ct2ct2ct_{2}, and the first event happened when t2=t1=0t2=t1=0t_{2} = t_{1} = 0, an observer in the frame of reference (1) could only perceive these two events as simultaneous if it were traveling at the speed of light?
What I tought was

Let coordinates of two evens be $(t_1,x_1)$ and $(t_2,x_2)$ in the original frame of reference, they are $(t'_1,x'_1)$ and $(t'_2,x'_2)$ in Lorentz transformed frame transformed by
x_i'=\gamma(x_i-\beta ct_i)
ct'_i=\gamma(ct_i-\beta x_i),where i=1,2. Then in order to meet your interest let times be $t_1=t_2$ so that the two events are simultaneous in the original FR. You will easily find that $t'_1 \neq t'_2$ unless $x_1 =x_2$, i.e. not two but single event.

 

[FactChecker]

Maybe I am misunderstanding the problem. Actually, any non-zero relative speed will do. Just have one reference frame move relative to the other and pick two events that are simultaneous in the first frame and are separated in the direction of motion. They will not be simultaneous in the other reference frame.

 

[Ibix]

Maybe I am misunderstanding the problem.

The events are specified to be null separated.

This is not stated in the initial "summary" paragraph. I think this initial summary is a new feature of PF5, and I haven't got used to it yet.

 

[PeterDonis]

You could consider a standard Lorentz boost in the $x$-direction of frame $1$ with a speed $v$.

But such a boost is not well-defined if $v = c$, so there is no way to apply this method to the scenario described in the OP.

A frame traveling at speed of light is not inertial

It would be better to say that there is no such thing at all as "a frame traveling at the speed of light". You can have a coordinate chart in which a photon's worldline has all coordinates constant except one; but there is no way to set up a "frame" that corresponds to such a chart, because a "frame" requires certain properties that such a coordinate chart does not have.

 

[kent davidge]

But such a boost is not well-defined if $v = c$, so there is no way to apply this method to the scenario described in the OP.

I don't understand you here. No boost behaves well if $v = c$ and no boost can be applied to the case described in the op. So what's the problem with my particular boost?

 

[PeterDonis]

No boost behaves well if $v = c$ and no boost can be applied to the case described in the op. So what's the problem with my particular boost?

Um, what? If no boost is well defined for $v = c$, then your boost is not well defined for $v = c$. So that's the problem with your boost.

 

[kent davidge]

Um, what? If no boost is well defined for $v = c$, then your boost is not well defined for $v = c$. So that's the problem with your boost.

Yea, I mean, if no boost can do it, then it's not a problem that my boost can't. In other words, my boost can't do it just as any other boost can't do it.

Also, those relations are derived assuming that $v < c$, so one cannot blindly set $v = c$ up there.

 

[PeterDonis]

if no boost can do it, then it's not a problem that my boost can't.

Yes, it is, because you were trying to apply your boost to the problem at hand. It can't.

 

[PeterDonis]

those relations are derived assuming that $v < c$, so one cannot blindly set $v = c$ up there.

But the problem statement in the OP requires $v = c$. Or, to put it another way, the events described in the OP must be null separated. @Ibix already pointed that out in post #6 (and he was hinting at it in post #2).