I Relativity of simultaneity doubt

[PeterDonis]

My point was that all Lorentz boosts for which those given spacelike separated events are simultaneous actually "share" the same velocity in $z$ direction.

In the sense that they can have no velocity in the $z$ direction, yes. Isn't that obvious? If you boost in the $z$ direction, you obviously have a change in simultaneity in the $z$ direction. I'm not sure why you felt the need to point that out, or why you think it is important.

 

[cianfa72]

In the sense that they can have no velocity in the $z$ direction, yes. Isn't that obvious? If you boost in the $z$ direction, you obviously have a change in simultaneity in the $z$ direction.

Consider Minkowski spacetime in standard inertial coordinates and from event A take an event B spacelike separated from it. Consider the spacelike straight line connecting them and name it $z$. Take the 3D-space of directions orthogonal to it. Which is the family of Lorentz boosts such that events on $z$ axis are simultaneous ?

 

[PeterDonis]

Which is the family of Lorentz boosts such that events on $z$ axis are simultaneous ?

All of them that don't include a $z$ component, as I already said. So a boost in any direction in the $x y$ plane.

 

[cianfa72]

All of them that don't include a $z$ component, as I already said. So a boost in any direction in the $x y$ plane.

ok, a boost in any direction in the $(x,y)$ plane however with the appropriate velocity otherwise events along that given spacelike direction $z$ do not result as simultaneous.

 

[PeterDonis]

a boost in any direction in the $(x,y)$ plane

Yes.

however with the appropriate velocity otherwise events along that given spacelike direction $z$ do not result as simultaneous.

I have no idea what you mean by this. Any boost in any direction in the $xy$ plane at any velocity leaves any pair of events separated only along the $z$ direction (i.e., with the same $t$, $x$, and $y$ coordinates) simultaneous.

 

[cianfa72]

Any boost in any direction in the $xy$ plane at any velocity leaves any pair of events separated only along the $z$ direction (i.e., with the same $t$, $x$, and $y$ coordinates) simultaneous.

That's true. However the $z$ axis has been redefined (i.e. it is not the same as the $z$ axis in the initial inertial coordinate system) since it is aligned with the spacelike direction defined by [spacelike separated] events A and B (note that A and B may not be simultaneous in the given inertial frame and may not have $x=0, y=0$ as well).

Then, of course, if events A and B are simultaneous in the given initial inertial frame any boost at any velocity in any direction orthogonal to the spacelike "line" defined from them leaves those events simultaneous. Hence if the two simultaneous events A and B are separated only along the $z$ direction then any velocity in any direction in the $xy$ plane leaves them simultaneous.

 

[italicus]

However the z axis has been redefined (i.e. it is not the same as the z axis in the initial inertial coordinate system) since it is aligned with the spacelike direction defined by [spacelike separated] events A and B (note that A and B may not be simultaneous in the given inertial frame and may not have x=0,y=0 as well).

What do you mean? The names of coordinates don’t matter. Given the events A and B, call the space axis through them as you want . They are simultaneous for any observer whose timeline is “normal “ to that axis, in the sense of Minkowski.

 

[cianfa72]

What do you mean? The names of coordinates don’t matter. Given the events A and B, call the space axis through them as you want . They are simultaneous for any observer whose timeline is “normal “ to that axis, in the sense of Minkowski.

Yes, the Lorentz boost should be at any velocity in any spacelike direction that does not include the spacelike direction defined by A and B events.

 

[Dale]

it is not the same as the z axis in the initial inertial coordinate system

Sure, but it is still an inertial coordinate system. By the first postulate it is equally valid as the original and all of the laws of physics are the same there.

 

[cianfa72]

Sure, but it is still an inertial coordinate system. By the first postulate it is equally valid as the original and all of the laws of physics are the same there.

Sorry, maybe I can't explain my point. Let's take two events A and B spacelike separated even if not simultaneous in the global inertial frame $(t,x,y,z)$. Consider the 'new' $z$ axis (call it $z'$) along them: it is in general a linear combination of $t,x,y,z$ axis.

Any Lorentz boost at any velocity in a direction that is linear combination of $x,y$ axes (i.e. having $t=0,z=0$) does not bring simultaneous those given two events, right ?

 

[PeterDonis]

Let's take two events A and B spacelike separated even if not simultaneous in the global inertial frame $(t,x,y,z)$. Consider the 'new' $z$ axis (call it $z'$) along them: it is in general a linear combination of $t,x,y,z$ axis.

In which case we can say nothing whatever useful about Lorentz transformations, since you have allowed the new $z'$ axis to depend on all of the original ones.

Any Lorentz boost at any velocity in a direction that is linear combination of $x,y$ axes (i.e. having $t=0,z=0$) does not bring simultaneous those given two events, right ?

Since you have given us zero useful information about where the $z'$ axis points, obviously we can give zero useful information about this question.

 

[Dale]

Sorry, maybe I can't explain my point. Let's take two events A and B spacelike separated even if not simultaneous in the global inertial frame $(t,x,y,z)$. Consider the 'new' $z$ axis (call it $z'$) along them: it is in general a linear combination of $t,x,y,z$ axis.

Any Lorentz boost at any velocity in a direction that is linear combination of $x,y$ axes (i.e. having $t=0,z=0$) does not bring simultaneous those given two events, right ?

Sorry, you are losing me. If you have two spacelike separated events then there is always some (unprimed) reference frame where those two events are simultaneous and both are located on the $z$ axis. Starting from that frame you can perform various transformations which have the following effects (the primed coordinates are in the transformed frame).

You can spatially translate in $z$ and they remain simultaneous and both on the $z'$ axis.

You can spatially translate in $x$ and $y$ and they remain simultaneous but not on the $z'$ axis.

You can temporally translate in $t$ and they remain simultaneous and both on the $z'$ axis.

You can rotate about $z$ and they remain simultaneous and both on the $z'$ axis.

You can rotate about $x$ or $y$ and they remain simultaneous but not on the $z'$ axis.

You can boost along $z$ and they remain on the $z'$ axis but not simultaneous.

You can boost along $x$ or $y$ and they remain simultaneous but not on the $z'$ axis.

All of the possible transformations from one inertial frame to another are some composition of these above transformations.

 

[cianfa72]

Thank you @Dale now it makes sense to me !

 

[alexandrinushka]

Hey there @lukka98 this video has helped me a lot. I found it brief and very well made. Maybe it helps you as well

 

[David Lewis]

i think light will arrive at OT at the same time also in his frame of reference, if i don't observe that i think light speed is greater than c from B to OT and lower from A to OT,

You're correct. In OE's frame, the closing speed from B to OT is greater than c. Call the speed of the train v. In OE's frame, light reaches from B to OT at a rate of c+v.

 

[FactChecker]

Summary:: My question is about on of the "famous" Einstein's mental experiment: the train and the lightnings, i don't understand why the result is what it is, I thought for day but I don't understand this particular case, If you can help me to understand I'll very grateful.

View attachment 289031
(excuse me for my english, but I'm studying physics and I am not a native English speaker)
One observer OE, is on the ground, we take him as the fixed frame of reference.
The other OT is on the train that is moving relatively to the OE at a costant velocity ( they are both inertial frame of reference).
So, two lighning strike the two points A and B when the train is with back and end in the same points ( A and B) and the two observer are aligned.

Stop right here. If one observer thinks that the two ends are aligned at the same time, then the other observer must think that the two endpoints are aligned at different times. If it is simultaneous for one, it can not be simultaneous for the other. Is that fact included in your logic?

(In fact, if one observer thinks that the endpoints and the observers are all aligned at time t=0, then the other observer thinks that all three alignments happened at different times.)

 

[David Lewis]

OT observes the light from each lightning strike at different times because OT was no longer at a location equidistant from both lightning strikes at the time the light from both strikes reached OT.

Assume strikes A and B are simultaneous in your frame, and you are midway between the strikes when they occur, but you are closer to B and farther from A when the light reaches you. Would you conclude the strikes are not simultaneous in your frame?

 

[italicus]

@lukka98

Have a look at post #30, which refers to a diagram that I posted in another discussion , one year ago . Look at the diagram.
Both OE and OT know that c is the same in all inertial reference frames; OT is reached by the light coming from B before than the light coming from A; (of course OT uses his time t’) .
Therefore, he concludes that event B happened before event A , in his reference frame.
Einstein tells us that both OT and OE are right. Events A and B are simultaneous for OE, so they aren’t simultaneous for OT.

 

[alan123hk]

I think this Einstein's mental experiment is just saying that simultaneous events in one frame of reference become non-simultaneous events in another frame of reference, which is not a hard concept to understand and no big deal. Most importantly, the order of two things that are causally related will not be reversed for observers in different frames of reference. The theory of relativity is completely consistent with this causality law.

 

[Ibix]

which is not a hard concept to understand

An optimist, I see...

 

[alan123hk]

If two events are causally or relationally dependent, their order should be absolute and cannot be reversed. For example, harvesting cannot be earlier than sowing. We can briefly understand how this property is embodied in the theory of relativity through the method described below.

Assuming that in the reference frame S, event A occurs at time t1, event B occurs at time t2, and event B is later than event A, then t2>t1. According to the Lorentz transformation,

$$ t_2'-t_1'=\gamma \left[ \left(t_2-t_1\right)- \frac {v}{c^2} \left(x_2-x_1 \right) \right] $$
It can be seen from above equation that if $\left( t_2-t_1\right) > \frac {v}{c^2} \left(x_2-x_1 \right)$, then for the observer in reference S', $~t_2'>t_1'~$, still event A happens before event B.

But if $~ \left( t_2-t_1\right) < \frac {v}{c^2} \left(x_2-x_1 \right) ~$, then event B occurs before event A in reference frame S', so the order is reversed.

Since the sequence of two events with causal relationship cannot be reversed, the following relationship must be guaranteed.
$$ | t_2-t_1| > \frac {v}{c^2} |(x_2-x_1| ~~~~~~~~~~\Rightarrow~~~~~~~~~|\frac {x_2-x_1}{t_2-t_1}| < \frac {c^2}{v}$$
where $~u=|\frac {x_2-x_1}{t_2-t_1}|~$ is the speed of propagation between two causally related events, so we get the following formula,
$$ uv<C^2 $$
where v is the velocity of reference S' relative to reference S. Since the movement of a particle fixed in S' relative to the S can also represent a signal propagating in S, the above formula actually states that as long as the propagation speed of the signal is less than the speed of light, the sequence of causally related events will not be reversed.

As far as I know, humans have not yet found a particle or method that can actually transmit signals faster than the speed of light. But perhaps no one can guarantee that this theory will never be disproved in the future.

 

[vanhees71]

Two events $x,y \in \mathbb{R}^4$ can be causally connected if they are time-like or light-like separated, i.e., if $(x-y)^{\mu} (x-y)^{\nu} \eta_{\mu \nu} = x^0 y^0-\vec{x} \cdot \vec{y}\geq 0$. Then $\text{sign}(x^0-y^0) \text{sign} (x^{\prime 0}-y^{\prime 0})$ for all Lorentz transformations $x'=\Lambda x$, $y'=\Lambda y$.