Relativity: events simultaneous

perseus

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What would be the characteristics of a system of reference in which two events no are simultaneous in this new system of reference are simultaneous,? (Except that the second reference system traveling at the speed of light)
 
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jedidia

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I don't follow you at all. Could you try to be more precise? special relativity is confusing enough in clear language...
 

Face

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What would be the characteristics of a system of reference in which no two events are simultaneous in this simultaneous,? (Except that the second reference system traveling at the speed of light)

Something like the Lorentz-contracted train inside the barn with both doors closed "simultaneously"?
 

perseus

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Ejemplo
The effect of the relativity of simultaneity is for each observer to consider that a different set of events is simultaneous. The relativistic phase difference between clocks ("relativistic phase")
f5d2e812d650734fd9395bc4bbf3cb84.png
means that observers who are moving relative to each other have different sets of things that are simultaneous
Rel2.gif

My question is is there a condition for a single system of reference in which the two events are simultaneous?
 

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My question is is there a condition for a single system of reference in which the two events are simultaneous?

I don't get it. Do you mean the 2 events from your example (general declares war, fleet is underway), perhaps?

If so - since those 2 events are causally linked (fleet underway can only occur after war was declared) - I'd say having both simultaneously is only possible in the frame-of-reference of a light particle, i.e. one moving with light-speed. But that's just an educated guess, I don't have calculations to back it up...
 

Urwumpe

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My question is is there a condition for a single system of reference in which the two events are simultaneous?

The single system of reference is likely what makes it hard to understand, since you actually mean two observers in different frames of reference (one relatively stationary, another relatively moving to the reader)

Of course, if you think about it, two events simultaneous in one frame of reference must not be simultaneous in another frame of reference. If you think about the addition of velocity, there might still be events that can be simultaneous, if the velocity in question is c... but that's still a hard one to find an example.
 

perseus

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It's amazing to think that If Mr. car phoned to Mr violin, he could tell the future.
 

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It's amazing to think that If Mr. car phoned to Mr violin, he could tell the future.

His phone call would arrive a while later, since it is also an event subject to the whole relativity...
 

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The picture means something like that: we have two people, one standing on the Earth, one passing him by in a moving car. Let's say it happens on a Monday, 12:00. For the standing person (the "violin guy") what is happening simultaneously in the Andromeda galaxy is that some space general is planning an invasion. For the guy in the car, what is happening simultaneously is that the fleet is already sent and on their way to Earth. Now, according to the violin guy, this will happen some time later - let's say, on Tuesday.

Face said:
I don't get it. Do you mean the 2 events from your example (general declares war, fleet is underway), perhaps?
Those two events are separated by a timelike interval, so there will be no frame of reference in which they would be simultaneous.

perseus said:
It's amazing to think that If Mr. car phoned to Mr violin, he could tell the future.
Actually it's not true, since they have no way of knowing what's happening in the Andromeda galaxy at the moment it is happening. It might be happening simultaneously, but they still have to wait 2 million years to get any information about that. You might think that the car guy can get the information about the fleet a day earlier than the violin guy, but again it's not true, since the car guy will travel a significant distance during those 2 million years, and before the violin guy gets the information about the future from him, he will already be able to obtain it on his own.
 

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Those two events are separated by a timelike interval, so there will be no frame of reference in which they would be simultaneous.

So you disagree with the rest of my post there? I always thought that traveling at light-speed would make you see everything at once, so to say. Is this not the case? Because if it is so, that would be the only frame of reference to have them simultaneously, even if not very practical.
 

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So you disagree with the rest of my post there? I always thought that traveling at light-speed would make you see everything at once, so to say. Is this not the case? Because if it is so, that would be the only frame of reference to have them simultaneously, even if not very practical.
In essence, I indeed disagree. You are right though in saying that traveling at light-speed you would see everything at once, in a way. Frames of reference travelling at light-speed are not really practical, since they are degenerate - the x direction becomes equivalent with the t direction and there is no sensible flow of time. But yeah, you can say that time stops and everything happens at once - except that this "everything" is only one 3-dimensional slice, so pretty much like simultaneous events for other observers. The problem is, you never get beyond that one slice, because your time axis is in that slice - so some events never happen.

It's really hard to put into words, so let me use some math:
Let (t, x, y, z) be a coordinate system for some regular, massive observer (like our Mr. Violin).

Our observer O' will be travelling at v=c along the x axis.

Using the traditional Lorentz transformation will give us 0 in the denominator, so that's not really useful. If v<c, then we have:
[math]t' = \frac{t - \frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}[/math][math]x' = \frac{x - vt}{\sqrt{1-\frac{v^2}{c^2}}}[/math][math]y' = y[/math][math]z' = z[/math]
We can safely ignore y and z, since nothing is happening there, and focus on 2D spacetime.

For any observer, simultaneous events are those for which t = const, where t is relative to that observer. This means that for the moving observer simultaneous events will have t'=const:

[math]t' = \frac{t - \frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}} = \mathrm{const}[/math]
If v is less than c, the denominator is a nonzero constant and we can write:
[math]ct - \frac{v}{c}x = \mathrm{const}[/math]
So, let's extend that to v=c:
[math]ct - x = \mathrm{const}[/math]
On the other hand, the world line of our observer is x' = 0:
[math]x' = \frac{x - vt}{\sqrt{1-\frac{v^2}{c^2}}} = 0[/math]
That means [math]x - vt = 0[/math]
Once again let's extend that to v=c:
[math]x - ct = 0[/math]
That is a line for which [math]ct - x = 0 = \mathrm{const}[/math] - so all events on the world line of the observer are simultaneous.

But! That means that on the whole world line we have t' = 0 (if we assume it was 0 at some point, which we usually do). On the other hand, the world line should somehow mark the time direction of the observer. This means that for such an observer it really makes no sense to talk about t' not equal to 0.

Now, what about simultaneity? Our observer is always in t' = 0, so we only need to talk about those events, which have t' = 0 - so, [math]ct - x = 0[/math]. Events that have [math]ct - x \neq 0[/math] are not simultaneous to any event on the world line of the observer, and so the observer will never perceive them as simultaneous to his "now".

Actually, t' and x' don't even make a good coordinate system. They are both 0 on [math]ct - x = 0[/math] and undefined anywhere else. It is possible to introduce so-called null coordinates, which are good coordinates and one of them is along the world line of the observer, but then neither of them can be interpreted as time. Time just doesn't flow for an observer travelling at light-speed.

As an interesting fact, if we call such coordinates u and v, the metric (a kind of generalization for the Pythagorean theorem) becomes something like [math]\Delta s^2 = \Delta u \Delta v[/math] (as opposed to the usual [math]\Delta s^2 = c^2 \Delta t^2 - \Delta x^2[/math] in spacetime and [math]\Delta s^2 = \Delta x^2 + \Delta y^2[/math] on a regular plane).

I hope I managed to shed some light on the problem ;) If something is unclear, feel free to ask, I will try my best to explain.
 
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perseus

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Thanks, I think the above for you, is that due to conservation of the interval,


x² + y² + z² = c²t²

the temporal contracion produces or implicit a spatial distancing, which prevents the arrival of advance information event, the system where the event takes longer. is not it?
velocidadUniverso.jpg

But it could be a very curved or closed space?

If space was very curved time this approach would have no sense.
and that information would come very quickly

Wow, surprised, like relativity, compensates her own wiles :lol:
 
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Fizyk

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Thanks, I think the above for you, is that due to conservation of the interval,


x² + y² + z² = c²t²

the temporal contracion produces or implicit a spatial distancing, which prevents the arrival of advance information event, the system where the event takes longer. is not it?
I'm not really sure if I understand you correctly, but you don't actually need relativity to explain this.

Let's look at the situation from the point of view of the stationary observer (Mr Violin). When Mr Car passes him by, in the Andromeda galaxy the general makes some plans. A day later, the fleet is sent and a signal with this information starts travelling towards Earth. By then, the car already moved a bit (it was travelling in the direction of the galaxy for a day).

The signal takes about 2 million years to reach Earth. By then Mr Car is in deep space (we have to assume he was going in a straight line towards Andromeda, otherwise the example won't work). Since he is closer to Andromeda than Earth is, he received the signal a bit earlier than Earth (and this is pretty much what causes the problems with simultaneity, but you have to take time dilation and Lorentz contraction into account here, too) - but even if he resends it immediately to Mr Violin, the signal from Andromeda and the signal from Mr Car will reach him in the same moment. There is just no way Mr Car can send the signal such that Mr Violin receives it before the signal from Andromeda.
 
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