Re: relativity of simultaneity - real or perceived?

"Curious" <anthonyroseuk-curious@xxxxxxxxxxx> schreef in bericht
news:1116036993.415902.212980@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Does 'relativity of simultaneity' mean that an event which is
> simultaneous in one frame of reference, need not be simultaneous in
> another frame? Or do the two frames just APPEAR to have a different
> timing of events?
>
> In the common example of a fast train struck by lightning at both ends,
> simultaneously from the point of view of the embankment, and an
> observer on the train sees the lightning at different times, is that
> not merely a result of his *perception* of the events? If he were a
> scientist, could he not calculate the time taken for the light to
> travel and realise that the lightning did indeed strike at the same
> time?
>
> If two photons pass opposite ends of a box at the same time as measured
> by an observer at rest wrt to the box, could they ever be correctly
> calculated to be in the same positions at some point in time to any
> observer at any speed? Or would they be correctly calculated to pass
> the ends of the box at different times by that other observer?
>

Relativity of simultanitity can best be compared with the following
space-space example.
Imagine a tower and a straight road.
The road passes the tower at a shortest distance of 5 km.
You are standing on the road at the point where the road passes the tower at
this shortest distance.
You are lined up in the direction of the road, so the tower is 5 km beside
you.
Let's define your frame of reference:
you are standing at the origin;
1 km ahead of you on the road is (x, y) = (1, 0)
1 km behind you on the road is (-1, 0)
the tower at 5 km beside you is at (0, 5)
the tower is neither ahead nor behind you: the x-coordinate is 0.
Got the picture so far?

Relativity of simultanity is about how the time coordinate is measured in
different frames of reference.
So to complete the space-space example, another frame of reference is
needed.
Imagine a second straight road, which crosses the first road at the point
where you are standing, at a certain angle alpha (let's assume tan(alpha) =
3/4).
So, you are standing at the junction of the two roads with a frame of
reference which is aligned with the first road and for you the tower is at
(0, 5) i.e. 5 km "right" beside you.

But I could be standing on the same junction, lined up in the direction of
road 2.
Let's define my frame of reference:
I am standing at the origin;
1 km ahead of me on road 2 is (x', y') = (1, 0)
1 km behind me on road 2 is (-1, 0)
the tower is 4 km beside me at (5*sin(alpha), 5*cos(alpha)) = (3, 4)
the tower is not "right" beside me, but also 3 km ahead of me: the
x'-coordinate is not 0.
Got the picture so far?

This space-space example is analogue to space-time in the following way.
The x-coordinate, which indicates how far a point is ahead of (or behind)
you can be compared to time: "ahead" of you is in your future and "behind"
you is your past.
The same is true for the x'-coordinate in my frame of reference: a positive
value indicates an event in my future and a negative value indicates an
event in my past.
The (location of the) tower can be compared to an event: (0, 5) is now at 5
km distance for you.
In space-time this same event would be at (5*sinh(alpha), 5*cosh(alpha)),
i.e. not now for me.
Got the whole picture now?

Here's a bonus:
The two roads can be compared to our respective world lines.
Alpha, the angle between our respective roads can be compared to the "angle"
between our repective world lines: tanh(alpha) = ds/cdt = v/c, where v is
our relative velocity.

This also shows why c is the maximum relative velocity:
max(v/c) = max(tanh(alpha)) = tanh(infinity) = 1

Benno


.

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