3.1 Comparing Time for O and O' (Space-Time Diagrams)
Description
This article is from the Relativity and FTL Travel FAQ, by Jason W. Hinson jason@physicsguy.com with numerous contributions by others.
3.1 Comparing Time for O and O' (Space-Time Diagrams)
So, how do we show time dilation on our space-time diagram. Well, the
key to this can be found by expressing time dilation in the following way:
In the O observer's frame of reference, let the tick t1 of his clock be
simultaneous with the tick t1' of the O' observer's clock. Also, let the
tick t2 of his (the O observer's) clock be simultaneous with the t2 tick of
the O' observer's clock. Then, we would find that
(Eq 3:1)
t2'- t1'= (t2 - t1)/gamma
where gamma (as defined in Section 1.4) would be calculated using the
relative velocity of O and O'. What Equation 3:1 says is that in the O
observer's frame of reference, the difference in the ticks of the O'
observer's clock is smaller than the difference in the O observer's own
ticks by a factor of gamma. Thus, we see that in the frame of O, the O'
observer's clock is running slowly.
As an example, from here on we will consider the case where the
relative velocity is 0.6 c such that gamma = 1.25. Using an example like
this will make the procedure easier to understand for the reader; however,
remember that we could redo this whole process with any speed (calculating a
new gamma factor, drawing a different speed for the observers, drawing
appropriate lines of simultaneity, etc.).
Now, what if we let the t1 tick be the "zero" tick. That means that at
the origin, when both of our observers are right next to one another, t1 =
t1' = 0. So, both of the observers agree (because there is no separation
between them in space at the origin) that t1 and t1' are simultaneous, and
happen at t = t'= 0. However, after some time, there will be a tick (t2) on
the O observer's clock. In the frame of reference of O, that tick is
simultaneous with the tick t2' of the O' observer's clock. Since t1 = t1' =
0, and we are using gamma = 1.25, we know (from Equation 3:1) that
(Eq 3:2)
t2'- 0 = (t2 - 0)/1.25.
so
t2' = 0.8*t2
So, this says that in the frame of the O observer, the tick t2 of his
clock is simultaneous with the tick 0.8 t2 on the O' observer's clock. If we
draw a line of simultaneity in the O observer's frame of reference such that
it goes through the tick t2 of his clock, then it must also go through the
tick 0.8 t2 of the O' observer's clock. If we let t2 = 1 second, then we get
what is shown in Diagram 3-1. The distance from the origin, o, to the first
mark along t in that diagram is defined to be 1 second for our O observer.
Meanwhile, the distance from o to the "*" symbol along t' in that diagram is
0.8 second FOR THE O' OBSERVER. So, we begin to see that we can relate
distances in time along the axes of the different observers.
Diagram 3-1
t t'
^ /
| /
| /
| /
| /
| /
t = 1 + - - * - - - - - - - line of simultaneity
| / t' = 0.8 for O at t = 1
| /
| /
| /
|/
---------o----------------------> x
|
|
(Note: The line for t' only approximately represents an observer
moving at 0.6 c. It probably more closely represents 0.5 c, but
that's my ASCII for you. For our example, it _should_
represent an observer traveling at 0.6 c in the O observer's
frame of reference.)
This puts us on our way to understanding how, for example, different
lengths along t and t' relate to particular times on the clocks of the two
observers. Our next step to understanding this better will be to look at the
situation from the O' observer's frame of reference.
We have found what tick of the O' observer's clock is simultaneous with
the t = 1 tick of the O observer's clock
_in_the_O_observer's_frame_of_reference_. However, say we want to decide
what t' tick is simultaneous with the O observer's t = 1 tick
_in_the_O'_observer's_frame_of_reference_ (remember, the line of
simultaneity in Diagram 3-1 is only valid for the O observer's frame). To
figure this out, we need to draw a line of simultaneity in the O' observer's
frame of reference which passes through the event "the O observer's clock
ticks 1". When we do this, we want to note where that line passes the t'
axis, because that mark points out the tick on the O' observer's clock which
is simultaneous with O observer's t = 1 tick
_in_the_O'_observer's_frame_of_reference_. I have drawn this line in Diagram
3-2, but I have also left everything that was in Diagram 3-1.
Diagram 3-2
t t' line of simultaneity
^ / ' for O' at t' = 1.25
| / '
| / '
| / '
| % t' = 1.25
| ' /
t = 1 + - - * - - - - - - line of simultaneity
| / t' = 0.8 for O at t = 1
| /
| /
| /
|/
---------o----------------------> x
|
|
(Note: The line of simultaneity for O' is a rough approximation)
Now, note that I marked the "%" symbol in that diagram (where the line
of simultaneity for O'--which goes through t = 1--crosses the t' axes) as
the event t' = 1.25. But how did I know that? Well, because in the frame of
reference of O', it is the O observer who is moving at 0.6 c, and thus it is
the O observer who's clocks are running slowly by a factor of 1.25. So, in
the frame of O', the event "t = 1 at the O observer's position" must be
simultaneous with the event "t'= 1.25 at the O' observer's position." That
way, in the O' observer's frame, it will be the O observer's clock which is
running slowly by a factor of 1.25. In addition, if the diagram were drawn
carefully I could use the length from the origin to "*" (which I know is 0.8
seconds for the O' observer) to figure out how much time passes between the
origin and the "%" symbol for O'. Either way, I find the same thing.
In Diagram 3-2, one can begin to see the power of using space-time
diagrams to understand special relativity. Note that from one glance at that
diagram not only can we see that in the O observer's frame of reference the
O' observer's clock is running slow by a factor of 1.25 (i.e. the event "t =
1" is simultaneous with the event "t'= 0.8" in the O observer's frame) but
we also see that in the O' observer's frame it is the O observer's clock
which is running slow by a factor of 1.25 (i.e. the event "t = 1" is
simultaneous with the event "t'= 1.25" in the O' observer's frame). Thus, we
can see at once on this diagram that in each observer's own frame, the other
observer's clock is running slow. This happens to be one of the first, key
points to understanding the twin paradox (which will be discussed fully in
the next section).
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