SPACETIME
DIAGRAMS
One approach to understanding relativity is to represent events on a
diagram which shows the coordinates in two frames of reference which
are in motion relative to each other. To construct such a diagram,
start with a few basic ideas:
 Relativistic effects are not obvious at speeds much less than c,
so represent all velocities as β = v/c.
 All observers agree on the speed of light, so the speed of light
should be a line with slope=1 in both coordinate systems.
 An object at rest in a frame moving at velocity v with respect to
the “rest frame” will follow a path with slope β = v/c in the “rest
frame”.
These are sufficient to begin
constructing the diagram. In
black are the “rest frame”
coordinates and the light path
with slope 1.
To construct the axes for the
moving frame, the x axis is the
path for an object at rest in the
moving frame, so it will have
slope β on the diagram.
To get the time axis, use the fact that the light path must still have
slope =1 in the moving frame, so it must be “centered” in the red axes,
at equal angles from the time and x axes. This in turn implies that the
time axis will be tilted in from the stationary time axis by the same
angle that the x tilted up from the stationary x axis.
The next step is to calibrate the moving frame axes. Where to place
the marks for 1 meter of length and 1 light meter of time? Consider
two events located at (x1,t1) and (x2,t2) in the stationary frame. Define
the interval between them as ((x2-x1)2 – (t2-t1)2)1/2 = (Δx2 –Δt2)1/2.
This is analogous to the Euclidean distance between 2 points, but with
the minus sign in front of the time difference.
Let’s see how this quantity looks in the moving frame. For this, we
cannot simply use the time dilation and Lorentz contractions because
simultaneous events in one frame are not simultaneous in another. As an
example, consider observing the simultaneous explosions two stars going
nova. In one frame, the two stars are at rest and we only have a time
difference to measure. In a frame moving at β with respect to this
one, the two explosions are at different times as well as distances, so
we have a length difference and a time difference to consider. Not
only that, because of the nonsimultaneity, the locations are measured at
different times and we must resort to the full Lorentz transformation to
calculate the interval in the moving frame.
The Lorentz transformation, using γ = 1/√(1-β2) and β = v/c, is
xmov = γ(x - βct)
ctmov = γ(ct – βx)
So, in the moving frame,
Δxmov = x2mov – x1mov = γ(x1 – βct1) - γ(x2– βct2)
= γ( [x1-x2]– β[ct1 – ct2])
= γ(Δx - βΔct)
The Δt calculation is exactly the same with the x’s and t’s
reversed, which gives
Δtmov = γ(Δct - βΔx)
Putting these together gives:
(Δxmov2 – Δctmov2) = [γ2(Δx - βΔct)2 – γ2(Δct - βΔx)2]
= γ[(Δx2 -2 Δx βΔct + β2Δct2) - (Δct2 -2 Δct βΔx + β2Δx2 )]
= γ[(Δx2 + β2Δct2) - (Δct2 + β2Δx2 )]
= γ[Δx2 (1- β2) - Δct2 (1- β2)]
So
(Δxmov2 – Δctmov2) = Δx2 –Δct2
What we have found is an invariant. The interval defined as
Δx2 –Δct2 is the same in all frames of reference! The same will be true
of Δct2 –Δx2.
If we take Δx as the difference between x and the origin and Δct as
the difference from the origin to ct, then the equation for a hyperbola
passing through the point x=0, ct = ct will be
Δct2 –Δx2 = ct2.
For points on the x axis, we similarly get
Δx2 –Δct2 = x2.
We can use this to calibrate our moving axes by drawing hyperbolae
through the marks ct=1,2,3, etc. and placing marks ctmov = 1,2,3, etc
on the tilted axes where the hyperbolae cross them.
The grid lines in the moving frame are then drawn in parallel to the
axes:
USING THE DIAGRAM TO ANSWER QUESTIONS.
Twin Paradox:
Two identical twins start at a common place and time, at the origin of
coordinates. One of the two then leaves on a trip at 0.6c, travels out
for 20 light meters and then returns at -0.6c. During the trip, the
moving twin sends signals to the other twin each light meter by light
beam. These will be shown as lines on the diagram.
First, set up the diagram with the two sets of axes. Include the return
trip with the opposite slope.
Draw in the time lines for the
traveling twin, shown in red on
the right. Notice the gap in
the middle during
acceleration. This is what
fixes the difference in ages
of the twins; the one who
accelerated is the younger
one. Velocity is relative,
acceleration is not.
What we have shown here is
the timing from the moving
twin’s point of view. For the
20 time intervals measured by
the moving twin, the other twin measures about 24 intervals (of 0.2
light meters).
Let’s draw another diagram assuming
the moving twin sends a light pulse
back to the other twin each 0.4
light meters. Drawing these in black
gives the diagram at right.
In this depiction, the age difference
is seen as due to the extreme
slowing of the moving twin’s clock on
the outward journey. The more
frequent signals on the return are
nowhere near enough to compensate,
so the age difference is locked in.
The same difference in time
intervals is shown here.
Docking the Long Ships
An illustration of the Fitzgerald contraction, with a more modern bent,
is docking a 500 meter long space ship in a 300 meter long repair dock.
The ship is to fit entirely within the dock. What speed must the ship
be moving, relative to the dock, for this to work? Ignore the small
detail of not being able to stop in the repair dock and still fit.
As an example, the figure below shows a 3 meter stick in a moving
frame. In the stationary frame, the stick appears 2.5 meters long as
shown by the yellow dots where the stick crosses the stationary axis in
the diagram on the right.
The slope of the moving frame axes can be read from the stationary
frame axes. The slope of the t=0 axis of the moving frame is 1.1 light
seconds per second or, (1.1/2)c = 0.55c which is the velocity of the
moving frame.
As a check, Δxstat = √(1-β2) Δxmov = √(1-0.552) 3m = 2.505, not bad!
On the next page is a blank Minkowski diagram with the stationary axes
and hyperbolae on which to draw your moving frame axes and find the
velocity of the moving frame.