Spacetime diagrams and Bondi’s k-calculus
Two important consequences of Einstein’s 1905 paper on SR:
1. It was immediately apparent that physicists had been taking the assignment and
measurement of coordinates for granted when, in fact, it was important to do it
correctly, in a way that could be "operationally" verified by experiment.
For
instance, the assumption that time was universal, and that the time measured between
any two events was independent of the observer, turned out to be an unexamined
and incorrect idea.
Another was that a rod of length 1 meter in some inertial
frame would naturally be found to have the same length in any other such frame.
2. At the same time, it became clear that in thinking (and drawing pictures!)
of
objects, we should pay careful attention to the object’s extension in time as
well as space.
This leads to the subject of spacetime diagrams, to which we now
turn.
We first have to set the mathematical scene:
Units: The essential features of SR can be understood using a 2-dimensional spacetime
diagram.
Geometrically, this is a plane, but not, as we’ll see, the Euclidean plane.
One of the two dimensions corresponds to time and one to space.
if both time and space have the same units.
in seconds.
It makes for simplicity
Our convention is to measure everything
This means that we replace x, measured in centimeters, by
x̃ =
x
x cm
=
sec,
c cm/sec
c
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which means:
light.
x̃ is the time it takes to travel a distance x moving at the speed of
Having made this definition, we now remove the accent and just write x.
So
for example, an object moving at the speed of light will have an equation x = ±t,
and will have velocity v = ±1.
Notice that in these units, velocity is dimensionless.
The conventional units can always be recovered by replacing x and v with x/c and v/c.
We make the following assumptions:
(1) there exists an inertial observer (one for
whom Newton’s first law holds), (2) any other observer moving with constant velocity
relative to this one is likewise an inertial observer.
will be used synonymously with inertial observer.
In what follows, the word observer
We also assume that all our observers
are equipped with identical standard clocks and with signalling devices which can emit,
detect, and reflect light rays.
More about this shortly.
Points in a spacetime diagram are called events.
by two coordinates (ta , xa ).
An observer A will label an event
The observer has a world line which we label A, and which,
in his coordinates, is just the time axis.
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Points on A’s world line have the coordinates
(ta , 0).
The light rays, the world lines of photons or quanta of radiation, are drawn
with slopes of ±1.
Time increases as we move from the bottom to the top of the diagram.
Figure 1 shows the world lines of three observers, together with a number of light
rays.
The k-factor
Suppose A and B are two observers in relative motion, and that A emits two photons
T seconds apart.
If B is motion relative to A, then B will receive the two photons
T 0 seconds apart, where T 0 < T if their motion is toward one another and T 0 > T if
they’re moving apart.
In either case, there’s a number k such that T 0 = kT .
If B
reflects these two photons back to A, then A will receive them k 2 T seconds apart.
The k-factor between A and B is the same as that between B and A; k is also known as
the relativistic Doppler shift.
See figure 2.
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Exercises: (a) If 2 1-dimensional observers are in relative motion, then their paths
cross.
Prior to that event, they are approaching each other, and afterwards, receding
from each other.
The picture above illustrates the case when they’re receding.
By
extending the spacetime diagram into the past, show that the k-factor as they approach
each other is 1/k.
(b) If if there are 3 observers A, B, and C, with k-factors kab , kac , and kbc ,
then kac = kab kbc .
Coordinates
We are ready to introduce coordinates in a well-defined way.
Fix an observer A and
an event E. A measures the distance to E by using radar ranging :
he sends a photon
to E at time t1 on his clock; the photon is reflected at E and received back by A at
time t2 .
The round-trip distance travelled by the photon, moving at the speed of light,
is just c(t2 −t1 ).
This is twice the distance to E. Since c = 1 in our system of units,
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A infers that E is located at a distance of
xa (E) = (1/2)(t2 − t1 ).
The event E must also be assigned a time, and sensible choice is the midpoint of the
interval [t1 , t2 ]:
ta (E) = (1/2)(t2 + t1 ).
Note that, given the coordinates (ta , xa ), we have immediately
t1 = ta − xa .
t2 = ta + xa ,
(We’ll need this in the derivation of the Lorentz transformation later.)
Velocity and the k-factor
Not surprisingly, if A and B are in relative motion, their velocity and their k-factor
are related.
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If "we" are observer A and wish to find the velocity of observer B, we need to take
two events on B’s worldline, assign coordinates to them, and then compute ∆x/∆t.
It
doesn’t matter which two events we take - we’re really just computing the slope of
a line.
So we make an easy choice:
since the two are in relative motion, their worldlines
cross at a point we label O in the figure above, and which we take as the first event.
If we set A’s clock to t = 0 at O, then we know the coordinates of O; they are (0, 0).
Reading things off from figure 4, we see that the event E has the coordinates xa (E) =
(1/2)(k 2 T − T ), ta (E) = (1/2)(k 2 T + T ).
Given the coordinates of O, this means that
v = xa (E)/ta (E)
= (k 2 T − T )/(k 2 T + T )
= (k 2 − 1)/(k 2 + 1)
Exercises:
• Show that
k2 =
1+v
.
1−v
• What was the velocity of B relative to A at a time before the event O? Isn’t that
cute?
We can get all the signs right if we’re tricky like this, but remember
that xa has really been defined as a distance, and so sometimes the correct signs
have to be put in by hand.
• Use a spacetime diagram, similar triangles, etc.
to draw the lines ta = constant
for some observer A. On the same diagram, draw some of the lines tb = constant
for an observer B in relative motion to A. Two events E and F are simultaneous
to observer A if they take place at the same time ta .
Will they be simultaneous
for observer B? This is called the relativity of simultaneity.
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It is REAL!
• By drawing some more pictures convince yourself that if the events E and F are
simultaneous for A there exist observers B and C such that E precedes F according
to B and F precedes E according to C.
The composition of velocities
We are about to encounter our first strange result (or second, if you think the relativity
of simultaneity is strange).
simply add:
Remember that in Newtonian mechanics, relative velocities
if vab is the velocity of B relative to A, and vbc the velocity of C relative
to B, then
vac = vab + vbc .
If I’m walking at a speed of 5 mph toward the front of a train moving at 70 mph, then
my velocity as measured by someone standing by the tracks is 75 mph.
However, we already
know that if I’m on a rocket ship moving relative to observer A at v = c/2 and shine
my flashlight in the direction of motion, then the speed of the light rays as measured
by A is not c+c/2; it’s just c.
So the Newtonian addition of velocities won’t hold
in SR. What does happen can be computed from the k-factors:
We start with the fact (exercise above) that for 3 observers, kac = kab kbc .
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Using the
expression for v in terms of k, this means that
vac =
2 k2 − 1
kab
bc
2 k2 + 1
kab
bc
after some algebra . . .
=
vac =
(1 + vab )(1 + vbc ) − (1 − vab )(1 − vbc )
or
(1 + vab )(1 + vbc ) + (1 − vab )(1 − vbc )
vab + vbc
1 + vab vbc
Examples:
• Suppose vab = 0.9c (or 0.9 in our units), and vbc = 0.9 too.
vac =
Then
0.9 + 0.9
1.8
=
= 0.994
1 + 0.81
1.81
Exercise: Show that if vab < 1 and vbc < 1, then vac < 1. If you’re moving at
less than c, no matter how much acceleration you can muster, you’ll always be moving
at less thant c.
• On the other hand, suppose vab = vbc = 3 km/sec, (a fairly hefty speed which would
take you more than 1/4 of the way around the earth at the equator in an hour).
This is "just" 10−5 c, however, and we compute
vac =
10−5 + 10−5
,
1 + 10−10
which is 2×10−5 to many decimal places.
This is why NASA doesn’t use special
relativity when it computes trajectories to the moon.
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