The Trouble with Time Dilation
For people new to special relativity, and people trying to understand
it from simple "lay" explanations, "time dilation" can be a major
stumbling block. As generally "explained", time dilation means
time slows down for objects which are moving quickly.
But this explanation is somewhere between misleading and completely
Consider two spaceships. On Ship A, an astronaut watches Ship B
zoom by at 0.866 C, and observes that Ship B's clocks are running slow
by a factor of 2. That's the so-called "time dilation".
Unfortunately, B sees the same thing happening with Ship A's
clocks. What's even worse, an observer floating in space, moving
at the average
of the speeds of the two ships, will assure us
that the clocks on Ship A and Ship B are ticking at exactly the
So, what's "really" going on?
First, let's look at γ.
It's a useful value, and for
simple cases it is typically defined as:
We can then very
is the time
dilation factor!", and in fact that's just what is usually said.
But there's a big problem with this. Look at the part of the
Lorentz transform which gives us the time coordinate in a spaceship
moving at velocity v
along the X axis:
Do you see the problem? It's that second term. Look at the
derivative of τ with respect to t:
This is "time dilation": 1/γ < 1. It is the derivative of τ along
the spaceship is following.
This is time contraction
. It's the derivative in the
direction of time
rather than in the direction the spaceship is
traveling. It has a physical meaning, too: If a single
observer in A's frame watches a series
of clocks go by, where
each of them is moving at velocity v
and they're all
synchronized to B's clock in B's frame of reference, then that observer
will see the time shown on the B clocks advancing more quickly than
time in A's frame
. Turn it around, look at it from the point
of view of the line of clocks in B's frame which are passing the A
observer, and it's just "time dilation" of a single observer in A's
frame as viewed from B's
Confused? If you're not, perhaps you should read that again!
In other words, in the spaceship's frame, time is not "going slower"
than time in the stationary observer's frame in any global sense.
Rather, time along the path
the spaceship is following seems to
go more slowly.
Or we could say, along the path Ship B is following in A's frame of
reference, less time elapses for B than for A.
Or we could say, Ship B is taking a shortcut through A's
frame of reference.
And at the same time, Ship A is taking a shortcut through B's frame
Perhaps that seems confusing, and it's hard to quantify it. But
it's certainly closer to "the truth" than the bland assertion that B's
clock is running slower than A's!
Consider a revolving clock, moving on a circular path at v
Viewed from a point in the center of the circle, the clock is moving
continuously, and it "runs slow". Geometrically, from the point
of view of an observer in the center of the circle, it's just like the
case of Ship B passing ship A moving linearly.
But when we turn it around, and view the stationary observer in the
middle from the moving clock, we see the stationary observer's clock running
Why is that?
Consider carefully the coordinates of the moving clock. If we
want to keep the "apparent motion" of the observer in the center of the
circle fixed so that it moves along the X axis, then the moving
coordinate system must be rotating.
. Furthermore, at the
point in the middle of the circle, the coordinates are rotating at
exactly the right rate to keep the the stationary observer from moving
either way along the X axis.
In other words, in the rotating coordinates which keep the
instantaneous motion of the clock in the middle lined up with the X
axis of the revolving clock, that clock does not move through space
Since it's not moving, it's also not taking the "shortcut" the moving
ships could take, above, and instead, we just see the bare "time
contraction" term: The stationary clock goes faster, from either
point of view.
This is an incomplete description (you could turn it around and claim
the revolving clock must be time contracted, too) but I think it's a
step in the right direction. I will say a bit more about it when
we consider the revolving clock in detail.
Another View: "Time Dilation" as a
Consequence of the Metric
Everything I've said up to now has been just an intuitive
description. All intuitive descriptions of spacetime behavior are
incomplete, and may be misleading. However, there are also some
factual statements we can make about the behavior of the metric which
may help with understanding "time dilation".
The distance between two points -- or two events -- is determined by
the "metric". In the ordinary Cartesian space of our everyday
experience the metric is given by Pythagoras's theorem:
All points located a particular distance from the origin form a
circle. With the metric of special relativity, on the other hand,
a set of points which are equidistant from the origin forms a hyperbola
And that difference is the source of the "time dilation" effect.
The Arithmetic of the Metric
With 1 space dimension and 1 time dimension, in an inertial frame, the
experienced by an object that moves along a straight line between two
For two events at the same location in space
, that's just Δt.
But if the space coordinate changes, then the proper distance must decrease
The proper distance from the origin to a point on
the T axis is
-- the coordinate time. However, the farther an event is from the
T axis, the smaller
is the proper distance from the origin to
the event, because the Δx2
term is subtracted
from the square of the T distance in the metric.
This is completely contrary to our everyday experience.
normal "Cartesian" space, the length of the hypotenuse of a right
triangle at least as long as each of the other
sides. Furthermore, the distance between two points in our
everyday world can never be zero or negative, while it can be in SR
4-space. We summarize this
by saying the metric in Cartesian space is positive definite.
The metric in relativity is not. Attempts at applying everyday
intuition to relativistic situations tend to result in
frustration. Because, in special relativity:
The farther away an event is physically, the closer
it is in proper time.
The faster you go, the less (proper) distance you cover.
A Hyperbolic View of What's Happening
A set of points that are equidistant from the origin using the SR
metric forms a hyperbola.
A point which lies on the T axis represents an event in the rest
frame of the
origin. It's in the same spacial location as the origin.
set of points which are equidistant from the origin with such a point
forms a hyperbola which bends "up" -- toward higher T values -- as it
moves away from the T axis (see Figure 2). All points on the
hyperbola are the same proper distance, ΔT, from the origin. All
points on the yellow hyperbola are also equidistant from the origin;
their distance appears to be roughly (3/4)ΔT on the diagram. All
points on the
hyperbola are at a proper distance of roughly (1/2)ΔT from the
A "future" event which is distant in space
from the T axis lies
which intersects the T axis at a lower
T value. Such an
event is no "farther away" from the origin than a point on the T axis
which has a smaller time coordinate.
So by moving away from the T axis, an object moves onto a different
"closer") hyperbola of equidistant points. Objects which are
moving rapidly away from the T axis move to "closer" hyperbolas at the
same time they are
moving to larger T values; hence, they don't get very far from the
fast. If an object moving to the right in Figure 2 moves ΔX
while ΔT (coordinate) time passed, we can see from the diagram that
only about (1/2)ΔT amount of proper
time would pass for the
The Lorentz transform performs a hyperbolic
, which rotates the points through which a moving object
passes along their hyperbolas onto the T axis. For most of us,
this is an unfamiliar operation -- we're used to circular rotations,
and we're used to a metric where all points on a circle
equidistant from the origin.
Confusion Alert: Why does motion toward the T axis
cause time dilation?
The "proper time", or subjective elapsed time, which is experienced by
an object must be measured between two events experienced by the
object. In the above discussion, I tacitly assumed the object we
were discussing had started at the origin.
The time which passes for you must be measured from a moment (and
location) where you are, up to another moment (and location) where you
are. This is simple; even obvious. But the consequence of
this simple observation may be
To find the elapsed time experienced by an object which starts far from
the T axis and moves toward the axis, we need to measure the proper
time between the object's starting point
and its ending point
-- and its starting point wasn't at the origin. So, to find its
(proper, subjective) elapsed time, we need to draw a line from its
starting point -- off to one side -- to its ending point, and find the
length of that
line. And what we'll find if we go
through the exercise is that curves of equal distance from its starting
point form hyperbolas centered on its starting point.
So the result is just the same as if we moved the origin to the point
where the object started -- and in fact that's what we generally do, if
it's at all possible, because it simplifies the math without affecting
Page created in 2004. Minor update to diagram, 11/14/06