
A Derivation of the Lorentz Transforms

Since
no relativity website is complete without a derivation of the Lorentz
transforms, I’ve put together a simple one here. This derivation
is somewhat different from the one given in Einstein’s 1905
Electrodynamics paper. I’ve used a single rather simple gedanken
experiment, with a single light ray traveling one way, in order to
obtain the formula for γ. I then used some slightly
indirect mathematical reasoning to derive all the other values we
need to complete the transforms.
(Note
that, throughout this page, I’ve dropped my usual assumption that
c=1.)
Assumptions
Einstein
was very parsimonious with his assumptions. I’m going to be a bit more
liberal.
 (A1) The laws of physics are the same in all
inertial frames – there is no way to determine, from within an inertial
frame, how “fast” it is moving and there is no “distinguished” inertial
frame. This is, of course, the principle of relativity; it has a
number of immediate corollaries which we’ll list a little later.
 (A2) The speed of light is measured as the
constant c in every inertial frame. This can be viewed as a direct
consequence of (A1) but it’s clearer to call it out as an explicit
assumption.
 (A3) The coordinate systems of any two
inertial frames are related by a linear transformation. This is
vital; without it we can’t proceed. It can probably be justified using
(A1) but again, I’m just going to assume it with no further
comment.
Corollaries to Assumption (A1)
The
relativity of physical laws has any number of easy corollaries. I’m
going to list a few which we need to keep in mind. In each case,
assume we have two inertial frames, Frame 1 and Frame 2, and that
they’re in motion relative to each other.
 (C1) Observers in each frame see the same
“effects” when viewing the other frame. This is somewhat vague, but
it means, basically, that time dilation and length contraction of the
“other” frame must appear the same regardless of which frame we’re in.
 (C2) If frame 2 appears to be traveling at
velocity v when viewed from frame 1, then frame 1 appears to be
traveling at velocity v when viewed from frame 2.
 (C3) Along a line which is perpendicular to
the line of motion of frame 2 relative to frame 1, distances will be
measured identically in each frame. If this were not true it would
provide an asymmetry between the frames, which would allow us to
determine which frame was “moving faster” and, ultimately, what frame
among all possible inertial frames is “at rest”.
 (C4) If two events lie on a line which is
perpendicular to the line of motion of frame 2 relative to frame 1, and
those two events are simultaneous in either frame, then they are
simultaneous in both frames. Like (C3), this is necessary
to avoid introducing a “preferred” frame.
Synchronization of Clocks
If two nonaccelerating
clocks, which we call 1 and 2, are stationary with respect to each
other, then the distance between them must be constant, and they can
be synchronized easily. Measure the distance between them with a
tape measure. Call the distance δ. Then an observer at clock 2 can
look at clock 1 (with a telescope, if necessary), and note the time. To
that time, add δ/c, the transit time of the light which came from
clock 1 (by assumption
A2). Set clock 2 to that
time. The two
clocks must now be synchronized.
More than two clocks in
the same inertial frame can be synchronized the same way. Use tape
measures or rulers to determine the spatial coordinates of all the
clocks, then compute their distances using trigonometry. Nominate a
“master” clock. Synchronize each clock with the master, as
described above. The clocks must all be synchronized with each
other as well, as we can see from simple trigonometry and assumption
A2.
The Gedanken Experiment
We
have two inertial frames, which we’ll call
S and
M.
Frame
M is moving along the
x
axis of frame
S at
(positive) velocity
v.
Since
we know that measurements along the
y
and
z axes are the
same in the two frames, we will, for the most part, ignore those
coordinates and just talk about the
x
coordinate and the time coordinate.
S,
which we’ll also refer to as the “stationary” frame, will be
given Latin coordinates, (x, t). Frame
M, which we’ll also
refer to as the “moving” frame, will be given Greek coordinates,
(ξ, τ).
At
time t = τ = 0, a flash of light is emitted from the origin. In the
moving frame
M, the flash moves directly up the Y axis
(perpendicular to the line of motion) and arrives at observer A,
located on the Y axis in that frame, at time ∆τ.
In
frame
S, the flash took time ∆t to get from the origin to A,
and, since A was moving, the flash appeared to move
diagonally
in frame
S. In that time, A moved v∆t units to the right,
and the flash traveled c∆t units.
See
Figure 1. The flash travels on the path
shown in red in frame
S,
and the path shown in blue in frame
M. Since the distance
traveled by the flash in the moving frame
M
lies along the Y axis, that distance will be the same in both
frames. Therefore, the distances, as laid out in Figure 1, are all
correct as viewed from frame
S. In particular, the “vertical”
distance from the origin to observer A must be c∆τ in the
stationary frame, as well as the moving frame.
Figure 1:
The Goal
Two of the space
coordinates – y and z – are unaffected by velocity along the x
axis. So, our goal is to find the linear transformation which
carries coordinates (t,x) in frame
S to coordinates (τ,ξ) in
frame
M.
or,
more simply, we want to find two linear equations:
For
that, we’ll need to find each of the partial derivatives, but we’re
actually going to start by finding something easier: The rate at
which time is passing for observer A.
Derivation of dt/dτ
Refer
to
figure 1. From simple trigonometry, we can
see that
In
“stationary” frame S, we watch observer “A”, who is carrying
a clock, move to the right at velocity
v. How fast does the
clock seem to be ticking? Or, equivalently, how
fast is time passing
in frame S
relative to the rate at which that clock (in frame
M)
is ticking?
In
frame
S the time coordinate is t. In frame
M it’s τ.
We’ll call the ratio of how fast time passes in frame
S, to
how fast the moving clock ticks,
γ, which we
define
as:
At
this point, for convenience in the formulas, we also define
From
equation
(3) we then have
Since
the term on the right in (7) doesn’t depend on the length of ∆t,
we immediately see that, for observer A,
or
equivalently,
Finding ∂τ/∂t
Consider the observer
at A once more. Note that ξ(A) = 0, and is constant. At A, only τ
is changing:
But ∂t/∂τ is exactly
the rate of change in t as τ changes, with ξ held constant. Since
ξ
is constant for A, we must therefore have that, at A,
To put this another
way, suppose the observer at A passes a line of clocks, each
stationary in frame
S, and
compares each with his own clock. For A, τ is changing, ξ is not,
and the rate at which the time he observes on the
line of
clocks in
S relative
to his own clock will be ∂t/∂τ, and we have just determined that
it must be equal to γ.
But
the frames are symmetric. An observer in frame
S will see the
same effects, looking at frame
M, as an observer in frame
M
who observes frame
S, save that the velocity of the “other”
frame is negated. γ is independent of the sign of
v. So, an
observer stationary in frame
S watching a line of clocks in
frame
M must
also see the time on the
M clocks
changing at γ relative to his own clock, and we must therefore have
and
by equation
(8b),
Finding ∂τ/∂x and (finally!) τ(t,x)
Along
A’s path, we have
and
so, for A, equation
(2a) reduces to
But
for observer A, we also have
Plugging
(13) into (12) we obtain
which
we solve for ∂τ/∂x:
from
which we conclude
Derivation of ξ(t,x)
For observer A, x = vt,
and equation
(2b) reduces to
But for A, we also have
ξ=0 (and is constant). Combining that with (17) we see that
Combining v=dx/dt (for
A), and dt/dτ=γ, we obtain
But since ξ is
identically 0 for A, and only τ is varying, we must also have
As noted, frames
S
and
M are symmetric. If we switch from A, who is viewing frame
S
from frame
M, to an
observer in frame
S
watching frame
M, the
only change is that
v
is negated. So, identical reasoning to that which led us to equation
(20) (but with the sign of
v switched)
must lead us to conclude
Plugging
(21) into
(18) we obtain
And
finally, combining
(21),
(23),
and
(2b), we obtain
With
(16), we now have the “forward” Lorentz transform, to go from a
“stationary” frame to one moving to the right at velocity
v:
The Inverse Transform
The frames are
symmetric. If we shift our point of view from
S
to
M all that changes
is the sign on
v. So, we can immediately write down the inverse
transform, just by
reversing the velocity:
Let’s
check to see if it really is the inverse, by multiplying (25) and (26)
together:
Sure enough, their product is the identity. Transforming into
another frame and then transforming back gets you back where you
started.
The Minkowski Metric
The Minkowski (or
Lorentz) metric measures the familiar “interval” of relativity,
which counts distances in space and time as having opposite signs:
We would like to
confirm that the Lorentz transform which we’ve just derived
preserves the Minkowski metric.
In matrix form, the
metric is,
To apply it to two
column vectors, we multiply the first into it on the left
(transposed), and the other on the right. For example:
A little fiddling with
eq. (30) shows first that, when we change coordinates, the metric
tensor itself must transform as,
where the metric is multiplied on the left and right by the
inverse
transform (
eq 26). When we cast
(31) into matrix form and compare it again with (30) we realize that
the term on the left must be
transposed. So, we have,
And the Minkowski
metric is indeed preserved by a Lorentz transformation, in
consequence of which all “intervals” will be the same for all
inertial observers, regardless of their relative velocities.
Page created on 8/8/05, and last updated 10/31/06 (minor correction to a
formula)