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 non-accelerating 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)