by Steve Bryson
There are many different of ways at arriving at the Lorentz Transformation. This is because the Lorentz Transformation can be thought of as coming from any one of the following equivalent statements:
•The speed of light is the same for all observers
•Spacetime is a 4 dimensional space with distance formula given by s2 = t2-x2
•Spacetime is a 4 -dimensional hyperbolic space
The derivation from the first principle can be found in Einstein's popular book on relativity (see book list) as an appendix. This is the classical approach and requires nothing more than a knowledge of algebra.
The derivation from the second principle is very easy and only requires a working knowledge of high school algebra. This is presented in detail in this handout.
The third derivation is, I feel, the deepest conceptually, but it requires knowledge of hyperbolic trigonometry. I give an introduction to this in this handout. It is certainly the way I think about it.
The following will perhaps look pretty hard, but that is because I am going to show every step one step at a time. I promise that for this section I will use nothing but high school algebra to get from one step to another, and I will describe the operation at each step.
First, consider two coordinate systems in relative motion with velocity v with coordinates t,x and t',x' and coinciding origins. Here the t,x system in defined as motionless, and the t',x' system is in relative motion with velocity v.
Look at an event with coordinates t', x' = 0:
If we are given the t' coordinates of the event (we know that the x' coordinate =0), how do we find the x and t coordinates of the event. We use the fact that the spacetime distance is independent of coordinates, which is to say that the spacetime distance in the two coordinate systems is the same.
The spacetime distance in this case is:
or
.
Take square root of both sides:
Factor out a (ct)2:
but x/t = velocity v, so isolate the x/t:
Solve this for t and find that
This is the formula for the t coordinate in the case described above. Now x=vt (as v=x/t), so
Thus the event at x'=0 and time t' will have coordinates
Thus something that happens at t'=2 seconds in the moving coordinate system will happen at some t seconds (where t > 2seconds) in the nonmoving coordinate systems. It will also have a non-zero x coordinate. Notice that even though the point in the moving system had x' coordinate =0, there is a nonzero x coordinate for the event. This is because the Lorentz Transformation mixes the coordinate axes.
The more general case: Consider now an event that has t' and x' coordinates both nonzero:
We ask the same question: What are the x and t coordinates of the event? We start by assuming that the Lorentz transformation is linear in x' and t' as all good rotations should be. This means that the transformation is of the form
t = Dt'+Ax'
x = Ct'+Bx'
What we need to do is determine what A, B,C and D are (we'll find that they are functions of velocity). But we already know what D and C are from the simpler case above, for they assumed that x'=0. Thus we know that the Lorentz transformation is of the form
So now we need to find A and B. We do this by again using the fact that the spacetime distance of the event from the origin is the same in both coordinate systems:
c2t2-x2=c2t'2-x'2
First we compute c2t2-x2 in the transformed coordinates: we substitute the above Lorentz transformation expressions for t and x into the spacetime distance:
What we want to do with this is compare it with c2t'2-x'2. As it should be equal, we will be able to determine A and B. We do this by pulling the x' and t' terms out of the above formula and comparing coefficients. First we expand the squares:
Then collect the like terms:
Now factor out the terms with t' and x' in them:
Simplify and note that this must be equal to c2t'2-x'2:
Now we demand that whatever multiplies the t' on the left hand side of the equal sign is equal to whatever multiplies the t' on the right hand side of the equal sign. Similarly for the x' and the t'x' term. Thus there are three cases:
t' case: we have c2 multiplying the t' on both sides, so we do not learn anything here.
t'x' case: we have no t'x' term on the left side at all, so whatever multiplies the t'x' term on the right side must be equal to zero:
We solve this for A in terms of B:
We will use this in a minute.
x' case: we have a -1 multiplying the x' on the left side, so whatever multiplies the x' on the right side must be equal to -1:
Now substitute the formula 
from above:
And solve for B:
Now plug this back into the formula for A:
We now know A and B. We can plug this into our starting formulas for the Lorentz Transformation
to get
This is the complete Lorentz transformation. Note that if something has a length (x) of 1 in the nonmoving coordinate system, then it's length in the moving coordinate system (x') would be shorter. Note that even though for a length there is no t part (t=0), there will be a t' part from the mixing of space and time by the Lorentz transformation. This would be seen as a small rotation of the object being measured along an axis perpendicular to the direction of motion and the direction measured.