We want to transform coordinates x,y,z,ct in one isor K to coordinates x',y',z',ct' in another isor that moves with u/c wrt to K.
To simplify things, we take the relative velocity V between K and K' to be along x, so that y'=y and z'=z.
In classical mechanics, t'=t. Take for example that for t=0 we have x'=x. Then, after time t, the moving isor is at a distance Vt, which means that in general, x' = x + V·t. This, with t'=t, y'=y and z'=z are the Galileo transf.
For Lorentz transf we still have y'=y and z'=z if we transform along x. However, t'=t cannot be true any more.
Remember the analogy of the interval with the space interval in Euclidean geometry. A rotation in 3D space keeps the distance between two events invariant.
Here, we want the "distance" between two events invariant as well, but instance of Euclidean distance we have the interval Δs.
So what we need is to transform like a rotation, but a rotation that is adapted to our new geometry, which is 4D and with some weird signs.
When we consider rotations, we decompose them into the numbers of possible planes we have. In 3d, these planes are xy, yz and zx, which means 3 numbers are needed for a rotation in xyz. In four dimensions we have xy, yz, zx but also xt, yt and zt, so we have 6 numbers in total. However, we are interested here in the last three, since the first three are just the usual 3D rotations.
Let's study the rotation on the xt plane. This rotation leaves y and z unchanged. It must deal only with x and t in such a way that c²t'² - x'² = c²t² - x². This is analogue to having x'² + y'² = x² + y² for a rotation in the xy plane. In such case, we would write
x = x'·cosθ - y'·sinθ y = x'·sinθ + y'·sinθ
for a finite angle θ rotation about the origin. Positive θ means rotating counter-clockwise.
In the case of xt we need almost the same expression, but one that can give us the proper signs. We know that
sin²θ + cos²θ = 1 cosh²θ - sinh²θ =1
where sinh and cosh are hyperbolic trigonometric functions. This is all we need to write
x = x'·coshψ + ct'·sinhψ ct = x'·sinhψ + ct'·coshψ,
where now ψ is the rotated angle. It is easy to verify that c²t²-x² = c²t'²-x'². But we don't understand the meaning of this angle ψ and we want to relate it with the relative speed of the frames, u.
Consider the motion of the origin of the K' system, which always is x'=0. Then we get
x = ct'·sinhψ ct = ct'·coshψ,
so dividing them we get x/(ct) = u/c = tanhψ, which is the relation we wanted.
Now, we know that once we have a tangent, we easily have the other two trig functions, sin and cos:
sinθ = tanθ / (1 + tanθ²)¹⸍² cosθ = 1 / (1 + tanθ²)¹⸍².
To translate to hyperbolic functions is easy, since
sinθ = -i·sinh(i·θ) cosθ = cosh(i·θ) tanθ = -i·tanh(i·θ).
So we can get
sinhψ = tanψ / (1 - tanh²ψ)¹⸍² coshψ = 1 / (1 - tanh²ψ)¹⸍².
Here we have used ψ= iθ, and now, as tanhψ = u/c,
sinh(ψ) = (u/c) / (1 - (u/c)²)¹⸍² cosh(ψ) = (u/c) / (1 - (u/c)²)¹⸍² .
Finally, we can write our Lorentz transf as
x = ( x' + (u/c)ct') / (1 - (u/c)²)¹⸍² ct = (ct' + (u/c)x') / (1 - (u/c)²)¹⸍² .
Notice how I prefer to use ct instead of t and always have V and c as u/c.
Lorentz transf are of absolute importance.
If you want to transform x to x' instead, just change V to -V and swap x and x' and also t and t'.
Let's derive now two formulas that are of essential importance as well. Both are consequences of Lorentz transf.
Imagine a rod at rest seen by K, parallel to the x axis. Its ends, x1 and x2 are at a distance Δx = x2 - x1 according to what K measures.
We want to know what K' things of the length of this rod. For K', the length is Δx' = x2'-x1'. We want, of course, positions x1' and x2' to be measured at the same time ct'. This means that
x₁ = ( x₁' + (u/c)ct') / (1 - (u/c)²)¹⸍² x₂ = ( x₂' + (u/c)ct') / (1 - (u/c)²)¹⸍² Δx = Δx' / (1 - (u/c)²)¹⸍².
In other words, K' sees the rod smaller than K. We call this Lorentz contraction. The *proper length* is what the rod measures for an observer at rest wrt the rod itself. Any observer not at rest wrt to the rod will see the rod with a smaller length. If proper length is l₀ and length is l,
l = l₀·(1 - (u/c)²)¹⸍².
If instead of a rod we had a cube, only the length for which we have motion is contracted. In this case, Δy and Δz are invariant, so the volume also transforms as
V = V₀·(1 - (u/c)²)¹⸍²,
where V is the letter we use for volume here. My convention for v's is the following:
We can now ask a similar question, this time related to time. Suppose a clock travels with K'. We take two events occurring in the same x',y',z' coordinates but at different times t₁' and t₂'. So K' measures the time interval as Δt' = t₂' - t₁'. Now we transform to t₁ and t₂ as
ct₁ = (ct₁' + (u/c)x') / (1 - (u/c)²)¹⸍² ct₂ = (ct₂' + (u/c)x') / (1 - (u/c)²)¹⸍² Δt = Δt' / (1 - (u/c)²)¹⸍².
We call the time measured by the observer moving with the clock as proper time, as we already know. And, to every other moving observer, the time given by the clock will be longer. This is called time dilation.
Another important observation: two successive Galilean transf are commutative. However, two Lorentz transf are not commutative. This is reasonable if you see how Galilean transf are translations, while Lorentz transf are rotations.