Lorentz Transformation and Lorentz Boosts [Special Relativity Special #7]

Alright, now I’m going to wrap up the Lorentz transformation for real and be done with it!!!! As I mentioned before, what we’re about to pull off is this:

A ct–x spacetime diagram with an event P

In $x$–$ct$ spacetime, event $P$ would be written as $(ct, x)$, right!?!??!

Event P on the unit hyperbola at hyperbolic angle φ₀ from the ct axis, with its ct and x coordinates marked cosh and sinh

And using the hyperbolic angle, we got as far as writing its coordinates simply, like this. Here we start on the future unit hyperbola; for a general timelike point, multiply both coordinates by its interval scale. The final linear boost matrix also applies to spacelike and null events.

The same picture with the blue ct′ and x′ axes of a moving frame added

But wait, for a frame in relative motion like this one, event $P$ — how are we supposed to give it $ct'$, $x'$ coordinates!!!

Can’t we pull something off with the hyperbolic angle phi?!?! That’s exactly what we’re going to do now!!!

The ct′ axis tilted by θ from ct; P sits at hyperbolic angle φ from ct′ and φ₀ from ct

So now the problem boils down to this!!! With the picture drawn out like this!!!

we can say this!!! The hyperbolic angle is proportional to the area of a hyperbolic sector, so adjacent angles add up!!! Here $\theta$ is rapidity, $\tanh\theta=v/c$, not the ordinary angle you measure with a protractor on the page; that drawn tilt has $\tan\alpha=v/c$.

And as for the definitions of the hyperbolic functions, if you look at a general math book, there are formulas like these!!!

(Oh but, it’s been ages since I graduated from high school,,,, hyperbolics aren’t taught in high school, right?!?!)

Hyperbolic addition formulas for sinh, cosh and tanh of φ₁ ± φ₂

There are formulas like these, see???? The derivation is simple… so I’ll just put them to use!!! (Switch to exponentials and the derivation really is simple!)

Ordinary perpendicular axes: P dropped onto both axes at right angles, with the angle φ₀ from the vertical axis

If you think carefully about writing the position of point $P$ with hyperbolic functions, we drop it onto the axes at 90 degrees like this and can call its coordinates $\cosh\Phi$ and $\sinh\Phi$.

P dropped onto two tilted axes along red dashed lines with right-angle marks; the feet give cosh φ and sinh φ

So in this kind of spacetime, it can be drawn like this. The red dashed lines and the black solid lines are at 90 degrees — spacetime’s kind of 90 degrees, the perpendicular we redefined last time!!!

Now we’re all set to work out the Lorentz transformation.

The O′ frame drawn in the O frame: red ct′ and x′ axes, each tilted by θ, and event P at hyperbolic angle φ₀ from ct

Let’s draw the spacetime of $O'$, moving with relative velocity $v$ with respect to $O$, like this, and hmm~~~ the info about $v$ should be packed into $\theta$, riiight?!?

Now let’s first write event $P$ in the black frame, and then write it in the red coordinate system.

P: ct = cosh φ₀, x = sinh φ₀

In the red frame: ct′ = cosh(φ₀ − θ) = ct cosh θ − x sinh θ and x′ = sinh(φ₀ − θ) = −ct sinh θ + x cosh θ

Using the fact that cosh is an even function and sinh is an odd function, you can derive it even more easily!

Now if we write those two equations as a matrix, we get the transformation!!!

And

if we plug $x' = 0$ into this equation, the equation is then about $x' = 0$ — in other words, out comes an equation for the relative velocity $v$. Let’s do it!!!

Setting x′ = 0: x cosh θ = ct sinh θ, so x/t = c tanh θ, that is v = c tanh θ

We can make it into this, oh, buuut

what if we mash it up together with this identity?

Dividing by cosh²θ and using tanh θ = v/c: cosh²θ = 1/(1 − (v/c)²) = γ²

We write the hyperbolic cosine with gamma like this, and the hyperbolic sine is

Then sinh²θ = cosh²θ − 1 = γ² − 1 = γ²(v/c)², so sinh θ = γ(v/c)

So now the matrix doesn’t need an angle anymore!!!

The boost matrix with γ on the diagonal and −γ(v/c) off the diagonal

If we include the $y$, $z$ axes too!!!?!?! (But these are $y$, $z$ axes that don’t get involved. Meaning it’s not general — even though the $y$, $z$ axes are there, we’re only looking along the $x$ direction.)

The 4×4 boost matrix acting on (ct, x, y, z), leaving y and z alone

Not a fan of this?!?! Then if we just list it all out as equations — typing that into the equation editor is way too much of a pain, so please… bear with me…

The transformation written out by hand: ct′ = cγ(t − vx/c²), x′ = γ(x − vt), y′ = y, z′ = z

This kind of transformation is called the Lorentz transformation~~~

…is right! But let me be more specific: what we just did was a special case, since we fixed the $y$, $z$ directions and only looked at relative velocity along the $x$ direction.

More specifically, it’s a Lorentz boost along $x$ — one particular Lorentz transformation~

Lorentz transformations form the broader family: they preserve the spacetime interval and include spatial rotations as well as boosts. So calling our $x$ boost a Lorentz transformation is perfectly correct — now we just know its more specific name too!

(Still, let’s at least know that such a term exists~ and let’s moooove on! hehehehe hehe)

P.S. note: seen from a frame, a frame moving at v has its ct′ and x′ axes each tilted by the same θ, which keeps the speed of light c the same in every frame

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

In the second copy of the transformation, the $x$ line is $x' = \gamma(x - vt)$, with a prime on the left. And in the hyperbolic addition formulas, the cosh line reads $\cosh(\phi_1 \pm \phi_2) = \cosh\phi_1\cosh\phi_2 \pm \sinh\phi_1\sinh\phi_2$ (the first factor lost its subscript).

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