At the beginning of the course, we looked at writing down a rotation of a coordinate system in terms of matrix operations.

Our Lorentz transformations can be written in terms of $\gamma$ and a new parameter $\eta=V/c$ as... $$x_{ {\cal m}} = \gamma(x_{ {\cal b}}-Vt_{ {\cal b}},$$ $$y_{ {\cal m}} = y_{ {\cal b}},$$ $$z_{ {\cal m}} = z_{ {\cal b}},$$ $$t_{ {\cal m}} = \gamma(t_{ {\cal b}}-Vx_{ {\cal b}}/c^2).$$

These equations look very similar to the rotations of coordinate systems. Let's make it look more that way by floating the idea of a four-dimensional vector: $bbb(x) = [(x_1),(x_2),(x_3),(x_4)]=[(x),(y),(z),(ct)]=(\myv x, ct)$.

Instead of just $t$ as the fourth component, we want that component to have the same units as the other components.

The Lorentz transformations can be written in terms of these coordinates and a new parameter $\eta=V/c$ as... $$x_1' = \gammax_1-\gamma \eta x_4,$$ $$x_2' = x_2$$, $$x_3' = x_3,$$ $$x_4' = \gamma x_4-\gamma \eta x_1.$$

Or, in matrix form: $$x' = \Lambda x,$$

where $$\Lambda = [(\gamma,0,0,-\gamma \eta),(0,1,0,0),(0,0,1,0),(-\gamma \eta,0,0,\gamma)].$$

In the same way that we defined a vector (well, now a 3-vector) to be a quantity that transformed under rotations like the displacement vector, we can define a four-vector:

In each inertial frame, a four-vector is specified by a set of four numbers $q=(q_1,q_2,q_3,q_4)$, such that the values in two frames are related by $q' = \Lambda q$, where $\Lambda $ is the Lorentz transformation connecting the two reference frames.

Any single quantity that is invariant under rotations alone is called a rotational scalar or three-scalar. For example, rotational scalars include:

  • the mass $m$ of a particle.
  • the time $t$.

Any single quantity that is invariant under a Lorentz transformation is called a Lorentz scalar or four-scalar. The time $t$ (related to coordinate $x_4$) is apparently not a four-scalar. But some quantities that are four-scalars include:

  • the speed of light $c$.
  • the rest mass $m$ of a particle,
  • the charge $q$ of a particle.

The invariant scalar product

Rotations in 3-dimensions do not change the length of a vector.

Let $\myv A$ be a vector in one coordinate system, and the same vector is represented as $\myv A'$ in a rotated coordinate system, then... $$|\myv A|^2 = \myv A*\myv A = A_1^2+A_2^2+A_3^2=|\myv A'|^2.$$

That is to say, $|\myv A|^2$ is a rotational scalar.

The dot product of this vector with itself $(\myv a +\myv b)*(\myv a+\myv b)$ is also a length squared, that is a rotational scalar, implying: $$\myv a*myva + 2myva*myvb+myvb*myvb=\myv a'*myva' + 2myva'*myvb'+myvb'*myvb'.$$

But, we know that $\myv a*myva=\myv a'*\myv a'$, and similarly for $\myv b$, so apparently the 3-space dot product, $$myva*myvb=\myv a'*\myv b',$$

must be a scalar invariant too.

Invariant scalar product in 4-space

This quantity $s=x_1^2+x_2^2+x_3^2-x_4^2 = r^2-c^2t^2$ turns out to be invariant under Lorentz transformations, as we shall now show: $$\begineq s' &=& x_1'^2+x_2'^2+x_3'^2-x_4'^2\\ &=&\gamma^2(x_1-\eta x_4)^2+x_2^2+x_3^2-\gamma^2(-\eta x_1+x_4)^2\\ &=&\gamma^2(1-\eta^2)x_1^2 +x_2^2+x_3^2-\gamma^2(1-\eta^2)x_4^2\\ &=&x_1^2+x_2^2+x_3^2-x_4^2 = s.\\ \endeq $$

This looks like a length-squared. But notice that $s$ can be positive or negative.

By the same line of argument that led from an invariant length squared to an invariant dot product in 3-space, we can define this quantity as an invariant scalar product of two 4-space vectors:

$x*y = x_1y_1+x_2y_2+x_3y_3-x_4y_4$.

We can think of flashing a light bulb at the origin $\myv x=0$ at time $t=0$. The light spreads out with speed $c$ in all directions, so that the wavefront is at $r^2=c^2t^2$. The whole set of wavefront positions can be denoted in terms of the 4-space dot product $x*x=0$.

Since this scalar product is invariant, in another reference frame $x'*x'=0$ also. That is, light is also spreading out to fill a sphere as well with speed $c$. So this invariant scalar product can be thought of as a consequence of the invariance of the speed of light in all reference frames.

We can divide 4-space into a couple of distinct regions:

Minkowski diagrams

4-space is 3 spatial dimensions and one proportional to time. Instead of talking about a point in this space as a "position", it's more appropriate to talk about one point as an "event" that has both a location and a time.

A Minkowski diagram depicts one (or more) space dimensions, and the time dimension. If the time dimension $ct$ runs vertically, and one of the spatial dimensions runs to the right, and both have the same scale, light follows a path on this diagram at a 45 degree angle.

Any material object must take a path (a "world line") with a slope always greater than 1.

If your position is $x_1=0$ at $t=0$, then the positions in 4-space that you might aspire to reach at a later time by hopping a train with some speed $V<c$ lie inside a hyper-cone as schematically depicted in two dimensions here as "Your future". This is also called your forward light cone.

They all have $t>0$ and obey the inequality:

$x_1^2+x_2^2+x_3^2 < c^2t^2$.

That is,

$s=x_1^2+x_2^2+x_3^2+x_4^2 < 0$.

Because this is a Lorentz-invariant scalar, any train (all reference frames) agree that $s'<0$ which means that they also think $r^2 < c^2t'^2>0$ in your forward light cone. That is to say, all possible trains agree that the points in your future life cone really are going to happen at a time which is *after* the flash. The displacement vector from the origin to a point in the future light cone is said to be a time-like interval.

By a similar argument, all observers agree that any of the events in the area designated as "your past" would have happened before the flash.

This is not the case for points in the area labelled "present". None of these points could be reached in a train (travelling less than the speed of light). 4-vectors from the origin to points in this region are called space-like. In fact events in this region, depending on which reference frame we took, might happen:

  • At an earlier time,
  • at a later time,
  • at the same time.

If all intervals were like this, our notion of one thing causing another--e.g. an explosion being followed by the collapse of a building--would be problematic. But the time-like intervals rescue our idea of causality: An event at the origin can "cause" things in the forward light cone to happen, because all observers agree on the time ordering of events.

Image credits

Race Gentry