# let’s try something easier: Special Relativity

Here’s a way of explaining special relativity which I wish I had run across at a much younger age. Thanks to author Greg Egan for finally getting this through my dense headbone. For the first time, I kind of feel like all that stuff about time dilation and increased mass and so forth makes some sense.

Let’s briefly review what the claims of special relativity are:

- As an object moves faster, its length contracts
- Also, time passes more slowly for that object
- These effects are relative to viewpoint: to that object, it’s you who are shorter and slower
- Whether two distant events are simultaneous depends on the observer’s velocity
- Nothing can move faster than light
- If it tries, it just gets heavier
- And mass is equivalent to energy by the relation
*E*=*mc*².

So here’s the idea:

Spacetime can be treated as four dimensional, with three dimensions in space
which are all at right angles to each other, and one dimension of time which is
at right angles to all three of the dimensions of space. Distances in space
and time can be expressed in comparable units by stating them as fractions of
the speed of light (*c*), so one light-second of distance corresponds to
one second of duration. So far, nothing we haven’t heard before...

The gotcha with this four dimensional scheme (at least, in the way I can best
understand it) is that distances in space are real numbers, but distances in
time are imaginary numbers. You can treat one second of time mathematically
as being like one light-second of space (about three quarters of the distance to
the moon) multiplied by *i*, the square root of negative
one. And here’s what that implies:

When you combine a distance in space with a duration in time, the total amount
of spacetime traversed can be found with Pythagoras’s theorem, because the time
dimension is at right angles to any spatial movement. Now in space,
Pythagoras says that if you go three paces left and four paces north, your
distance from the starting position is five paces, because 3² +
4² = 5². And this same rule applies to spacetime, but the twist is
that, because distances in time are imaginary numbers, the square of such a
number is negative. And this means that if you visualize the spatial
movement, the time required, and the diagonal path combining them as a right
triangle, you end up with a situation where the hypotenuse is *shorter*
than at least one of the sides — a situation impossible to draw on
paper. For example, if you travel a distance of three light-seconds from
where I’m standing, in a duration (from my viewpoint) of five seconds, then the
total spacetime distance you cover comes out Pythagoreanically as follows: the
square of three light-seconds distance is 9, and the square of five seconds of
time is –25 (because as a distance that duration is the imaginary number
5*i*), and the sum of those is –16. The square root of –16
is 4*i*, or a duration of four seconds of time.

And hey, there’s your time dilation, as advertised: while I lived five seconds, you only lived four. If you turned around and came back at that same colossal speed, we’d meet again with my watch showing ten seconds gone and yours showing only eight.

(You can also arbitrarily choose to consider time as real and distance as imaginary, and it works out the same way. But that would be kind of silly — of the two, I would have to say space is the more real.)

This is supposed to be all relative. How does it look from your travelling viewpoint? In your frame of reference during the outbound leg, you sit still for four seconds, and I recede from you at three fifths of the speed of light. So the way you see it, we don’t end up separated by 3 light-seconds, but only by 2.4 light seconds. And bam, there’s your foreshortening of distance — the Lorentz contraction.

Now what about the return phase? Well, you can’t just change horses and
drag the viewpoint of reference back with you. Everything is relative, but
not that relative. To finish working this out, we have to stick with the
same frame of reference: the continuation of your outbound path. From that
viewpoint, I continue receding at 0.6 *c*, and you turn around
and chase after me. How fast? Well, let’s figure out where the
endpoint is in this view. We know that the total spacetime length of my
path is ten seconds. That works out to me receding over a distance of 7.5
light seconds for a time of 12.5 seconds (7.5² + 12.5*i*² = 56.25 – 156.25 = –100 = (10*i*)²).

Your return trip has to meet that point setting out four seconds later than my
path did. How long is that return journey? It has to cover 7.5
light-seconds in 8.5 seconds. 8.5*i*² is –72.25, and 56.25
– 72.25 is –16, or 4*i*². Your catch-up journey takes you
four seconds, just as it did from the first viewpoint — the total spacetime
distance does not vary, though the spatial and temporal components of it do.

Note that what we previously considered the midpoint of the journey is now less than a third of the way through it. Though your own clock on the return trip showed just four more seconds gone, from the outbound viewpoint that we started with, your return took 8.5 seconds. At the time I’m halfway through waiting for you to come back, you’ve already been underway for 2.25 seconds, whereas from my view you are just turning around at that time. Poof, there’s your nonsimultaneity: events that happened at the same moment from my view happen at quite different times from the outbound viewpoint.

If you pick the return journey as your viewpoint, you get those same numbers swapped end for end: you rush to the turnaround point in 8.5 seconds and then wait four seconds for me to catch up.

How fast is the return trip? 7.5 light-seconds in 8.5 seconds is about 88%
of lightspeed. But from my viewpoint, you decelerated by
0.6*c*, then piled another 0.6*c* on top of
that. Somehow, 0.6 + 0.6 adds up not to 1.2, but to 0.88235...abracadabra,
there’s your limitation on velocity. If you add speed to a fast moving
object, it doesn’t keep getting faster without limit: the faster it’s already
going, the less its speed increases with any further acceleration.

But the momentum and kinetic energy increases with every push (here’s nothing relative about the conservation laws for those quantities) and since momentum is mass times velocity, if velocity can’t increase then mass must be larger instead. Alakazam, there’s the relativistic mass increase.

(Some say that we should not speak of the object having increased mass — that this is in some sense not real mass. Einstein himself recommended just treating the momentum and energy as the well-defined values, and not trying to define what “mass” means in that context.)

At this point we’re not far from *E* =
*mc*², the formula which says that energy weighs something — it
has inertia and gravity. But to get to that last step is tricky, and I
think the above covers enough ground to get the idea across. I’m not going
to work out every detail.

The final important point to note is that Einstein did not just pull an imaginary-numbered time dimension out if his ass and then work through the consequences: what he did (building on the work of Lorentz) was to take the observed fact that the speed of light never varied with one’s point of view, and work out what the nature of time must be in order for that to be the case.

So you could say that the fundamental formula for special relativity, from which the rest follows, is

ΔS² = Δx² + Δy² + Δz² –cΔt²

where *x*, *y*, and *z* are the three
coordinates in space, *t* is the coordinate in time, Δ means the
difference between the coordinates of two locations, and Δ*S* is the resulting “spacetime interval”.

Alternatively, for two points in spacetime (events) denoted *a*
and *b*, some people prefer to express each coordinate explicitly
as a difference, like instead of Δ*x* say
(*x _{b}* –

*x*).

_{a}This formula does count distances in time as imaginary relative to distances in space. In essence, the theory of special relativity consists of the assertion that the spacetime distance measured by this interval is the value which remains unchanged regardless of an observer’s frame of reference.

But this formula wasn’t written by Einstein. It from Hermann Minkowski, a former teacher of Einstein’s whose response to the special relativity theory was to reframe it in terms of geometry. People trying to solve special relativity problems often do so using Minkowski Diagrams based on this formula. It’s usually considered a better way to understand the theory than Einstein’s original approach was. And yeah, it’s what works for me.

Apparently general relativity can also be boiled down to a single one-line formula, but only if you take pages of explanation to define the terms in it. Many of the letters stand for complex entities known as tensors. If you expand those terms you get ten partial differential equations. Here is the tensor form of it:

Let’s take a little look at how to use special relativity in practice. To
generalize from the examples above and give yourself practical formulae, you
need to use the dilation factor, commonly called gamma
(*γ*). This is much easier to define and describe if you go
by its reciprocal, which is sometimes called alpha (*α*), though
not usually. This reciprocal is a number between zero and one — it’s one
when the subject in question is stationary and approaches zero as the subject’s
velocity approaches lightspeed. The relation is that the speed (as a
fraction of c) squared, plus alpha squared, equals one. So if you graph
it, you get a quarter circle — a nice clear geometric relation. If you
graph gamma itself, you get a curve which looks kind of circular at first and
then shoots up to infinity, with no intuitively obvious shape to it. So
the formula is *γ* = 1 ∕
√1 –
|*v*|² ∕ *c*², where |*v*|
is the object’s speed (the magnitude of its velocity vector).

The rate which time progresses in the moving frame of reference is the
stationary rate divided by gamma. The foreshortened length along the axis
of movement is the stationary length divided by gamma. The apparent mass
is the rest mass (*m*) times gamma.

If people ever really do fly around in relativistic spaceships, they might want
to make use of a “false velocity”, which is the real velocity times
gamma. In these terms acceleration adds up in a linear way: like, if you
accelerate at 1/400 *c* per day for 800 days, your false speed is
2*c*, and the distance you cover across the galaxy is, as measured by your
local clocks but by the galaxy’s stationary ideas of distance, two light days
per day. Framing it this way makes sense in human terms because you’ll be
looking at a map of stars which says your destination is some number of light
years away, and you’ll be looking at your local calendar and wanting to know
how many days of your own time it’ll take to get there.

This fake speed gives you nice linear predictions of how fast you will cover
ground between two landmarks, given how much you accelerated. It correctly
states various properties such as your momentum (which is fake velocity times
rest mass)... but it doesn’t work for kinetic energy: if you use that fake
speed and your rest mass in the old ½*mv*² formula, you overestimate your
kinetic energy, but if you plug in your real speed, you underestimate
it. The real formula is just relativistic mass minus rest mass, expressed
as energy, or (*γ* – 1)*mc*². This
approximates ½*mv*² at lower speeds, but there’s no
clean way to express it in terms of fake speed. Unlike other equations
which still work if you define the terms right, the Newtonian kinetic energy
formula is actually wrong in its overall shape. It’s quite close at low
speeds (the error is barely one percent at 15% of lightspeed) and you can get
even closer by using the relativistic mass, i.e.
½*γmv*², but that is still different from
(*γ* – 1)*mc*².

To make it work, the “½” in that second approximate formula would have to be
replaced with a special correction factor: at a tenth of lightspeed it’s 0.501,
at half of lightspeed it’s 0.536, at three quarters it’s 0.602, at nine tenths
it’s 0.696, at 99% it’s 0.876, and at lightspeed (for massless particles) it’s
1. This is why I was never able to see how kinetic energy made sense when
I tried to grapple with relativity in the past — for slow particles kinetic
energy is half of momentum times speed, but for the fastest ones it’s momentum
times speed with no ½. So it wasn’t that I was failing at math: the
½*mv*² formula really just doesn’t apply. For a
while I thought I could make it work by integrating the change of apparent mass
over different velocities, but it didn't work out.