Supersonic Man

May 17, 2016

let’s try something easier: special relativity

Filed under: Hobbyism and Nerdry — Supersonic Man @ 5:21 pm

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.

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 I 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 5i), and the sum of those is -16.  The square root of -16 is 4i, 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 silly — of the two, space is definitely 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 drag the viewpoint 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 a distance of 7.5 light seconds over a time of 12.5 seconds (7.5² + 12.5i² = 56.25 – 156.25 = -100 = 10i²).

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.  8.5i² is -72.25, and 56.25 – 72.25 is -16, or 4i².  Four seconds, just as it was 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 just over a third of the way through it.  Poof, there’s your nonsimultaneity: events that happened at the same moment from my view happen 2.25 seconds apart from yours.

(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 0.6c, then piled another 0.6c 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 tines 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 = mc², the formula which says that energy weighs something — it has inertia and gravity.  But I think that 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.



  1. 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. 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. 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. The formula is γ = 1 ∕ √1 – |v|²∕c², 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 mass is the rest mass (m) times gamma.

    If people ever really do fly around in relativistic spaceships, they’ll probably 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 ½mv² 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)mc². This approximates ½mv² at lower speeds, but there’s no clean way to express it in terms of fake speed.

    Comment by Supersonic Man — May 24, 2016 @ 12:44 pm | Reply

    • Yeah, unlike other equations which still work if you define the terms right, the Newtonian kinetic energy formula ½mv² is actually wrong in its overall shape. It’s 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. ½γmv², but that is still different from (γ-1)mc².

      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 ½mv² formula really just doesn’t apply.

      Comment by Supersonic Man — May 24, 2016 @ 11:41 pm | Reply

      • For a while I was trying to make the old ½mv² formula work by integrating over a varying m, but it still doesn’t fly. The basic idea was to allow for the fact that the kinetic energy it’s already got adds inertia to any further acceleration. But the way that energy goes to infinity as v approaches c isn’t fully accounted for by that — the cumulative effect of inertia feedback pushes the energy higher, but not towards infinity.

        Comment by Supersonic Man — July 24, 2016 @ 12:26 pm

  2. 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 ΔS 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 Δx say (xb – xa).

    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 this interval measures a combined spacetime distance 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.
    the Einstein Field Equation

    Comment by Supersonic Man — June 23, 2016 @ 9:14 am | Reply

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