Given the way objects move, they seem to be going all the same "velocity" so to speak, that velocity being the speed of light. Except, velocity is displacement/time, so if something goes faster, the displacement is equally balanced by the reciprocally changing time factor so that it always balances out to 299,792,458 m/s. Well, that's the way it appears to be. The faster something displaces, the slower time moves for it; the slower something displaces, the faster time moves for it. These two factors always balance out to equal $c$.

Why is this? Is there something obvious and mathematical I'm missing about it, or is this a little deeper? Explanations perhaps? Deeper insight on this factor? It seems very interesting.

Why is it that spacetime is always $c$? Is it simply what we use to observe it, and anything we observe it with is always traveling at $c$....jeez it's so freaking weird.

I could be completely wrong though; please clarify if so, but honestly, $E = \frac{mc^2} {\sqrt{1-\frac{v^2}{c^2}}}$ – it's freakin' weird.

  • $\begingroup$ Yes, as a keen popular-science reader, I can confirm your intuition is totally correct! To explain poorly, in a soundbite, "we're moving through spacetime at c". If you're "moving" (in the commonplace sense) it's like the angle of your line tilts more upwards. You can only either be flat or straight up or in between. (Now there's precision in science for you.) I can direct you to a somewhat unusual book that explains this very well indeed: purchase amazon.fr/gp/product/B00JH02G88 Hope it helps! $\endgroup$
    – Fattie
    Jul 22, 2014 at 9:28
  • $\begingroup$ "Deeper insight on this factor?" It's amazing that the whole mysteries-of-time-and-space thing comes down to pythogras' theorem .. that's where the squares come from on the diagonals. $\endgroup$
    – Fattie
    Jul 22, 2014 at 9:30
  • $\begingroup$ @JoeBlow: I haven't read Andrew Thomas' third book but in his second book (about gravity) he comprehensively misunderstands the Big Bang, which doesn't bode well. $\endgroup$ Jul 22, 2014 at 9:49
  • $\begingroup$ Hi JR. Thanks a lot for that comment; Did you happen to read his first book. Funnily enough I was going to ask on here, essentially: "Do real physicists think Andrew Thomas is plain whacky??" My take: IMO, setting aside the "possibly whacky theories" part of Andrew Thomas. As a connoisseur of Popular Science [Particularly Cosmology, Mysteries Of Space And Time, Etc] - funnily enough I feel his general explanations of relativity, quantum matters, and the rest -- are about the best out there. Every popular science book includes the obligatory Explanation of Modern Physics.... $\endgroup$
    – Fattie
    Jul 22, 2014 at 10:27
  • $\begingroup$ ....for me, his are outstanding: clear, clear-minded, crisp ... Dawkins-like. Again, this is very strictly completely setting aside the "whacky theories" aspect, which, essentially, I'm not important enough to comment on. Thanks again. $\endgroup$
    – Fattie
    Jul 22, 2014 at 10:29

1 Answer 1


Actually you're quite correct, though possibly not in the way you expected. Ordinary velocity isn't an invariant because obviously different observers moving at different speeds will measure different velocities. However there is an invariant form of velocity called the four velocity that is an invariant under special relativistic (i.e. Lorentz) transformations. The magnitude of the four velocity is always $c$ - the speed of light.

The reason the magnitude of the four velocity is always $c$ is because it measures the speed at which we move through time as well as the speed with which we move through space. This may seem strange because surely the speed we move through time is always one second per second - and indeed it is as long as you're measuring your own velocity in your own inertial frame. However if you're measuring the velocity of someone moving relative to you then time dilation will slow their clocks. If $t$ is their time and $\tau$ is your time this means $dt/d\tau$ is not equal to one second per second i.e. the rate at which they move through time has changed.

The end result is that if someone is moving relative to you, i.e. $dx/d\tau \ne 0$, then $dt/d\tau < 1$ and the two effects balance out to keep the magnitude of the four velocity at $c$.

  • $\begingroup$ I have seen people getting confused about this. The magnitude of the four velocity of any massive object is c (the speed of light). But the magnitude of the four velocity of light is zero. $\endgroup$
    – MBN
    Jul 22, 2014 at 10:07
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    $\begingroup$ @MBN: is the four velocity of light well defined? I've had the impression it isn't because you can't differentiate wrt the proper time. $\endgroup$ Jul 22, 2014 at 10:11
  • $\begingroup$ I see. That's a good point. I thought that the four velocity is the tangent vector whatever the parametrization of the world line may be. And the normalization for timelike curves is just a matter of choice. I guess I didn't know the definition of four velocity. $\endgroup$
    – MBN
    Jul 22, 2014 at 10:51
  • $\begingroup$ @JohnRennie a strange question: Is 'invarient' a correct way to spell what I know as 'invariAnt'? $\endgroup$
    – Danu
    Jul 22, 2014 at 12:29
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    $\begingroup$ @Danu: damn, I always get that spelling wrong. Thanks :-) $\endgroup$ Jul 22, 2014 at 13:58

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