This is not some kind of metaphysical, philosophical or metaphorical statement, it is actually true. It is an implication of special relativity that I think is pretty interesting and pretty cool, and I hope you'll think so too. It will take a little while to get to the point here, though.

We normally think of space and time being distinct things, independent and fundamentally different in a way that we can't really describe. We think of space being three-dimensional (x, y, and z), and time being one-dimensional (t). We describe an object by its spatial coordinates at a point in time, and then it may be at another set of coordinates at another point in time. We could say that at t=1 an object is at 〈1,2,3〉, and at t=2 it is at 〈2,4,6〉. It seems like a small step to just bring the time dimension inside our space vector and make a spacetime vector, 〈t,x,y,z〉, but when this is done, the time coordinate must be multiplied by c (the speed of light) and i (square root of -1, this comes up later) so that the elements of the vector have the same units (distance). This makes the spacetime vector 〈ict,x,y,z〉, and we denote this vector by R .

If we want to look at the "speed" at which something moves through spacetime, it will be the derivative of R with respect to "proper time", denoted by τ (Greek letter tau). For some background on this, the special relativity node has some good information, but the bottom line is if person 1 is standing still and watches person 2 move past at a constant speed on a train, they see person 2's ticking more slowly than their own. Person 2, who is on the train, will see their clock ticking normally, but will see person 1's clock ticking slowly. How slowly the two clocks appear to tick depends on how quickly the train is moving; the faster it is moving, the more slowly the clocks will appear to tick. Proper time is time it takes for one tick of either person's clock from their own perspective, and if τ is person 2's proper time, then person 1 sees person 2's clock taking time t to tick, which is related to τ through γ (Greek letter gamma) as such: t = τ × γ. Equivalently, τ = t/γ. Gamma depends on person 2's speed relative to person 1. This relationship between τ and t makes it far easier to find a meaningful derivative of R , since we can turn the derivative with respect to an external proper time into a derivative with respect to the observed time of the object.

So after that digression, we can find the "speed" of the spacetime vector with respect to proper time, which is denoted by U . We differentiate the spacetime vector with respect to proper time:

U = d R /dτ

= d R /d(t/γ)

= γ × d R /dt

= γ〈ic, dx/dt , dy/dt , dz/dt〉

= 〈γic, γ(dx/dt) , γ(dy/dt) , γ(dz/dt)〉.

This is the spacetime "speed" in vector form, but the magnitude of this vector is a little different from the magnitude of a spatial 3-vector. The magnitude of a spatial vector 〈x,y,z〉 is found by s 2 = x 2 + y 2 + z 2 . This is a geometric invariant, which basically means that you can rotate or move this vector however you like, or change the coordinate system, and its magnitude (length in this case) will stay the same. Similarly, our spacetime 4-vector 〈ct,x,y,z〉 should be invariant through a change in inertial frames of reference, so that all observers will record the same magnitude of the vector, even if they see the components differently. The magnitude of the 4-vector is given by:

s 2 = (ict) 2 + x 2 + y 2 + z 2

= -(ct) 2 + x 2 + y 2 + z 2

It is conventional to reverse the sign of the right hand sign, making it s 2 = (ct) 2 - x 2 - y 2 - z 2 . This is also a geometric invariant, and the same form of the magnitude will be applicable to the derivative of the vector.

Now, the definition of γ is (1 - (v/c) 2 ) -1/2 = c(c 2 - v 2 ) -1/2 , so γ 2 = c 2 (c 2 - v 2 ) -1 , v being the speed of the object being observed.

So, finally, we can get an expression for the magnitude of the derivative of the spacetime vector, denoted by s:

s 2 = (γc) 2 - (γ(dx/dt)) 2 - (γ(dy/dt)) 2 - (γ(dz/dt)) 2

= γ 2 (c 2 - (dx/dt) 2 - (dy/dt) 2 - (dz/dt) 2 )

= γ 2 (c 2 - v 2 ), since v 2 = (dx/dt) 2 + (dy/dt) 2 + (dz/dt) 2 .

= (c 2 - v 2 ) × c 2 (c 2 - v 2 ) -1 , by substituting in the expression for γ.

= c 2

Therefore s = c, regardless of where you are or how quickly you are moving†.

Special relativity, which has been verified experimentally and is 100% bona fide serious business, predicts that when an object is moving more quickly, outside observers will see time moving more slowly for that object. Strange things happen at high speeds, where space and time stop being independent and become mixed up together, and each influences the other in weird ways. So I interpret this result as saying that when your spatial speed increases and your time speed decreases, they are compensating for each other so that their sum remains the same, and your speed in spacetime is always at its maximum no matter what you do. I don't know enough to give a more satisfying or in-depth analysis of what all this means, I just think it gives a bit of insight into how special relativity works.

† Technically this should say that s = ±c, but I'm not really sure what that would mean. It might have something to do with the symmetry of spacetime, but I don't know nearly enough about that to talk about it, so I've just taken the cowardly way out and only mentioned it in this tiny text at the bottom.

Many thanks to Gorgonzola for correcting me on the geometric invariant, to abiessu for proof-reading it for me, and to raincomplex for correcting my spelling.

This writeup is based on the special relativity module of a physics course I did last year, which I enjoyed greatly. Still, I'm not a scientist, so if anyone out there see a mistake or anything here, please let me know, I don't want to be misinforming anyone.