Is the magnitude of the four-velocity vector in spacetime arbitrary?

Question

I know that the squared magnitude of the four-velocity vector is plus or minus $c^2$, but I’m a little confused on whether or not this vector has been normalised arbitrarily, since it is often claimed that we “move through spacetime at the speed of light”, and I have heard conflicting answers on this.

Also, depending on the sign convention of the metric the squared magnitude can be $-c^2$, meaning the magnitude would be $ic$ and not $c$ (if I am correct). So it seems to me that the four velocity magnitude is just arbitrary and has no physical meaning.

So I am wondering if saying that we “move through spacetime at the speed of light” is something genuinely derived or just a matter of definition, or perhaps not even accurate at all.

Answer 1

i am wondering if saying that we “move through spacetime at the speed of light” is something genuinely derived or just a matter of definition

I would say that the truth is somewhere in the middle. The four-velocity is defined as $$\frac{dx^{\mu}(\tau)}{d\tau}$$ We can look at this expression from the perspective of Newtonian physics. The top is a change in position and the bottom is a change in time, so a change in position divided by a change in time is a velocity. Since these are relativistic quantities it makes sense to think of this expression as the relativistic generalization of velocity.

We can also look at this expression from the perspective of geometry. Since $d\tau=\sqrt{dt^2-(dx^2+dy^2+dz^2)/c}$ it can also be looked at as the relativistic generalization of an arc length. Geometrically when you divide a change in position by a change in arc length you get a unit tangent vector.

So what we naturally think of as the relativistic generalization of velocity is also a relativistic generalization of a unit tangent vector. These properties are then perhaps a bit unsurprising. It is a unit vector, so of course the length is always the same, and it should be unsurprising that the length of the unit vector is c given how frequently we set c=1.

So overall it is a straightforward consequence of the mathematical framework. The four-velocity is a unit vector and unit vectors have unit length. It certainly strengthens the argument for considering relativity in terms of geometry, but the fact that a unit vector has unit length is otherwise not as insightful as some pop-science authors want to make it seem. The important geometrical insight is typically given short shrift in such works.

Answer 2

The four velocity is defined as tangent vector to a curve you track as you are moving through spacetime (i.e. into the future). This makes sense, because this vector tells you where in spacetime are you moving next, which is the same meaning as velocity has in Newtonian physics.

The curve is moreover parametrized by your own clocks and this parametrization tells you the magnitude of the tangent vector. Now, because of principle of relativity the clocks ticks at any place on the trajectory and for all clocks working on the same principle must all be equivalent. Therefore the squares of all 4-velocities, which are invariants, should also be equivalent. The value might depend on some conventions (like units of measurement, signture of metric) but once it is fixed, it must be same for all of the objects living in spacetime that have clocks based on the same principle - which is every massive object.

For the light, there are no such clocks and thus the magnitude can be different.