# Question about the norm of the four-velocity being equal to $c$

On the way to the Einstein equation we derived the four-velocity: $$u^\mu=(c,v^k)$$ with $$v^k$$ being the 3-velocity, which can can be very low ($$|v|<). However, the square of the four velocity is $$\eta_{\mu\nu}u^\mu u^\nu=\gamma^2(c^2-v^1v^1-v^2v^2-v^3v^3).$$ [We use the diag($$+---$$) Minkowski metric.] $$\eta_{\mu\nu}u^\mu u^\nu=\gamma^2(c^2-v^2)$$ $$v=|v|$$ ($$|v|<). $$\eta_{\mu\nu}u^\mu u^\nu=\frac{1}{1-\frac{v^2}{c^2}}(c^2-v^2)=c^2$$ Since it is a tensor equation it is true for all moving systems, for example also for the driver of a car which moves at 10km per hour. How can it be grasped that this driver has a 4-speed of $$c$$?

• The four-velocity does not have any natural normalization. The condition $u^\alpha u_\alpha = c^2$ is a normalization choice. Commented Nov 8, 2023 at 18:26

One easy way of grasping it is to say that an observer at rest moves forward along the time direction at lightspeed, while the driver tilts their velocity vector so it has a spatial component. It turns out that four-accelerations are spacelike and orthogonal to the four-velocity, allowing objects to stay on-shell and the velocity to remain $$c$$.