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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|<<c$). 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|<<c$). $$\eta_{\mu\nu}u^\mu u^\nu=\frac{1}{1-\frac{v^2}{c^2}}(c^2-v^2)=c^2$$ 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 $c$?

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

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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 $c$.

...except that saying a velocity through time has a certain value is pretty incorrect! Velocities as we know them are distance traversed per unit of time, so speaking of a velocity through time makes violence to how we normally use the word. One way out that is philosophically and mathematically fine is to just say that 4-velocities are not at all like our everyday velocities and should be treated as a very different kind of object that just happens to have an ordinary velocity as a part of itself (plus the time component, which is something else).

In the spacetime picture worldlines are parametrized by the proper time, and the 4-velocity is the change in position per unit of proper time. That makes the constancy of 4-velocity easier to grasp: it is like how for a curve in 3D space parametrized by arc length the tangent vector is constant in length but changes direction.

So the driver is just driving along a straight line in spacetime that is slightly tilted compared to an observer at relative rest. The 4-velocity points along this line, and has 3D components corresponding to the spatial velocity and a time component linked to the time dilation.

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  • $\begingroup$ Hi, Anders, thanks for your answer. The last two and a half sections of your answer solved the problem. (To the admins: It says to avoid comments like "Thanks" - how can I be polite without a thanks to the person who answered my question?? Am I supposed to be rude?) $\endgroup$
    – Fuzzy
    Commented Nov 8, 2023 at 14:10
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    $\begingroup$ @Fuzzy - Thanks! I think the standard approach is just upvotes if the answer is useful. The format of the site does not lend itself to the politeness we would have if we had a free-form conversation. $\endgroup$ Commented Nov 8, 2023 at 14:19

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