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Too many text books (in fact, all of them that I've found including 'Gravity'), just throw out the term Four Velocity without digging into what it means, exactly. I understand $\frac{dx}{dt}$, but I don't understand how you can take the derivative of time against time, $\frac{dt}{dt}$. I mean, that's 1, isn't it?

So looking at the symbols a little closer, it appears that the components are actually $$\frac{dx}{d\tau}.$$ That is, it's the derivative of normal space to proper time. So then first component of the 4-Velocity vector is:$$\frac{dt}{d\tau}$$I'm guessing that's the ratio of the observer's time to the proper time?

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    $\begingroup$ I would like to suggest taking some to to consider the full implication of the phrase "time $t$ is a coordinate in SR". While time $t$ is a (universal) parameter in Newtonian mechanics, proper time $\tau$ (along a world line) is a parameter in relativistic mechanics. $\endgroup$
    – Alfred Centauri
    Commented Jun 14, 2018 at 1:37
  • $\begingroup$ You might want to indicate which Gravity book you're reading, it's not a very specific name. $\endgroup$
    – Javier
    Commented Jun 15, 2018 at 18:12

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That's right, but you can also think of the four velocity as just the velocity vector with a special parameter. A trajectory in spacetime is an assignment of a spacetime point $x^\mu(\tau)$ (remember this is $(ct, x, y, z)$) for each proper time $\tau$. The four velocity is just the derivative of this, that is, the velocity vector: $u^\mu = dx^\mu/d\tau = (d(ct)t/d\tau, dx/d\tau, dy/d\tau, dz/d\tau)$.

Its first component $u^0 = c dt/d\tau$ measures the rate of change of coordinate time as a function of proper time, and it is always greater than or equal to 1.

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  • $\begingroup$ Isn't the first component $\frac{d(ict)t}{dr}$? $\endgroup$
    – user32023
    Commented Jun 14, 2018 at 9:45
  • $\begingroup$ @MikeDoonsebury It is if you use the convention where the first coordinate is imaginary time, but no one does that anymore. We prefer to directly say that the interval is $s^2 = -t^2 + x^2 + y^2 + z^2$ instead of using imaginary numbers to get that minus sign. $\endgroup$
    – Javier
    Commented Jun 14, 2018 at 16:23
  • $\begingroup$ How does just changing the sign on a square change the physical reality? I've never understood why the square of spacial distances add to the total distance and time distances subtract. $\endgroup$
    – user32023
    Commented Jun 14, 2018 at 18:37
  • $\begingroup$ @MikeDoonsebury you're basically asking me to explain the mathematical foundations of special relativity, which certainly won't fit in this comment; consult any textbook on the subject. The simple fact is that Lorentz transformations leave $s^2$ invariant and, conversely, those transformations that leave $s^2$ invariant are exactly the Lorentz transformations. $\endgroup$
    – Javier
    Commented Jun 14, 2018 at 18:48
  • $\begingroup$ @MikeDoonsebury trying to understand a new physical model in the setting of the established theory doesn't always make sense. Instead, fully embrace the new theory as a mathematical model, and then ask the question of - how the old familiar setting of Newtonian mechanics arises in a certain limit. Asking why a postulate of Special Relativity is what it is doesn't really carry a lot of meaning - it just is, and the justification is that it simply works. $\endgroup$
    – Prof. Legolasov
    Commented Jun 15, 2018 at 7:22

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