How is four-velocity automatically normalized? This is page 31 from Sean Carroll's Spacetime and Geometry. There is a line above eq (1.100) on this page which says that "the four-velocity is automatically normalized". This absolute normalization is a reflection of the fact that the four-velocity is not velocity through space,which can of course take on different magnitude, but a velocity through spacetime through which one travels at the same speed.

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*What is the absolute normalization here?


*How four-velocity is automatically normalized?


*Is four velocity always a unity?

 A: Written out, the four-velocity's four components are $\frac{dt}{d\tau}$ and the 3-vector $\frac{d\vec{x}}{d\tau}$. Note that $\frac{dt}{d\tau}=\gamma$, and $\frac{dx_i}{d\tau}=\frac{dx_i}{dt}\frac{dt}{d\tau}=\beta_i\gamma$, where we set $c=1$ as usual.
When we take the "length" of this vector under the Minkowski metric $\eta_{\mu\nu}$, we get (using the signature $(-,+,+,+)$ as Carroll uses here):
\begin{align}
\eta_{\mu\nu}U^\mu U^\nu&= -\left(\frac{dt}{d\tau}\right)^2+\left(\frac{d\vec{x}}{d\tau}\right)^2\\
&=-\gamma^2+(\beta_x^2+\beta_y^2+\beta_z^2)\gamma^2\\
&=-\gamma^2(1-\beta^2)\\
&=-1
\end{align}
where the last line is derived from the fact that $\gamma=\frac{1}{\sqrt{1-\beta^2}}$. So the "length" of this 4-vector under the Minkowski metric is always $-1$ under these conventions. (If you were to use the other convention for defining the metric, with signature $(+,-,-,-)$, then the "length" of this 4-vector would be $1$). 
It is "automatically" normalized because its normalization comes directly from its definition. The relationship between proper time and time in other frames guarantees that when you change frames, any change in the four-velocity's length from time dilation is exactly cancelled by an equal and opposite change in the four-velocity's length due to length contraction.
