# How is four-velocity automatically normalized?

This is a page from Sean Carroll's Spacetime and Geometry.There is a line in 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.

1. What is the absolute normalization here?

2. How four-velocity is automatically normalized?

3. Is four velocity always a unity?

• He's using natural units here, normally that $1$ would be a $c^2$ – Craig Jan 17 '19 at 15:08
• Possible duplicates: physics.stackexchange.com/q/410603/2451 and links therein. – Qmechanic Jan 17 '19 at 15:12
• Please don't post questions using images of text. It breaks search functionality and adaptive technology for the blind. – user4552 Jan 17 '19 at 15:57

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 $$1$$).