Q:

4-vector velocity to Newtonian?

The four-vector condition for a particle free of forces is:
$\frac{du}{dτ} = 0$
and the equivalence of this to the statement of newton's first law follows from the expression for four-velocity:

Then I came across this part in my lecture notes:

Now this got me curious. How exactly does one get to the Newtonian statement i.e. what algebra steps are involved? Thank you to anyone who can enlighten me~

A:

First consider the spatial component of the equation above, you get
$$
\frac{d}{d\tau} \left( \gamma(v)\overrightarrow{v} \right) = 0
$$
But $\gamma(v) = \frac{dt}{d\tau}$ so you have
$$
\frac{d^2 t}{d\tau^2} v + \gamma(v) \frac{d\, \overrightarrow{v}}{d\tau} = 0
$$
Now since
$$
\frac{d \, u}{d\tau} = 0 \implies \frac{du^0}{d\tau} = \frac{d^2 t}{d\tau^2} = 0
$$
we finally get
$$
\frac{d \, \overrightarrow{v}}{d\tau} = 0 \implies \frac{d \, \overrightarrow{v}}{dt} = 0 
$$
Which is your result. Of course, we should remember that $\gamma(v) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

