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:

enter image description here

Then I came across this part in my lecture notes: enter image description here

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~


1 Answer 1


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 $\gamma(v) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$


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