2
$\begingroup$

Ordinarily in Newtonian physics, velocity is defined as $${\vec v}=\frac{d{\vec x}(t)}{dt}$$ where we use the coordinates of an observer and the universal time $t$.

When we dive into special relativity, we define the spatial components of the four-velocity as $$v^i=\frac{dx^i(\tau)}{d\tau}$$ instead of $$v^i=\frac{dx^i}{dt}.$$

  • But in the usual definition (second equation), are we not using a hybrid system of spacetime coordinates? I mean, while $x^i(\tau)$ are the coordinates measured by the stationary observer, $\tau$ is the proper time of the moving observer.
Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Would you prefer that we use $$U^i = \frac{\text{d}x^i}{\text{d}s},$$ where $s$ is the invariant interval and agreed upon by everyone? ;) $\endgroup$
    – Philip
    Commented Nov 21, 2020 at 20:41

1 Answer 1

Reset to default
2
$\begingroup$

In Newtonian mechanics, $t$ is the time for the object that is moving, as it is in SR. The difference is that any inertial frame believes that it is also their own time. It is an approximation that is corrected by SR.

Only after that correction some important quantities are conserved. For example, momentum before and after a collision, as observed by different inertial frames. If defined as $m\frac{d\mathbf x}{dt}$, it is not conserved (we think it is because discrepancy for usual velocities is completely neglible).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.