1
$\begingroup$

In special theory of relativity, why is 4-velocity defined as:

$$ u^\mu = \frac{dx^\mu}{d\tau} $$

and not as $$ u^\mu = \frac{dx^\mu}{dt} $$

where ${\tau}$ is proper-time and t is time in some other observers' frame of reference. What does derivation with respect to invariant (like proper-time here) have to do with contravariant components of the vector? I'd prefer intuitive explanation rather than strict mathematical.

$\endgroup$
3
  • 5
    $\begingroup$ The latter is not a 4-vector. To butcher the math, you are "dividing" a 4-vector with a non-invariant quantity, $dt$. $\endgroup$
    – knzhou
    Jul 1, 2016 at 0:29
  • $\begingroup$ What's wrong with dividing the 4-vector with non-invariant quantity? Why that doesn't make it another 4-vector? $\endgroup$
    – matori82
    Jul 1, 2016 at 12:35
  • $\begingroup$ The length of the vector can change under a Lorentz transformation. $\endgroup$
    – knzhou
    Jul 2, 2016 at 0:13

3 Answers 3

1
$\begingroup$

Proper time is the reference time all observers can agree on. Makes for calculating things much much easier.

Since tau is attached to the particle the space differentials are zero

Imagine we used some other reference time (which you are more than welcome to do). Then the spacetime line element would have dx, dy, and dz not equal to zero.

Furthermore using a different frame would mean we would have to transform everything from one frame to another. Yikes!

Proper time is simply chosen to avoid all these extra nastynesses.

Hope that helps

$\endgroup$
1
  • $\begingroup$ Why would we "have to transform everything from one frame to another" in the first place? Basically my question is why is it an issue if the derivative is observer dependent? $\endgroup$
    – mdcq
    Feb 12, 2018 at 23:23
1
$\begingroup$

4-vectors have invariant length as defined by $$\vec{v}\cdot \vec{v} = g_{ab}v^a v^b.$$

Coordinate velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}t}$ does not have this property. Proper velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}\tau}$ does. Coordinate velocity is defined. It's not a 4-vector, so it's not that useful.

$\endgroup$
1
  • $\begingroup$ Can you show, why it wouldn't be a four vector? And why is it only useful as a four vector? $\endgroup$
    – mdcq
    Feb 12, 2018 at 23:23
0
$\begingroup$

Given a path $x(s)$ on a manifold, the velocity with respect to that path is defined taking derivatives with respect to the invariant parameter the path is described with, on the manifold (the arc length). In general relativity such parameter is the proper length $s$ (or proper time $\tau = \gamma s$).

$\endgroup$

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