1
$\begingroup$

The usual definition for the four velocity of a worldline parameterized by proper time, $x(\tau)$, is simply $$U^{\mu} = \frac{dx^{\mu}}{d\tau}.$$

However, I've read in many places that the four velocity is only defined for a time-like worldline. Why can it not be defined for a null or space-like worldline?

$\endgroup$

4 Answers 4

3
$\begingroup$

First of all, there's no such thing as a spacelike world line. A world line is a succession of causally connected events, and you can't have causal connection between two events that are spatially separated. It's the line that traces a particle's motion through space time. If it could connect spatially separated events, then the effect could precede the cause.

You can think about it another way: consider two events of the world line, A and B, with their respective light cones. If B occurs some time later than A, then B's future lightcone is completely contained inside A's (and viceversa, A's past lightcone is contained inside B's past cone).

Same thing (not coincidental) happens with proper time, it's only defined for timelike separations. That's because the proper time between two intervals is the time measured in a frame of reference where the two events happen in the same place. That means that in that frame those two events have zero spatial separation and are therefore time separated. And given that the interval is Lorentz invariant, they are time separated in every frame of reference.

So, without a world line and without a notion of proper time, you can't have a 4-velocity.

$\endgroup$
3
$\begingroup$

The other answers are not wrong; a spacelike (or null) curve doesn't have a proper time, so you can't define the four velocity by $U^\mu = dx^\mu/d\tau$. I think your intuition is saying that you should still be able to define some kind of velocity, and that's right: simply parametrize your curve with some parameter $\eta$, and $V^\mu = dx^\mu/d\eta$ is your tangent vector.

The problem with this is that you can't normalize it. Requiring that $U^\mu U_\mu = -1$ is one way of defining what we mean by proper time. But if the curve is spacelike (or null), then $V^\mu V_\mu$ is positive (or zero), so you can't normalize to $-1$. This is of course the same as saying that such curves have no proper time.

$\endgroup$
1
$\begingroup$

Because you need clocks and rulers, basically. If you try the same measuring technique with space like paths, you will end up measuring proper distance rather than proper time, so velocity is out, with no derivatives with respect to proper time available to you.

For lightlike paths, proper time does not make any sense, as we understand it, $ds^2$ = 0 and we use a different, completely arbitrary variable, (an affine parameter) with no connection with proper or coordinate time.

enter image description here

The blue vertical line corresponds to an inertial observer measuring a coordinate time interval $t$ between events $E_1$ and $E_2$. The red curve corresponds to a clock measuring its proper time interval $\tau $ that passes  between the same two events.

$\endgroup$
0
$\begingroup$

The idea of a world-line was first introduced by Minkowski in his 1908 lecture:

Let the variations dx, dy, dz of the space co-ordinates of this substantial point correspond to a time element dt. Then we obtain, as an image, so to speak, of the everlasting career of the substantial point, a curve in the world, a world-line, the points of which can be referred unequivocally to the parameter t...

Trying to define a four-velocity for a space-like world-line can't be done here because it would mean the substantial point moving faster than the universal limiting velocity c in every frame.

Forgetting world-lines of substantial points, you can invent a quantity analogous to four-velocity by differentiating a space-like interval by a proper space parameter instead of a proper time parameter. But it's difficult to see how a proper space parameter can be usefully applied to anything.

$\endgroup$

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.