4-Velocity and World Line Parameterization

[pmr]

I would like to think that I understand how 4-velocity is defined in special relativity. It makes sense to me that one takes a differential segment of the world line $(dt, d\vec x)$ and translates it into a frame in which its spatial component disappears, leaving only $(d\tau, 0) = (dt/\gamma, 0)$, where $d\tau$ is referred to as the "proper time." One then divides the differential world line segment by the proper time, and the 4-velocity pops out:

$$\vec U = {1 \over d\tau}(dt, d\vec x) = {\gamma \over dt}(dt, d\vec x) = \gamma(1, \vec v)$$

The magnitude of the 4-velocity is then frame invariant, massless particles can't have 4-velocities, etc..., etc...

What I don't understand is this: The 4-velocity involves the terms $dt/d\tau$, $dx/d\tau$, $dy/d\tau$, $dz/d\tau$. Now, these are derivatives. Every other time I've seen a derivative anywhere in calculus or physics it has been the case that the thing being differentiated is a function that can be parameterized by the specified variable. Yet in this case I can't make sense of the idea that the world line is parameterized the proper time $\tau$. To begin with, the proper time for a given differential world line segment is only well defined in the momentarily co-moving reference frame (MCRF). Every differential world line segment will have a different MCRF, and this makes the proper time feel ephemeral, strange, and not globally persistent enough to parameterize the world line as a whole.

Could anyone help clear up my confusion here? Should I even be trying to think of the world lines as parameterized entities? And if not, then how can we possibly take the derivatives of an entity that isn't even a function (in the sense that functions are nothing if not "things which are parameterized").

 

[Nugatory]

You can interpret $\tau$ as the distance along the world line (from some arbitrarily chosen zero point on the world line).

Now there's no problem writing $x$, $y$, $z$, and $t$ as functions of $\tau$ and evaluating the derivatives that yield the four-velocity vector. Choosing a reference frame is equivalent to selecting specific functions of $\tau$ to define your coordinates, and a momentarily comoving inertial frame is one in which $dx/d\tau$, $dy/d\tau$, and $dz/d\tau$ are zero at the point of interest.

From this point of view, it's the coordinates $x$, $y$, $z$, and $t$ that should feel ephemeral and not especially globally significant - they're just the result of a more or less arbitrary choice of functions mapping points on the world line to numbers. The proper time $\tau$, on the other hand, has a real and frame-independent physical significance: it's the time recorded by a clock moving along the world line.

This perspective may feel a bit strange when you're still learning special relativity and concentrating on the coordinates that an observer naturally assigns to events - but it will serve you well when you get to general relativity where there are no global inertial frames.

 

[Orodruin]

Just to make a comparison to curves in Euclidean spaces, it is perfectly fine to parametrize such a curve by using the curve length (again from an arbitrary point on the curve). If I make a crooked path in the forest and tell you to follow the curve for 5 km, it will lead you to a specific point in the forest. It is also a parametrization that is invariant regardless of what coordinate system you put on top of the forest as a whole. Same thing with a world line and the proper time. If I have a world line and start from the (arbitrary) zero point and tell you to find the event on the world line such that a clock following the world line progresses an hour, then this is a specific event in space-time. Again the parametrization is independent of the coordinate system that we put on space time (e.g., what inertial frame we choose) as we will always end up at the same event.

 

[pervect]

What I don't understand is this: The 4-velocity involves the terms $dt/d\tau$, $dx/d\tau$, $dy/d\tau$, $dz/d\tau$. Now, these are derivatives. Every other time I've seen a derivative anywhere in calculus or physics it has been the case that the thing being differentiated is a function that can be parameterized by the specified variable. Yet in this case I can't make sense of the idea that the world line is parameterized the proper time $\tau$. To begin with, the proper time for a given differential world line segment is only well defined in the momentarily co-moving reference frame (MCRF). Every differential world line segment will have a different MCRF, and this makes the proper time feel ephemeral, strange, and not globally persistent enough to parameterize the world line as a whole.

I'm not sure why you think this. The proper time is not only defined for all observers, but it's a world scalar, It's definition is ndependent of any specific coordinates, and the value of proper time between any two points on a worldline is something that is also an invariant, i.e. it's not only defined for all observers, but its value is independent of the choice of reference frame.

This is why one differentiate with respect to proper time rather than coordinate time, differentiation by a world scalar gives a result that's not tied to any specific coordinate system. It's often called a geometric object (the simplest sort of object, a scalar. The 4-velocity is another geomteric object.

To calculate the proper time in the flat space-time of SR, you just evaluate
$\sqrt{(c dt)^2 - dx^2 - dy^2 - dz^2}$

Of course, in the MCIRF, the calculation is simpler, because dx=dy=dz=0. This is not required to calculate the proper time, however, it's still a geometric world scalar that's defined for all observers when you use the full formula.

Could anyone help clear up my confusion here? Should I even be trying to think of the world lines as parameterized entities? And if not, then how can we possibly take the derivatives of an entity that isn't even a function (in the sense that functions are nothing if not "things which are parameterized").

Yes, the usual parameterization of a worldline used to compute 4-velocities is
$t(\tau), x(\tau), y(\tau), z(\tau)$

Sometimes you get a wordline parameerized by some other parameter than $\tau$. Say it is, for example $T$. Then you can take

$(c \, dt/dT, dx/dT, dy/T, dz/dT)$ and multiply it by a scalar so it has the correct length. This scalar will be $dT / d\tau$ You can also use the fact that the "length" of the 4-velocity vector must have a magnitude of +1 or -1 depending on your sign conventions, rather than calculating $dT / d\tau$.

 

[stevendaryl]

What I don't understand is this: The 4-velocity involves the terms $dt/d\tau$, $dx/d\tau$, $dy/d\tau$, $dz/d\tau$. Now, these are derivatives. Every other time I've seen a derivative anywhere in calculus or physics it has been the case that the thing being differentiated is a function that can be parameterized by the specified variable.

I sort of understand why it seems strange to you, but mathematically speaking, if we have a way to associate a unique value of $x$ to each value of $\tau$, then we implicitly have a way to view $x$ as a function of $\tau$.

The more general notion is "parametrized path" which is not confined to discussions of relativity. Anytime you have some N-dimensional space, and you have a path, or curve, through that space, you can describe the path by giving N functions of a real-valued parameter $s$: $x^j(s)$. What is $s$? It's completely arbitrary. Just pick a point along the path, and call that $s=0$. Then just let $s$ increase smoothly and monotonically (meaning it always increases, and never decreases or remains the same) as you go along the path. Then with respect to that parameter, we can define a kind of "velocity":

$v^i = \dfrac{dx^i}{ds}$

However, even though you can parametrize a path by any parameter you like, the most convenient way is to let $s$ be the distance along the path. I understand that it seems circular to think of $x^i$ as a function of the distance $s$ along the path, because you have to have a path first before you can figure out $s$. But mathematically, it doesn't matter whether you have the path first, and then calculate $s$. If there is exactly one value of $x^i$ associated with each value of $s$, then you can view $x^i$ as a function of $s$.