In Hobson et al, General Relativity: An Introduction for Physicists (pg. 15), it was said that it is natural to describe the worldline of a massive particle by giving the four coordinates $(t,x,y,z)$ as functions of the proper time $\tau$, i.e. $(t(\tau),x(\tau),y(\tau),z(\tau))$.
Why did the author feel that it was necessary to specify that it is a massive particle? Isn't this four-dimensional way of describing a worldline applicable to any particle, regardless of whether it is massive or not?
The question here, by the way, is not "Why use proper time at all?", but "Why is the qualifier 'massive' important?" It is this second question I am going to answer.
There is a colloquial understanding that "a photon does not experience time". While this is often contested typically with recourse to some argument about how you cannot define a Lorentz frame comoving with a photon, there is a much more airtight way to formalize this in mathematics that captures very well what I think most people have in mind when they say this. And it's this:
The proper time between any two points along a light like worldline is always zero.
From this it follows that no light-like object can change or modify in any way as seen by an outside observer, which is a pretty good definition, I'd say, of "having your clock stopped" or "not experiencing time".
Returning to the question, then, if you want to parameterize a world line by proper time, that should mean that each space-time point threupon should get its own proper time coordinate different from other such space-time points on the line. And massless particles, including but not limited to photons, follow light-like world lines. Then, because the proper time between any two points is zero, there is in effect only one proper time for all of them: zero, so how can it function usefully as a coordinate that informatively distinguishes between them?
In more formal terms, the mapping $E(\tau)$ which parameterizes the worldline events in the life of the massless particle would have to be "infinitely ill-defined" at the sole point $\tau = 0$, mapping it to everything on it, and be fully undefined for any other value of $\tau$, which isn't really very useful as a "parameterization".
And to top it all off, if you wanna plow through and take that seriously regardless, you'll have to describe the map's horribly and hgeepy multi-valued "output" at $\tau = 0$ through some other means anyways, so you gain precisely nothing by this approach.
Parametrizing the worldline of a particle by the proper time $ \tau $ is the pseudo-Riemannian equivalent of parametrizing a curve in $ \mathbb R^n $ by its arc length parameter. However, there is an additional complication introduced by the fact that the metric tensor of a general pseudo-Riemannian manifold is not positive definite, unlike the inner product of $ \mathbb R^n $, so it's not always possible to normalize the norm (in special relativity, this is the arc length parameter $ ds $) of the derivative of your parametrization to $ 1 $.
For a massless particle in special relativity, its path $ \gamma : \mathbb R \to M $ in the spacetime manifold $ M $ has the property that $ g(\dot \gamma, \dot \gamma) = 0 $ (this is equivalent to the particle travelling at the speed of light), so there's no possible way to normalize $ \gamma $ by a reparametrization so as to make it have norm $ 1 $.
It seems to me like no answer yet has addressed the point:
Why does a massless particle follow a lightlike trajectory?
The thing is, relativistic four-momentum for a massive particle is defined as $p^\mu = m u^\mu$, where $u^\mu$ is the particle' four-velocity. If we parametrize by the proper time, the four-velocity always has unit norm (depending on your metric convention, either + or -1; I'll choose + here), so we have that the square modulus of the momentum is given by $p^2 = m^2$.
This is all well and good if $m \neq 0$, but what happens if we set $m=0$? This approach breaks down, since we would be defining $p^\mu = 0 u^\mu$, that is, massless particles would all have zero energy and zero momentum... But we know that photons have energy and momentum, so we must approach it another way.
So, we throw away the assumption of the parameterization of the trajectory by proper time: this way, we can have a lightlike trajectory, whose four-velocity has norm 0, so that the condition $p^\mu p_\mu = 0$ makes sense if we define $p^\mu = \hbar k^\mu$, where $k$ is now the wavevector, which obeys $k^\mu k_\mu = 0$.
This allows us to use the four-momentum for both massive and massless particles, even though in the latter case its definition through $u^\mu = \mathrm{d} x^\mu / \mathrm{d} \tau$ does not make sense anymore. We can parameterize the curve described by the particle through any parameter $\lambda$ we want, and the vector $\mathrm{d} x^\mu / \mathrm{d} \lambda$ will always have 0 norm.
