For a mass point in a local intertial system on which no forces act, we have that $$ \frac{\partial^2\xi}{d\tau^2}=0$$ where $\tau$, the proper time, is defined through $ds = c d\tau$ and the $\xi$ denote the Minkowski coordinates in the local intertial system.
I have read that for photons the proper time can not be identified with the $\tau$ above. Why is that the case? In particular, why is $ds=0$ for light? (mathematically, and intuitively)
Imagine computing the interval, $$ \Delta s^2 = (c\Delta t)^2 - (\Delta x^2 + \Delta y^2 + \Delta z^2), $$ between two events on the worldlines of a massless particle and of a massive particle. This interval will have the same value not matter which inertial reference frame is used to compute it. For the massive particle, we can simplify the computation by choosing to do the calculation in its rest frame. There the space part of the interval is zero, and we have $$ \Delta s^2 = (c\Delta t_\text{rest})^2 - (\text{zero}) $$ which you used to define the proper time, $\tau$, between the two events.
For the massless particle, that doesn't work: no reference frame exists where the massless particle is at rest. Instead, the massless particle is observed to travel at speed $c$ in all reference frames. The only freedom we have is to orient the velocity along one axis, so that we can say e.g. \begin{align} \Delta x &= c\Delta t & \Delta y &= \Delta z = 0 \end{align} For this case the interval is $$ \Delta s^2 = (c\Delta t)^2 - (c\Delta t)^2 = 0 $$ which I think is the observation you were hoping would be explained here.
If you consider a massive particle moving nearer and nearer to the speed of light, as some sort of a limiting process, what you find is that its spacetime interval between two spatial locations gets shorter and shorter as the speed approaches $c$. Sometimes one finds this information elided to a statement like photons experience zero proper time elapse as they travel between any two points. Such a simplification may or may not be useful, depending on your circumstances.
The definition of the proper time interval, for a particle with an arbitrary motion $ \vec{v} $ in a reference frame $ R $ between times $ t_1 $ and $ t_2 $ in this reference frame, is: $$ \Delta\tau=\tau_2-\tau_1=\int_{t_1}^{t_2}d\tau=\int_{t_1}^{t_2}\sqrt{1-\frac{\vec{v(t)}^2}{c^2}}dt $$ So, if the particle is a photon, $ \vec{v(t)}^2=c^2 $ which leads to $ d\tau=0 $ and $\Delta\tau=0 $: the proper time that elapses between any two events on the world line of a photon is zero.
An other demonstration:
if $ ds $ is the elementary interval of length, the elementary interval of proper time $ d\tau $ can be defined as $$ d\tau=\frac{ds}{c}\ \ \ \ [1] $$ Considering the elementary interval of length squared as $ ds^2=c^2dt^2-dx^2-dy^2-dz^2=c^2dt^2-\vec{dr}^2 $,
for a photon you have $ |\frac{dr}{dt}|=c $, which leads to $ ds^2=0 $, and then with $ [1] \Rightarrow d\tau=0 $.
Hoping to have answered your question,
Best regards.
There is an intuitive physical reason why "photons can't have time": clocks are necessarily massive systems. Why is this? Because a clock is a local energy reservoir and some physical mechanism that disperses this energy in a well defined way. "Good clocks" do this in equal proportions in equal clock intervals, but there is nothing wrong with stochastic clocks like radioactive decay, either. Energy-mass equivalence then requires that this localized energy reservoir has to have a non-zero mass property.
