# Proper time for a light particle

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.
• That's one way to think of it. Another way is to remember (or re-derive) that the interval $\Delta s^2$ is unchanged by boosts; if you have a light-like interval in one reference frame, it is light-like in all reference frames. – rob Aug 8 '17 at 20:38