Let's suppose we have a spaceship with the exact speed of light. If a traveller takes this spaceship to go to proxima centauri (approximately 4 years light away from Earth) and come back, we (as observers on Earth) will see the ship coming back after approximately 8 years.
But how much time would have passed for the traveller on the ship? How can be this calculated with a formula?
The quantity that tells you what time an observer travelling along a path $\gamma : [t_0,t_1] \rightarrow \mathbb{R}^4$ experiences is the proper time
$$ \tau = \int_\gamma \sqrt{\mathrm{d}x_\mu\mathrm{d}x^\mu}$$
Assuming flat spacetime, i.e Minkowski metric/special relativity, this reduces to
$$ \tau = \int_\gamma\sqrt{\mathrm{d}t^2 - \frac{1}{c^2}\mathrm{d}x^i\mathrm{d}x_i} =\int_{t_0}^{t_1} \sqrt{1 - \frac{\vec{v}(t)^2}{c^2}}\mathrm{d}t $$
For $\vec{v} = c$, i.e an object travelling with the speed of light, this is $\int_{t_0}^{t_1} 0 \mathrm{d}t = 0$, so anyone hypothetically travelling with the speed of light will indeed not experience any time at all.
By plugging in other values for the travelling speed $\vec{v}$, you are able to calculate the experienced time for arbitary travellers.
