# Question involving special relativity

I'm starting to study the special theory of relativity, so the concepts and the 'methods' to solve problems are still not very clear in my mind. The question is basically:

A person is running towards a tunnel (which has length $$L$$) with velocity $$v$$, at the moment she's about to enter, a photon is emitted on the other side of the tunnel. when the photon and the person encounter each other, the length traveled by the person was $$f$$. Find $$f$$ in two different ways: working on the reference frame of the person and working in the reference frame of the tunnel.

I know I have two simultaneous events (in the reference frame of the tunnel): the entry of the person on the tunnel and the photon emission, right? but I have a third one: it occurs after some time interval in the same point in space (where the person encounter with the photon), so here's all I could do:

working in the reference frame of the tunnel: the simultaneous event ((L,0)) as viewed by the tunnel occurs in $$x=\gamma(L + vt')={\gamma}L$$. The encounter of the person with the photon ((L-f, t')) as viewed by the tunnel in $$x=\gamma(L-f + vt')$$ (or should it be ct?). I'm really confused of how to start this. I know the right thing to do is using Lorentz transformations, but nothing clear comes to my mind.

• Please define your quantities clearly. The simultaneous what occurs at $x_0$? What are the primed and unprimed coordinates, how is $t'$ defined? Are you aware that simultaneity is relative (frame-dependent)?
– Puk
Oct 28, 2020 at 22:18
• I am, that's why i said i was working on the reference frame of the tunnel to consider simultaneity. my edit should've made it clear. Oct 28, 2020 at 22:24
• drawing a spacetime diagram usually helps Oct 30, 2020 at 8:51

In the tunnel frame, everything works out as in Newtonian mechanics. If the two meet at time $$t$$, then the person travels a distance $$vt = f$$, and the photon a distance $$ct$$.
As we have $$L = vt + ct$$, we get $$f = vt = \frac{v}{c-v}L$$.