Simultaneity in Special Relativity (conflicting times in thought experiment) I'm learning Simultaneous Events in Special Relativity. Lets say I have a light emitter in the center of the train; it shoots photons horizontally to each end of the train. The train moves with constant speed $V$.
From the point of view of someone inside the train, its length is $l$ then a photon arrives at each end in $t_b = \frac{l}{2c} = t_f$ time. So it arrives at the same time to both ends - as expected.
However, if I'm watching this from the outside, then from my point of view I see the photon going to the back of the train described by $\frac{l}{2}-Vt_b^{'} = ct_b^{'}$ and the photon hits the front of the train described by equation $\frac{l}{2} + Vt_f^{'} = ct_f^{'}$. With some algebra manipulation in order to get the time I have
$$t_b^{'} = \frac{l}{2(c+v)}$$
$$t_f^{'} = \frac{l}{2(c-v)}$$
Both observers experience the events at different times, as the Theory of Special Relativity predicts. But how is Lorentz Transformation applied here? Shouldn't $t_b^{'} = \frac{t_b}{\gamma}  = \frac{l}{2c}\sqrt{1-\frac{v^2}{c^2}} \neq \frac{l}{2(c+v)}$. What am I doing wrong?
 A: 
What am I doing wrong?

In your analysis of the situation in the embankment view you said that the position of the front of the train is given by $\frac{l}{2}+V t'$ but because of length contraction the correct equation is $\frac{l}{2} \sqrt{1-\frac{v^2}{c^2}}+ V t'$. If you start with that then you will get agreement between the two approaches.
It is always best to use the full Lorentz transform. It will correctly account for length contraction, time dilation, and the relativity of simultaneity. Using the "shortcut" formulas is not recommended. In situations where the "shortcut" formulas apply the Lorentz transform will automatically simplify, and by consistently using the Lorentz transform you will avoid mistakes from misusing the "shortcut" formulas.
A: Both equations are incorrect. The stationary observer on the ground does not see the proper length of the train $l$ but $l/\gamma$.
Then you must Lorentz transform your space and time coordinates in the train, x=l/2, and t=l/2c to the time coordinate in the platform. Then both results coincide
$t'_f=\frac{l}{2c}\sqrt{\frac{c+v}{c-v}}$
