# Error in Derivation of Time Dilation Formula [closed]

Suppose a train is travelling forwards with speed $$v$$ as perceived in the inertial reference frame of the railroad. Suppose moreover that there is a lightbulb at the leftmost-end of the train, and that it emits a ray of light that hits the rightmost-end of the train. If the length of the train in its rest frame is $$\ell$$, then in its rest frame the time $$t'$$ taken for the light ray to reach the rightmost-end of the train should be: $$t'=\frac{\ell}{c}$$ On the other hand, in the rest frame of the railroad, taking the Lorentz-Fitzgerald contraction of the train along its direction of motion into account, the time $$t$$ that the light ray appears to take to hit the opposite end should satisfy: $$ct=\frac{\ell}{\gamma_v}+vt$$ However, when combining these equations together, I get: $$t=\sqrt{\frac{c+v}{c-v}}t'$$ rather than the usual time dilation formula $$t=\gamma_vt'$$. Can someone explain the error in this derivation?

• What is your question? Apr 24, 2022 at 14:46
• sorry, I accidentally posted before I had asked my question Apr 24, 2022 at 14:49
• Your equation c*t=... is not valid, it says the speed of light is (c-v), if you write it in the form $ct-vt=l/\gamma$ Apr 24, 2022 at 15:00
• As I mention here, you can't just use the $\gamma$ separately on distance & time and combine them, you need to use the full Lorentz transformation to transform $(t, x)$ coordinates between frames. Apr 24, 2022 at 15:03
• Even when I apply the full Lorentz transformation, I get the same answer as above: $$ct'=x'=\ell$$ so $$ct=\gamma_v(ct'+\beta_vx')=\gamma_v(1+\beta_v)ct'=c\sqrt{\frac{c+v}{c-v}}t'$$ Hence $$t=\gamma_v(1+\beta_v)t'=\sqrt{\frac{c+v}{c-v}}t'$$ Apr 24, 2022 at 19:01