Suppose there is a bus which travels at constant speed $v=0.9c$, relative to my friend Eric who is standing still on the ground.
I'm exactly in the middle of the bus, at distance $d$ form both ends, and at the same time $t=0$ I shoot a light beam in both the directions, one toward the head of the bus and the other toward the back! enter image description here

Since I'm beginning to study special relativity, my intuition is that:

  • Since I'm on the Bus, I will see the 2 beam of lights reaching the 2 opposite sides of the bus at the same moment $t=d/c$.

  • Eric instead will see the one touching the back of the bus first at time $t=d/(c+0.9c)$ and the one touching the head of the bus after $t=(d+ 0.9c*t)/c=d/c+0.9t=10d/c$.

Are my intuitions right?

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    $\begingroup$ I don't understand why pepole downvote without explanation, I put effort to make questions that I think are useful and it's clear that I spent some time preparing it. Please downvote with explanation so that I can edit the question $\endgroup$ – Gabriele Scarlatti May 25 '17 at 9:54
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    $\begingroup$ Upvoted purely for the artwork $\endgroup$ – Spine Feast May 25 '17 at 10:53
  • $\begingroup$ The only mistake is to assume that the observer on the platform also sees that the train has a half-width $d$. Due to length contraction it will instead be $\gamma d$, where $\gamma=1/\sqrt{1-(v/c)^2}$ $\endgroup$ – user126422 May 25 '17 at 13:24
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    $\begingroup$ @Willy Billy Williams : "...it will instead be $\;d/\gamma$, where $\gamma=\cdots$" $\endgroup$ – Frobenius May 31 '17 at 16:41
  • $\begingroup$ @frobenius yes! $\endgroup$ – user126422 May 31 '17 at 18:54

I think it's better to define clearly frames, space-time events with their coordinates and use the Lorentz transformation to find the relations between them. I believe that to derive results intuitively using length contraction and/or time dilation is not so safe.

enter image description here

So, let the primed $\;S'\;$ denote the frame of the bus moving with speed $\;v=0.9\,c\;$ as in Figure 1. We have 3 space-time events $\;\mathrm{A,B,C}\;$ :
\begin{align} \mathrm{A} & =\text{shooting light beams to both directions from the middle of the bus} \tag{01.A}\\ \mathrm{B} & =\text{the backwards light beam strikes the back of the bus} \tag{01.B}\\ \mathrm{C} & =\text{the forwards light beam strikes the front of the bus} \tag{01.C} \end{align}

enter image description here

Also, let the unprimed $\;S\;$ denote the rest frame of Eric as in Figure 2 (1) .

The space-time coordinates of events in the primed frame $\;S'\;$ of the bus are : \begin{align} \left(x'_\mathrm{A},t'_\mathrm{A}\right) & =\left(0,0\right) \tag{02.A}\\ \left(x'_\mathrm{B},t'_\mathrm{B}\right) & =\left(-d,d/c \right) \tag{02.B}\\ \left(x'_\mathrm{C},t'_\mathrm{C}\right) & =\left(+d,d/c \right) \tag{02.C} \end{align} Events $\;\mathrm{B,C}\;$ are simultaneous in the system of the bus.

The space-time coordinates of these events in the unprimed Eric's frame $\;S\;$ are : \begin{align} \left(x_\mathrm{A},t_\mathrm{A}\right) & =\left(0,0\right) \tag{03.A}\\ \left(x_\mathrm{B},t_\mathrm{B}\right) & =\left(???,??? \right) \tag{03.B}\\ \left(x_\mathrm{C},t_\mathrm{C}\right) & =\left(???,??? \right) \tag{03.C} \end{align}

We suppose that at the shooting time moment ($t'_\mathrm{A}=0$) Eric is standing opposite to the middle and outside the bus and is setting there the origin of his space-time events $\left(x_\mathrm{A},t_\mathrm{A}\right) =\left(0,0\right)$.

Now, the Lorentz transformation between frames$\;S,S'\;$ expressed with differences is \begin{align} \Delta x & =\gamma\left(\Delta x'+v\,\Delta t'\right) \tag{04.1}\\ \Delta t & =\gamma\left(\Delta t'+\dfrac{\,v\,}{c^2}\Delta x'\right) \tag{04.2} \end{align} So \begin{align} \Delta x_\mathrm{BA} & =\gamma\left(\Delta x'_\mathrm{BA}+v\,\Delta t'_\mathrm{BA}\right) \Longrightarrow x_\mathrm{B}-x_\mathrm{A}=\gamma\left[\left(x'_\mathrm{B}-x'_\mathrm{A}\right)+v\,\left(t'_\mathrm{B}-t'_\mathrm{A}\right)\right] \Longrightarrow \nonumber\\ x_\mathrm{B} & =\gamma \left(-d+v\,d/c\right)=-\dfrac{c}{c+v}\dfrac{d}{\gamma}=-\sqrt{\dfrac{c-v}{c+v}}\,d=-\sqrt{\dfrac{1}{19}}\,d\approx -0.2294\,d \tag{05.1}\\ \Delta t_\mathrm{BA} & =\gamma\left(\Delta t'_\mathrm{BA}+\dfrac{\,v\,}{c^2}\Delta x'_\mathrm{BA}\right)\Longrightarrow t_\mathrm{B}-t_\mathrm{A}=\gamma\left[\left(t'_\mathrm{B}-t'_\mathrm{A}\right)+\dfrac{\,v\,}{c^2}\,\left(x'_\mathrm{B}-x'_\mathrm{A}\right)\right] \Longrightarrow \nonumber\\ t_\mathrm{B} & =\gamma \left(\dfrac{\,d\,}{c}-\dfrac{\,v\,}{c^2}\,d\right)=\dfrac{1}{c+v}\dfrac{d}{\gamma}=\sqrt{\dfrac{c-v}{c+v}}\,\dfrac{\,d\,}{c}=\sqrt{\dfrac{1}{19}}\,\dfrac{\,d\,}{c}\approx 0.2294\,\dfrac{\,d\,}{c} \tag{05.2} \end{align} and on the same footing \begin{align} \Delta x_\mathrm{CA} & =\gamma\left(\Delta x'_\mathrm{CA}+v\,\Delta t'_\mathrm{CA}\right) \Longrightarrow x_\mathrm{C}-x_\mathrm{A}=\gamma\left[\left(x'_\mathrm{C}-x'_\mathrm{A}\right)+v\,\left(t'_\mathrm{C}-t'_\mathrm{A}\right)\right] \Longrightarrow \nonumber\\ x_\mathrm{C} & =\gamma \left(d+v\,d/c\right)=\dfrac{c}{c-v}\dfrac{d}{\gamma}=\sqrt{\dfrac{c+v}{c-v}}\,d=\sqrt{19}\,d\approx 4.3589\,d \tag{06.1}\\ \Delta t_\mathrm{CA} & =\gamma\left(\Delta t'_\mathrm{CA}+\dfrac{\,v\,}{c^2}\Delta x'_\mathrm{CA}\right)\Longrightarrow t_\mathrm{C}-t_\mathrm{A}=\gamma\left[\left(t'_\mathrm{C}-t'_\mathrm{A}\right)+\dfrac{\,v\,}{c^2}\,\left(x'_\mathrm{C}-x'_\mathrm{A}\right)\right] \Longrightarrow \nonumber\\ t_\mathrm{C} & =\gamma \left(\dfrac{\,d\,}{c}+\dfrac{\,v\,}{c^2}\,d\right)=\dfrac{1}{c-v}\dfrac{d}{\gamma}=\sqrt{\dfrac{c+v}{c-v}}\,\dfrac{\,d\,}{c}=\sqrt{19}\,\dfrac{\,d\,}{c}\approx 4.3589\,\dfrac{\,d\,}{c} \tag{06.2} \end{align}

For a space-time diagram, see Figure 3.

enter image description here

(1) For convenience Figure 2 has been drawn with scale but with $\;v/c=0.60 (\gamma=1.25)\;$ instead of $\;v/c=0.90 (\gamma=1/\sqrt{0.19}\approx 2.2942)\;$ of the question.

(2)Kinematics: (a) If two cars 1 and 2 at a distance $\;s\;$ apart start running towards to each other with speeds $\;v_{1}\;$ and $\;v_{2}\;$ respectively, then they will meet each other after time \begin{equation} \Delta t =\dfrac{s}{v_{1}+v_{2}} \tag{a.1} \end{equation} at a distance \begin{equation} s_{1}=\dfrac{v_{1}}{v_{1}+v_{2}}\;s \tag{a.2} \end{equation} from the starting point of car 1. (b) If two cars 1 and 2 at a distance $\;s\;$ apart start running with speeds $\;v_{1}\;$ and $\;v_{2}\;$ respectively,where $\;v_{1}>v_{2}\;$, so that the faster car 1 is "hunting" the other car 2, then car 1 will "catch" car 2 after time \begin{equation} \Delta t=\dfrac{s}{v_{1}-v_{2}} \tag{b.1} \end{equation} at a distance \begin{equation} s_{1}=\dfrac{v_{1}}{v_{1}-v_{2}}\;s \tag{b.2} \end{equation} from the starting point of car 1.With $\;v_{1}=c$, $\;v_{2}=v\;$ but $\;s=d/\gamma$=the length-contracted bus half-width $\;d\;$, equations (a.2),(a.1),(b.2) and (b.1) yield equations (5.1),(5.2),(6.1) and (6.2) respectively (see also data in Figure 2).

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    $\begingroup$ @Gabriele Scarlatti $$ \dfrac{c}{c\pm v}\dfrac{d}{\gamma}=\dfrac{c}{c\pm v}\dfrac{d}{\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}}=\sqrt{\dfrac{c^2-v^2}{(c\pm v)^2}}\,d=\sqrt{\dfrac{c\mp v}{c\pm v}}\,d $$ $\endgroup$ – Frobenius Jun 1 '17 at 8:18
  • $\begingroup$ I was wrong ahahah, I double checked and then deleted the comment, thank you very much $\endgroup$ – Gabriele Scarlatti Jun 1 '17 at 10:38

Thank you very much for asking me to revise my answer, as I was actually initially incorrect; I didn't think about my result carefully enough. Let me try again.

Your intuition is nearly correct. The only difference is that there's this special relativistic weirdness called length contraction. So we anticipate that the result for the time in the unprimed frame that the light hits the right wall of the bus is $$t_R = \frac{1}{c-v}\frac{d}{\gamma}$$ and the time the light hits the left wall of the bus is $$t_L = \frac{1}{c+v}\frac{d}{\gamma},$$ where those additional $1/\gamma$'s account for the length contraction.

Now let's actually derive the result to check our intuition.

Let's write down equations in the unprimed ($v=0$) frame for the motion of the two ends of the bus $x_L(t)$ and $x_R(t)$ as well as equations for the motion of the two light rays $\ell_L(t)$ and $\ell_R(t)$. What you're looking for is the time when the light ray moving right intersects with the right end of the bus (i.e. for $t_R$ such that $x(t_R)=\ell_R(t_R)$) and the time when the light ray moving left intersects with the left end of the bus (i.e. for $t_L$ such that $x(t_L)=\ell_L(t_L)$).

Let's write down the coordinates of the right side of the bus, the left side of the bus, and you in your primed frame: $$(t'=0,x'=d)$$ $$(t'=0,x'=-d)$$ $$(t'=0,x'=0).$$ We may use the standard Lorentz transformation laws $$x'=\gamma(x-vt)\\t'=\gamma(t-vx/c^2) \\ x=\gamma(x'+vt') \\ t=\gamma(t+vx/c^2)$$ to determine the locations of these coordinates in the unprimed frame: $$(t=\gamma v \frac{d}{c^2},x=\gamma d)$$ $$(t=-\gamma v \frac{d}{c^2},x=-\gamma d)$$ $$(t=0,x=0).$$

Let's work out in detail the position of the right hand side of the bus in the unprimed frame. Since the bus is moving with a constant velocity, we must have $$x_R(t)=vt+c_R,$$ where $c_R$ is some constant. We can find $x_0$ from $$x_R(t=\gamma v \frac{d}{c^2}) = v(\gamma v \frac{d}{c^2})+c_R = \gamma d,$$ which means that $$c_R = \gamma d(1-\frac{v^2}{c^2}) = \frac{d}{\gamma}.$$

At the same time, we may write down the equations that describe the right-moving light in the unprimed frame: $$\ell_R(t)=ct.$$

Solving for $t_R$ by setting $x_R(t_R)=\ell(t_R)$, we find $$vt_R+\frac{d}{\gamma}=ct_R$$ implies that $$t_R=\frac{1}{c-v}\frac{d}{\gamma}.$$

Going through similar steps for the left side of the bus, we find $$x_L(t)=vt-\frac{d}{\gamma}$$ and $$t_L=\frac{1}{c+v}\frac{d}{\gamma}.$$

  • $\begingroup$ Hi, thank you for answering! I see that you know much more than me on this topic, but I cann0t understand very much of what you have written. Maybe I'm looking for a more intuitive explanation, I'd love if you'd try to add a more intuitive one! Thank you very much! $\endgroup$ – Gabriele Scarlatti May 25 '17 at 11:28
  • $\begingroup$ Hi @GabrieleScarlatti, I edited my answer above. I hope you find it satisfactory. $\endgroup$ – WAH May 25 '17 at 13:34
  • $\begingroup$ Whoops! Yes, one should take d => 2d everywhere above. $\endgroup$ – WAH May 25 '17 at 19:38
  • $\begingroup$ Note that the above solution now has the correct factors of 2. Thanks to @Frobenius for making the necessary edits. $\endgroup$ – WAH May 27 '17 at 10:56
  • $\begingroup$ @WAH May be you know it, but parentheses are of fixed size using "(" and ")",see equation (1), and of adjustable size to their content using "\left(" and "\right)", see equation (2) : \begin{equation} (1-\dfrac{v^2}{c^2}) \tag{1} \end{equation} \begin{equation} \left(1-\dfrac{v^2}{c^2}\right) \;,\quad \left(a-\dfrac{b^2}{\dfrac{\rho}{\sigma}}\right) \tag{2} \end{equation} $\endgroup$ – Frobenius Jun 1 '17 at 21:45

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