A thought experiment on simultaneity

[member 743765]

Consider two mirrors opposite to eachothers at the mid point between them two photons simultaneously emitted each one towards one of the mirrors.
relative to observer at rest to the mirrors at the mid point he will see the photons reflected simultaneously but I think a moving observer won't observe them simultaneously my question is why would one mirror reflect the photon earlier than the other relative to a moving observer?

 

[Ibix]

You have described a setup with two mirrors moving at speed zero and added an observer moving with speed $v$. How would this observer describe the setup if he regards himself as stationary?

 

[member 743765]

You have described a setup with two mirrors moving at speed zero and added an observer moving with speed $v$. How would this observer describe the setup if he regards himself as stationary?

So one of the mirrors is moving away from him and hence the photon needs more time to reach it than the photon of the other mirror?

 

[Ibix]

Correct, and the other mirror is moving towards him. So the reflection events are not simultaneous according to him.

 

[PeroK]

So one of the mirrors is moving away from him and hence the photon needs more time to reach it than the photon of the other mirror?

Indeed, this scenario is probably the simplest to demonstrate the relativity of simultaneity associated with light having the same measured speed in both frames.

Let's consider the frame of reference in which the mirrors are moving. Let's say the mirrors are travelling with speed $v$ (assume this is to the right). The two photons are moving to the left and right at speed $c$ (as measured in this frame). The source of light is moving in this frame and remains at the midpoint between the mirrors. Let's say that (in this frame), it is a distance $D$ from both mirrors.

This implies (by the definition of velocity/speed), that the photon moving to the left will reach the mirror after $\Delta t_1 = \frac{D}{c+v}$. In this frame of reference the light and the mirror are moving towards each other, so their combined separation speed is $c + v$. For example, if $v = \frac c 2$, then the light will move two-thirds of the distance $D$, and the mirror will move one-third of the distance $D$ in the given time. That must be the case if the light is measured (in this frame) to have speed $c$.

Likewise, the photon moving to the right will take time $\Delta t_2 = \frac{D}{c - v}$ to reach the mirror. For example, if $v = \frac c 2$, then the mirror will move a distance $D$ and the light move a distance $2D$ before the light reaches the mirror. Again, this is consistent with the light moving at $c$.

The important point, however, is that in the rest frame of the source and mirrors, the measured speed of light is also $c$ - for both photons. And the mirrors are at rest. So, the light reaches both mirrors at the same time (as measured in this frame).

It's this transformation between reference frames that is different in SR from what we are used to in classical, Newtonian mechanics.

 

[thomsj4]

Imagine enclosing the entire setup in a spacetime “drum” whose boundary is formed by all possible light paths between the midpoint and the two mirrors. Let the distance between the mirrors be $L$, and place them so that in the rest frame $S$ the left mirror is at $-\tfrac{L}{2}$ and the right mirror is at $+\tfrac{L}{2}$. In $S$, two photons are emitted from $x=0$ at $t=0$. Each photon follows a null worldline, so in $S$ one photon strikes the left mirror when


$$


\Delta s^2 = c^2 \Delta t^2 - \Delta x^2 = 0\implies c^2t_L^2 = \Bigl(\tfrac{L}{2}\Bigr)^2 \implies t_L = \tfrac{L}{2c},


$$


and similarly the other photon strikes the right mirror at


$$


t_R = \tfrac{L}{2c}.


$$


Hence, $t_L = t_R$ in $S$, so the reflections are simultaneous there.


To see why a moving observer sees them as non-simultaneous, perform a Lorentz boost of velocity $v$ along the $+x$-axis. Label the new frame $S’$. The Lorentz transformation for time is


$$


t’ = \gamma\Bigl(t - \tfrac{v x}{c^2}\Bigr),


\quad \gamma = \tfrac{1}{\sqrt{1-\tfrac{v^2}{c^2}}}.


$$


The reflection event on the left mirror in $S$ has coordinates $\big(t, x\big) = \big(\tfrac{L}{2c}, -\tfrac{L}{2}\big)$. Under the boost its time becomes


$$


t’_L = \gamma\Bigl(\tfrac{L}{2c} - \tfrac{v\left(-\tfrac{L}{2}\right)}{c^2}\Bigr)


= \gamma \Bigl(\tfrac{L}{2c} + \tfrac{v L}{2 c^2}\Bigr)


= \tfrac{\gamma L}{2c}\Bigl(1 + \tfrac{v}{c}\Bigr).


$$


Meanwhile the reflection on the right mirror in $S$ has $\big(t,x\big) = \big(\tfrac{L}{2c}, +\tfrac{L}{2}\big)$, transforming to


$$


t’_R = \gamma\Bigl(\tfrac{L}{2c} - \tfrac{v\left(\tfrac{L}{2}\right)}{c^2}\Bigr)


= \tfrac{\gamma L}{2c}\Bigl(1 - \tfrac{v}{c}\Bigr).


$$


Because


$$


t’_L - t’_R = \tfrac{\gamma L}{2c}\Bigl[\bigl(1 + \tfrac{v}{c}\bigr) - \bigl(1 - \tfrac{v}{c}\bigr)\Bigr]


= \tfrac{\gamma L}{c},\tfrac{v}{c}


\neq 0,


$$


the two reflections do not occur at the same $t’$. This is not due to any trick of signal travel times; it is a direct consequence of light-speed constancy and the Minkowski structure of spacetime. Each observer measures light moving at speed $c$, but in the moving frame the spatial location of one mirror is “closer” in time than the other because the offset $\tfrac{v x}{c^2}$ in the Lorentz transformation skews simultaneity. In essence, the observer in $S’$ “slants” through the spacetime drum’s boundary of light paths, causing them to register one reflection before the other.

 

[PeroK]

To see why a moving observer sees them as non-simultaneous, perform a Lorentz boost of velocity $v$ along the $+x$-axis. Label the new frame $S’$. The Lorentz transformation for time is

If you have already developed the Lorentz Transformation, the RoS is something significantly more basic. The original thought experiment does this assuming nothing more than the invariance of $c$. That's from first principles.