# How is the speed of light constant in all directions for all observers?

Order of Events:

1. Pulse - A single pulse of light is emitted from the light towards the mirror
2. Reflect - The pulse hits the mirror and is reflected back towards the light
3. Return - The pulse returns to the light.

Observers:

1. BoxGuy - An observer on the boxcar
2. PlatGirl - An observer on the platform

Question:

With the above configuration, how can the speed of light be constant for both observers in both directions?

Analysis:

Assuming the speed of light is constant for BoxGuy relative to himself, the time between Pulse and Reflect is equal to the time between Reflect and Return. This is because the distance the light travels relative to him is d in both cases.

With the same assumptions for PlatGirl, the time between Pulse and Reflect is less than the time between Reflect and Return. This is because the mirror will travel 2 * d on the away trip (because when light has traveled 2 * d, the mirror will be d farther to the left, so both the mirror and pulse will be in the same location), but only 2/3 * d on the return trip (using similar logic).

Assuming that the light pulse is in the same location for all observers at any given moment, Pulse has to occur simultaneously for both BoxGuy and PlatGirl, Reflect has to occur simultaneously for BoxGuy and PlatGirl, and Return has to occur simultaneously for BoxGuy and PlatGirl.

Finally, if we try to figure out the relative passage of time for BoxGuy and PlatGirl with the above, we get that time travels faster for PlatGirl than for BoxGuy during Pulse-Reflect. This is because light travels farther for her (2*d) than him (d) during that time. With similar logic, we get that time travels slower for PlatGirl than for BoxGuy during Reflect-Return.

The last conclusions do not make sense, since the coming or going of a beam of light should not affect the relative time-lapse for two observers. For example, if this were the case what would happen if another pulse was emitted the moment the first pulse is reflected? Time cannot move faster AND slower for both of them.

Thus, either the speed of light is not constant, the same light beam can simultaneously be in different locations at once for different observers, or there is another flaw in the analysis.

Which is it and why?

Notes:

• As mentioned by other users, d will be shorter for PlatGirl than for BoxBoy according to SR. However, the duration of Pulse-Reflect is still shorter than Reflect-Return for PlatGirl, and the durations are equal for BoxBoy.
• In response to my question on Mark's answer, we can use the Lorentz Transform to calculate PlatGirl's space-time coordinate for BoxGuy's Reflect observed event, which happens at (d,d/c) in his frame of reference:

$\lambda = (1/\sqrt{1-.5^2}) = (1/\sqrt{.75}) = \sqrt{4}/\sqrt{3} = \frac{2\sqrt{3}}{3}$
$t' = \lambda (t - vx/c^2) = \lambda (d/c - (-.5) * d/c) = \frac{2\sqrt{3}}{3}*(1.5d/c) = \sqrt{3}d/c$ $x' = \lambda (x - vt) = \lambda (d + .5c * d/c) = \frac{2\sqrt{3}}{3}*1.5d = \sqrt{3}d$

• Similarly for (0, 2d/c):

$t' = \lambda (t - vx/c^2) = \frac{2\sqrt{3}}{3} (2d/c) = \frac{4\sqrt{3}}{3} d/c$ $x' = \lambda (x - vt) = \frac{2\sqrt{3}}{3} (.5c * 2d/c) = \frac{2\sqrt{3}}{3}d$

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The thing is that the constantcy of the speed of light is both (a) a consequence of the symmetries of Maxwell's equations and (b) an experimentaly measured fact (since 1887 and many times since then). In anycase in your list of possibilities you left out "the distance between points may be different for different observers" – dmckee Jan 10 at 21:47

The problem is in a misunderstanding of "simultaneous".

"Simultaneous" refers to two different events that occur at the same time in some particular reference frame, but you're applying it to the same event in two different frames. So it doesn't make sense to say "Pulse has to occur simultaneously for both BoxGuy and PlatGirl." That's a single event - it can't be simultaneous all by itself, even when observed by two different people.

You could, if you want, set the origins of the coordinate systems they are using so PlatGirl and BoxGuy assign the same time coordinate to Pulse. If you do, they will not assign the same time coordinate to Reflect. The time between the events Pulse and Reflect is different in different frames.

Additionally, PlatGirl and BoxGuy will not agree on the length of the boxcar. Your calculation assumes they both measure the length to be $d$, but actually PlatGirl will observe the boxcar to be Lorentz-contracted.

One way to analyze your scenario is to set up coordinate systems $S$ for the boxcar and $S'$ for the platform. We set (x,t) = (0,0) = Pulse in both systems.

In frame $S$ (box), the coordinates are:

Pulse: (0,0) Reflect: (d,d/c) Return: (0,2d/c)

In frame $S'$ (platform), the coordinates are:

Pulse: (0,0)

Reflect: $(\sqrt{3}d,\sqrt{3}d/c)$

Return: $(\frac{2\sqrt{3}}{3} d, \frac{4\sqrt{3}}{3} d/c)$

You can verify that in both frames, light moves outward at speed $c$ and returns at speed $-c$

In reply to your edit, yes the durations from Pulse to Reflect and Reflect to Return are the same for BoxGuy and different for PlatGirl. That is just a fact. That's how it is. Notice, though, that the spatial separations are also different. For BoxGuy, these events the same distance apart. For PlatGirl, they are different distances apart. What's the same between frames is the interval $\Delta x^2 - \Delta t^2$.

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 Ok, so we set the space-time-origin (0,0) to Pulse at the Light for both BoxGuy and PlatGirl. The space-origin "moves" with the Light for BoxGuy, but is fixed relative to the platform for PlatGirl. At (d,d/c) and (0, 2d/c) for BoxGuy, Reflect and Return happen respectively. What is the formula for translating these space-time coordinates directly back to PlatGirl's coordinate system? – Briguy37 Jan 11 at 17:07 Nevermind, I've updated my question notes with the Lorenz Transform. From this, it appears time observed at a point in space is warped according to the direction of relative travel. For example, at the time of the Pulse, BoxGuy's mirror is at $(d,0)$, which maps to PlatGirl's $(\frac{2\sqrt{3}}{3}d, \frac{\sqrt{3}}{3}d/c)$. Thus, PlatGirl observes a current version of BoxGuy (for them both), but BoxGuy's "current" mirror is a future version of PlatGirl's "current" mirror. By symmetry, if the mirror and origin were reversed, PlatGirl's mirror would be the "future" one. Is this correct? – Briguy37 Jan 11 at 20:39 That's correct at the moment they're passing each other, yes. – Mark Eichenlaub Jan 11 at 20:46 Very interesting, thanks much for the response! – Briguy37 Jan 11 at 20:53