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I'm struggling with an introductory example of special relativity. We haven't done the math yet so I would like an explanation based only on the fact that the speed of light is the same in every inertial frame.

An airplane travels east with a certain speed. There is a "clock" at both ends of the airplane. If there is a flash in the middle of the airplane and we're in the airplane's inertial frame, both clocks will register the flash at the same time, say 3.

But if we are an observer standing on the ground, in our inertial frame the light will reach the clock at the end of the airplane faster than the clock in the front. So the clock in the back may reads 3 when the clock in the front only reads 1, for example.

Question: Let's say the airplane has a speed v. The light's speed as it travels in the inertial frame of the observer towards the back will c+v, and the speed as it travels towards the front will be c-v. Let's say the distance from the middle to the back and front respectively is L. Then the time it takes for the light to travel from the middle to the back is L/(c+v) and to the front L/(c-v).

Shouldn't these be the same? If not, why can we say that the clocks will read 3 when the light hits them, no matter what frame is used? This is the argument the book uses to say that the clock must be less than 3, i.e. 1, in the front when the light hits the clock in the back in the observer's inertial frame.

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You started by sentences that indicated that you realize that the speed of light is $c$ regardless of the speed of the source and the observer. However, you quickly wrote another sentence that indicated that the conclusion about knowledge in the previous sentence was an illusion:

The light's speed as it travels in the inertial frame of the observer towards the back will $c+v$, and the speed as it travels towards the front will be $c-v$.

No. This is not how it works according to special theory of relativity. The speed of light is always $c$, it is never $c+v$ or $c-v$. Because the light from the middle of the aircraft gets to the rear end of the aircraft earlier (because the rear end is moving towards the light while the front end tries to escape it and the light has the same speed in both directions), it just proves another fact of relativity, namely the "relativity of simultaneity". Two events that are simultaneous according to the aircraft's inertial system, $t_1=t_2$, are not simultaneous according to the inertial system attached to the Earth's surface: $t'_1\neq t'_2$.

Objects are contracted in the direction of the motion. The front-rear length of an object moving by speed $v$ is $$ L = L_0 \cdot \sqrt{1-v^2/c^2}$$ where $L$ is the length of the object in its own rest frame. The formula to add velocities that replaces $u+v$ is $$V_{\rm total} = \frac{u+v}{1+uv/c^2}$$ See two derivations of this formula.

Check that with $u=c$, you get $V_{\rm total}=v$: the speed of light "added" to anything else in the relativistic way is still the speed of light. All these things may be derived from a more general Lorentz transformation that maps the coordinates $(t,x,y,z)$ from one inertial system to another.

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