I am told that the weak equivalent principle, that $m_i=m_g$ (inertial and gravitational masses are equivalent) is equivalent to the statement that in a small system you can't tell whether you are in a uniform gravitational field, or in an equivalent accelerating frame. My question is about light: Special relativity doesn't say anything about how light falls in a gravitational field, so if you are in a rocket ship accelerating at $g$ why should it be true that when you shine a light in the rocket ship, you see the same thing happening as if you shined a light on earth? (to clarify, if you shot a bullet in the rocket ship, I understand why the trajectory of the bullet would look the same as if the bullet were shot on Earth, but this is because special relativity (or classical mechanics) tells us what the trajectory of a bullet looks like on Earth).
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Your statement of the weak equivalence principle (WEP) is fine with me, but I think you're not applying it in the correct context. Special relativity, as mentioned by @Jerry Schirmer, says that light travels along null geodesics. This meaning that it travels at $c$ since for a null geodesic the spacetime interval is null (zero). $$ds^2 = - c^2dt^2 + dx^2 = 0 \implies \frac{dx}{dt} =c $$ In a locally flat space we can use special relativity to say that light will travel in this way. Hope this clears things up some! |
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