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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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special relativity and general relativity both make predictions for the motion of light--it travels along null geodesics. –  Jerry Schirmer Jan 10 '13 at 6:56
Re "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?" ... some things aren't the same - the speed of light for instance, as discussed in this answer –  twistor59 Jan 10 '13 at 8:53

2 Answers 2

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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You didn't at all explain how the weak equivalence principle would tell how light would bend in an accelerated reference frame or gravitational field. –  FrankH Mar 6 '13 at 6:59
The question is almost a non-question. So I wasn't sure what to put as a response. The weak equivalence principle posits that all bodies have the same "gravitational charge". Test particles of differing masses thus respond identically in a given field. Since this response (at least locally) is simply an acceleration we cannot distinguish the effect of the gravitational field from some other acceleration. Since light would bend in order to reach you in an accelerated frame, the same is true in a gravitational field. Thus redshift, gravitational lensing etc. –  user2053414 Mar 6 '13 at 7:18
I would think he nullified his own question by bringing gravitation into a SR problem. And I have to ask about the null geodesics. Using GR to derive SR? –  C. Towne Springer Apr 7 at 4:58

Think of it this way:

Imagine you are in a rocket ship that has transparent portholes (very small ones) diametrically opposed on the body of the ship, and the ship is accelerating at g. At some instant during flight, a laser beam pulse enters one of the portholes, and as soon as that event occurs, it is detected and another laser apparatus emits a laser pulse that crosses the cabin of the rocket ship at the same instant, alongside the one that entered the porthole.

Is there any reason that the laser pulse emitted from inside the cabin should not be deflected downward any less than the one that came through the porthole? Make the rocket portholes be separated as far apart as it takes in order for the effect to be measurable.

This thought experiment even works if the laser was fired in some other direction (like from a spot on the ceiling of the rocket ship). Additionally, clocks inside the rocket ship also slow down exactly the same amount they would on the surface of the Earth.

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