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I'm just reading a book about gravity. An example it gives is a spaceship accelerating. A beam of light travelling at right angles to the direction of movement of the spaceship enters it via a small pinhole. The book states that the beam of light would appear to bend to an observer within the spaceship. Intuitively this makes sense - if the ship were travelling at $0.5\,c$, for example, and was $10\,m$ across (from the pinhole to the opposing wall) then I'd expect the light to hit the wall $5\,m$ further back down (assuming the ship is accelerating "up") from the pinhole - the wall would have moved 5m further in the time it takes the light beam to cross the interior to the opposing wall

I think I'm fundamentally misunderstanding though. The book states this is down to the acceleration of the ship and goes on to talk about how using equivalence the same bending would be a result of gravity. In my description above though, it's the velocity that causes the apparent bending, and the same would be visible if the ship was maintaining a constant $0.5\,c$. I suspect that if this were the case (constant $0.5\,c$) there would be no bending at all, but I'm not understanding why

Can anyone enlighten me?

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Hi Kevin, and welcome to Physics Stack Exchange! Despite appearances, your question doesn't have all that much to do with gravity, so I added some other relevant tags. –  David Z Mar 27 '13 at 13:28
    
Thank you David. You're right of course. My aim is a better understanding of gravity/relativity, which is what made me choose gravity, but this question is quite early in the journey, I think –  Kevin O'Donovan Mar 27 '13 at 15:27
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It's not the velocity that causes the bending, it's the acceleration! If the spaceship were moving by any but constant velocity in the absence of a gravitational field, the path of the photons would be straight, wouldn't it? The motion of any uniformly moving object (or photon) always looks straight in any other inertial (uniformly moving) frame. The path gets curved, "parabolic", just because the velocity isn't constant.

The equivalence principle equates the situations of an accelerated spaceship with a static spaceship sitting in the gravitational field, with the correspondence $a\sim g$. So if the path of the photon is curved in the accelerated spaceship, it should be curved in the gravitational field (but with no extra motion), too, although I am not quite sure whether this simple argument produces the right factor of two that a naive Newtonian "attraction acting on light" misses relatively to the correct general relativistic calculation.

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I think that's where I'm struggling. I'm an engineer rather than a physicist, so this isn't my domain. I think I'm being extremely naïve. If the photon enters through the pinhole the ship will have moved on from its original position by the time it hits the wall, displacing the point of impact. Ah, I may have missed the point. In the presence of acceleration the path traced to get to that point would be a curve. Without acceleration there would still be a displacement but the path would not be a curve. Is that right, or am I still missing the point. –  Kevin O'Donovan Mar 27 '13 at 15:25
    
It's one of the points, there are also other points - so far, you haven't gotten to the key point involving the equivalence principle. Could you please be more specific about which sentence you don't understand? Otherwise it is pretty much impossible to deduce what you're struggling with, it remains an esoteric purely internal psychological struggle no one else can understand. –  Luboš Motl Mar 28 '13 at 6:49
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Kevin you are quite right in your assumption, although these are two different situations. If the ship was just travelling at a speed, the beam would be displaced travelling across the ship, so in the ship's frame of reference it would appear travelling at an angle, as if it was shone this way - this is called aberration of light.

If the ship is accelerating, the displacement taking place would have a more complex form, a curve, corresponding to the ship's motion. But obviously in the external observer's reference frame the beam would still be travelling straight.

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