# Equivalence priciple, light bend

by tomz
Tags: gtr
 P: 35 in my text book it say that a frame which move at constant velocity in deep space and a frame under free fall in uniform gravitational field will be Equivalent. There is no experiement that observer can do to see whether he is in which of the two frame. But it later say that light bent in gravitation field. So is that means observer can use a beam of light to see whether he is under free fall or in deep space? It is contract to what Equivalence priciple says? thank you
 Sci Advisor P: 2,801 The equivalence principle works only "locally". This means that it works on scales where you can't see the global curvature of the space. Definitely, there is a difference between an accelerated reference frame and a gravitational field; the gravitational field must die off as you get very far from it. But locally, they are equivalent.
 P: 3,187 From the Equivalence principle you should expect that light will similarly bend in a (hypothetical) uniform gravitational field. As a result, if you are in free fall in such a field you should not notice any bending relative to your falling system, so that it looks to you (by only looking "inside") exactly the same as if you are not falling. And light is indeed deflected by gravitational fields. Perhaps that book doesn't sufficiently elaborate. 1. If you accelerate in a rocket without gravitation then light will bend relative to your rocket and you will feel the force of acceleration. Equivalence principle: 2. If you feel the same force due to gravitation (neglecting that it is not perfectly uniform) then light will bend just the same. 1+2. If you feel no force then the light will not bend relative to your rocket. It's similar to if you are falling and a stone is falling next to you: the stone will not fall relative to you.
 Sci Advisor PF Gold P: 5,059 Equivalence priciple, light bend Actually, as I read it, the OP is citing the more modern formulations of EP, rather than Einstein's elevator. What is described is local lorentz invariance. However, the answer is still the same. If your free fall 'lab' is big enough to include detectable curvature, it can be distinguished from uniform motion in empty space - it is too big to be local. You don't need light - just measure that two objects at rest in 'space' in your lab move towards each other over time (beyond what might be expected from their mutual attraction). This would be due to their geodesics converging due to curvature.
P: 8,544
 Quote by harrylin From the Equivalence principle you should expect that light will similarly bend in a (hypothetical) uniform gravitational field. As a result, if you are in free fall in such a field you should not notice any bending relative to your falling system, so that it looks to you (by only looking "inside") exactly the same as if you are not falling. And light is indeed deflected by gravitational fields. Perhaps that book doesn't sufficiently elaborate. 1. If you accelerate in a rocket without gravitation then light will bend relative to your rocket and you will feel the force of acceleration. Equivalence principle: 2. If you feel the same force due to gravitation (neglecting that it is not perfectly uniform) then light will bend just the same. 1+2. If you feel no force then the light will not bend relative to your rocket. It's similar to if you are falling and a stone is falling next to you: the stone will not fall relative to you.
To add a bit: So 1 + EP implies that light will bend in a gravitational field.

However, since this bending of light is predicted by the EP, it is local bending.

The famous Eddington sorts of experiments detect global=EP+curvature light bending.

Nordstrom's theory and GR both obey the EP, and thus predict local light bending. However, due to different predictions on curvature and light, the global light bending in Nordstrom's theory is zero, but in GR it is twice the EP prediction.

http://www.einstein-online.info/spot...nce_deflection
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,502 The equivalence principle requires that light bend in a gravitational field. Imagine falling freely in a gravitational field in a small room. Light comes through a small hole in one wall and illuminates a spot on the other wall. Since the room is falling freely, it will accelerate during the time the light takes to cross the room. If the light did not "fall" in the gravitational field in exactly way the room does, it would hit the opposite wall higher than if the room were not falling. Because light does bend downward, it goes downward as much as the room does and falls on then opposite wall exactly where it would if there were no gravity.
PF Gold
P: 5,059
 Quote by HallsofIvy The equivalence principle requires that light bend in a gravitational field. Imagine falling freely in a gravitational field in a small room. Light comes through a small hole in one wall and illuminates a spot on the other wall. Since the room is falling freely, it will accelerate during the time the light takes to cross the room. If the light did not "fall" in the gravitational field in exactly way the room does, it would hit the opposite wall higher than if the room were not falling. Because light does bend downward, it goes downward as much as the room does and falls on then opposite wall exactly where it would if there were no gravity.
Ah, yes, and following this you can specify when your 'lab' is too big to be local: if it is impossible to aim two laser beams so as to maintain exactly the same distance between them over the length of the lab.
P: 3,967
 Quote by tomz in my text book it say that a frame which move at constant velocity in deep space and a frame under free fall in uniform gravitational field will be Equivalent. There is no experiement that observer can do to see whether he is in which of the two frame.
Light will not bend in a free falling lab just as it does not bend in a constant velocity lab in deep space.
 Quote by tomz But it later say that light bent in gravitation field.
Perhaps the book should of made it clear that light will also bend in an accelerating lab in deep space. If the acceleration due to gravity and the artificial acceleration of the lab in deep space are the same, then the lab will bend by the same amount in both cases.

 Quote by tomz So is that means observer can use a beam of light to see whether he is under free fall or in deep space? It is contract to what Equivalence priciple says?
Hopefully you can now see that the bending of a beam of light will not determine if a lab is in free fall or in deep space and so no contradiction with the Equivalence Principle (EP). Also as others have mentioned if the lab is very large, tidal effects may be detected which will differentiate between the two cases, but the EP is defined as the case where the lab is sufficiently small or local that tidal effects are not noticeable.

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