michel123456 Posted May 17, 2010 Posted May 17, 2010 from http://en.wikipedia.org/wiki/Equivalence_principle#Development_of_gravitation_theory The equivalence principle proper was introduced by Albert Einstein in 1907, when he observed that the acceleration of bodies towards the center of the Earth at a rate of 1g (g = 9.81 m/s2 being a standard reference of gravitational acceleration at the Earth's surface) is equivalent to the acceleration of an inertially moving body that would be observed on a rocket in free space being accelerated at a rate of 1g. Einstein stated it thus: "we [...] assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system." (Einstein 1907). That is, being at rest on the surface of the Earth is equivalent to being inside a spaceship (far from any sources of gravity) that is being accelerated by its engines. My comment below. There is a difference: _when a spaceship is accelerated, it is accelerated in the direction of its path. When you are in upon the Earth, what is the direction of acceleration? .... In the rocket case, or in the case of any accelerated body, there is a well determined direction. From left to right, upward, downward, etc. Anyway, along the object's trajectory. In the Earth case, the direction of the acceleration (namely Gravity) is not along any Earth's path. The direction is not from left to right, nor upwards or downwards, the direction is from the outside to the inside. That is a clear difference. Now. We know that gravity exerts a (pseudo) force oriented from the outside to the inside. And we know that this force is exactly the same as a force exerted upon a massive body under acceleration. So, if I take a massive body, and I blow it in an accelerated way, I will recreate a kind of gravity. But, if I take a massive body, and do absolutely nothing, I will observe a kind of gravity, already there. Not "a kind of gravity", but "Gravity". As if the massive object was already blowned in an accelerated way, and me stupid not noticing anything.
swansont Posted May 17, 2010 Posted May 17, 2010 Freefall in a uniform gravitational field is an inertial frame. When you are at rest in that field, what you feel is the force that keeps you from undergoing a net acceleration. By Newton's laws, that force must be in the opposite direction from the force of gravity, and equal in magnitude. The equivalence principle is saying you can't tell the difference between standing on a surface, feeling that acceleration (1g), or being a container in deep space, accelerating at 1g.
michel123456 Posted May 19, 2010 Author Posted May 19, 2010 We are in deep space. And we are accelerating at 1g= 9.81 m/s2. The problem is not that we are accelerating. The problem is the direction of the acceleration. Lets take the example of the container moving linearly. If we were born upon such a spaceship* feeling a force that pushes our bodies against the back of our seat, we could deduce from the laws of physics that the entire spaceship is displacing under acceleration. Everybody on the spaceship would measure the same acceleration, and the vector would point into the opposite direction of the displacement. In our common understanding, all directions, as measured by all the occupants astronauts, will be parallel. In our case, as occupants of the spaceship Earth, we are measuring such an acceleration. So we can conclude very positively that we are moving under acceleration. The weird thing is that the directions of all vectors as measured by all the occupants are not parallel. The directions are radial from the outside to the inside. *the Earth is a spaceship.
PSM5 Posted May 19, 2010 Posted May 19, 2010 The equivalence principle applies only to point masses. What if the accelerating rocket were rotating? What would be it's direction? The milky way galaxy rotates, so how can you know that the rocket is not rotating?
michel123456 Posted May 19, 2010 Author Posted May 19, 2010 I posed moving linearly. Let's suppose the rocket is rotating. How could it manage to get an acceleration corresponding to a force from the outside to the inside (centripetal)? Take in mind that the answer must correspond to the Earth's hypothetical movement, but also to any material body, since all material bodies are the source of some gravitational field.
swansont Posted May 19, 2010 Posted May 19, 2010 We are in deep space. And we are accelerating at 1g= 9.81 m/s2. The problem is not that we are accelerating. The problem is the direction of the acceleration. Lets take the example of the container moving linearly. If we were born upon such a spaceship* feeling a force that pushes our bodies against the back of our seat, we could deduce from the laws of physics that the entire spaceship is displacing under acceleration. Everybody on the spaceship would measure the same acceleration, and the vector would point into the opposite direction of the displacement. In our common understanding, all directions, as measured by all the occupants astronauts, will be parallel. In our case, as occupants of the spaceship Earth, we are measuring such an acceleration. So we can conclude very positively that we are moving under acceleration. The weird thing is that the directions of all vectors as measured by all the occupants are not parallel. The directions are radial from the outside to the inside. *the Earth is a spaceship. The earth is a sphere. It cannot be considered a uniform gravitational field for macroscopic objects.
PSM5 Posted May 20, 2010 Posted May 20, 2010 Let's suppose the rocket is rotating.How could it manage to get an acceleration corresponding to a force from the outside to the inside (centripetal)? Take in mind that the answer must correspond to the Earth's hypothetical movement' date=' but also to any material body, since all material bodies are the source of some gravitational field.[/quote'] I wasn't trying to answer a question because I do not completely understand what your point is. But I did find your post interesting and that is why I commented. A rotating accelerating rocket will have a force that is opposite from that of gravity. But they do share something in common. The direction of their forces both converge at a center.Just thought it was an interesting observation. So, if I take a massive body, and I blow it in an accelerated way, I will recreate a kind of gravity. That's the part I don't understand.
michel123456 Posted May 20, 2010 Author Posted May 20, 2010 Say you are standing on the surface of a huge hollow sphere of mass approximatively zero (for simplicity). If I blow into the sphere in such a way that the sphere expand, you will feel a push. I can technically blow the sphere at a rate of 9.81 m/s2=g. Of course we know that the Earth is not expanding is such a way. And even if the Earth was actually expanding, what about all the other objects around us? Gravity is not a property of the Earth only, it is a property of all massive objects. Or, putting it another way, if I dare to consider that the Earth could expand in such a way, I should also consider that all objects around me are expanding in the same way too. Including myself. Of course, that sounds totally crazy, because we are not observing such an expansion happening. All that we are observing is a force that is equal to an acceleration.
PSM5 Posted May 21, 2010 Posted May 21, 2010 Okay, I think I'm understanding your thought experiment now. I'm just not sure how it relates to the equivalence principle proper. Your OP claim that there is a difference between the acceleration of an object in free fall and an accelerating rocket is true for macroscopic objects. From the same Wikipedia page that you linked: So the original equivalence principle, as described by Einstein, concluded that free-fall and inertial motion were physically equivalent. This form of the equivalence principle can be stated as follows. An observer in a windowless room cannot distinguish between being on the surface of the Earth, and being in a spaceship in deep space accelerating at 1g. This is not strictly true, because massive bodies give rise to tidal effects (caused by variations in the strength and direction of the gravitational field) which are absent from an accelerating spaceship in deep space. I have read some text that claim the equivalence principle applies only to point masses. But that was not part of Einstein's original.
michel123456 Posted May 21, 2010 Author Posted May 21, 2010 Okay, I think I'm understanding your thought experiment now. I'm just not sure how it relates to the equivalence principle proper. Your OP claim that there is a difference between the acceleration of an object in free fall and an accelerating rocket is true for macroscopic objects. From the same Wikipedia page that you linked: I have read some text that claim the equivalence principle applies only to point masses. But that was not part of Einstein's original. The difference I want to point out is about direction. When there is a slight difference between 2 phenomenas, there are 2 ways of reasonning: 1. to emphasize the difference in order to prove that the 2 phenomenas are completely different. In this case, you can say A=B but A has another physical nature of B, due to the mentionned "slight difference". That's another way of saying "physically equivalent". 2. to disform everything in order to prove that the 2 phenomenas are exactly the same. In this case, you can say that A=B, point. They are exactly the one and same phenomena, they are "physically the same", and the slight difference is caused by [fill in your hypothesis].
PaulS1950 Posted May 21, 2010 Posted May 21, 2010 If I understand what you are suggesting, and I may not be; On the inside of an expanding sphere (that is stationary), it (the wall of the sphere) would be falling away from you as it is inflated. On the outside of the expanding sphere you would feel the expansion of the sphere in much the same way that gravity would affect you. If the sphere was accelerating it would depend on where in or on the sphere you were positioned as to the affect that the combination of the acceleration and expansion would be experienced by you. Paul, the 60 year old student
michel123456 Posted May 21, 2010 Author Posted May 21, 2010 You have understood very well. In fact, i was proposing to stand on the external part of the surface. And it works only under acceleration. An expanding sphere at regular rate don't product any more effect than moving through space at regular speed: impossible to detect. If you prefer, you can change the words "expanding sphere" by the words "scale factor".
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