I would like to preface this by saying that this isn't necessarily meant to be a full question, though it may become that, but it rather a confirmation of my understanding of a concept.

Newton would say that an individual standing on the earth represents an Inertial frame of Reference. In this scenario, if an object, let's say an apple, is released above the ground, the observer would suggest that the apple accelerated towards the ground due to gravity.

Einstein on the other hand would suggest that an individual standing on the earth is NOT an Inertial frame of reference, and that the apple itself represents an inertial frame of reference. More broadly, an inertial frame of reference is one in which an object has no forces acting on it, such as an object in deep space, or in free fall, not being PULLED by gravity, but rather following geodesic a through curved spacetime, just as the apple is doing as it follows a geodesic a towards the Earth.

If we accept this to be the Inertial frame of reference, the logical conclusion is that an apple located above the Earth does not accelerate towards Earth, but rather, from the Inertial frame of reference, from which we measure true acceleration, the Earth is accelerating upwards (radially outwards).

My understanding of the radially outwards acceleration, which is the concept that I would like to be confirmed or denied, is as follows:

Due to the curvature of spacetime, the 'future' of the Earth in spacetime is bent due to it's own matter, inwardly, or to put it simply, it should be collapsing into itself. However, due to the rigidity (may not be the correct term, but I assume it is evident what I mean... Can clarify) of the planet, it does not collapse inward. This causes the Earth to be, relative to the inertial frame, accelerating away from it's future path through spacetime, at 9.8m/s/s, exactly the rate at which it would collapse inward if this wasn't the case.

**Note: I apologize for the rather long-winded question, but this was a very broad concept for me, and I'm afraid that my education isn't refined enough to be able to concisely explain this.

My thanks for clarification on this matter.

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    $\begingroup$ The surface of Earth is not a inertial system and Newton's laws are not valid around here, so Newton would never have said that. $\endgroup$ – CuriousOne Jun 28 '16 at 4:45

Newton would not say that an individual standing on the Earth represents an inertial frame of reference. An inertial frame is one in which Newton first law applies i.e. an object moves in a straight line at constant speed. Since the observer dropping the apple observers the apple to accelerate that observer's frame is non-inertial.

However you are correct, or rather almost correct, to say that the apple's frame is inertial. If you were falling alongside that apple (ignoring air resistance) then you would observe the apple remaining stationary beside you. A good example of this is the astronauts in the International Space Station. They are falling freely so, as countless videos show, anything they release remains next to them, or moves away at constant velocity.

I say almost correct because in general relativity inertial frames are normally only locally inertial. The apple appears to be stationary relative to you, but actually it is slowly approaching you because you and the apple are both moving along different lines towards the centre of the Earth. The farther away from you the apple is the greater will be this relative motion.

But I'm guessing that your real interest is in describing the acceleration of the Earth's surface, and you are quite correct that it is accelerating. Technically its proper acceleration is non-zero. The proper acceleration is calculated using a somewhat complicated equation but basically it is the acceleration relative to a freely falling object, and as you say relative to the apple the surface of the Earth is accelerating outwards at $9.81$ m/s$^2$.

Another convenient definition is that the proper acceleration is the acceleration felt by an observer in their rest frame. For example I weigh 68 kg and right now in my rest frame I'm experiencing a force of $9.81 \times 68$ Newtons, which is just the upwards force exerted on my by my chair. My proper acceleration acceleration is just this force divided by my mass i.e. $9.81$ m/s$^2$.

Likewise the surface of the Earth is being pushed up by the rock/whatever immediately under the surface, which is being pushed up by the stuff immediately under it and so on down to the centre of the Earth (where the gravity is zero so in fact the centre of the Earth is approximately an inertial frame).

It seems weird to be talking about something that is stationary as accelerating, but this is normal in general relativity. In everyday life when we talk about acceleration we normally mean coordinate acceleration i.e. if we measure the position of an object using some coordinate system (e.g. latitude, longitude and altitude) then acceleration means the position is changing as measured using those coordinates. However GR tells us that the latitude/longitude/altitude are actually accelerating coordinates not inertial ones, and to be stationary in an accelerating coordinate system is to be accelerating in an inertial coordinate system.

If you're interested in What is the weight equation through general relativity? there is a detailed calculation of the proper acceleration in a spherical gravitational field like the Earth's. However you might find this a bit hard going if you're new to GR.

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    $\begingroup$ I was under the impression that Newton would consider the frame of reference of, not necessarily a person on Earth's surface, but Earth's surface itself (I apologize for this being unclear) would be considered to be inertial because it wouldn't be accelerating, since it would be 'stationary', as it wouldn't be falling towards the Earth. While I may need to have this clarified, you're right that the crux of my question was on the radial acceleration of the Earth. Thank you for your very clear answer. $\endgroup$ – Kieran Moynihan Jun 28 '16 at 5:39

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