Needles to say I am a visitor here. I do not belong to the science world;)
But I have read both of these things before:
I don't think these can both be true. I'd appreciate any clarification - thank you.
The difficulty here is one of definitions.
At the kinds of energies that apply to falling apples on Earth you can do physics correctly in either view. We teach the former in physics 101, but the latter has a pretty strong claim to being more fundamental.
To explain this we need to examine exactly what we mean by acceleration in general relativity, and this will take a while. But bear with me and we’ll do it step by step.
Let’s suppose it’s you rather than an apple that is falling. So I’m standing on the Earth’s surface and you’re falling freely towards me. We’ll ignore air resistance so you have an acceleration equal to Earth’s gravity.
To measure acceleration we have to measure how position changes with time, so I’ll choose some coordinates with myself stationary at the origin. Then I can measure distances with my rulers and times with my clock, so I can measure the change in your height $r$ with time $t$ and calculate your acceleration:
$$ a = \frac{d^2r}{dt^2} = g \tag{1} $$
And what I’ll find is that your acceleration is equal to the gravitational acceleration $g$ or about 9.81 m/s$^2$. So in my coordinates I am stationary and you are accelerating downwards towards me.
But you also have a ruler and clock, and you can also use these to measure distances and times. So you also choose a coordinate system with yourself stationary at the origin, then use this to measure the distance to me $r’$ and the time $t’$. Then you calculate my acceleration $a’$ using:
$$ a’ = \frac{d^2r’}{dt’^2} = -g \tag{2} $$
And what you find is that I am accelerating upwards towards you with an acceleration of $-g$, which is exactly the opposite of what I found. So who is correct?
Well what general relativity tells is is that both you and I are correct. That is because what we have measured is called the coordinate acceleration and the value of the coordinate acceleration depends on what coordinates are being used to measure it. The coordinate acceleration has no special significance in GR.
But in GR there is a form of acceleration that all observers will agree on, and this is called the proper acceleration. Relativity describes spacetime as a four dimensional manifold, and vectors like acceleration have four components. So we call them four-vectors, and the acceleration is a four-vector called the four-acceleration. The proper acceleration is just the magnitude of the four-acceleration.
In Newtonian mechanics the (coordinate) acceleration is given by the equation we used above:
$$ a^\alpha = \frac{d^2r^\alpha}{dt^2} \tag{3} $$
Note that I’m using index notation here. The acceleration $\mathbf a$ and the position $\mathbf r$ are three-vectors that we can write as $(a^0, a^1, a^2)$ and $(r^0, r^1, r^2)$. Here $a^0$ means the $x$ component of $\mathbf a$, $a^1$ means the $y$ component and $a^2$ means the $z$ component. So in the equation above the value of $\alpha $ ranges from $0$ to $2$. This is just a compact way of writing the three equations for the three components of the acceleration.
Anyhow, in general relativity the four-acceleration is given by a similar equation but it has an extra term that is related to the curvature of spacetime:
$$ a^\alpha = \frac{d^2x^\alpha}{d\tau^2} + \Gamma^\alpha_{\,\,\mu\nu}\frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} \tag{4} $$
The first term looks similar to the Newtonian equation (3) (there are some subtle differences but we’ll gloss over these) but now we have a second term involving the symbols $\Gamma^\alpha_{\,\,\mu\nu}$. These symbols are called Christoffel symbols and they tells about the spacetime curvature.
Assuming you’ve made it this far, let’s go back to where we started and look at the four-accelerations of me standing on the ground and you falling from the sky. When I calculate my four-acceleration the first term goes to zero so my four acceleration is:
$$ a^\alpha_\text{me} = \Gamma^\alpha_{\,\,\mu\nu}\frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} $$
So my four-acceleration is not zero. In fact my acceleration is equal to the gravitational acceleration, which makes sense because that’s exactly what gravitational acceleration is. It’s the contribution to the four-acceleration caused by the curvature of spacetime. If you're interested the calculation of my proper acceleration is described in detail in What is the weight equation through general relativity?.
To calculate your acceleration we need to know that the motion of a freely falling object is described by the geodesic equation:
$$ \frac{d^2x^\alpha}{d\tau^2} = -\Gamma^\alpha_{\,\,\mu\nu}\frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} $$
And, perhaps surprisingly if we use this to substitute for $ d^2x^\alpha/d\tau^2$ in equation (4) we get the result for your four-acceleration:
$$ a^\alpha_\text{you} = 0 $$
So your four-acceleration is zero.
And finally we can make sense of the statement in your question:
the apple falls to ground because the ground is rushing up to meet the apple (which is actually suspended in space) because of Earth's acceleration through space.
The Earth’s surface is accelerating upwards in the sense that its proper acceleration is not zero, and both I and you/the apple will agree on this despite our different coordinate systems.
In statement 2) you make (which is a Newtonian vision), you assume the earth accelerates through space in the direction towards the apple. But the direction of the earth's acceleration is always directed towards the sun (ignoring the other celestial bodies in our solar system). The apple though will be attracted to the sun in the same way as the earth is, so it can't "hang" motionless somewhere in space trough which the earth passes during her motion. But if the apple would hang still somehow, then it still wouldn't accelerate towards the earth, because the earth is moving through space with constant velocity. So the apple would "fall" towards the earth with constant velocity. It would fall perpendicular towards earth if it would be hanging around (as a figure of speech) exactly in the center of the "cylinder" the earth traces out while moving trough space and would hit the earth at an angle between 0 (if it would hang out at the edge of the cylinder) and 90 degrees (as said, when it would hang around at the centre of the cylinder), and wouldn't hit the earth all.
Statement 2) comes much closer to what's going on. The apple isn't pushed (or pulled) though (which actually happens according to this old theory of gravitation, but please don't think this model shows you what's happening) precisely because the curvature of spacetime. An electron (in a non-curved spacetime) would "feel" a push by a negative charge distribution, which is essentially what makes gravity different than the three basic forces (the weak, the strong and the electromagnetic interaction). If you would find yourself in a closed space together with the apple it would seem like you weren't being accelerated at all (this is, in fact, an approximation, since if you were in the same closed space near a black hole this argument wouldn't hold because of your size, because tidal forces would make it feel very unpleasant for you, to say the least; only very, very locally, if you had a sub-atomic size, it would seem you were feeling no pull).
So essentially, statement 2) comes closest to what's going on.
Number 1 is correct. Number 2 is incorrect. The apple has the same velocity as the Earth since it grew on earth. Also, if the earth were moving into the apple you wouldn't see the apple accelerating unless the Earth is constantly linearly accelerating at 9.8m/s^2, which is impossible.
