Baseem T. answered • 04/12/19
Experienced UC Berkeley Graduate (Maths, Chemistry, Physics)
The simplest answer is because Einstein's general theory of relativity (GTR or GR) must reduce to Newtonian gravity in the classical limit of small masses and slow velocities. Newtonian gravity isn't wrong, per se, it just does not accurately describe physics in the regime of the very large and very fast— that's within the purview of GR, which seeks to generalize Newtonian gravity, not undermine it. In grade school, you will tend to investigate the motion of relatively slow objects in small regions of spacetime, and according to the equivalence principle, the effects imparted by the curvature of spacetime in such situations become indistinguishable from being in an accelerating reference frame, which is just a short Lorentz boost away from us appearing to be subject to a downward force that we call gravity.
A Quick Overview from General Relativity (Extremely truncated for brevity) [Feel free to skip!]
Take an object that's moving through the vacuum of space, far from the influence of any gravitational sources. According to Newton's second law, you would predict that, in the absence of any net forces, the object's momentum does not change with time, and that it should continue to travel in a straight line through space. This makes sense if spacetime is Minkowskian, or flat. However, if your spacetime is curved, like the surface of a sphere, "straight lines" are going to look different (for instance, all lines of longitude are "straight lines"), and differential geometry (and especially tensor calculus) lets us use parallel transport to generalize the concept of straight lines to paths called geodesics. Thus, we can generalize Newton's first and second laws to obtain the geodesic equations.
Now suppose that our object is in freefall (meaning there are no non-gravitational forces acting on it) in the vicinity of a massive body. According to GR, this massive body will curve spacetime, and the Einstein field equations are the second-order differential equations for the different components of the metric tensor (which is a mathematical object that lets you describe how to measure "distances" in space and time).
This object will travel along a geodesic, as explained earlier, because it does not have any forces acting on it, and therefore experiences no net acceleration (the magnitude of its four-acceleration is zero), because gravity is a fictitious force that arises as an artifact of being in an accelerating reference frame. However, because of the curvature of spacetime, this geodesic may not look like your traditional idea of a straight line. One must always bear in mind that even though this geodesic is a "straight line" in 4D spacetime, it very well might appear to be a curved path in the embedded 3D space (much like how the orbit of the moon is nearly circular in 3D space, but is a straight line in 4D spacetime).
How does this relate to the question?
Well, if we take the Einstein field equations (which give us a whole set of geodesics in spacetime) and the geodesic equations (which give us the specific geodesic along which the object will travel) and make the following assumptions (the Newtonian or classical limit):
- Spacetime is static (not changing in time)
- Spacetime is approximately flat (the weak-field approximation) [which can be done by assuming that the mass of the massive body is sufficiently small, or by looking at a small enough region of spacetime]
- Motion through space is sufficiently slow (i.e. the magnitude of the three-velocity is much less than the "speed" of light |v| << c)
then one can show that the results reduce to what you would expect from classical mechanics and Newtonian gravity. In fact, the factor of 8πG/c4 sitting in front of the stress-energy-momentum tensor in the Einstein field equations is there to ensure that GR agrees with Newtonian gravity in the classical limit.
The Important Section / Conclusion!
The truth of the matter is that for the applications that you are likely to use in grade school, spacetime is extremely flat, and there's no need to invoke the Einstein field equations if the accuracy you'll gain is far, far smaller than the errors in any of your measuring devices. Newtonian gravity was good enough to get us to the moon and back, but not good enough to predict the precession in Mercury's perihelion (because the planet is so close to the Sun that the curvature of spacetime is far more significant) or the bending of light due to gravitational lensing. In the regime of the very massive, very fast, and very large, treating gravity as a force will begin to introduce large errors.
To give one final visual example of this, think about the paths that radially infalling particles would take as they accelerate toward the Earth— they're basically the paths/geodesics defined by the classical gravitational field lines that resemble the spokes of a bicycle wheel. If you have two particles falling toward the Earth along two adjacent field lines, they'll gradually get closer and closer to one another as they fall over a large distance (because the curvature of spacetime is too significant to ignore over large distances). If one particle looks at another, they'll notice this, and might think that there's a force being imparted on one of them. This is the phenomenon of geodesic deviation. However, over a sufficiently small patch of spacetime, the two trajectories look perfectly parallel, and if the two particles were enclosed in a kind of falling elevator with each other, they wouldn't see either of them moving, and would conclude that there's no force acting on them—this is the concept of an inertial frame. This notion of inertial frames works when you're looking at a small enough region of spacetime such that the curvature is negligible, which is the case for most physics problems that you deal with early in life.
And this is why it's acceptable to treat gravity as a force in the classical regime, but one should always remember that it's a crutch that doesn't apply in the relativistic regime.
