How does relativity explain gravity, without assuming gravity [duplicate]

Question

I have seen the "objects pull down on space-time" explanations, but they assume a "pull down" force themselves. Could anyone explain the space-time explanation without assuming gravity in the first place?

Answer 1

Why don't you want to assume gravity? Gravity it is an experimental fact, a starting point for doing physics. General Relativy is a geometrical theory of gravity, built on the basis of Special Relativity and always having in mind that it should recover the non-relativistic Newtonian theory of the gravitational field.

The "pull down" is a deviation of the flat, Minkowski spacetime, governed by the Einstein Field equations. And in my opinion a not-so-good analogy because it is rather difficult to imagine how to pull down time.

Answer 2

but they assume a "pull down" force themselves.

The images of flat sheets "pulled down" where the planets are do not reflect the fact that the curvature of spacetime is an intrinsic curvature that is measured by geodesic deviation.

What has been done, in order to help visualize the spatial curvature, is to take a two dimensional spatial slice and then embed that into a fictional, flat 3D space where the intrinsic curvature of the slice is represented as an extrinsically curved 2D surface.

A good example of how this is done for a spherically symmetric static star can be found in the book "Gravitation" on page 613:

Therefore, depict 3-space only as it is at one time, t = constant. Moreover, at any one time the space itself has spherical symmetry. Consequently, one slice through the center, $r=0$, that divides the space symmetrically into two halves (for example, the equatorial slice, $\theta = \pi/2$ ) has the same 2-geometry as any other such slice (any selected angle of tilt, at any azimuth) through the center. Therefore limit attention to the 2-geometry of the equatorial slice. The geometry on this slice is described by the line element

$$ds^2 = [1-2m(r)/r]^{-1}dr^2 + r^2d\phi^2.$$

Now one may embed this two-dimensional curved-space geometry in the flat geometry of a Euclidean three-dimensional manifold.

Read more using Google Books.

Answer 3

Massive objects distort spacetime, as described by the Einstein Field Equations. In turn, this causes particles to accelerate: the GR equivalent of $\mathbf{F}=m\mathbf{a}$ are the geodesic equations: $$ \frac{\text{d}^2x^\alpha}{\text{d}\lambda^2} + \Gamma^{\alpha}_{\mu\nu}\frac{\text{d}x^\mu}{\text{d}\lambda}\frac{\text{d}x^\nu}{\text{d}\lambda} = 0,\qquad \alpha=0,\ldots 4, $$ with $$ \Gamma^{\alpha}_{\mu\nu} = \frac{1}{2}g^{\alpha\beta} \left(\frac{\partial g_{\beta\mu}}{\partial x^\nu} + \frac{\partial g_{\beta\nu}}{\partial x^\mu} - \frac{\partial g_{\mu\nu}}{\partial x^\beta}\right), $$ the so-called Christoffel symbols, and $g_{\mu\nu}(x^\alpha)$ the spacetime metric. In the absence of matter, $g_{\mu\nu}(x^\alpha)$ is constant, so that the geodesic equations reduce to $$ \frac{\text{d}^2x^\alpha}{\text{d}\lambda^2} = 0, $$ which describe a constant motion, as expected.