From my (limited) understanding of general relativity, most of what we experience as gravity is a result of the distortion of the temporal dimension, and not the spatial dimensions. Therefore, most of the spacetime curvature caused by the earth (and most astronomic objects, with the exception of maybe black holes) occurs along the temporal dimension, with very little on the spatial dimensions. This is why the bent sheet analogy is misleading, if I am not mistaken. Why is this so? Why aren't all four dimensions distorted equally, or the spatial dimensions distorted more than the temporal?
[Physics] Why does mass bend the temporal dimension more than the spatial dimensions of spacetime
general-relativitygravityspacetime
Best Answer
This is kind of true, but also kind of not true. The coordinate acceleration of a freely moving object is given by the geodesic equation:
$$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu{}_{\alpha\beta} u^\alpha u^\beta $$
This is not as scary as it seems at first glance. The left side is basically just the acceleration we measure for the falling object. On the right side the symbol $\Gamma^\mu{}_{\alpha\beta}$ are the Christoffel symbols that describe the curvature of spacetime, while the symbols $u^\alpha$ are the four-velocity.
The four velocity is given by:
$$ \mathbf u = \left( c\frac{dt}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau} \right) $$
where $dx/d\tau$ etc are basically the spatial velocity of the particle, and $dt/d\tau$ is the velocity through time. By the velocity through time I mean the rate we move through time i.e. one second per second if we are stationary.
The key thing we need to note is the factor of $c$ in $c~dt/d\tau$. Suppose I am moving along the $x$ axis at 100 m/s, which is quite fast by everyday standards, then my my four-velocity will be approximately;
$$ \mathbf u \approx (c, 100, 0, 0) $$
i.e. the time component $c$ is around $10^6$ times greater than the space component. So to a good approximation at everyday speeds we are only moving through time and the space components of the four-velocity can be ignored. Then the geodesic equation simplifies to:
$$ {d^2 x^\mu \over d\tau^2} \approx - \Gamma^\mu{}_{tt} c^2 $$
So of all the Christofel symbols only the four symbols $\Gamma^\mu{}_{tt}$ matter. This is what is meant by the statement that only the curvature in time matters.
The point of all this is that it is not true to say that the time dimension is curved more or less than the spatial dimensions. However at everyday speeds it is true to say that gravity is mostly due to the time curvature because we are moving along the time axis around a factor of $c$ faster than we are moving along the space axes.
For a more popular science level explanation of how the speed affects the "force" see my answer to Why does the speed of an object affect its path if gravity is warped spacetime?