Why does mass bend the temporal dimension more than the spatial dimensions of spacetime?

Question

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?

Answer 1

most of what we experience as gravity is a result of the distortion of the temporal dimension, and not the spatial dimensions.

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?

Answer 2

It is the result of our choice of time and spatial coordinates. For practical reasons, we use seconds and meters, so the accelerations of gravity in the earth and solar system are far from negligible.

Suppose that our perception of time were such that we used microseconds instead of seconds, while keeping meters for spatial distances. Everything would seem to be almost at rest, because our daily objects barely moves on a scale of meters and micoseconds. And if things were viewed as at rest, the situation would be similar to a flat Minkowskian spacetime both in temporal and spatial dimension.

Answer 3

There is actually no such thing as curvature in one dimension, so the premise of the question is based on a misunderstanding. When we talk about curvature in general relativity, we mean intrinsic curvature, such as the curvature of a basketball that can be detected by a bug that never leaves the surface of the basketball and can't conceive of a third spatial dimension. The bug can detect phenomena like the fact that the angles of a triangle add up to more than 180 degrees. Intrinsic curvature can't exist for a one-dimensional curve, e.g., a circle has no intrinsic curvature. For these reasons, curvature always involves at least two dimensions.

At a fancier mathematical level, we can see this because the Riemann curvature tensor is antisymmetric, but an antisymmetric tensor in one dimensions is zero.

Answer 4

Why aren't all four dimensions distorted equally

Although spacetime is often used term one should clearly understand that space and time aren't interchangeable one into another; similarly when they talk about curvature of space-time it's by definition curvature of space relatively to time.

Why aren't all four dimensions distorted equally

If they would have been equally distorting you won't notice any "distortion" being an observer belonging to the same "equally distorted space-time".