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?

  • $\begingroup$ If you know some calculus, you might like this gentle introduction to GR: hepweb.ucsd.edu/ph110b/110b_notes/node72.html $\endgroup$
    – PM 2Ring
    Commented Dec 18, 2021 at 7:12
  • $\begingroup$ The bent sheet analogy is mistaken because it is embedded in space and its in this space it is bent. However, spacetime has nothing to bend in. Still, its a theorem of Whitney that we can embed any smooth manifold into some flat space. And a theorem of Nash that if the manifold is Riemannian we can do this isometrically. Spacetime manifolds aren't Riemannian, they are Lorentzian. However, a generalisation of Nash's theorem says semi-Riemannian manifolds can also embed isometrically (apparently the proof is simpler) and hence Lorentzian manifolds, and hence spacetime can too ... $\endgroup$ Commented Dec 18, 2021 at 7:24
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    $\begingroup$ ... and so the analogy is not as mistaken as all that! $\endgroup$ Commented Dec 18, 2021 at 7:25
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    $\begingroup$ Glibly: Because relatively speaking, we are moving through much more time than space. $\endgroup$ Commented Dec 18, 2021 at 14:08
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    $\begingroup$ I asked a related question a while ago: How much of gravity is caused by time dilation? (possibly a duplicate). Quoting from the accepted answer: "Most everyday objects on Earth [...] have a much larger temporal component of velocity than spatial component of velocity" $\endgroup$
    – jng224
    Commented Dec 18, 2021 at 14:35

4 Answers 4


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?

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    $\begingroup$ I am no mathematician, but I almost understand this. I see, at least, that if movement on the x-axis is, say, .5c, then the relativistic effect (?) of that comes into play because we are no longer at "everyday speeds." I upvoted this answer although the math is otherwise beyond me. $\endgroup$
    – Wastrel
    Commented Dec 18, 2021 at 15:54
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    $\begingroup$ @Wastrel yes, exactly. If we are moving at speeds approaching $c$ then the curvature in the time dimension no longer dominates and the motion would be determined by the spatial curvature as well. $\endgroup$ Commented Dec 18, 2021 at 15:57

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.

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    $\begingroup$ This is the answer to OP's question as stated. The only way to say one is "more" than the other is with numbers in terms of an arbitrary choice of units. $\endgroup$ Commented Dec 19, 2021 at 13:20
  • $\begingroup$ That is a very interesting observation! But in reality our units are adequate to the speeds we travel in the time dimension (fast, therefore coarse granularity) and the spatial dimension (slow, hence fine granularity). If we were huge and fast our units would look different, and we'd be affected more by the spatial distortion. $\endgroup$ Commented Dec 20, 2021 at 3:31

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.


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".


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