Why would spacetime curvature cause gravity given that the value of gravitational time dilation is so small?

Question

In this question, mpv provides the clearest explanation of the operation of gravity in his answer:

The apple moving first only in the time direction (i.e. at rest in space) starts accelerating in space thanks to the curvature (the "mixing" of the space and time axes) - the velocity in time becomes velocity in space. The acceleration happens because the time flows slower when the gravitational potential is decreasing. Apple is moving deeper into the gravitational field, thus its velocity in the "time direction" is changing (as time gets slower and slower). The four-velocity is conserved (always equal to the speed of light), so the object must accelerate in space.

How can the difference in gravitational potential have an effect when the calculated values of gravitational time dilation and space contraction are so very small? If you take the clock out of a geostationary orbit satellite and put it on the ground it will run slower but only a very little amount slower. And over small length scales this is even less significant.

Secondly does this imply that gravity in Einstein's relativity requires time to pass to operate?

Answer 1

To understand this consider the analogy of two cars driving north from the equator as discussed in the question: Why does the speed of an object affect its path if gravity is warped spacetime? The relevant diagram from that question is:

Due to the curvature of the Earth's surface the two cars converge even though they started out parallel, and this is what happens in general relativity, except that there we need to treat time as a dimension as well.

If you look at the diagram it should be obvious that the faster the cars drive north the faster they will converge i.e. the large the apparent force between them. In fact as discussed in Geodesic devation on a two sphere if the cars start a distance $d$ apart their separation at time $t$ is given by:

$$ s(t) = d \cos\left(\frac{vt}{r} \right) $$

and a quick differentiation later we find the acceleration $d^2s/dt^2$ is proportional to $v^2$. So the apparent force between the two cars is proportional to the square of their speed northwards.

In general relativity we find for a stationary object the four acceleration is given by:

$$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} u^\alpha u^\beta $$

See How does "curved space" explain gravitational attraction? for where this equation comes from. The quantity $u$ is the four velocity, so once again we have a quadratic dependence of the acceleration on velocity just as in the case of the 2-sphere above.

And this neatly explains why even a small curvature causes large accelerations. It's because the magnitude of the four velocity is always equal to $c$, so in the equation for the acceleration we are multiplying the (small) curvature term $\Gamma^\mu_{\alpha\beta}$ by the (very large) term $c^2$. So even at the Earth's surface, where the spacetime curvature is barely measurable, we still get an acceleration of $9.81~\textrm{m}^2$ because we are multiplying that small curvature by $9 \times 10^{18}$.

As for time passing, no it does not imply time flows. In GR the trajectory of an object is a line in a 4D manifold i.e. the set of all possible positions in spacetime of the object. While humans perceive the position on that line to be changing with time there is no equivalent concept of flow in GR. For more on this see What is time, does it flow, and if so what defines its direction?

Answer 2

The resulting gravitational potential (GPE per unit mass) is also "small", compared with $c^2$... but in SI units, it may still be quite large. It's like asking how a speed can be measurably large when it's expected to be small compared with $c$.

does this imply that gravity in Einstein's relativity requires time to pass to operate?

It does in Newtonian physics too; in both cases we want to explain an acceleration, which is a second-order time derivative.

Answer 3

I find it helpful to think of gravity like a low pressure system in the weather. You don't have to consider the difference between the rate of time at altitude and at the earth surface. All you need to know is that the "pressure" of time immediately below the object is slightly lower than the pressure of time immediately above the object. So the object wants to move in that direction, ie downwards.

Think of why a helium balloon rises in the air. It doesn't know that the air pressure 100 metres high is different from that at the surface. All it knows is that the air pressure immediately above it is a tiny bit lower than the air pressure below it. So it moves in that direction - upwards.

So it is only the difference between this pressure immediately above and immediately below the object that matters in the process of falling.