# How does curved spacetime cause motion revisited

There was a previous question titled "Why would spacetime curvature cause gravity?" asked March 10, 2014.

The answer given was essentially that since the time component of an object in curved space slows that automatically has to generate a motion in space so the total spacetime velocity is still c.

So far so good but there seems to be a problem. Suppose another object is sitting on the surface of the earth. Its clock is also slowed due to curved space but there is NO motion in space.

1. So how can the object still have a total spacetime velocity of c if its coordinate time is slowed but it is not moving in space?

2. It doesn’t seem like the answer to the original question works if the slowing of coordinate time has to produce motion in space because there isn’t any motion in space in this case but there is still slowing of coordinate time. Is this correct?

3. In light of this how does curved spacetime cause motion? Is there some other explanation that works or am I missing something?

Thanks, Edgar

• Note that in (1), the object's world line is accelerated, i.e., an accelerometer attached to the object reads non-zero acceleration. If the surface of the Earth were not there, the object would free-fall (accelerometer reads zero) 'downward'. Indeed, relative to a free-falling observer, the object is accelerating. – Alfred Centauri Nov 28 '15 at 2:38
• Think of it this way. Objects in space follow geodesic paths. By virtue of the fact that spacetime is curved, objects must follow a geodesic path which results in motion. – Horus Nov 28 '15 at 12:27

Consider the example of the Schwarzschild metric. If we take an observer who is at fixed $r$, $\theta$ and $\phi$ the metric simplifies to:

$$d\tau^2 = \left( 1 - \frac{2M}{r}\right) dt^2$$

the four-velocity is $dx^\alpha/d\tau$, so in this case we find:

$$\mathbf{u} = \left( \frac{1}{\sqrt{1-2M/r}}, 0, 0, 0 \right)$$

The norm of the four-velocity is given by:

$$u = \sqrt{g_{\alpha\beta}x^\alpha x^\beta}$$

So in this case we find:

$$u = u^0 \sqrt{g_{00}} = \frac{1}{\sqrt{1-2M/r}} \sqrt{1-2M/r} = 1$$

So the norm of the four velocity is indeed $c$ even in this example of a curved spacetime.

When moving from special relativity to general relativity it's easy to forget that computing the norm of a four-vector requires the metric. This is true in special relativity as well of course, but since in SR the entries in the metric all have modulus unity it's easy to overlook it.

Time duration does not cause motion. If you had a spherical shell of matter then the inside has time dilation compared to distant outside, but there is no motion caused in the inside.

On the other hand your example is also totally wrong. An inertially moving object is in freefall and the non inertial earth based coordinate system would say it is falling down.

In fact, an object sitting on the surface of the earth is accelerated upwards in every inertial frame because of the pressure of the ground below it. It sinks into the earth until the pressure is enough to accelerate it upwards at 1 g.

The word General in General Relativity refers to the fact that there are not global inertial frames.