0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Alfred Centauri Nov 28 '15 at 2:38
  • $\begingroup$ 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. $\endgroup$ – Horus Nov 28 '15 at 12:27
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$
  1. The "total spacetime velocity" refers to the four-velocity (c, 0, 0 , 0) where even though the object is still in space, it is moving through time. The spacetime coordinates are given by (ct, x, y, z) differentiating which we get the four velocity, and since the object is at the same pt. in space, the four-velocity is c.

  2. & 3. Let us take the spherically symmetric solution of Einstein's field equation, also called the Schwarzchild solution described by the metric

$$ ds^2 = (1-2GM/r)dt^2 - (1-2GM/r)^{-1}dr^2 - r^2d\Omega^2 $$

Here we want to find see how the time interval emitted by a stationary clock at $r_1 $ changes at some arbitrary location $r_2. $ Since the proper time interval $\delta \tau$ is invariant, we have

$$ {\delta \tau}^2 = (1-2GM/r_1){\delta t_1}^2 = (1-2GM/r_2)\delta{t_2}^2$$

where $\delta t_1$ and $\delta t_2$ are the time intervals measured at $r_1$ and $r_2$ respectively. The consequences from the above relation can be worked out if $r_1 < r_2$ or vice versa. So gravitational time dilation can occur even if things aren't in motion.

$\endgroup$
  • $\begingroup$ Respectfully, this doesn't seem to answer any of the questions I asked. It merely restates the obvious fact that gravitational time dilation can occur even if things aren't in motion. My basic question is why does spacetime curvature produce motion in space? And if everything always travels at c through spacetime how does that work in a gravitational field when there is no motion in space even though time is slowed? $\endgroup$ – Edgar Owen Nov 28 '15 at 1:12
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.