Are time and gravity affected when at rest compared to free fall?

A falling object moves along a geodesic path ('straight path') in spacetime. When it comes to rest it now follows a 'curved path' through spacetime. Is the passage of time and force of gravity fundamentally affected by this difference?

To be really clear, assume one object is falling , at height x from the ground and another is hovering at height x. Do they experience the same gravitational pull? Will they age at the same rate?

I feel I have mangled the correct terminology, forgive me. Thanks!

• Just to note, when speaking about relativity "at rest" and "in free fall" are generally used interchangeably, since they both refer to geodesic paths. If you're standing on the surface of a planet, you're being accelerated by the surface and thus no longer on a geodesic path -- which nonetheless, confusingly, is the state called "at rest" in Newtonian language. – Asher Jan 14 '15 at 0:41
• @Asher, no free fall is not the same as being at rest. There is no absolute notion of rest in SR/GR whereas free fall is absolute. Obviously any state of rest is dependent on the reference frame. In GR the closest thing to an absolute rest frame is that defined by the timelike Killing field in a stationary space-time. In the OP's case standing on the surface of the Earth would certainly count as being at rest in the above sense. – FenderLesPaul Jan 14 '15 at 1:07
• @KieranPC, a falling object experiences no gravitational force locally. So did you mean something other than gravitational pull? – FenderLesPaul Jan 14 '15 at 1:09
• @FenderLesPaul I had only considered the case of an inertial frame. My mistake. – Asher Jan 14 '15 at 1:38
• What does it mean to ask "Will they age at the same rate?" Presumably this means that they cross paths twice, and you want to know whether each will have aged the same amount between those two crossings. But to answer the questions, we need to know why they crossed paths twice --- as you've described it, it sounds like they'll only cross once. The answer to your question depends on what causes the second crossing. – WillO Jan 14 '15 at 3:22

Since from your previous questions you're obviously interested in learning how this is done I'll go into the detail of the calculation. Note that a lot of what follows can be found in existing answers, but I'll tailor this answer specifically at you.

Time dilation is calculated by calculating the proper time change, $d\tau$, using the expression:

$$c^2d\tau^2 = -g_{ab}dx^adx^b \tag{1}$$

where $g_{ab}$ is the metric tensor. The reason we can use this to calculate the time dilation is that the proper time is an invariant i.e. all observers will agree on its value. To illustrate how we do this let's consider the simple example of an astronaut moving at velocity $v$ in flat spacetime. In this case the metric is just the Minkowski metric, and equation (1) simplifies to:

$$c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 \tag{2}$$

First we do the calculation in the astronaut's rest frame. In that frame the astronaut isn't moving so $dx = dy = dz = 0$, and the proper time as observed by the astronaut is just:

$$d\tau_{astronaut} = dt$$

Now let us here on earth calculate the proper time. We'll arrange our coordinates so the astronaut is moving along the $x$ axis, so $dy = dz = 0$. In that case equation (2) becomes:

$$c^2d\tau^2 = c^2dt^2 - dx^2$$

To procede we have note that if the astronaut is moving at velocity $v$ that means $dx/dt = v$, because that's what we mean by velocity. So $dx = vdt$. Put this into our equation and we get:

$$c^2d\tau^2 = c^2dt^2 - (vdt)^2$$

which rearranges to:

$$d\tau_{Earth} = \sqrt{1 - \frac{v^2}{c^2}}dt$$

Because the proper time is an invariant both we and the astronaut must have calculated the same value i.e. $d\tau_{Earth} = d\tau_{astronaut}$, and if we substitute for $d\tau_{Earth}$ in the equation above we get :

$$\frac{d\tau_{astronaut}}{dt} = \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{\gamma}$$

where $\gamma$ is the Lorentz factor. But the left hand side is just the variation of the astronaut's time with our time - in other words it's the time dilation. And the equation is just the standard expression for time dilation in special relativity that we teach to all students of SR.

The point of all this is that we can use exactly the same procedure to work out the time dilation in gravitational fields. Let's take the gravitational field of a spherically symmetric body, which is given by the Schwarzschild metric:

$$c^2d\tau^2 = c^2\left(1-\frac{2GM}{r c^2}\right)dt^2 - \left(1-\frac{2GM}{r c^2}\right)^{-1}dr^2 - r^2 (d\theta^2 + sin^2\theta d\phi^2) \tag{3}$$

This is very similar to the equation (2) that we used in flat spacetime, except that the coefficients for $dt$ etc are now functions of distance, and we do the calculation in exactly the same way. Let's start with calculating the time dilation for a stationary astronaut at a distance $r$. Because the astronaut is stationary we have $dr = d\theta = d\phi = 0$, and equation (3) simplifies to:

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

and this time we get:

$$\frac{d\tau}{dt} = \sqrt{1-\frac{2GM}{r c^2}} = \sqrt{1-\frac{r_s}{r}} \tag{4}$$

where $r_s$ is the Schwarzschild radius. And that's it - calculating the time dilation for a stationary observer in a gravitational field is as simple as that. You'll find this expression in any introductory text on GR.

But the real point of your question (finally we get to it!) is what happens if our observer in the gravitational field is moving? Well, let's assume they are moving in a radial; direction at velocity $v$, so just as in the flat space case we have $dr = vdt$ and $d\theta = d\phi = 0$. We substitute this into equation (3) to get:

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

which rearranges to:

$$\frac{d\tau}{dt} = \sqrt{1-\frac{r_s}{r} - \frac{v^2/c^2}{1-\frac{r_s}{r}}} \tag{5}$$

And once again, it's as simple as that. If you compare this result with equation (4) you'll see the time dilation for a moving object is different because we have an extra term $\frac{v^2/c^2}{1-\frac{r_s}{r}}$ in the square root.

One last sanity check: what happens if we go an infinite distance away from the gravitating object so $r \rightarrow \infty$? Well if $r \rightarrow \infty$ then $r_s/r \rightarrow 0$ and equation (5) becomes:

$$\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{\gamma}$$

which is exactly what we calculated for flat spacetime.

Because in relativity simultaneity is relative, there is no way of telling which object experiences a slower rate of time. But if you take two paths joining two spacetime points $A$ and $B$, then you can compare time rates along this paths.

They aren't equal. Suppose the first path is a geodesic and the second one is not and forget for a moment about the global topological issues. Then there is a simple argument: because the geodesic equation can be derived by variation of the self-time action for the particle, it is less for path $A$ than for $B$.