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Is gravitational time dilation caused by gravity, or is it an effect of the inertial force caused by gravity?

Is gravitational time dilation fundamentally different from time dilation due to acceleration, are they the same but examples of different configurations?

Could you recreate the same kind of time dilation without gravity using centrifugal force?

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Bummer that the asker never came back to "accept" @johnrennie 's Answer. (Hey Jay, if you see this notification, recognise the work that went into answering by clicking on the check mark under the voting arrows at the top of the answer!) – JDługosz Jan 4 '15 at 8:59

No, gravitational time dilation is no different to other forms of time dilation. They all stem from the invariance of the line element.

If we choose some coordinates, $x^i$, then the line element is given by:

$$ ds^2 = g_{ab}dx^adx^b \tag{1} $$

where the matrix $g_{ab}$ is called the metric tensor. In both GR and SR the line element is an invariant, that is all observers in all coordinate systems will calculate the same value for $ds$.

Suppose I'm using some set of coordinates $(t, x, y, z)$ to calculate your line element using equation (1). We'll stick to SR for now, where $g$ is just the Minkowski metric, so I get (I'm pulling the usual trick of setting $c = 1$):

$$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$

Now suppose you're doing the same calculation in your rest frame coordinates $(t', x', y', z')$. By definition, in your rest frame $dx' = dy' = dz' = 0$, so you would calculate:

$$ ds^2 = -dt'^2 \tag{2} $$

Since we must both agree on the value of $ds^2$ we can equate the right hand sides of equations (1) and (2) to get:

$$ -dt^2 + dx^2 + dy^2 + dz^2 = -dt'^2 $$

If any of $dx$, $dy$ or $dz$ are non-zero, i.e. if you're moving in any way in my coordinate system this means that:

$$ dt \ne dt' $$

and therefore our measurements of elapsed time will not match. This is why we get time dilation. In introductory works on SR you'll see time dilation calculated using various arrangements of light beams and mirrors, but this is the fundamental reason it occurs.

I've used the example of SR above because the metric tensor is diagonal and all the elements are $-1$ or $1$, so it's easy to write out the expression for $ds^2$. In GR the metric may not be diagonal (it's often possible to choose coordinates where it is) and the values of the elements in the metric will typically be functions of position. However the working is exactly the same. We'd end up concluding that $dt \ne dt'$ in exactly the same way.

Since you specifically asked about time dilation and centrifugal force, let's do the calculation explicitly. Suppose you're whirling about a pivot with velocity $v$ at a radius $r$ and I'm watching you from the pivot. I'm going to measure your position using polar coordinates $(t, r, \theta,\phi)$, and in polar coordinates the line interval is given by (I'm leaving $c$ in the equation this time):

$$ ds^2 = -c^2dt^2 + dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2) $$

Note that this is just the flat space, i.e. Minkowski metric, in polar coordinates. We're using the flat space metric because there are no masses around to curve spacetime (we'll assume you and I have been on a diet :-). We can choose our axes so you are rotating in the plane $\theta = \pi/2$, and you're moving at constant radius so both $dr$ and $d\theta$ are zero. The metric simplifies to:

$$ ds^2 = -c^2dt^2 + r^2d\phi^2 $$

We can simplify this further because in my frame you're moving at velocity $v$ so $d\theta$ is given by:

$$ d\phi= \frac{v}{r} dt $$

and therefore:

$$ ds^2 = -c^2dt^2 + v^2dt^2 = (v^2 - c^2)dt^2 $$

In your frame you're at rest, so $ds^2 = -c^2dt'^2$, and equating this to my value for $ds^2$ gives:

$$ -c^2dt'^2 = (v^2 - c^2)dt^2 $$


$$ dt'^2 = (1 - \frac{v^2}{c^2})dt^2 $$


$$ dt' = dt \sqrt{1 - \tfrac{v^2}{c^2}} = \frac{dt}{\gamma} $$

which you should immediately recognise as the usual expression for time dilation in SR. Note that the centripetal force/acceleration does not appear in this expression. The time dilation is just due to our relative velocities and not to your acceleration towards the pivot.

Finally, since I did say there was no difference between gravitational and other forms of time dilation I should justify this by proving that the special relativity calculation above works in the same way for combined gravitational and speed related time dilation. Specifically we'll calculate the time dilation for an object in orbit around a black hole. This turns out to be straightforward, showing how powerful this technique is. All we need to know is that the metric for a black hole is:

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

We proceed as before setting $dr = d\theta = 0$ and $\theta = \pi/2$ to get:

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

The orbital velocity is:

$$ v = \sqrt{\frac{GM}{r}} $$

and as before we can rewrite $d\phi$ as:

$$ d\phi = \frac{v}{r}dt = \frac{\sqrt{GM/r}}{r} dt $$

and substituting this in our metric gives:

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

As before, in the rest frame of the orbiting body we have $ds^2 = -c^2dt'^2$, and equating this to the above value for $ds^2$ gives:

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

which simplifies to:

$$ dt' = \sqrt{1-\frac{3GM}{c^2r}}dt = \sqrt{1-\frac{3r_s}{2r}}dt $$

where $r_s$ is the Schwarzschild radius: $r_s = 2GM/c^2$.

And, reassuringly, this is exactly the result Wikipedia gives for the time dilation of an object in a circular orbit.

This is the point I want you to take away. Once you understand the basic principle that the line element is an invariant you can use this to calculate the time dilation for any object, whether in a gravitational field or not, and whether moving or not. In fact, as I've just demonstrated, understanding this basic principle opens the door to understanding general relativity as well as special relativity. That's how important it is!

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Sure, circular observers observe time dilation. Consider the worldline of a circular observer in flat spacetime. We know, right off hand, relative to a "stationary" observer, that the spatial coordinates of such an observer will be

$$\begin{align} x&= r\cos\left(\omega\,\tau\right)\\ y&= r\sin\left(\omega\,\tau\right) \end{align}$$

for some parameter $\tau$. Let's just call it the proper time.

Then, remembering that $-1 = -{\dot t}^{2} + {\dot x}^{2} +{\dot y}^{2} + {\dot z}^{2}$, for this path, we have:

$$\begin{align} -1 &= -{\dot t}^{2} + r^{2}\omega^{2}\sin^{2}\left(\omega\,\tau\right) + r^{2}\omega^{2}\cos^{2}\left(\omega\,\tau\right)\\ 1&= {\dot t}^{2} - r^{2}\omega^{2}\\ \dot t &= \sqrt{1 + r^{2}\omega^{2}}\\ t &= \tau\sqrt{1 + r^{2}\omega^{2}} \end{align}$$

Compare this to the GR formula for gravitational time dialation (at a stationary point) $t = \tau/\sqrt{1-\frac{2GM}{rc^{2}}}$

Thus, if you are orbiting in a circle at radius $R$ in empty space, you have the equivalent time dilation to being held stationary at a radius $r$ from a gravitational body of mass $M$ if you are rotating with angular speed:

$$\begin{align} 1 + R^{2}\omega^{2} &= \frac{rc^{2}}{rc^{2} - 2GM}\\ R^{2}\omega^{2} &= \frac{2GM}{rc^{2} - 2GM}\\ \omega &= \frac{1}{R}\sqrt{\frac{2GM}{rc^{2} - 2GM}} \end{align}$$

How "equivalent" this is is obviously subject to debate, since everything else would feel different. But you can certainly get the same time dilation kinetically.

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Gravity is a inertial force where reference frame is Spacetime. So, Gravitational Time Dilation and Time Dilation due to accelerated frame are both same.

And yes, you can recreate same kind of Time Dilation using centrifugal force.

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