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?


5 Answers 5


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 $d\phi$ 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!

  • 1
    $\begingroup$ I hate to ask about a fairly old answer, but you mention "In both GR and SR the line element is an invariant"...are there theories where the line element isn't taken to be an invariant? $\endgroup$ Feb 18, 2018 at 18:01
  • $\begingroup$ @danielunderwood An invariant interval is the definition of a metric space (spacetime), on which relativity is based. A non-invariant interval may be possible in more general theories, such as where spacetime is viewed as a projection. Hopefully someone with an expertise in a String or M-Theory can provide a clarification. $\endgroup$
    – safesphere
    Nov 25, 2018 at 19:05
  • $\begingroup$ Don't get me wrong, I liked the answer, but your basic argument is based on the models we have; I would expect you to mentioned some experimental observations, and physical arguments also, not just doing some calculations in model of the universe that we currently have. $\endgroup$
    – Our
    Mar 3, 2019 at 4:51
  • $\begingroup$ Where does $v = \sqrt{GM/r}$ come from? $\endgroup$
    – ProfRob
    Feb 15, 2021 at 23:57
  • $\begingroup$ @ProfRob I have to admit that I took a short cut there. That equation is correct - it's a quirk of the Schwarzschild coordinates that they give the same result as Newtonian gravity - but deriving it is a lot of work. $\endgroup$ Feb 16, 2021 at 6:13

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.


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.

  • $\begingroup$ what do you mean by "Gravity is a inertial force where reference frame is Spacetime" ? $\endgroup$
    – Our
    Mar 3, 2019 at 4:53
  • 1
    $\begingroup$ It's General Theory of Relativity. Gravity is actually a fictional force. Gravitational acceleration is actually result of local Spacetime geodesic. $\endgroup$ Mar 3, 2019 at 9:14

Well, the answer is "no" time dilation is always the same effect and is due to velocity! Indeed, when an object is located in a gravitational field it is falling. Even when you sit on your chair you are falling in the Earth's gravitational field, otherwise you would float in the air as in the ISS! Let's equate the factors for time dilation of special and of general relativity, where $R_S$ is the Schwarzschild radius: \begin{equation} \Delta t'=\frac{\Delta t}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{\Delta t}{\sqrt{1-\frac{R_{S}}{r}}}\label{eq:time dilation compared-1-1} \end{equation} hence $\frac{v^{2}}{c^{2}}=\frac{R_{S}}{r}\Longrightarrow v^{2}=\frac{2GM}{r}\label{eq:time dilation compared-2-1}$ and $v=\sqrt{2rg}=\sqrt{2V}$, where $V$ is the gravitational potential. We therefore put in evidence the relationship that links velocity and gravity in relativistic time dilation (both in SR and GR): \begin{equation} v=\sqrt{2V} \end{equation} Thinking of time dilation as always due to velocity also explains why in artificial gravity (e.g. in a hypothetical vertically accelerating space elevator) we would experience both the sensation of natural gravity "and" time dilation, though no natural gravity is present.
If you love gravity and relativity, I would add that curved spacetime produces quantitatively exact results but "qualitatively" (I think) it is a wrong concept. There is no curved spacetime but a fluid quantum vacuum. The relativistic effects are the same: https://hal.archives-ouvertes.fr/hal-01423134v6


Interesting thread, imagine the following test concerning time dilation.

  1. Two synchronized atomic clocks, one on earth, the other one in a vehicle going to the moon with astronauts. The astronauts report their clock status to earth.

  2. Four measuring points, comparing the clocks:

    A. Before the trip starts (showing the same time)

    B. After the full acceleration, starting cruising in space to the moon.

    C. Before launching the brakes (before decelerating) when coming close to the moon.

    D. After landing on the moon.

I am 99,9% convinced that at point B there will be a significant time difference between the clocks. Between point B and C basically nothing happens (no time dilation). After point D, another significant time difference has occured.

In other words, I would eat my hat if time dilation is not always caused by a force (acceleration, deceleration, being close to massive matter) acting on matter, and never can happen when just cruising in "empty" space. Muons, yes, they live longer because of the deceleration caused by the friction in the earth atmosphere (which has the same time dilation effect as acceleration).

The spacetime concept works as a purely mathematical framework, but is like focusing on a shadow in a room with a physical moving object.

Time dilation is a physical phenomenon which involves a force field that slows down the particles of a system. When particles interactions take place with lower frequency, time is effectively slowed down. The mechanics of this is unknown. Time emerges from this effect, it is not "floating around" in space as a dimension.

  • $\begingroup$ "In other words, I would eat my hat if time dilation is not always caused by a force (acceleration, deceleration, being close to massive matter) acting on matter..." Consider a case where 2 rockets go out & back, accelerating and decelerating exactly the same at their turnaround points, but traveling different total distances. Are you claiming that their time dilations would be the same? $\endgroup$
    – D. Halsey
    Oct 31, 2018 at 0:09
  • 2
    $\begingroup$ Time dilation is a physical phenomenon which involves a force field that slows down the particles of a system. No, this is a completely incorrect description. $\endgroup$
    – user4552
    Oct 31, 2018 at 2:20
  • $\begingroup$ D.Halsey 1. With your 2 rockets, make the case simpler, just let the rockets fall back to earth and crash on earth after accelerating and then cruising in space for a while. So, it's obvious that a rocket travelling a longer distance must accelerate more (longer acceleration or more forceful acceleration), when leaving earth. Therefore, time dilations will (must) be different. Then your case can be complicated with decelerations, turn-arounds, other gravitational forces etc etc. Doesn't matter, same principle. SR is a special case of GR, it's not GR + SR = time dilation for moving objects. $\endgroup$
    – user211380
    Nov 1, 2018 at 18:26
  • $\begingroup$ You should eat your hat. SR alone tells us that relative velocities absent any acceleration actually causes clocks to tick slower. Read up on the twin paradox, which can be constructed such that no accelerations are involved at all. The traveling twin/clock really will be younger upon return simply due to traveling out and back at a high relative velocity to the stay-at-home twin/clock. $\endgroup$ Dec 31, 2021 at 7:17

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.