In his book, Relativity: The Special and General Theory, Einstein claimed that the clocks located on a rotating disc run slower solely due to their tangential velocities as appear in the Lorentz factor, no matter how much acceleration they take. He then replaces potential per unit mass with velocity square $(r^2\omega^2)$:

If we represent the difference of potential of the centrifugal force between the position of the clock and the centre of the disc by $\phi$ , i.e. the work, considered negatively, which must be performed on the unit of mass against the centrifugal force in order to transport it from the position of the clock on the rotating disc to the centre of the disc, then we have


However, I cannot really understand why the centrifugal acceleration does not affect clocks at all. Assume we have two concentric rings one with a large radius and the other with a very small one. If the rings rotate at the same tangential velocity, according to Einstein, the clocks run slower at the same rate as measured by an inertial observer at rest with respect to the plate's center. However, according to the centrifugal acceleration formula:


the clock on the ring with a smaller radius experiences much more acceleration than one located on that with a larger radius. How can it be possible that such a large centrifugal force/acceleration, which can easily mash the nearer clock to the center of rotation (if the radius is small enough), is ineffective in altering time rates? (Forget about the viewpoint of the rotating observers.)

Remember that if the radius approaches zero, the centrifugal acceleration tends to infinity, yet the tangential velocity can remain unchanged. It is really hard for me to understand why an infinite acceleration/force cannot affect clock rates!

  • $\begingroup$ Have you tried just calculating the time dilation factor and seeing what happens? $\endgroup$
    – knzhou
    Nov 10, 2019 at 8:54
  • $\begingroup$ Yes, the time dilation factor just depends on velocity rather than acceleration. $\endgroup$ Nov 10, 2019 at 8:57
  • $\begingroup$ @MohammadJavanshiry Well then, there you have it - the time dilation is different because the velocity is different. Are you looking instead for an explanation why time dilation depends on velocity and not acceleration? $\endgroup$ Nov 10, 2019 at 9:26
  • $\begingroup$ @probably_someone Yes. It is slightly awkward to me why the acceleration has no place in the Lorentz factor at least for when an inertial observer investigates the behavior of accelerated objects. $\endgroup$ Nov 10, 2019 at 9:31
  • $\begingroup$ It seems to me that you're asking "why doesn't acceleration affect time dilation in addition to the time dilation caused by velocity?". Is that correct? $\endgroup$
    – PM 2Ring
    Nov 10, 2019 at 11:40

5 Answers 5


Your question has the same ring as, say, the following question:

"We know that everything slows down and stops, unless you keep pushing. Push an object across the floor, it keeps moving. Stop pushing: it stops. It is really hard for me to understand how satelites in orbit keep moving without anything pushing them around."

In your case what you are stuck on is a wrong assumption (a wrong assumption about time dilation.)

I do grant you that your wrong assumption is very tempting. Many go down that road.

To address the wrong assumption let me take the Twin scenario as example.

As is often done, I will call them Alice and Bob.
Alice goes on a journey, Bob stays put.
Alice makes her far away U-turn, and when Alice and Bob rejoin the amount of elapsed proper time for Alice is less than the amount of elapsed proper time for Bob.

Both Alice and Bob are scientists with full understanding of time dilation, so given the travel plan they know in advance how large the difference in elapsed proper time will be.

The way to calculate the difference in amount of elapsed proper time is to apply the Minkowski metric. Bob has traveled forward in time along the shortest path; he hasn't moved. Alice has not taken the shortest path: in addition to travelling forward in time she has traveled spatial distance. To calculate the difference in elapsed proper time you evaluate one thing and one thing only: the difference in spatial distance traveled.

The shape of the journey that Alice made does not matter. Several other siblings can make all kinds of journeys, if all those journeys are in the end the same spatial distance (compared to each other) then for all those travellers the same amount of proper time will have elapsed.

The fact that the shape of the journey doesn't matter follows from this: as you evaluate the total spatial distance traveled (applying the Minkowski metric) the shape of the journey drops out of the calculation.

Of course, the acceleration is necessary; Alice has to make that U-turn, that's the only way to rejoin Bob. But whether the U-turn is sharp (pulling a lot of G's) or gradual (low G-load), that doesn't matter. The journey that Alice makes can be a zigzag course, pulling G's all the time; it doesn't matter, only the difference in spatial distance travelled counts.

Yes, that is very counter-intuitive.
Superficially you might expect that the difference in spatial distance travelled is irrelevant, it looks so passive.
By contrast, the acceleration is violent, surely that's where it happens.

In actual fact the difference of elapsed proper time is described by the Minkowski metric. The Minkowski metric is necessary, and sufficient.

To learn the logical implications of the Minkowski metric is to learn Special Relativity.

  • $\begingroup$ Repeating the twin paradox scenario cannot provide me with a clear answer to my question. Why this very counter-intuitive phenomenon (as mentioned by you) is assumed to be a good description of nature? $\endgroup$ Nov 10, 2019 at 10:41
  • 1
    $\begingroup$ @MohammadJavanshiry Because we have observed its effects everywhere, and never once have we observed anything counter to its predictions. One of the most famous examples is the fact that we see muons in particle detectors at Earth's surface. Muons have a lifetime of 2 microseconds, which means it would naively be expected to travel around $c\times 2\times 10^{-6}\approx 300$ meters before decaying (we've also never seen faster-than-light muons). They're produced at the top of the atmosphere, tens of kilometers up, so the fact that we see them at sea level is because of time dilation. $\endgroup$ Nov 10, 2019 at 12:06
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    $\begingroup$ @probably_someone Muon decay does not relate to the rotating disc. I am not questioning special relativity for inertial frames, but rather I am confused about applying SR to accelerated objects as measured by inertial observers. Moreover, every precise experiment is performed within an error range. It is possible that a novel theory or an alternative to SR that corrects SR deficiencies falls within the obtained error range too. Besides, I am really doubtful if there has been carried out any experiment in high centrifugal accelerations showing that relativity is not flawed in this case. $\endgroup$ Nov 10, 2019 at 12:27
  • 1
    $\begingroup$ @MohammadJavanshiry 1) Particles don't all have a nonzero diameter. For instance, everything we measure about leptons is consistent with them being point particles (i.e. having zero diameter). 2) If you can come up with a framework surrounding that "acceleration effect" that both explains all previous experimental data and predicts future results better than special relativity, congratulations on your Nobel Prize. Until then, we have absolutely no reason to believe that such an effect exists. Conclusions in science require the support of experimental evidence. $\endgroup$ Nov 10, 2019 at 17:26
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    $\begingroup$ @MohammadJavanshiry Regarding centripetal acceleration, we have tested to much higher accelerations than the LHC, in the early days of accelerator physics. For example, the Cambridge Electron Accelerator was 236 ft (72 m) in diameter and accelerated electrons ($m=511$ keV) to an energy of 6 GeV, which puts them at a velocity of $0.99995c$, making for a centripetal acceleration of $1.2\times 10^{15}$ m/s$^{2}$. But probably whatever number I give you, your response will be "yeah, but what if it's higher than that?", which is why I say: give us a reason to test that hypothesis (i.e. see above). $\endgroup$ Nov 10, 2019 at 17:49

I would not expect centripetal acceleration to have any effect. The motivation for SR was the principle that the speed of light should be the same for all observers. The Lorentz transform explains how one can account for the effects of relative motion between observers, ie the speed and direction of the motion. Acceleration is simply a change either in the speed or the direction of a motion, so it simply requires you to plug different values into the transformation equations. At any instant an accelerating body has a definite speed and a definite direction to its motion, so at that instant the Lorentz transform adequately quantifies all relativistic effects. The magnitude of those effects changes from instant to instant as the velocity of the body changes, but we have no reason to suppose that the nature of the effects would change.

  • $\begingroup$ Then do you mean that even at infinite accelerations, the clocks easily comply with Lorentz transformation, whereas defining a clock at infinite accelerations can even be impossible?! (as a matter of logic) $\endgroup$ Nov 11, 2019 at 10:21
  • $\begingroup$ Hi Mohammad, I wouldn't like to guess what would happen at an infinite acceleration. Out of interest, why do you say that defining a clock is impossible under those circumstances? $\endgroup$ Nov 11, 2019 at 11:27
  • $\begingroup$ Hello Marco, I think even if it is possible to define a clock, it is absurd $-$ to me $-$ if this infinite acceleration does not affect the clock (time rate, that is.). $\endgroup$ Nov 11, 2019 at 11:36
  • $\begingroup$ Hi Mohammed. I'm interested in knowing why you think it would have an effect, so please share your thoughts on that. In particular, do you imagine that the effect would be proportional to the magnitude of the acceleration? If so, what might the constant of proportionality be? $\endgroup$ Nov 11, 2019 at 13:48
  • $\begingroup$ I am afraid that I cannot present my thoughts in detail because, first, I have not reached a certain deduction on it yet, and second, I am trying to write an article about it. However, my initial calculations show that there can be proportionality to acceleration. $\endgroup$ Nov 11, 2019 at 18:19

In special relativity time dilation is not affected by body's acceleration, however in general relativity - it is, because of equivalence principle (accelerating reference frame is indistinguishable from non-accelerating frame in gravity field). Gravitational time dilation is defined through Schwarzschild metrics:

$$ t_{0}=t_{f}{\sqrt {1-{\frac {2GM}{rc^{2}}}}} $$

Using Newton's gravitation and second laws, one can re-write equation above as :

$$ t_0 =t_f\sqrt{1-2\frac{GMmr}{r^2c^2m}} \\=t_f\sqrt{1-2\frac{F_{_G}r}{mc^2}} \\=t_f\sqrt{1-2a\frac{r}{c^2}} \\ =t_f\sqrt{1 - \frac{a}{a_s}} $$

where $a_s$ is Schwarzschild acceleration and is defined as :

$$ a_s = \frac{c^2}{2r} $$

  • $\begingroup$ Potential per unit mass or $2GM/r$ has a velocity square dimension. It is not important how you rewrite equations as a function of acceleration. Moreover, according to my own example of the rotating rings, it is evident that velocity determines the time dilation as mentioned by Einstein himself. $\endgroup$ Nov 11, 2019 at 9:26
  • $\begingroup$ Einstein also introduced equivalence principle which you completely ignore here $\endgroup$ Nov 11, 2019 at 9:32
  • $\begingroup$ My example has nothing to do with EEP; it is clear, and Einstein's position towards it is clear too. $\endgroup$ Nov 11, 2019 at 9:36
  • $\begingroup$ Seems you don't get it. If equivalence principle is right, then time dilation experienced near massive objects (like black hole) should be the same experienced by objects simply moving at extra-ordinary accelerations. Otherwise Einstein's EP has no point and/or is faulty $\endgroup$ Nov 11, 2019 at 9:44
  • $\begingroup$ Please focus on my example and show me how acceleration is inserted into the time dilation equation from the viewpoint of the inertial observer at rest WRT the rings. $\endgroup$ Nov 11, 2019 at 9:48

for a rotate coordinate system with the constant angular velocity $\omega$ the coordinates are:

$$x=x'\cos(\omega\,t')-y'\sin(\omega\,t')$$ $$y=x'\sin(\omega\,t')+y'\cos(\omega\,t')$$ $$z=z'$$ $$t=t'$$

with $ds^2=\eta_{\mu\nu}\,dx^\mu\,dx^\nu=c^2\,dt^2-dx^2-dy^2-dz^2$

we get the metric

$$G=\left[ \begin {array}{cccc} \left( -{x'}^{2}-{y'}^{2} \right) {\omega} ^{2}+{c}^{2}&\omega\,y'&-\omega\,x'&0\\ \omega\,y'&-1&0 &0\\-\omega\,x'&0&-1&0\\ 0&0&0&-1 \end {array} \right] $$


$dx'=dy'=dz'=0$ we get

$$d\tau=\frac{ds_{\text{clock}}}{c}=\sqrt{G_{00}}\,dt'= \sqrt{1-\frac{\omega^2\,(x'^2+y'^2)}{c^2}}\,dt'$$

so $d\tau$ is a function of the clock distance from the center $x'^2+y'^2=r^2$ and the angular velocity. with $\omega=\frac{v}{r}=\frac{v}{\sqrt{x'^2+y'^2}}$ we get for $d\tau$


  • $\begingroup$ As you see, there is no place for acceleration in the time dilation formula! $\endgroup$ Nov 11, 2019 at 7:28
  • $\begingroup$ But $\omega ^{2}\cdot r^{2}=\left( \omega ^{2}\cdot r\right) \cdot r$ so you have the zentrifugal force in this equation? $\endgroup$
    – Eli
    Nov 11, 2019 at 9:09
  • $\begingroup$ It's not important how you rewrite velocity as a function of acceleration. In any form, there is no pure dependence of time dilation on acceleration, and it is velocity that plays a decisive role. $\endgroup$ Nov 11, 2019 at 9:16
  • $\begingroup$ I know this but the acceleration is canceled the force per mass is still there $\endgroup$
    – Eli
    Nov 11, 2019 at 11:16
  • $\begingroup$ I edited your answer two hours ago, you had made a mistake in the last equation. Have you seen my edit? $\endgroup$ Nov 11, 2019 at 11:27

In Special Relativity, it is assumed that only relative velocity has an effect on the ideal-clock rate, as read from another inertial observer. It is implicitely postulated that the clock's acceleration has no effect, or else you would need some universal acceleration to compare, like there's an universal constant $c$ to compare velocities. On dimensional grounds, you would need to introduce some acceleration parameter $g$ of some sort into the Lorentz transformation, a kind of universal constant (the maximal or minimal acceleration available in the universe?).

It is assumed that this $g = 0$ in SR.

In the past, some authors have already created extended theories which go above SR, but empirical data have ruled out most of these theories. For example, you may be interested in doubly special relativity:


From dimensional analysis, these theories could introduce some acceleration constant that could have an effect on clock rates.

  • $\begingroup$ Defining a universal acceleration is not necessary. My calculation shows that a term $\frac{a^{3/2}\sqrt{d}}{c^2\omega}$, as well as the clock's tangential velocity, can play a decisive role in time dilation for accelerated particles, where $a$ is the centrifugal acceleration, $\omega$ is the angular velocity, and $d$ is the effective length of the rotating clock. I am not so sure about the mentioned term because I am still working on it. $\endgroup$ Nov 11, 2019 at 20:04
  • $\begingroup$ @MohammadJavanshiry, if such an effect would have been there, don't you think that it would have already been noticed or found before, by the thousands of physicists and mathematicians that worked hard on SR since it was created by Einstein? There is none! So you're making a mistake in your analysis. $\endgroup$
    – Cham
    Nov 11, 2019 at 20:43
  • $\begingroup$ don't you think that it would have already been noticed or found before No, I do not. Because proving such an effect is not that easy. It needs the definition of observer to be revisited fundamentally. You have to know about equal observers, which has a long story behind it, to make a step forward. It is unlikely that anyone has heard the story before! $\endgroup$ Nov 11, 2019 at 21:17
  • $\begingroup$ I just thought that "cats" can better understand "lion kids", but it seems that I was wrong ;) $\endgroup$ Nov 12, 2019 at 16:10

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