so let's say we have a Ferris wheel of diameter 100 m. To my understanding, the equivalence principle states that an object under acceleration experiences the same effects as an object in a gravitational field. Furthermore, gravitational time dilation exists, so I would think an object with enough centripetal acceleration from a Ferris wheel could slow down someone's aging. If I wanted to slow down someone's aging by a factor of 100 relative to a normal person on the surface of the Earth, how fast would a 100 m diameter Ferris wheel have to be spinning?

  • $\begingroup$ Ain Dubai ferris wheel has a diameter of 250 meter, you can start from there if you want $\endgroup$ Aug 26, 2022 at 6:03

4 Answers 4


Technically yes (assuming an indestructible Ferris wheel). However, long before you achieved a significant relativistic velocity, they'd die instantly from the centripetal force crushing them flatter than the flattest pancake against the floor of the Ferris wheel car. Setting the wheel's tangent velocity to a mere $1000m/s$ (about the speed of a rifle bullet) would make the felt acceleration 2000 times that of Earth's gravity, and would have a similar outcome to dropping a tank on their head. Such an acceleration corresponds to a time dilation factor of about $1\times 10^{-6}$, that is, for every million seconds experienced on the ground, a linearly accelerated passenger would experience 999,999 seconds - although making the acceleration circular makes the problem much more complicated and I don't know how to solve it. Accelerated frames are hard enough when they don't keep coming back to comoving with the same inertial frame.

Although they're already very, very dead, we could spin up our indestructible Ferris wheel yet further. At around a tangent velocity of $10000m/s$ (still long before we reach relativistic velocities or a significant accelerated-frame time dilation factor) the very flat pancake that used to be the person whose life you wanted to prolong would be vaporized by air ignited by the shockwave of the Ferris Wheel car plus person pancake ripping through it at the speed of a falling meteor.

This might make reducing their rate of aging by another few seconds per million somewhat less useful.

Suppose the Ferris Wheel cars are airtight (and indestructible). The contents are now stratified with a hot dilute gas of person molecules at the radially inward side, increasing in density to a hot liquid person puddle at the bottom. The whole container is glowing white hot. We continue to spin up the Ferris Wheel, desperate to slow the rate at which the molecules age.

Suppose we spin up the Ferris Wheel instantly (never-mind how) to a tangent velocity of a significant fraction of the speed of light. Now at least we are getting close to serious time dilation, although nothing close to a factor of 100. This, however, is as far as we can get. At this point, the indestructible Ferris Wheel is dumping so much energy into the air that the air has ignited into a plasma with effects similar to a small nuclear weapon. The unimaginable torque applied by the wheel on its foundations instantly rips it free of the Earth, but there's no stopping so much rotational kinetic energy now that we have gotten it started. The wheel tips and slams into the earth. It is launched into the sky at many times escape velocity, never to return. Behind it, the foundation, the city it was in, and a significant part of the Earth's crust vanishes in a brilliant flash of light. The shockwave crosses the globe, making the asteroid that killed the dinosaurs look like a cheap firecracker. On the bright side, global warming is no longer an issue.

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    $\begingroup$ Bravo - a brilliant answer !! $\endgroup$
    – gandalf61
    Aug 21, 2022 at 13:06
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    $\begingroup$ This is definitely inspired by What If style from XKCD, haha $\endgroup$
    – justhalf
    Aug 21, 2022 at 13:41
  • $\begingroup$ doesn't it depend on the diameter of the wheel? $\endgroup$
    – njzk2
    Aug 21, 2022 at 21:17
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    $\begingroup$ +1 awesome. A much more reasonable and practical solution would be to spin the Ferris wheel around a black hole $\endgroup$
    – bobflux
    Aug 21, 2022 at 21:55
  • $\begingroup$ @njzk2 yes. The diameter of the wheel was specified as 100m in the OP. $a=v^2 / r$ for centripetal acceleration, tangent velocity, radius. $\endgroup$
    – g s
    Aug 21, 2022 at 22:04

First, I'll try an intuitive answer and then the mathematical answer.

Imagine the stationary observer in the center of the ferris wheel at the origin of a Cartesian coordinate system. For each sufficiently small duration in time, the tangent velocity vector of the rotating observer is arbitrarily near to constant and the time dilation formula of special relativity applies. The person riding the ferris wheel is constantly moving by the observer at the same speed with a velocity vector perpendicular to the line connecting them.

Note that $|v|=\omega r$, and

$$100 = \frac{1}{\sqrt{1-\frac{|v|^2}{c^2}}}.$$

Solve for $|v|$.

This follows from the tensor character of $g_{\mu\nu}$.

The transformation from the rotating frame is given by


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

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

Calculate $g'_{\mu\nu}$ in the rotating frame. You only need to calculate $g_{t't'}$.

$$d\tau^2=\left(1- \frac{{\omega^2(x'^2+y'^2)}}{c^2}\right)dt'^2 + \text{other terms} $$

The $\text{other terms}$ in the metric are zero in the rotating frame.

$$\frac{1}{100}=\sqrt{\frac{d\tau^2}{dt^2}}=\sqrt{1-\frac{\omega^2 r^2}{c^2}}=\sqrt{1-\frac{|v|^2}{c^2}}$$

So $100 = \frac{1}{\sqrt{1-{|v|^2}/{c^2}}}$ as expected.

Note that, at the origin, $c d\tau=dt$ and time dilation increases outward from the origin.

For details see A Short Course in General Relativity. Nightingale & Foster. Page 64. Longman, 1979. (This is an early edition. The later editions of this book are not as good.)



Similar Questions.

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Metric coefficients in rotating coordinates

Rotating Observers in Special Relativity: Coriolis-like effect?

How do you get time dilation from $g_{00}$ in general and from the Schwarzchild metric in particular?


Yes, but it is because of the velocity.

Particle accelerators do this all the time with elementary particles. Muons have a half life of $2.2$ microseconds before they decay. This means if they were going near the speed of light, they could travel an average distance of $660$ meters without relativistic time dilation.

But in the Large Hadron Collider at $0.999999...$ c, time dilation makes them last much longer. They make thousands of trips around the 26.7 kilometer circumference.

But a tiny $100$ m ferris wheel would require spinning them at more than $10$ billion revolutions per second. The g force would be preposterous.

Keep in mind that time dilation works at all speeds. It is just too tiny to notice at ordinary speeds. A person on a regular ferris wheel will age perhaps a nanosecond or so less that a person waiting on the ground.

I am not sure about the acceleration of a ferris wheel producing a time dilation.

See Why can't I do this to get infinite energy? for an explanation of why acceration and gravity are the same.

Notice that an acceration in the direction of a beam of light produces a change in velocity, which produces a doppler shift. This leads to time slowing.

In a Ferris wheel, acceleration is perpendicular to the velocity. No change in speed and no doppler shift, so no change in time?

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    $\begingroup$ The time experienced by an object is, letting $t,x,y,z$ be some inertial coordinates, $\Delta\tau=\int_{t_0}^{t_1}\sqrt{1-\frac1{c^2}\left(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2\right)}\,dt.$ This is not sensitive to acceleration, only the (magnitude of the) velocity. In terms of the equivalence principle, gravitational time dilation only makes sense between two different heights in the same field. It's hard to apply directly when you have one person not accelerating and one who is, and especially hard when there is complicated motion. $\endgroup$
    – HTNW
    Aug 21, 2022 at 5:17
  • $\begingroup$ $0.999999... c$: that's just ridiculous. What do you mean by $0.999999...$? You try to make it sound authoritative by mentioning the 26.7-km circumference, but it's too late for that! $\endgroup$
    – TonyK
    Aug 21, 2022 at 21:56
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    $\begingroup$ @TonyK - The Large Hadron Collider accelerates particles to extremely close to the speed of light, but I didn't know how close. I didn't know how many $9$'s to use. But I have now looked it up here CERN-BROCHURE-2017-002-Eng. At top energy, particles go 0.999999991 times the speed of light. I just quoted the circumference I read on the internet. $\endgroup$
    – mmesser314
    Aug 22, 2022 at 3:34
  • $\begingroup$ That's better! Thank you. $\endgroup$
    – TonyK
    Aug 22, 2022 at 10:12

All these answers, but no answer! Since I didn't see any numeric answers but wanted one, here it is. It is probably based on flawed logic, since the circular motion makes it screwy. (I'll just let the comments cover that using Cunningham's law.) To get a 100 to 1 time you need to go .99995c or 299777.47 km/s. An object travelling in a 100m diameter circle will go that fast if it makes 954221.32 rotations per second. And yes, it will have a similar effect to summoning a hefty blob of antimatter in that location.

  • $\begingroup$ i tried to solve my formula to compare. i usually make mistakes with numbers even whenusing a calculator. $100=1/\sqrt(1-[v^2]/[299792458]^2)$ for $v$. $v = (449688687 \sqrt(1111))/50$ m/s $=2997 77468 m/s~=~ .99995c$ , $\omega=v/R=v/50~=~6(10)^6$ rotations per second. $\endgroup$
    – jelly ears
    Aug 22, 2022 at 6:00

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