With today's ultracentrifuge technology, they can spin so fast that the sample can be subjected to accelerations of up to 2 millions Gs. That is equivalent to two solar masses. Has someone tried to measure the time dilation in a radioactive sample? How calculate that time dilation respect to the time outside the ultracentrifuge, for example one week. I think that that time dilation will be significant enough to be measured.


8 Answers 8


This is really the same as a couple of the other answers, but I note that in the comments to those answers you are insistent that your experiment is a test of general relativity. However this is not the case. As long as spacetime is flat the experiment can be analysed using special relativity, and in this answer I shall explain why.

It's commonly believed that special relativity cannot be used for accelerating frames, but this is wholly false. Special relativity only fails when spacetime is not flat, i.e. when the metric that describes the spacetime is not the Minkowski metric.

The analysis I'll give here originally formed part of my answer to Is gravitational time dilation different from other forms of time dilation?, but I'll repeat it here since it is the core issure in your question.

In the centrifuge the observer is rotating about the pivot with some velocity $v$ at some radius $r$. We are watching the observer from the laboratory frame, and we measure position of the observer using polar coordinates $(t, r, \theta,\phi)$. Since spacetime is flat the line interval is given by the Minkowski metric, and in polar coordinates the Minkowski metric is:

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

We can choose our axes so the rotating observer is rotating in the plane $\theta = \pi/2$, and since it is moving at constant radius 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 the laboratory frame the rotating observer is moving at velocity $v$ so $d\phi$ is given by:

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

and therefore our equation for the line element becomes:

$$ ds^2 = -c^2dt^2 + v^2dt^2 = (v^2 - c^2)dt^2 \tag{1} $$

Now we switch to the frame of the rotating observer. In their frame they are at rest, so the value of the line element they measure is simply:

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

where I'm using the primed coordinate $t'$ to distinguish the time measured by the rotating observer from the time we measure $t$.

The fundamental symmetry of special relativity is that all observers agree on the value of the line element $ds$, so our value given by equation (1) and the rotating observer's value given by equation (2) must be the same. If we equate equations (1) and (2) we get:

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

and rearranging this gives:

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


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

which you should immediately recognise as the usual expression for time dilation in SR.

So the time dilation for the rotating observer is given by the same function as for an observer moving in a straight line at constant speed. This is why it's perfectly valid for the other answers to calculate time dilation using the normal special relativity formula. The centripetal force/acceleration does not appear in this expression and general relativity is not required.

  • $\begingroup$ Please read my comment to @Physicist137 and physics.stackexchange.com/questions/87673/… $\endgroup$ Nov 24, 2014 at 11:10
  • $\begingroup$ @CarlosFreites: which comment? If you mean your second comment asking about the three scenarios then I think that's a different question to the one you asked. $\endgroup$ Nov 24, 2014 at 11:13
  • $\begingroup$ @JohnRennie Yes John, the imaginary elevator experiment one. $\endgroup$ Nov 24, 2014 at 20:35
  • $\begingroup$ Ah, ds is the thing to agree on, not necessarily dt. So we can have observer in frame 1 at proper time T see the clock in frame 2 as being behind his, which implies that the odometers differ too. $\endgroup$
    – JDługosz
    Jan 4, 2015 at 8:40
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    $\begingroup$ @doetoe: No. The computation is the same in GR. In fact SR is just a specific case of GR where the metric is the Minkowski metric. In a curved spacetime the method for the calculation would be the same but the metric would be more complicated. Actually this is an important point since if you understand the calculation I describe you basically understand how to use metrics in GR, so GR isn't as hard as you think! $\endgroup$ Apr 5, 2017 at 15:15

Short answer: I'm afraid this is not a test of general relativity. I'll tell you why. I'll try to keep simple.

You may use special relativity when your frame of reference is inertial. Let's say you see a Ultra-centrifuge spinning. You are experiencing no gravity at all (Earth's gravity is negligible for time dilation effects). You are experiencing no non-inertial forces (Earth spin and Coriolis force are in a very small scale for time dilation). Therefore, you are (approximately) a valid inertial frame of reference, and therefore, you can use special relativity for the spinning radioactive sample.

But, let's move to the reference frame of the sample. On there, the sample is experiencing 2 million Gs of non-inertial forces. It is cleary a non-inertial frame of reference. Thus, you cannot use special relativity here. You must use general relativity.

However, both observers, you and the sample, must agree and came up with the same results. Since it is far easier to treat the problem using special relativity, we can do so, using your inertial frame of reference. Let's calculate the $\gamma$-factor: $$ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1 - \frac{gr}{c^2}}}, \quad\quad g = \frac{v^2}{r} $$

I'm keeping the calculations very simple so you can understand. This is not right, but likely holds as good approximation. The non-inertial acceleration as you said has the value: $g = 10^6m/s^2$. I'll exagerate the radius: $r=1m$. Therefore, the gamma factor: $$ \gamma\approx\frac{1}{\sqrt{1-10^{-10}}}\approx 1 $$

Therefore, the time dilation: $\Delta t' = \gamma\Delta t$, is negligible, in your small experiment.

It was a nice ideia. It would work with atomic clocks. For instance, take a look in the Hafele–Keating experiment.

  • $\begingroup$ This is exactly what I was trying to get across in my answer. I'm glad someone agrees. $\endgroup$
    – HDE 226868
    Nov 24, 2014 at 1:14
  • $\begingroup$ First you are agree with me that special relativity effect are negligible and because that, general relativity must be used. But then you use special relativity formulas to conclude that general relativity effects are negligible. Feel free to use the differential geometry tools that is necessary to handle general relativity in order to get the formula to calculate the time dilation that is applicable. On the other hand, my experiment is a direct test for General Relativity. $\endgroup$ Nov 24, 2014 at 6:24
  • $\begingroup$ OK. Let's do the imaginary elevator experiment. You are inside a close elevator with no windows. Suppose that in Earth your weight is 1 Kg and that your weight inside that elevator is 2000000 Kg. Can you differentiate whether: 1) Your elevator is immobile over the surface of a planet with the mass of 2 suns with the same radius of the Earth. 2) Your elevator is being accelerated with a propulsion of 20000000 Gs in the empty space. 3) Your elevator is inside an ultracentrifuge that spin so fast that makes you weight 20000000 Kg. $\endgroup$ Nov 24, 2014 at 10:47
  • $\begingroup$ Please notice that by imaginary means things like: a) You can forget you can be crushed by your own weight. b) You can no get that acceleration because you would soon be at c speed. b) The ultracentrifuge radius is enough big that the whole elevator feel the same acceleration. d) You are unable to detect anything outside the elevator. Apart from that, you have all kind of equipment to measure the field that makes you weight 20000000 Kg. $\endgroup$ Nov 24, 2014 at 10:48
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    $\begingroup$ @CarlosFreites: re your last comment: the equivalence principle tells us it is impossible to distinguish acceleration from gravity in a local measurement. However over any non-zero distance you can tell your options 1, 2 and 3 apart because the tidal variation of the force will be different in the three cases. $\endgroup$ Nov 24, 2014 at 18:42

You recently commented

I think the special relativity influence is absolutely negligible.

It's exactly the opposite. It's the influence from general relativity that's absolutely negligible.

Time dilation is a prediction of both general and special relativity. In general relativity, it's caused by an object being near a massive body. In special relativity, it's caused by an object moving very quickly. This is the time dilation that you're talking about. In an ultracentrifuge, the object is being spun around a center point very quickly. Any time dilation would be a result of its enormous speed. There is no giant bit of matter causing time dilation.

The acceleration mentioned is most likely the centripetal acceleration the object experiences. Centripetal acceleration is always towards the axis of revolution, so the object moves at a constant speed.

  • $\begingroup$ In general relativity, it's caused by an object being near a massive body. Exactly, like a body with two sun masses. What "feels" that object that makes its time dilated? It is a strong gravitational field that is locally identical (according to general relativity) to an acceleration of 2000000 Gs (the mass of the sun is 1000000 times the mass of the earth). The tangential speed in the ultracentrifuge is negligible respect to c. $\endgroup$ Nov 23, 2014 at 21:17
  • $\begingroup$ @CarlosFreites Acceleration does not necessarily come from gravity. It comes from a force; gravity is just one example of a force. $\endgroup$
    – HDE 226868
    Nov 23, 2014 at 21:18
  • $\begingroup$ That is right but one fundamental aspect of general relativity is that locally you can not differentiate if you are in an accelerated system or in presence of a gravitational field. $\endgroup$ Nov 23, 2014 at 22:46
  • $\begingroup$ @CarlosFreites That doesn't mean that all accelerations are due to gravity. $\endgroup$
    – HDE 226868
    Nov 23, 2014 at 22:51
  • $\begingroup$ All accelerations behave locally as gravity. $\endgroup$ Nov 27, 2014 at 5:53

A good start here would be to compute the time dilation effect expected from a centrifuge operating with a million Gs. With $\frac{v^2}{r} = 10^6 g$ I would assume something like a radius of 10 cm, in which case $v\approx 1000$ m/s. We know gamma is $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$, which means that $gamma-1$ is about $6\cdot 10^{-12}$. Using radioactive decay directly then is probably not such a great measure of this then since you'd need greater than ${10^{12}}^2$ counts to have a precision necessary to measure the dilation. That would be a pretty hot radioactive emitter...

On the other hand, if you're willing to use something other than counting the half-life of a radioactive substance, then you might consider using mossbauer absorption, where the resonance of a photon absorption in a nuclear transition is so sharp that very slight deviations can be measured. Indeed, in mossbauer systems, velocities of $10^{-2}$ cm/s can be sufficient to measure the deviation from a resonant system. There is too much detail in that computation to reproduce here, so better to read a simple explanation of it in Experiments in Modern Physics by Melissinos and Napolitano, Chapter 9.3.

Granted, this doesn't really test General Relativity directly, so the short answer is, no. Wonderful idea though!

  • $\begingroup$ Yes, 1000 m/s = 1 Km/s is totally negligible respect to c (300000 Km/s) this mean that at this speed the time dilation is negligible for special relativity effects. On the other hand, this is a direct test for General Relativity. $\endgroup$ Nov 23, 2014 at 23:08
  • $\begingroup$ @CarlosFreites Just use an atomic clock and hold on a different elevated position. $\endgroup$
    – Markoul11
    Apr 3, 2023 at 14:41
  • $\begingroup$ That is not the point. The point is to check acceleration has identical effect that gravitational mass. $\endgroup$ Apr 4, 2023 at 15:36

The previous answers are correct and even an acceleration $2 x 10^6g$ centrifuge at $1m$ radius will generate a relative compared to the speed of light, tiny constant tangential speed of $υ=1.41 Km/s$ thus just only $0.00047$% of the speed of light $c$ in the vacuum.

Therefore, the SR time dilation effect value calculation which depends not on acceleration but exclusively on the scalar value of velocity thus speed and not any vector direction, is calculated as tiny and negligible in the OP's given example.

$$ \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\approx 1 $$

However, with that said in particle physics high energy experiments circular rotation time dilation effect is significant. For example in the 2021 muon g-2 Fermilab anomalous magnetic moment experiment published results here the SR time dilation correction in the cyclotron was $γ=29.3$ (see eq.1 page 3). Time inside the cyclotron passed $x29.3$ slower than outside in the lab.

Of course the muons were accelerated inside the cyclotron in this experiment at a constant speed $υ$ very close to the speed of light $c$.


It is a nice very idea for an experiment, but I don't think that radioactive decay would be an accurate enough 'clock' to use because generally with these types of measurements very small differences in time are detected - generally atomic clocks are used to measure the time in these experiments. With the radioactive decay process it is random decay and there will always be some uncetainty in the activity of the sample after the experiment.

I also wonder whether the centrifuge is equivalent to a gravitational field; according to this reference it is equivalent. The time dilation would be most noticeable if the sample was rotated at 'relativistic speeds'. If the speed is not so fast we would need more accurate time measurement.

  • $\begingroup$ I think the special relativity influence is absolutely negligible. Yes inertial mass is equal to gravitational mass that is a fundamental aspect of general relativity. $\endgroup$ Nov 23, 2014 at 20:24
  • $\begingroup$ @CarlosFreites this question is about general relativity where there is time dilation in frames that are being accelerated. I agree with you that the effect will be very small, but it is an interesting question and if an atomic clock could be used to measure time it would probably show an effect - see the link for experiments where general relativity has been observed with atomic clocks. I don't think it is realistic here to centrifuge an atomic clock, but the idea of radioactive decay is interesting as a way of measuring time. $\endgroup$
    – tom
    Nov 23, 2014 at 21:30

I am not a physicist (my background is biotech) but from a non-technical perspective I believe that since the centrifuge has no gravitational field (it operates in the same part of earths gravitational field as the observer) then the only time dilation that might be measured would be due to the velocity of the test subject rather than the g-forces created by bending the subjects trajectory into a tight curve. In other words since both the subject and the observer are both in the exact same point of the exact same gravity well, there is no gravitational time dilation, but since the centrifuge also creates speed, that might create some time dilation.

  • $\begingroup$ Hello Coral. One of the more amazing postulates of General relativity is that you can not, locally differentiate, a gravitational field from an accelerated referential system. That is if you are inside a elevator without any contact with the outside world, no matter what experiment you do, you can not say if you are in the earth with no acceleration at all or if you are in the space accelerating at 9.81 m/(seg^2). $\endgroup$ May 16, 2015 at 5:53

I put a cesium clock in a centrifuge for 24 hours, got 45.9 microseconds first relative to gps satellite. 20,200km up Centrifuge spun at 2 G's it was 91.8 microseconds. You can suck on numbers all you want, but experiments don't lie.

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    $\begingroup$ You put a cesium clock in a centrifuge? How did you manage that? Are you publishing this anywhere? $\endgroup$
    – Kyle Kanos
    May 11, 2015 at 21:21
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    $\begingroup$ And how do those experimental values correlate to general relativity expectations? By the way, the tone of your last sentence is not helping get your point across. $\endgroup$
    – Jon Custer
    May 11, 2015 at 22:17

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