Can a ultracentrifuge be used to test general relativity?

Question

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.

Answer 1

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:

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

then:

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

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.

Answer 2

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.

Answer 3

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.

Answer 4

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!

Answer 5

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 $c$.

Answer 6

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.

Answer 7

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.

Answer 8

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.