# Are today's chip scale atomic clocks accurate enough to conduct my own test of time dilation?

Regarding the Symmetricom SA.45s Quantum™ Chip Scale Atomic Clock, is it accurate enough to test time dilation if I place one at sea level, and one on a mountain? It's accurate to 3.0⋅10−10 per month.

• This experiment with an atomic clock on a mountain was done: leapsecond.com/great2005 – Uwe Aug 9 '16 at 7:57
• What about the idea of flying around the world on commercial? That is easily accessible to amateurs for only a few thousand $. Would these devices be sensitive enough for that sort of test?? – Fattie Aug 9 '16 at 13:47 • @Uwe It was done with a state of the art research grade clock. This question is about if the relatively affordable models available commercially are sensitive enough to start replicating the original experiments. – Dan Neely Aug 9 '16 at 14:11 • They want to mark the word "Quantum" as their own brand? (On the side, unless you're affiliated with them you aren't expected to maintain their branding. Indeed it even risks conveying that you are affiliated.) – user10851 Aug 9 '16 at 19:26 • @ChrisWhite No affiliations here. Just as happy to talk about whatever other brands are available, especially if somebody knows of a more accurate chip scale atomic clock? Who is the field leader in this tech? – Steve Aug 10 '16 at 3:18 ## 4 Answers According to my calculations, no. Let's treat the earth as perfectly spherical, then the relative error at which time passes for the observer on the ground$r_1=R_E$and on the mountain of height$h$,$r_2=R_E+h$, is $$\frac{t_2-t_1}{t_1}=\frac{t_2}{t_1}-1 = \frac{\sqrt{1-r_s/r_2}}{\sqrt{1-r_s/r_1}}-1$$ where$r_s=2GM_E/c^2$. Here is a log-plot of this error versus the mountain height$h$: It would seem that the time dilation error, even on Mount Everest, would be at least a few orders of magnitude too small for this device to resolve. • @Steve For$h=7$km, the time dilation error is$7.64\times 10^{-13}$. – lemon Aug 9 '16 at 7:47 • Let's treat the Earth as a perfectly spherical frictionless cow :) – Lightness Races with Monica Aug 9 '16 at 15:04 • @LightnessRacesinOrbit It can't be a cow; cows have fluid inside. Let's treat the cow as a perfectly uniform ideal solid. – wizzwizz4 Aug 9 '16 at 17:33 • @LightnessRacesinOrbit Well, it's moooving at 30 km/sec.... – Loren Pechtel Aug 9 '16 at 18:00 • Somewhat related: With super precise: optically linked clocks, even 25m of elevation distance is detectable, and it would be noticeable down to 0.5m of elevation difference. – Johnny Aug 10 '16 at 0:36 When gravity isn't very strong, i.e. as long as we aren't near any black holes, The relative time dilation between two points is related to the difference in their gravivitational potential energy by the weak field equation: $$\frac{dt_a}{dt_b} = \sqrt{1 + \frac{2\Delta\Phi_{ab}}{c^2}} \tag{1}$$ where$\Phi$is the Newtonian gravitational potential energy per unit mass. Near a spherical mass$M$this is given by: $$\Phi = -\frac{GM}{r}$$ or if the distance is small the difference is just: $$\Delta\Phi \approx gh$$ where$h$is the vertical distance between the points$A$and$B$. Since we're just trying to get a rough idea of whether the clocks will work lets use this approximation for$\Delta\Phi$in which case our equation (1) becomes: $$\frac{dt_a}{dt_b} \approx \sqrt{1 + \frac{2gh}{c^2}} \tag{2}$$ and since$gh/c^2$is small we can use a binomial expansion to expand the square root to give. $$\frac{dt_a}{dt_b} \approx 1 + \frac{gh}{c^2} \tag{3}$$ Now I'm not totally sure what the cited accuracy of$3 \times 10^{-10}$per month means, but I'm guessing that it means the ratio of the measured time to the real time over one month would be$1 \pm 3 \times 10^{-10}$. In that case our calculation is now very easy, because it means that for the gravitational time dilation to be detectable: $$3 \times 10^{-10} \lt \left|\frac{gh}{c^2}\right|$$ or: $$\left|h\right| \gt 3 \times 10^{-10} \frac{c^2}{g}$$ And feeding in the numbers gives me: $$\left|h\right| \gt 917\,\text{km}$$ Actually this height is large enough that our approximation of using$\Delta\Phi=gh$breaks down, and when you do the calculation properly you find the minimum value for the height difference$h$is around$5,000$km. But either way your chip scale atomic clock isn't useful for measuring time dilation unless you can find an exceedingly high mountain. For some perspective it's worth comparing this result to the well established time dilation experienced by GPS satellites. The gravitational time dilation experienced by a GPS satellite is about$47\mu\text{s}$per day, with is a drift of about$5\times 10^{-10}$, so it is just detectable by your chip. However GPS satellites orbit at a height of about$20,000$km. • Thank you, super impressed. I'll have to buy a better clock. I would require a clock capable of measuring time dilation separated by an altitude of 7km. Are you able to calculate my minimum sensitivity requirement plz? I wonder how much such a clock would cost. – Steve Aug 9 '16 at 7:41 • @Steve: to find the required stability for 7km just feed$h=7000$into equation 3 and you get a required stability of around$5 \times 10^{-13}$, which is about three orders of magnitude better than the clock on a chip. The atomic clocks used for international timekeeping have a stability of better than$10^{-14}$so they could easily do it. I'm afraid I have no idea what atomic clocks cost. I did try Googling but didn't find anything useful. – John Rennie Aug 9 '16 at 8:06 • @Steve: For a useful stability in that range (you actually want a good deal better to not get lost in the noise) you would at least need an active hydrogen maser, starting at around \$250k (and are sold by microsemi/symmetricom too). A non fountain Cs clock you can get on ebay starting at \$4k or so, but it is just a bit too inaccurate. – PlasmaHH Aug 9 '16 at 8:18 • ... and then it bears mention that if you want to do this at 7000m above sea level, then you also need to consider exactly what sort of environmental conditions you have, how stably your clock will run if e.g. the power is provided by a jumpy generator, and how much equipment you're going to need to lug all the way to wherever you want to run the experiment. – Emilio Pisanty Aug 10 '16 at 2:08 • @EmilioPisanty thank you. Apparently these units are small, lightweight, energy efficient, fairly resistant to heat influences, self contained. Sounds like bumbling amateurs will soon be able to track time dilation while out on their weekend picnic. A new pass time, forgive the pun – Steve Aug 10 '16 at 3:35 Regarding the Symmetricom SA.45s Quantum™ Chip Scale Atomic Clock, is it accurate enough to test time dilation if I place one at sea level, and one on a mountain? Its accurate to 3.0E-10 per month Yes, but you will have to conduct the experiment over a longer period of time, and using a high mountain so the error in the clock is swamped by the effects of time dilation. The atomic clock you are looking at has an accuracy of ±5.0E-11. Here's an example of someone with an accuracy only two orders of magnitude higher measuring the time dilation occurring at a height of only 2km: http://www.leapsecond.com/great2016a/index.htm This example shows six clocks being used, three at the top of the mountain, three at the bottom, over a period of 24 hours, with a height of 2km. By using several clocks you increase redundancy in case equipment failure occurs, but you also increase accuracy - assuming the error is distributed (and you can't simply assume that, but it may be good enough for your purposes) then the more clocks you use in your calculations, the more accuracy you gain. You may have to use your less accurate clocks over a much longer period of time, and at a much higher height, given your larger error, but they are accurate enough to measure time dilation empirically. The original experiment resulted in some conference talks and papers: http://leapsecond.com/great2005/ • AdamDavis Thank you. That sounds promising. Can you please confirm the accuracy is 5.0E-11? I got my number from the video presentation, but cant say I did much research. If this is the correct accuracy, I would like to consider editing my question to reflect this value. @John Rennie how do you feel about me updating the sensitivity value in the question, because obviously it has implications for your answer? Would you then want to edit your calculations? What do you think we should do please? What would you like to do Lemon, if the sensitivity is to be revised? – Steve Aug 10 '16 at 3:27 • You have to think carefully about systematic vs random errors. You can't improve systematic ones by averaging over a long time. The chip published specs give an overall error, which should (but often people miss some systematic errors in the assessment) include both. Once you identify the systematic errors you can often reduce them by clever experiment design. When you get close to the error bounds, it is hard. – Ross Millikan Aug 10 '16 at 4:44 • @Steve The manufacturer's Quantum SA.45s web page makes the accuracy claim. You'll want to read through all the literature there to understand the accuracy claim, though. It does provide a few different accuracy measures which may help you understand whether you can design a time dilation experiment with it successfully. – Adam Davis Aug 10 '16 at 14:40 • @RossMillikan Correct. I haven't looked too far into the part's data sheet and FAQ, but this presentation shows that the error improves over longer periods of time. I expect that the systemic error is still large enough that it has to be accounted for, but if Steve discusses his application with Microsemi they can probably verify whether the device can be used and how best to create the experiment. They are also skilled in taking data from multiple clocks in a precise time-scale system to improve the overall accuracy of measurement. – Adam Davis Aug 10 '16 at 14:58 • @AdamDavis yeah, most aging affects occur near beginning or end of lifetime, so if you burn the clock in for enough hours after buying it you can probably get into a more-stable-than-average region. A two-clock experiment with crossover should also be able to correct for some of the systematic errors as far as I can see. In fact, a lot hinges on the experimental design :) – hobbs Aug 10 '16 at 20:32 As I understand the specification of the accuracy of the mentioned clock, the actual accuracy is$\lambda t =\frac{3.0 e^{-10}}{30x24x60x60}$=$1.15 e^{-16}\$. If this is correct, you should be able to detect the time dilation at much lower altitude than has been calculated (in the neighborhood of 0.5 meter).
As you can see, the result depends on what the accuracy specification means.

• Thank you. I have received such a wide variety of advice and opinion, from here and Cosmoquest and Quora. Frankly I'm at a loss. My not having the qualified math to work this out on my own, or discriminate other peoples opinions. – Steve Aug 18 '16 at 9:24
• Steve, it is very simple, (1) buy (borrow/rent) 2 clocks, (2) set both to the same time, and (3) take one to the mountains. When you return, compare the clocks and see if there is a difference. – Guill Aug 18 '16 at 22:08
• If you have a problem setting both clocks to the "same" time, just find the starting difference and compare it to the final difference. Good luck. – Guill Aug 18 '16 at 22:19

## protected by Qmechanic♦Aug 9 '16 at 9:41

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