The tidal range of a perfectly fluid inertialess ocean on the Earth (taking into account lunar tides only) is approximately half a metre: this is the range between "high" and "low" points of an equipotential surface.

The question therefore arises: what is the tidal range in a swimming pool, and can one deduce the presence of the Moon and the period of its orbit without going outdoors?

Tidal range: Since the water in a swimming pool doesn't have anywhere to go, it will be in the same place at high tide and at low tide. However, half way between high and low tide, one end of the swimming pool will be nearest the high point of the equipotential surface and the other end will be nearest the low point.

The circumference of the Earth is about $4\times{10^7}$ metres. Therefore the height in metres (relative to the mean) of the equipotential surface at a distance $x$ metres along the great circle through the high and low points is $$0.25\cos{\frac{2\pi x}{4\times 10^7}}$$

Differentiating, this gives a tilt of $-1.25\times{10^{-8}}\pi\sin{\frac{2\pi x}{4\times 10^7}}$ metres per metre. Half way between the high and low points, the sine is 1, so we get $±1.25\pi\times{10^{-8}}$ metres per metre as the maximum observable tilt.

For a 25m swimming pool, this means that the tilt between one end of the pool and the other will vary between -981nm and +981nm: a total range over the course of 6¼ hours of just under 2μm.

This at least sounds measurable.

Measurability: The range in question is about 4 wavelengths of green light, so interferometric techniques are appropriate.

There are many sources of variation, so the signal to be found is well below the noise floor; but on the other hand we have 12½ hours to integrate over.

Here are the error sources I have thought of. Are there others?

  • Noise: This is audible noise, noise from distant traffic, and the occasional earthquake.
  • Evaporation: The pool evaporates at approximately 0.1μm per second. This is actually beneficial because it produces a steadily moving pattern of fringes in the interferometer.
  • Thermal expansion: A very approximate figure for a 4-metre-deep pool is that its water level rises by about 100μm/°C. This is the biggest source of variation (0.01°C equals 1μm) but (a) the temperature can be measured and corrected for; (b) temperature variations are likely to have a 24-hour period rather than a 12½-hour one; and (c) it is not the expansion of the water as a whole that matters but its differential effect between one end of the pool and the other, and that is much smaller.

All in all, then, the experiment seems at least doable, and likely to provide plenty of practice in identifying and allowing for experimental error.

Is it doable, or have I missed something?

Or, for that matter, has it already been done?

  • $\begingroup$ the speed of light is smaller in water, so doing this by measuring the effective path length of the light by reflecting from mirrors fixed to the bottom of the pool will help avoid having to do something at the surface of the water, which seems tricky to me. $\endgroup$
    – Gremlin
    May 26, 2016 at 14:34
  • $\begingroup$ At first sight this seemed good but now I'm not so sure, because the effect of a 1mm change in path length would be 2000 wavelengths in my setup, while in yours it would be the difference between the number of wavelengths in 1mm of water versus 1mm of air – which is about a quarter of that, I think. Easier experimental setup versus larger effect to measure? Tricky: it could go either way! $\endgroup$ May 26, 2016 at 14:43
  • $\begingroup$ I think that thermal expansion is less of a factor than you think. Even if one end was quite a bit warmer than the other, flow within the pool would prevent the warmer end from being deeper than the cool end. However, that flow would add to the noise, causing waves and ripples. $\endgroup$
    – James
    May 26, 2016 at 18:06
  • 1
    $\begingroup$ Speaking as an experimentalist with over 25 years of experience, if I got to this stage in my pre-experimental, back-of-the-envelope calculations of the feasibility of the experiment that you just did, then I would conclude that it's probably not the best way to detect the presence of the moon by measuring tidal forces. The signal you are looking for is very small, and there are too many significant competing noise/background sources. Why do you use a swimming pool of water as a starting point for your 'sensor'? Perhaps there are better gravitational sensors to base an experiment on? $\endgroup$
    – user93237
    May 26, 2016 at 18:17

1 Answer 1


Theoretically there is nothing wrong with your experiment. But I doubt you could succeed.

Firstly, how would you prevent air currents from creating water waves that spoil your measurement?

A light source at one end of the pool will heat up the water and cause turbulence. This could spoil your measurement because of scintillation of the light beam in the water. This effect would be reduced if you're reflecting off the top of the water and not through it.

How will you protect your setup from microscopic temperature fluctuations? A small change in the temperature of the mirror/laser support will change the distance between mirror and laser, thereby spoiling your measurement. Over 12 hours, I predict that keeping things temperature stable is hopeless.

But these are only technical problems which could, in principle, be solved.


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