(I'm not a physicist & I hope this question isn't too basic/lame)

The question says it all & assuming:

  • No atmosphere
  • No Earth rotation
  • Not a single object in the ocean(s)
  • Ignore all other disturbances (earthquakes, etc.)
  • The ocean is 'perfectly' still

If (due to-me-unknown reasons) the level wouldn't rise, how big an object had to be, so the sea/ocean level rise would be measurable?

  • $\begingroup$ Why do you think it shouldn't rise? About the measurability, it really depends on what are you using to measure! But it should be some percent of ocean's volume. $\endgroup$ – Ali Jul 18 '13 at 8:23
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    $\begingroup$ A rise would happen in principle but it would be practically unmeasurable. Here is a similar question tackled over at xkcd that might give you a feel for the numbers. $\endgroup$ – Michael Brown Jul 18 '13 at 8:23
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    $\begingroup$ It will rise, unless you are floating on the ocean (swimming or on a boat). So it depends on whether you are an object :-) But it may take some time to rise (wave propagation). How small is still measurable ? $\endgroup$ – babou Jul 18 '13 at 9:38
  • $\begingroup$ If we talk of minute effects on abstract liquids in an otherwise smoothly evolving world following with infinite precision the laws of fluid statics without disturbance, the sea level may ultimately seem to go down where you stand on shore, while it will rise on most of Earth. But the truth is that with just a normal finger, the effect would be many orders of magnitude too small to have meaning, let alone be measurable. $\endgroup$ – babou Jul 19 '13 at 11:40
  • $\begingroup$ One of my interest in SE and similar sites is to understand the sociology and the motivations of users. I assume you ask the questions because you are interested in the answer, but I am interested in understanding on what basis you choose the answer you decide to accept. Please forgive me if you consider this as too inquisitive. $\endgroup$ – babou Sep 5 '13 at 22:48

a rough, back of the envelope computation you can do: Find out the surface area of the earth in your proposed thought experiment, find out the volume of a finger. Compare the number of water molecules in each case.

Now look at the compressibilty of water (its not 0) and determine whether it would be less energy to push all the water molecules on earth up a small amount or to compress a small amount. I'm assuming the region of compression would decay as you moved further away from the perturbation (so your finger doesn't do work against the whole world's water body, just a local region.

I presume it would be more likely to compress locally than move globally. Doing work against water means not only work against gravity, but work also against surface tension.

  • $\begingroup$ Can you give some more details to explain how "the region of compression would decay as you moved further away ..." ? I am not sure I understand how this decay operates. Is that the compression created by the finger doing work against the water? $\endgroup$ – babou Jul 20 '13 at 22:57
  • $\begingroup$ A formal treatment would rely on Navier Stokes and volume viscocity, but just thinking of the bulk modulus for simplicity as you draw large concentric rings around the point where you put your finger in. The expression (dV/V) dV is change in volume, V is total volume. As you consider larger and larger concentric circles, dV stays the same, but V is getting larger, so the ratio dV/V is getting smaller, thus dP (change in pressure) is getting smaller as you go further away from the perturbation point. $\endgroup$ – Xurtio Jul 21 '13 at 18:39
  • $\begingroup$ I see... I was wondering because you stated that you "assumed the question was about the steady state solution". But Navier-Stokes is about fluid dynamics. So I did not see how there could be a decaying compression. $\endgroup$ – babou Jul 21 '13 at 20:51
  • $\begingroup$ I mentioned only spatial decay, not temporal decay. $\endgroup$ – Xurtio Jul 21 '13 at 22:42
  • $\begingroup$ I am trying to understand the relation between the spatial decay of compression and Pascal's law in steady state. Can you say a bit more on how this decay is evaluated ? $\endgroup$ – babou Aug 31 '13 at 20:31

It is important to note that, even if the analysis in Xurtio's answer does predict an increase in the water level on the other side of the Earth, this increase will take a finite time to take effect, because the disturbance will travel at a finite speed. This is because the disturbance will travel in the form of a surface wave, and these are comparatively slow. The fastest are tsunamis, which have a huge wavelength, but even they travel at about jet-plane speed and would thus take about a day to go round the Earth. Smaller waves are typically slower and would thus take longer. Aditionally, the relaxation time (the time needed for the disturbance to die down and the new level to be established) would be a largeish multiple of the round-trip time of the wave.

In short: the Earth is big!

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    $\begingroup$ True, I assumed the question was about the steady state solution after all the transients have passed. $\endgroup$ – Xurtio Jul 18 '13 at 15:00

Ultimately, the sea would rise on the other side of Earth, but maybe not where you stand, assuming your finger is big enough.

The first approximation to answering this question is that, as long as you are not floating on the ocean, directly or in a boat, the effect is identical to increasing the volume of water by an amount V equal to the immerged volume of finger. But it will take some time for the effect to be felt around the world, as it must propagate.

However, is there any effect worth considering ?

How much of a rise for a normal finger ?

Is there a significant effect. Let's do the back of the envelope calculation. We will assume that the volume of the finger is about 10cc (cubic centimeters), i.e. $10^{-5}\ m^2$. The see surface of Earth is 361 millions square kilometers, i.e., $3.61\ 10^{14}\ m^2$. Assuming you have an ideal incompressible liquid with no granularity (which is true as a first approximation), we want to know how much the level rises. We only have to divide the added volume by the total surface, i.e.: $\frac{10^{-5}}{3.61\ 10^{14}}\ =\ 1/3.61\ 10^{-19}\ =\ 2.8\ 10^{-20}\ m$. That is not very much, but what is it to a water molecule ?

We know that one mole of water contains $6\ 10^{23}$ molecules (Avogadro's number) and weights approximately 18g (16 for oxygen and twice 1 for hydrogen), i.e. has a volume of 18cc. We only have to divide this number by Avogadro's to get the average volume taken by one molecule: $18\ 10^{-6}\ /\ 6\ 10^{23}\ =\ 3\ 10^{-29}\ =\ 30\ 10^{-30}\ m^3$. Assuming the volume taken by each molecule is a small cube, to simplify computation (we look only for orders of magnitude), we simply take the cubic root, which give a height of $3.1\ 10^{-10}\ m$.

So the rise to be expected is at best about one tenth of a billionth of the size of a water molecule. Is that measurable ? I am not sure we can measure anything that small. The only devices I can think of are interferometers. But I doubt. It would probably require frequencies far too high and energetic (this is really beyond my competence). Anyway it would not make sense for measuring the rise of water level as the frontier between water and air cannot be defined with that much precision, even when water is absolutely still.

Now, considering real liquids and other real phenomena is a waste of time given all the approximation, and the infinitesimal character of what might happen, even compared to brownian motion of molecules.

If you want anything significant, you need a much bigger finger. Let's try to do that.

The big finger case

There is more to this problem. Since you take this as an abstract problem, one may assume that an abstract finger can have any mass and volume (nail included). One may also assume that since you listed out parameters that should be ignored, all others are fair game. I will come back to the standard size finger in the end.

Suppose you have a huge finger, with a mass in trillions of tons or more (that is when, in reality, the point I am making may becomes weakly significant, I think), or if you prefer a volume in thousands of cubic kilometers (just a cube 10km on each side). An even bigger finger (25 millions km$^3$) is currently sitting on the south pole, and is likely to be dipped in the ocean this century (see link below).

Then other things happen that may be measurable. The continents are themselves floating on the Earth mantle, which has a mass density around D=4 kg/dm$^3$.

Dipping your huge finger (volume V) in the ocean will ligthen the continental plate you are sitting on by a weight equivalent to the volume V of water (of your finger). Hence the plate will rise (just give it time, plated things are slow, as we learned from turtles). However, since the mantle has a greater density D than water (D is actually also the density ratio), it will rise only by a volume V/D, which is about V/4. But one datum is missing: what is the surface of the plate your are sitting on.

All I know is that plates are generally smaller than oceanic surface, smaller than 1/4 of that surface: there are 8 major plates, and the ocean are 72% of the planet surface. So I would think that, while people on the other side of the planet will experience a rise in water level. People on your tectonic plate will experience, in due time, a fall in the water level, which is really due to the fact that your plate - the ground you stand on - is rising.

The reasoning above uses very larges actual figures, taken from the real Earth, to give some reality to the analysis. As I said, it may well happen sooner than we should wish, with these figures, as discussed in an answer to another question.

To conclude

But if we come back to the original abstract small finger in an abstract smooth world obeying without fuss the laws of fluid statics, the same reasoning can apply. And since the continental plate rise is in the same order of magnitude as the see level rise, there is no reason not to take it into account. Hence the above conclusion is also true in the case of the small finger, absurdly non significant as it may be.


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