Suppose I do two experiments to find the triple point of water, one in zero-g and one on Earth. On Earth, water in the liquid or solid phase has less gravitational potential per unit mass than water in the gas phase. Therefore, the solid and liquid phases should be favored slightly more on Earth than in zero-g.

In a back-of-the-envelope calculation, how does the temperature of the triple-point of water depend on the gravitational acceleration and, if necessary, on the mass of water and volume and shape of container?

Edit Let's say I have a box in zero-g. The box is one meter on a side. It has nothing in it but water. Its temperature and pressure are just right so that it's at the triple point. All the water and ice and steam are floating around the box because it's zero-g.

Now I turn on gravity. The liquid water and ice fall to the bottom of the box, but the average height of the steam remains almost half a meter above the bottom of the box. So when gravity got turned on, the potential energy of the ice and liquid water went down significantly, but the potential energy of the steam didn't. Doesn't this mean that once gravity is turned on, water molecules would rather be part of the ice or liquid phase so that they can have lower energy? Wouldn't we no longer be at the triple point?

Several people have posted saying the answer is "no". I don't disbelieve that. Maybe the answer is just "no". I don't understand why the answer is no. Answers such as "No, because gravity doesn't affect the triple point," or "No, because the triple point only depends on pressure and temperature" simply restate the answer "no" with more words.

  • $\begingroup$ The correction is likely vanishing except in the case of very strong fields as in the vicinity of a black hole. In that limit, however, this is an interesting and relevant question. $\endgroup$
    – user346
    Dec 27, 2010 at 9:40
  • 2
    $\begingroup$ Why would the temperature depend on g and not the pressure? The triple point is a (p,T) point. $\endgroup$
    – Sklivvz
    Dec 27, 2010 at 17:07
  • $\begingroup$ Changing the pressure (i.e. by adding gravity) means we move away from the triple point. What actually happens could depend on the initial density of the water in the box. Note that this is not fixed by the pressure because at coexistence, different average densities can occur with the same pressure (simply different amounts of the coexisting phases). Worth thinking about... $\endgroup$
    – Greg P
    Dec 28, 2010 at 3:47
  • $\begingroup$ @Greg P Thanks. On rereading your answer, I see that you got what I was asking more than I realized. $\endgroup$ Dec 28, 2010 at 6:15
  • 1
    $\begingroup$ I think a good starting point is to understand what happens to liquid-vapor coexistence: web.ist.utl.pt/berberan/data/79.pdf $\endgroup$
    – Greg P
    Dec 28, 2010 at 17:35

5 Answers 5


This is perhaps similar to what mbq meant, but I will elaborate. The T-p phase diagram of water tells us, for a given temperature and pressure, what phase we will get if we have a bunch of that substance. If I apply different pressures to a bottle of water, I am moving around in the p-direction of the T-p plane. I am not changing the pressure of the triple point of water, just changing the pressure of that particular bottle of water! Similarly, if a tank of water is in a gravitational field, it affects the pressure. In fact, it leads to different pressures at different locations of the tank. It could lead some parts of the tank to freeze, for example. But it does not in any way change the triple point of water itself, which is an intrinsic property of that substance. So I would say that the question is ill-posed. It might be better to ask: what will happen to a tank of water at a given temperature and density if we now apply a gravitational field?

  • $\begingroup$ Yeah, I agree this is the correct answer. The questioner's demand for a more involved answer doesn't make one possible. It's a reasonable question to ask, but the answer is simply that the triple point of water is intrinsic to the material. It makes no more sense to ask for a better explanation that it does to ask why the temperature on Jupiter doesn't make my rice crispies taste better. $\endgroup$
    – Colin K
    Dec 28, 2010 at 2:01
  • $\begingroup$ @Colin The triple point is not some fundamental thing. It is not an axiom that the triple point is intrinsic to the material. It comes from more basic physics. Thus, one can explain why certain phenomena do or do not come into play when considering it. $\endgroup$ Dec 28, 2010 at 7:12
  • $\begingroup$ @Colin: I agree with @Mark that it's by no means intrinsic to a single water molecule (or to what do you refer by intrinsic?). If you start with some model in statistical physics then you can (in principle) compute its triple point (or more generally multiple-phases coexistence points). But this will in general depend on all interactions between all parts of the system and triple point is a property of a complete system, not just one molecule. In particular, it's a property also of the external fields and the way they couple to the molecules. $\endgroup$
    – Marek
    Dec 28, 2010 at 13:06
  • $\begingroup$ @Marek @Mark: By intrinsic I mean it's a property of the material. It doesn't change due to ambient gravitational or other fields (unless we're talking about very strong fields which can distort atomic structures). As Greg P wrote, it does not depend on "all interactions between all parts of a system." If the water at the top of a tall tank is at its triple point, the water at the bottom of the tank doesn't care. All that matters is the local temp and pressure. Gravity or other affects may impact these variables, but they remain the only things which affect the triple point. $\endgroup$
    – Colin K
    Dec 28, 2010 at 14:16
  • $\begingroup$ @Colin: I don't understand why temperature and pressure should be the only things that affect the triple point. Triple point is characterized geometrically as unique point where phases coexist. It doesn't matter whether you draw it on T-p diagram or p-V diagram or any other sort of diagram. Only thing that matters is what kind of couplings can change it. And it's not clear to me that gravitational coupling doesn't belong to those valid couplings. I can definitely imagine e.g. some lattice models whose phase properties do depend on the presence/absence of other interactions. $\endgroup$
    – Marek
    Dec 28, 2010 at 14:20

To expand on mpq and Greg's remarks.

A triple point is a unique (pressure,temperature) pair for each material (and where appropriate group of phases) where three phases of matter can coexist in equilibrium.

If you are not at the right pressure, there is no temperature where this occurs, and if you are not at the right temperature there is no pressure where this occurs.

In that sense the question is simply based on a misconception.

Several posters have tried to address the effects of changing gravitation on the locally experienced pressure due to a atmosphere column. That's fine as far as it goes, but it doesn't really go to the question.

Diddling the local gravity will change the pressure naturally experienced due the local atmosphere but will not affect the parameters of the triple point.

  • $\begingroup$ If I assume the normal sort of geometry, the gas phase will be upgravity of the liquid and solid phases. I think this means that we can include the extra gravitational potential energy of the gas molecules to about a mean free path, as Mark wants to. This wouldn't be much, as a MFP is going to be rather small (at least for say water/steam) probably only a few microns. But I'm not even sure if that is right, does the potential a mean free path above the liquid/gas interface matter, or just the potential right at the interface? I'm not sure. $\endgroup$ Dec 28, 2010 at 2:02

There is no difference; phase transitions does not change gravitational potential.

  • $\begingroup$ What kind of answer is this? $\endgroup$
    – Marek
    Dec 27, 2010 at 10:17
  • 1
    $\begingroup$ @Marek Concise (-; $\endgroup$
    – user68
    Dec 27, 2010 at 10:32
  • $\begingroup$ This answer is simply wrong I'm afraid. The gravitational potential affects the internal pressure of the fluid, and thus the triple point temperature. $\endgroup$
    – Noldorin
    Dec 27, 2010 at 14:59
  • $\begingroup$ Changing the pressure also affects the pressure ;-) Do we then say that changing the pressure therefore affects the 'pressure of the triple point'? The question itself does not make sense to me. $\endgroup$
    – Greg P
    Dec 27, 2010 at 19:33
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    $\begingroup$ @Noldorin: You might want to look again at the meaning of "triple point". It is usually defined as the point in the P-T plane where three phases can co-exist. Note that that is a fixed pressure and a fixed temperature. Changing the gas pressure does not change where the triple point is. Merely whether you need a pressure vessel to achieve it. To elaborate you can't get water vapor, liquid water and ice I to coexist in equilibrium at 1 atmosphere because the triple point occurs at a much lower pressure. Period. $\endgroup$ Dec 27, 2010 at 20:23

I don't see how to derive triple point directly, but let me talk about something similar and see where it leads.

For ideal gas atmosphere you'll find out that $g$ works as an inverse temperature. That is, $0K$ is the same as infinitely strong gravity because both keep the gas at the ground.

The real gases exhibit phase transitions, so we should promote ideal gas to something like van der Waals gas. This should not change the above correspondence of $\beta = {1 \over k_B T}$ and $g$ much. Therefore, I would imagine that one would get a phase transition at high $g$ with constant $\beta$. Or in other words, that in the presence of gravitational field, higher temperatures are enough for the phase transition. Or in other words yet, we have another coupling $g$ that tells system to cluster besides the usual thermal coupling $\beta$.

Now, this carries over (by a huge and completely unjustified extrapolation) to the triple point of water. I'll try to provide some calculations later.

  • $\begingroup$ excuse my ignorance - what is $\beta$? $\endgroup$ Dec 27, 2010 at 10:25
  • $\begingroup$ @Mark: sorry, I didn't realize this wasn't universally known. I'll edit the info into my answer. $\endgroup$
    – Marek
    Dec 27, 2010 at 10:34
  • $\begingroup$ How can g have the same effect as a change in temperature? It has a position-dependent effect. In a gravitational field, the density will depend on location - in contrast to the situation with a different temperature. $\endgroup$
    – Greg P
    Dec 27, 2010 at 19:35
  • $\begingroup$ @Greg: I think Marek is getting at the idea that the scale height of an atmosphere can be changed by dialing either the gravity or the temperature. Which is true but beside the point. $\endgroup$ Dec 27, 2010 at 20:45
  • $\begingroup$ @dmckee: yes, that's what I was getting at, but I realize now that I missed the point completely. $\endgroup$
    – Marek
    Dec 28, 2010 at 0:24

Back of the envelope calculation:

  • gravity will manifest itself by increasing the pressure by a factor of $mg/A$ where A is the surface of the container with water at the triple point. So basically the effect is shifting the triple point phase diagram down. So the triple point would be at the same temperature with a lower pressure.

$$p_{measured} = p_{gas} + p_{gravity}$$

  • Assuming a regular shape we can say that $p_{gravity}=\rho g h$ where $\rho$ is the density of the gas at our triple point pressure and h is the typical height of a column of gas.

$$p_{gravity}=\rho g h$$

  • to calculate $\rho$ we can use the ideal gas law:

$$\rho = \frac{Nm_w}{V} = \frac{p_{gas}m_w}{kT}$$

  • we can now calculate what fraction of pressure is due to gravity

$$\frac{p_{gravity}}{p_{gas}}=\rho g h/p_{gas}=\frac{m_w g h}{kT}$$

  • using the following values we can calculate this ratio

$m_w = 3 \times 10^{-26} \mathrm{kg}$

$g = 9.81 \mathrm{m/s^{2}}$

$h = 1 \mathrm{m}$

$k = 1.38 \times 10^{-23} \mathrm{J/K}$

$T = 273.16 K$


This means that you need a value for g one thousand times stronger to see a large effect.

  • $\begingroup$ A bigger value of g, or a container a kilometer high... $\endgroup$
    – Chad Orzel
    Dec 27, 2010 at 16:39
  • $\begingroup$ The kilometre high container would need to have constant pressure though. $\endgroup$
    – Sklivvz
    Dec 27, 2010 at 16:44
  • $\begingroup$ If $T$ was lowered though, you'd see a bigger change. Nice calculation ! $\endgroup$
    – user346
    Dec 27, 2010 at 17:44
  • $\begingroup$ Would you care to elaborate on what you mean by $p_{gas}$ and $p_{gravity}$? My immediate instinct is the call shenanigans, but I'm willing to be convinced that I've missed something. $\endgroup$ Dec 27, 2010 at 20:35
  • $\begingroup$ To expand on my comment, I think you might be re-inventing the distinction between gauge pressure and absolute pressure. $\endgroup$ Dec 27, 2010 at 20:55

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