How can one determine the height of a liquid droplet on a flat horizontal surface if we know the necessary physical properties of the different materials? The acceleration due to gravity is $g$. I want to look at all cases where the contact angle (between liquid surface and ground) is $0 < \alpha < 180^\circ$.

There may not be an exact answer, but I am looking for a good approximation. Here are some information, which may or may not be helpful in solving this problem:

There are three interfaces: liquid-gas, liquid-solid, solid-gas. If their respective surface tensions are $\gamma_{l-g}$, $\gamma_{l-s}$, and $\gamma_{s-g}$, then we can show (using force balance or energy conservation) that: $$\gamma_{l-g}\cos\alpha+\gamma_{l-s}=\gamma_{s-g}$$

What I tried was using this information, I was able to determine the contact angle $\alpha$. Then I tried expressing the energy in terms of the given variables, the volume $V$ and the height $h$. For a large $h$, the surface area would be small, reducing the energy stored in the surface tension, but the gravitational potential energy would increase. Similarly for a small $h$, the opposite would be true. If I am somehow able to express the surface area and location of the center of mass as a function of these mentioned variables, then I can find the value of $h$ that would minimize the energy.

Here are some general questions about the setup:

  • Will energy be stored in the solid-air interface? The liquid-air interface is straightforward, it's simply the area of the surface above the ground. The liquid-ground interface similarly would be the area in which the droplet is in contact with the ground. But is it possible (or is it even necessary) to consider the interface between the ground and the air in terms of energy?
  • How can I determine the surface area and the height of the center of mass? I do not think it is clear what shape this droplet will take. For certain values of $\alpha$, I could take it as a spherical cap, but it will not work for all values of $\alpha$ which I am interested in answering.
  • 1
    $\begingroup$ See my answer on physics.stackexchange.com/questions/509623/… $\endgroup$ Commented Apr 6, 2020 at 22:21
  • $\begingroup$ Thanks for linking me there, but I don't really think that helps answer the question here. You introduced a dimensionless factor that can qualitatively describe the behavior, but it doesn't take into account the other factors that I have mentioned. $\endgroup$
    – QiLin Xue
    Commented Apr 7, 2020 at 0:53
  • $\begingroup$ You did not indicate the gravity field, the angle of inclination of the plane relative to the vertical, and the location of the drop. The word ground in this context is misleading. $\endgroup$ Commented Apr 7, 2020 at 10:50
  • $\begingroup$ I have updated the details. $\endgroup$
    – QiLin Xue
    Commented Apr 7, 2020 at 17:15
  • 1
    $\begingroup$ A search on Google Scholar for "height of drop on surface" gives a host of links that may be of interest, including this first one: royalsocietypublishing.org/doi/abs/10.1098/rspa.1972.0159 onlinelibrary.wiley.com/doi/abs/10.1002/cjce.5450580203 $\endgroup$ Commented Apr 9, 2020 at 17:03

2 Answers 2


Here's my incomplete attempt, hopefully it will be helpful!

By symmetry, we know that the surface of the droplet must be rotationally symmetric about some axis perpendicular to the horizontal surface, so we can describe it as a solid of revolution about this axis. We can assume that the droplet only makes contact with the horizontal surface up to some maximum value $r_0$, and has a height $h_0$ at its centre. It is best to describe the surface using the radius as a function $r(h)$ of height, as describing the height as a function of radius $h(r)$ would not allow the description of droplets which overhang $r_0$. This is shown in the figure below. I've made the droplet wobbly because we don't yet know what shape it will take.

droplet on surface

The shape which surface will take is the one which minimises the total energy of the configuration, while keeping the volume of the drop constant. There are two contributions to the energy: the gravitiational potential energy, which depends only on the height of the centre of mass of the droplet; and the surface areas of the various interfaces, each multiplied by the constants $\gamma$. Let $S_{\mathrm{lg}}$ be the area of the curved surface of the drop, $S_{\mathrm{ls}}$ be its contact area with the solid plane, $S_{\mathrm{sg}}$ be the area of the solid surface in contact with the air, $V$ be the total volume of the drop, $\rho$ its density (assumed constant), and finally $\bar{h}$ the height of its centre of mass. Then the energy is $$ E =\gamma_{\mathrm{lg}}S_{\mathrm{lg}} + \gamma_{\mathrm{ls}}S_{\mathrm{ls}} + \gamma_{\mathrm{sg}}S_{\mathrm{sg}} +g V\rho\bar{h}. $$ The area of the entire plane surface is constant, so $S_{\mathrm{ls}} + S_{\mathrm{sg}} = \mathrm{const}$, so, to within a constant offset which doesn't matter, the energy is $$ E =\gamma_{\mathrm{lg}}S_{\mathrm{lg}} + (\gamma_{\mathrm{ls}} - \gamma_{\mathrm{sg}})S_{\mathrm{ls}} + g V\rho\bar{h}. $$ This answers the first of your questions: yes, in general energy is stored in the solid-gas interface, but it can be accounted for simply by changing the liquid-solid surface tension constant from $\gamma_{\mathrm{ls}}$ to the 'effective' surface tension constant $\gamma_{\mathrm{ls}} - \gamma_{\mathrm{sg}}$.

We now need to express the various terms in our expression for the energy as functionals of $r(h)$. It is not too difficult to see that \begin{align} \rho V\bar{h}[r(h)] & = \int_0^{h_0}\rho \pi r^2 h \,\text{d}h,\\ V[r(h)] & = \int_0^{h_0} \pi r^2 \,\text{d}h,\\ S_{\mathrm{lg}}[r(h)] &= \int_0^{h_0} 2\pi r\sqrt{1+(r')^2} \,\text{d}h,\\ S_{\mathrm{ls}}[r(h)] &= \pi r(0)^2 = \int_0^{h_0} \pi r^2 \delta(h) \,\text{d}h. \end{align} In the last line I have used the Dirac delta 'function' to express $S_{\mathrm{ls}}$ as an integral - I'm not sure if this is a sensible thing to do or not.

To minimise the energy functional while keeping the volume constant, we use Lagrange multipliers and so must minimise the functional \begin{align} I[r] &= E[r] +\lambda V[r]\\ &=\pi\int_0^{h_0} \underbrace{\left[2\gamma_{\mathrm{lg}}r\sqrt{1+(r')^2} +(\gamma_{\mathrm{ls}} - \gamma_{\mathrm{sg}})r^2 \delta(h) + g\rho r^2h +\lambda r^2\right]}_{\mathcal{L}(h, r, r')}\,\text{d}h. \end{align} Using the Euler-Lagrange equation $$ \frac{\partial\mathcal{L}}{\partial r} = \frac{\text{d}}{\text{d} h}\frac{\partial\mathcal{L}}{\partial r'} $$ gives us (after some simplification) $$ g\rho r h +\gamma_{\text{lg}}\sqrt{1+(r')^2} +(\gamma_{\text{ls}}-\gamma_{\text{sg}})r \delta +\lambda r = \frac{\gamma_{\text{lg}}}{(1+(r')^2)^{3/2}}((r')^2+(r')^4 +r''r). $$ In theory, solving this horrible looking differential equation should give the correct solution! Since there is a delta function, I'm not even sure if it's well-posed. I would be interested to hear if anyone knows how to proceed from here.

Note that setting $g =0$ in the above equation gives solutions of the form $r = \sqrt{\frac{1}{\lambda^2\gamma_{\text{lg}}^2}- (h-b)^2}$ for $h>0$, where $(h_0-b)^2=1/\lambda\gamma_{\text{lg}}$. This shows that when gravity is neglected the shape of the droplet is a sphere.

Once you have solved the above equation, you can use the constraint on the volume to determine the constant $\lambda$. The functional form of $r(h)$ would also allow you to find the wetting angle, using $\tan\alpha =r'(0) $.

Finally, we note that the endpoint $h_0$ is variable. This means that, in addition to the Euler-Lagrange equations, we require that $$ \left.\mathcal{L} - r'\frac{\partial\mathcal{L}}{\partial r'} \right|_{h=h_0} = 0 $$ (see e.g. here for a derivation). This condition says that $$ \left.2\gamma_{\mathrm{lg}}r\sqrt{1+(r')^2} +(\gamma_{\mathrm{ls}} - \gamma_{\mathrm{sg}})r^2 \delta(h) + g\rho r^2h +\lambda r^2 - r'\cdot 2\gamma_{\text{lg}}r'r\sqrt{1+(r')^2} \right|_{h=h_0} = 0 $$ which is true when $r(h_0) = 0$. This condition is therefore equivalent to the height of the droplet being $h_0$.

  • $\begingroup$ Thank you for your insightful answer! I will need some time to process everything as I am not familiar with Lagrange multipliers. I did take a look at a paper Jeffrey linked (royalsocietypublishing.org/doi/abs/10.1098/rspa.1972.0159) and the authors there derived $h^2=\frac{2\gamma_{lg}}{\rho g}(1-\cos\alpha)$. Although I tried simplifying your final conditions, I am unsure on how to deal with the delta function, and the other variables. In the paper linked, the authors modeled it as a very large drop, perhaps the same assumption can simplify things here? $\endgroup$
    – QiLin Xue
    Commented Apr 10, 2020 at 3:37

Here is another way to obtain the right equation to solve. Instead of using energy, the problem is more easily solved by realizing that we are in a static equilibrium situation. Thus, writing the balance of forces on a small element of droplet surface will allow us to obtain a differential equation that we can solve. This answer is mostly inspired by Wikipedia https://en.wikipedia.org/wiki/Young%E2%80%93Laplace_equation.

There is tension in the surface, which is curved, that is balanced by a pressure inside the droplet. More specifically, the equation to use is the Young–Laplace equation $\Delta p=\rho g h-\gamma \nabla \cdot \hat{n}$, $\Delta p$ being the pressure differential at the surface (called Laplace Pressure), $\rho$ the density of the confined liquid, h the depth in the liquid, g the gravitational acceleration, $\gamma$ the surface tension and $\hat{n}$ the normal vector on the surface.

For the axisymmetric situation (isotropic around the angular direction), we can use cylindrical coordinates to write an equation for r(z): $r''/{(1+r'^2)}^{3/2}-1/(r (1+r'^2)^{1/2})=z-\Delta p$ with pressures and lenghts renormalized to $(\gamma \rho g)^{1/2}$ and $(\gamma/( \rho g))^{1/2}$ respectively. This is a second order differential equation. If you know the radius at a certain z and the value of r' at this point, there is a unique solution that gives you the shape of the droplet.

In your case, you already determined the slope of the droplet at the point of contact with the solid surface (the contact angle). If you presume a given radius at the contact with the solid surface, you can get r(z). To get the volume, you can integrate over the surface delimited by r(z). This will depend on the radius of the droplet (in cylindrical coordinates) at the contact with the solid surface. I don't think there is a general analytical solution to your problem though. Note that there is no law to give $\Delta p$. It is a constant that is determined by existence of a solution.

The above equation gives you conditions under which you can neglect the hydrostatic pressure term: the pressure due to surface tension must be much larger than the hydrostatic pressure. This is valid at least for very small droplets.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.