# A question about the form of Laplace law

Sometimes, I find some websites writing laplace law as: $$$$p = \gamma\frac{1}{R}$$$$ and some $$$$p = \gamma\frac{2}{R}$$$$ I can understand that the second form is a result of spherical interfaces where curvatures are of same radii, but what about the first equation, i got confused, (knowing that the radius was not multiplied by 2 so that 2 disappeared after simplifying)? What is the interpretation of such law?

• Could you link one example of each? Apr 27, 2022 at 8:56
• @stafusa for example in this paper hal.sorbonne-universite.fr/hal-00783720/document equation (2.5), they included later that $\mu$ which is laplace pressure is $\gamma/r$ Apr 27, 2022 at 9:18

The Young–Laplace equation is usually written as

$$\Delta p = -\gamma\left(\frac{1}{R_1}+\frac{1}{R_2}\right), \label{1}\tag{1}$$

where $$\Delta p$$ is the Laplace pressure, the pressure difference across the fluid interface (the exterior pressure minus the interior pressure), $$\gamma$$ is the surface tension (or wall tension) [...] and $$R_{1}$$ and $$R_{2}$$ are the principal radii of curvature.

In the given source the equation without the factor $$2$$ appears in a section considering a 2D system:

For simplicity we adopt the two dimensional framework introduced in Rivetti and Neukirch (2012) where the liquid-vapor interface is a cylindrical arc

where $$R_2\to\infty$$, rendering Eq.(\ref{1}):

$$\Delta p = -\gamma\frac{1}{R_1}.$$

While in a (cylindrical) capillary the surface is a portion of a sphere, where $$R_1=R_2=R$$, so that $$(1/R_1+1/R_2)$$ becomes $$2/R$$ and Eq.(\ref{1}):

$$\Delta p = -\gamma\frac{2}{R}.$$

• Well in YOung-laplace equation, equation (1), you have a minus sign. In another link on wikipedia, it is positive. How to deal then with this? Jun 25, 2022 at 11:18
• @Aly It's just a matter of convention - of which direction is positive or whether $\Delta P$ stands for $P_{\text{out}}-P_\text{in}$ or $P_{\text{in}}-P_\text{out}$: in my link they define it as "exterior pressure minus the interior pressure", in yours, "$P_\text{inside}-P_\text{outside}$. Jun 25, 2022 at 11:43
• yeh exactly, I thought about this, but it was not clear enough in the website, I thought there was another explanation. Thanks! Jun 25, 2022 at 12:21

It's because there are two radii of curvature, and there are two radii because a liquid-air interface is two dimensional. We conventionally call the two radii $$R_1$$ and $$R_2$$, and the equation for the pressure difference across the interface is:

$$\Delta P = \gamma \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$$

If we consider a cylinder this is curved in only one direction, or more formally one of the radii is infinite so its reciprocal is zero. In that case we get:

$$\Delta P = \gamma \left( \frac{1}{R} + 0 \right) = \frac{\gamma}{R} \tag{1}$$

If we now consider a sphere it curves in both directions with equal radii of curvature $$R$$ so we get:

$$\Delta P = \gamma \left( \frac{1}{R} + \frac{1}{R} \right) = \frac{2\gamma}{R} \tag{2}$$

If we look at the diagram in the paper you cited:

the drop they are considering is a cylinder so equation (1) applies.