0
$\begingroup$

Sometimes, I find some websites writing laplace law as: \begin{equation} p = \gamma\frac{1}{R} \end{equation} and some \begin{equation} p = \gamma\frac{2}{R} \end{equation} 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?

$\endgroup$
2
  • 1
    $\begingroup$ Could you link one example of each? $\endgroup$
    – stafusa
    Apr 27, 2022 at 8:56
  • $\begingroup$ @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 $\gamma/r$ $\endgroup$
    – Remember
    Apr 27, 2022 at 9:18

2 Answers 2

1
$\begingroup$

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}):

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

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – Remember
    Jun 25, 2022 at 11:18
  • 1
    $\begingroup$ @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, "$P_\text{inside}-P_\text{outside}$. $\endgroup$
    – stafusa
    Jun 25, 2022 at 11:43
  • 1
    $\begingroup$ yeh exactly, I thought about this, but it was not clear enough in the website, I thought there was another explanation. Thanks! $\endgroup$
    – Remember
    Jun 25, 2022 at 12:21
1
$\begingroup$

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:

Cylindrical drop

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

$\endgroup$

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.