Can somebody derive the Young's equation for the contact angle and surface energy of a droplet. Young's equation


I've scratched my head for days finding how it exist.but couldn't find it anywhere on internet.please help:|

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Ideal_surface $\endgroup$
    – G. Smith
    Commented Jun 13, 2019 at 16:40
  • $\begingroup$ Thanks Smith it helped like jhon's answer below i kinda understood what the prove is but cannot figure out that on what object the force is acting that we are talking of. Somesort of boundary line is mentioned everywhere but forces act on real world objects not imaginary boundaries.may be boundary means some sort of edge of atoms.but whose atom solid liquid or gas and if it is liquid then how can solid vapour surface forces can act on liquid. $\endgroup$ Commented Jun 14, 2019 at 10:28

1 Answer 1


To try and make this clearer I've drawn a three dimensional picture of the drop:


The surface tension is the force per unit length acting normal to a line. Consider the small segment of the perimeter of the drop I have marked as $\ell$. We'll assume $\ell$ is small enough that it can be considered as straight. If the surface-vapour surface tension is $\gamma_{sv}$ then this produces a force on our line $\ell$ of $F = \gamma_{sv} \ell$ as shown by by the red line. So the total outwards force is:

$$ F_\text{out} = \gamma_{sv} \ell $$

Likewise inside the drop the surface tension between the solid and the liquid is pulling the drop inwards with a force $F = \gamma_{ls} \ell$. This is shown by the blue line.

And finally the surface tension at the liquid-vapour interface is pulling our line element with a force $F = \gamma_{lv} \ell$ and it is pulling upwards at an angle $\theta$ where $\theta$ is the contact angle. So the horizontal component of this force is pulling our line inwards with a force $F = \gamma_{lv} \ell \cos\theta$. This is shown by the green line. So the total inwards forces (the blue and green arrows) are:

$$ F_\text{in} = \gamma_{sl} \ell + \gamma_{lv} \ell \cos\theta $$

And the last step is just to say that if the drop is in equilibrium, i.e. it is neither spreading out not rolling up, then the inwards and outwards forces must be the same:

$$ \gamma_{sv} \ell = \gamma_{sl} \ell + \gamma_{lv} \ell \cos\theta $$

And just divide through by $\ell$ to get Young's equation:

$$ \gamma_{sv} = \gamma_{sl} + \gamma_{lv} \cos\theta $$

  • $\begingroup$ Force always act on some body.here I can not understand on what object the force is acting is it a line of liquid atoms or vapour atoms and if it is a line of liquid atoms then how the force of solid vapour surface tension is acting over it. $\endgroup$ Commented Jun 14, 2019 at 6:01
  • $\begingroup$ @gaurangagarwal The force is acting on the edge of the drop i.e. it is pulling the edge of the drop outwards or inwards. $\endgroup$ Commented Jun 14, 2019 at 6:18
  • $\begingroup$ Then how solid vapour surface tension is acting over it . I mean that solid vapour surface tension must work on solid/ vapour atoms not the liquid atoms(edge of the drop) $\endgroup$ Commented Jun 14, 2019 at 6:44
  • $\begingroup$ Is there something wrong with my statement? Please correct me if it is wrong:-( $\endgroup$ Commented Jun 14, 2019 at 10:32
  • 3
    $\begingroup$ @Buraian all surfaces have an associated interfacial energy i.e. an excess energy per unit area of the interface compared with no interface. Or to put it another way, if you take a bulk sample and divide it along a plane to create two surfaces this consumes energy. So for any surface it is energetically favourable for the surface to reduce its area. Hence the solid water surface wants to pull the liquid inwards and the solid air interface wants to pull the liquid outwards. $\endgroup$ Commented Jun 27, 2021 at 16:15

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.