In order to decide whether buoyancy is a conservative force, let us study do two thought experiments.

  1. For objects with density less than the density of fluid, no matter how deep we place the object below the surface of the fluid, it comes up with some kinetic energy. This indicates that the force due to buoyancy is a conservative force with some potential energy associated with it.
  2. However, for objects with density more than the density of fluid, no matter how deep we place the object below the surface of the fluid, it never comes up. This indicates that the force due to bouyancy is a non-conservative force.

Which of these two conclusions is correct?

  • $\begingroup$ I was ready to offer a bounty , unfortunately my my reputation is just 1 less than 75( the privilege to offer bounty) :) $\endgroup$ Dec 21, 2016 at 5:44
  • $\begingroup$ You have just posted the query; wait for at least two days before thinking about bounty. $\endgroup$
    – user36790
    Dec 21, 2016 at 5:45
  • $\begingroup$ Oh no, another "thoughtful" username... when will it end? :) $\endgroup$
    – anon01
    Dec 21, 2016 at 6:26
  • $\begingroup$ @ConfusinglyCuriousTheThird Whose name mine or MAFIA36790 ? $\endgroup$ Dec 21, 2016 at 6:28
  • $\begingroup$ yours! There are a few (frequenting) users here with names close enough to cause confusion... $\endgroup$
    – anon01
    Dec 21, 2016 at 6:29

1 Answer 1


For an object that is fully submerged the buoyant force is constant (in both magnitude and direction) exactly as the usual "near the surface of the Earth" approximation to gravity. That part is clearly conservative.

To figure the behavior of a object while partly submerged we much rely on Archimedes' principle, which tells us that the buoyant force is proportional to the weight of the volume then displaced. The act of submerging the body may require a different distance depending on the shape and orientation of the body, so we can be suspicious.

Take for a concrete example a non-spherical, ellipsoidal object. By symmetry and construction it will require the same average force to insert it into the fluid at any angle, but that force will be applied over a different distance depending on orientation. The act of submerging the object itself requires different amounts of work depending on the orientation of the object.

But before we say "Ah! Ha!" we need to consider that the starting and ending conditions for putting it end the long way are separated by a larger total distance than those for putting it in the short way. To know if the force is conservative over those paths we have to make the starting and ending points the same. Allow the starting point of the COM to be a distance $a$ over the fluid surface and the ending point to be a distance $a$ under the surface (where $a$ is the semi-major axis of the ellipse in some cross-section). The work done by gravity plus buoyancy on a vertical path between those points is $$W_a = 2a\left(\rho_o - \frac{1}{2} \rho_f \right)V_o g\;,$$ where $V_o$ is the volume of the object and the densities $\rho_{o,f}$ are for the object and the fluid respectively. Compare to lowering the object oriented so that the minor axis $2b$ is vertical. The work done is \begin{align*} W_b &= \left[(a-b) \rho_o V_o g\right] + \left[2b\left(\rho_o - \frac{1}{2} \rho_f\right)V_o g \right]+ \left[(a-b) \left(\rho_o - \rho_f\right)g V_o\right] \\ &= \left[ \left((a-b) \rho_o\right) + \left(2b\left(\rho_o - \frac{1}{2} \rho_f\right) \right)+ \left((a-b) \left(\rho_o - \rho_f\right)\right)\right]V_o g \\ &= \left[ \left[ (a-b) + 2b + (a-b) \right] \rho_o - \left(b + (a-b)\right)\rho_f\right]V_o g \\ &= \left[ 2a\rho_o - a\rho_f\right]V_o g \\ &= 2a\left( \rho_o - \frac{1}{2}\rho_f\right)V_o g \;, \end{align*} exactly as before.

This shows that at least for sufficiently symmetric objects and considering the combination of gravity and buoyancy even the dunking process is conservative. Explicit extension to arbitrary shapes requires that the summation done here be performed as an integral with careful attention to the endpoints and the completely out of water and completely underwater parts of the movement, but one might also appeal to intuition about having to move the same amount of fluid upward to achieve submersion.

  • $\begingroup$ Now that buoyant force is conservative for at least moderately symmetrical bodies , can we define a potential energy such that Work done by Buoyant force = - ( Change in potential energy) $\endgroup$ Dec 21, 2016 at 6:26
  • 1
    $\begingroup$ The question isn't "Can you define it?" but "Will it be sufficiently worthwhile to bother?", and because almost all fluid environments are heavily influenced by dissipative forces most people will probably answer "No." most of the time. The two important lessons here are the argument by analogy and the critical importance of paying attention to starting and ending points when asking about conservativeness. $\endgroup$ Dec 21, 2016 at 6:32
  • $\begingroup$ @dmckee nice answer. Ignoring the dissipative forces, and assuming a rigid object (e.g. for balloons this is not conservative), it seems like there should be stronger statements one could make. Something about smooth mapping of the buoyant force to the objects position and orientation or something. $\endgroup$
    – anon01
    Dec 21, 2016 at 6:41
  • $\begingroup$ @ConfusinglyCuriousTheThird I feel that you are right, but I am too tired tonight for anything but the kind of straight ahead grind you see here. $\endgroup$ Dec 21, 2016 at 7:03

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