# What is the direction of buoyancy in the bulk of liquid on earth?

I have a trouble when considering the direction of buoyancy in the bulk of liquid subjected to vertical temperature gradient. The liquid is heated from below or above that induces a natural convection in the bulk with a sufficient temperature difference, because the liquid density decreases with increasing temperature, see the following figures. Here, the bigger dot in the liquid denotes liquid particle of bigger density and the smaller one for lighter one. Taking the left figure as example, I understand that the lower liquid with less density will rise up to the surface where it is cooled down and the upper liquid with bigger density will go down to displace the lighter one then it is heated and rises further.

My confusion is that if considering the lighter liquid the buoyancy force is upwards, opposing the gravity and temperature gradient. On the other hand, for upper-layer liquid with bigger density the buoyancy force is downwards and along the gravity and temperature gradient. I determined these directions of buoyancy simply based on intuition from Newtonian mechanics.

My questions are:

1. What is the direction of the buoyant force on earth? Or how to identify its unique direction, if it is unambiguous, with respect to the temperature gradient and gravity directions.

2. What is the relation between the directions of the buoyant force and gravity as well as temperature gradient? The reason why I ask this question is that I am still thinking the direction of buoyant force is related to that of temperature gradient and gravity.

Thank you very much!

Let's assume that you have some reference density $\rho_0$. Then if $\rho=\rho_0$ everywhere and the fluid is at rest, $p = \rho_0 g z$. Now let's call this baseline pressure $p_0(z)$, without loss of generality we can write the pressure as $p = p' + p_0(z)$. If you rewrite your momentum equation with $p'$, you can group the $p_0$ term with the gravity force and obtain a combination that you'll call the buoyancy force: $$F_b = \rho(\vec{r}\;) \vec{g} - \nabla p_0(z) = (\rho - \rho_0) \vec{g}$$ You can see that depending of the sign of $\rho - \rho_0$, it will either be oriented up or downward (but it is always aligned with $\vec{g}$). Note also that this buoyancy force depends on your arbitrary choice of $\rho_0$, the pressure $p'$ will take up the difference if you make a different choice.