‘equilibrium’ means – no change in
motion. So, if a moving body is in equilibrium, it cannot have any
Either it is at rest or it moves at uniform velocity.
To understand ‘stability’
we can consider a ball at rest ( in equilibrium, obviously). We want to
stable its equilibrium is. Let us consider the ball in three different
situations as shown in the diagram.
the ball is kept on the top of an
inverted bowl. In case-B, the
ball is inside the bowl and in case-C,
is placed on a plane surface.
things are observed in these
- If we now
displace the ball slightly, in the case A, the ball displaces ifself
further. So, it is said to
be in unstable equilibrium.
- In case-B, the
ball, when displaced (slightly),
goes back to its previous position and so, this is the case of stable
- In case-C, the
ball, when displaced, neither can
goes back to the previous position, nor does it go away from that
location. So, we
call it to be in neutral
We may observe an
air parcel in the atmosphere. At a
certain position z = z0,it
is part of the atmosphere and its density is ρ0 . So, it’s
(the force due to gravity) per unit volume is ρ0g. This
air-parcel is now displaced to z = z1,
where the density of the surrounding is ρ1.
At this point, we should renew our knowledge about
buoyancy. Buoyancy is the
is applied by a fluid, on a body that is immersed in it. This buoynat
due to the fluid-pressure on the body.
The fluid-pressure on the top(downward) of the body is always less than
the lower-face (upward). The reason is simple, - the top side is always
smaller depth than the bottom. Due to this difference in pressure, the
force acts in the upward direction, trying to counterbalance the force
The force of
buoyancy, on a body in a fluid, is
equal to the weight of the fluid that is displaced by the body itself.
So, if the
parcel we are observing doesn’t mix with its surrounding, then the net
on it, per unit volume, is:
F = (ρo – ρ1)g.
stability S as,
S = (ρo- ρ1)/(zo – z1) when,(z – z1)-> 0
Comparing with the previous examples
stability, we can say that the parcel is :
- Stable if S > 0, or
Unstable if S
0, or negative.
if s = 0
Stability in a
This model is
more realistic because it fits the atmospheric properties better.
process is adiabatic, which ensures that
the parcel under consideration does not exchange heat with its
It can be derived
that for a polytropic atmosphere :
ρo / ρ1 = (po – p1)1/n
where, n is some
index. Now, the stability factor S
can be shown to be depending on the
relative value of n as compared to γ, the ratio of two specific heat of
fluid (here, air). So,
< γ Stable
as S > 0, or
> γ Unstable
as S<0, or
- If n
= γ Neutral as
S = 0.
: Phys-645, Fall-2007, UAF
for 'Home" - taken fromwww.animationlibrary.com