Stability
- I

**Equilibrium
and
stability:**

**In dynamics,
‘equilibrium’ means – no change in
motion. So, if a moving body is in equilibrium, it cannot have any
acceleration.
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
know, how
stable its equilibrium is. Let us consider the ball in three different
situations as shown in the diagram. **

**In case-A,
the ball is kept on the top of an
inverted bowl. In case-B, the
ball is inside the bowl and in case-C,
the ball
is placed on a plane surface.**

**Three different
things are observed in these
situations:**

**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
equilibrium*.

**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
equilibrium** .*

**Stability in
(static) atmosphere:**

**
**

**We may observe an
air parcel in the atmosphere. At a
certain position z = z**_{0},it
is part of the atmosphere and its density is ρ_{0} . So, it’s
weight
(the force due to gravity) per unit volume is ρ_{0}g. This
air-parcel is now displaced to z = z_{1,}
where the density of the surrounding is ρ_{1. }

**
At this point, we should renew our knowledge about
buoyancy. Buoyancy is the
force that
is applied by a fluid, on a body that is immersed in it. This buoynat
force develops
due to the fluid-pressure on the body.
The fluid-pressure on the top(downward) of the body is always less than
that on
the lower-face (upward). The reason is simple, - the top side is always
at a
smaller depth than the bottom. Due to this difference in pressure, the
buoyant
force acts in the upward direction, trying to counterbalance the force
of
gravity. **

**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
force F
on it, per unit volume, is:**

**
**

**
F = (ρ**_{o – }ρ_{1})g.

**We define
stability S as,**

**
S = (ρ**_{o-} ρ_{1)}/(z_{o} – z_{1}) when,(z – z_{1})-> **0 **

**or, S
= -(dρ / dz)**

Comparing with the previous examples
for explaining
stability, we can say that the parcel is :

**Stable if S > 0, or
positive.**
**Unstable if S
<
0, or negative.**
**Neutral
if s = 0**

# A
better atmospheric-model:

**Stability in a
polytropic atmosphere:**

**This model is
more realistic because it fits the atmospheric properties better.**

**Assumptions: the
process is adiabatic, which ensures that
the parcel under consideration does not exchange heat with its
surroundings.**

**It can be derived
that for a polytropic atmosphere :**

**
ρ**_{o }/ ρ_{1 } = (p_{o} – p_{1})^{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
the
fluid (here, air). So, **

**If
n
< γ Stable
as S > 0, or
position.**

**If
n
> γ Unstable
as S<0, or
negative.**

**If n
= γ Neutral as
S = 0.**

Tapas
bhattacharya

Web-project
: Phys-645, Fall-2007, UAF

Animation
for 'Home" - taken from*www.animationlibrary.com *