Imagine a point on the surface of the earth. Now consider unit vectors, one pointing South, one
pointing East and one pointing directly away from the center of the earth.
PICT
Denote the first as i, the second as j, and the third as k. If you are standing on the earth you
will consider these vectors as fixed, but of course they are not. As the earth turns, they change
direction and so each is in reality a function of t. Nevertheless, it is with respect to
these apparently fixed vectors that you wish to understand acceleration, velocities, and
displacements.
In general, let i^{∗},j^{∗},k^{∗} be the usual fixed vectors in space and let i
(t)
,j
(t)
,k
(t)
be an
orthonormal basis of vectors for each t, like the vectors described in the first paragraph.
It is assumed these vectors are C^{1} functions of t. Letting the positive x axis extend
in the direction of i
(t)
, the positive y axis extend in the direction of j
(t)
, and the
positive z axis extend in the direction of k
(t)
, yields a moving coordinate system.
Now let u be a vector and let t_{0} be some reference time. For example you could let
t_{0} = 0. Then define the components of u with respect to these vectors, i,j,k at time t_{0}
as
u ≡ u1i(t0)+ u2j(t0)+ u3k (t0).
Let u
(t)
be defined as the vector which has the same components with respect to i,j,k but at
time t. Thus
u(t) ≡ u1i(t) + u2j(t)+ u3k(t).
and the vector has changed although the components have not.
This is exactly the situation in the case of the apparently fixed basis vectors on the earth if u
is a position vector from the given spot on the earth’s surface to a point regarded as fixed with the
earth due to its keeping the same coordinates relative to the coordinate axes which
are fixed with the earth. Now define a linear transformation Q
(t)
mapping ℝ^{3} to ℝ^{3}
by
Q (t)u ≡ u1i(t)+ u2j(t)+ u3k (t)
where
u ≡ u1i(t0)+ u2j(t0)+ u3k (t0)
Thus letting v be a vector defined in the same manner as u and α,β, scalars,
^{T} = I. This proves the
first part of the lemma.
It follows from the product rule, Lemma 2.8.2 that
′ T ′ T
Q (t) Q(t) + Q (t)Q (t) = 0
and so
( )
Q ′(t)Q (t)T = − Q′(t)Q(t)T T . (2.28)
(2.28)
From the definition, Q
(t)
u = u
(t)
,
◜--=◞u◟---◝
u′(t) = Q′(t) u =Q ′(t)Q(t)T u (t).
Then writing the matrix of Q^{′}
(t)
Q
(t)
^{T} with respect to fixed in space orthonormal basis vectors,
i^{∗},j^{∗},k^{∗}, where these are the usual basis vectors for ℝ^{3}, it follows from 2.28 that the matrix of
Q^{′}
′ ′ ′ ′ ′ ′ ′ ′
v = R + x i+y j+ z k+ Ω × rB = R + x i+ yj+ z k+ Ω× (xi+ yj+ zk).
Now consider the acceleration. Quantities which are relative to the moving coordinate system
and quantities which are relative to a fixed coordinate system are distinguished by using the
subscript B on those relative to the moving coordinate system.
Ω ×vB
′ ′′ ′′ ′′ ′′ ◜′′---◞′◟′---′◝′ ′
a = v = R + x i+ y j+ z k+ x i+ y j+ z k + Ω ×rB
The acceleration a_{B} is that perceived by an observer who is moving with the moving coordinate
system and for whom the moving coordinate system is fixed. The term Ω×
(Ω× rB)
is called the
centripetal acceleration. Solving for a_{B},
aB = a − R ′′ − Ω ′ × rB − 2Ω × vB − Ω× (Ω× rB ). (2.30)
(2.30)
Here the term −
(Ω× (Ω × rB))
is called the centrifugal acceleration, it being an acceleration felt
by the observer relative to the moving coordinate system which he regards as fixed, and the term
−2Ω × v_{B} is called the Coriolis acceleration, an acceleration experienced by the observer as he
moves relative to the moving coordinate system. The mass multiplied by the Coriolis acceleration
defines the Coriolis force.
There is a ride found in some amusement parks in which the victims stand next to a circular
wall covered with a carpet or some rough material. Then the whole circular room begins to revolve
faster and faster. At some point, the bottom drops out and the victims are held in place by
friction. The force they feel is called centrifugal force and it causes centrifugal acceleration. It is
not necessary to move relative to coordinates fixed with the revolving wall in order
to feel this force and it is pretty predictable. However, if the nauseated victim moves
relative to the rotating wall, he will feel the effects of the Coriolis force and this force is
really strange. The difference between these forces is that the Coriolis force is caused
by movement relative to the moving coordinate system and the centrifugal force is
not.