Download Rotating Frames and Infinitesimal Rotations in Physics and more Study notes Mechanics in PDF only on Docsity!
Lecture 4 – Non-Inertial Reference Frames (Chapter 2 in F&W)
Now we want to provide a brief overview of the properties of relevant accelerating
frames. Of special interest are rotating frames, i.e ., transformed from an inertial
frame by a rotation that is varying with time. We have already mentioned one
specific example in Lecture 1. Let us define two sets of right-handed Cartesian unit
vectors (defined by the right-hand-rule or the antisymmetric tensor ε jkl
), one set in an
inertial frame (and therefore constant in time) and one set in the rotating frame,
inertial: ˆ ˆ ˆ, ˆ,
ˆ ˆ ˆ ˆ rotating: ,.
k j k jkl l
k j k jkl l
e e e e
e e e e
( 4. 1 )
For now the two reference frames share a common origin. Thus the position vector
of a point particle in the two frames is expressed as
3 3
1 1
ˆ ˆ.
j j j j
j j
r r e r e
( 4. 2 )
The velocity is then
3 3 3
1 1 1
ˆ ˆ ˆ.
j j j j j j
j j j
r r e r e r e
( 4. 3 )
Clearly we want to derive some handy notation for the time variations of the rotating
unit vectors. As suggested in the previous discussion of the rotation group we can
express the infinitesimal change in the unit vectors in terms of an infinitesimal 3x
matrix ( i.e ., a representation of the sum of the generators with appropriate
infinitesimal rotation angles as coefficients),
j jk k
de d e
( 4. 4 )
Now we need to understand the properties of jk
d
. Since the infinitesimal rotation
of a vector (about one end of the vector) produces a change that is perpendicular to
the original vector, we have
0 0, all ,
k k kk
de e d k
( 4. 5 )
i.e .,
jk
d is purely off-diagonal. Next we know that the unit vectors stay orthogonal
as they rotate (this is an orthogonal transformation after all), which means that
j k j k j k
j k j k
jl l k j kl l
jk kj
jk kj
j k e e e e d e e
de e e de
d e e e d e
d d
d d
( 4. 6 )
Thus jk
d
is an antisymmetric matrix. As expected from our earlier discussion of
SO(3), such a 3x3 antisymmetric matrix has just 3 independent components. As
suggested above we can think of the various components of the rotation as occurring
in the various planes, i.e ., as tensor components. For the special case of 3-dimensions
we can equivalently think of the components of the rotation as rotations about the
corresponding (unique) axes orthogonal to each plane. (This does not work in more
than 3 dimensions.) This suggests that we can convert from the matrix representation
of the infinitesimal rotation in Eq. ( 4. 6 ) to a vector representation. We define (the
signs come from the right-hand-rule) vector components for the infinitesimal rotation
parameters via
12 3 23 1 31 2
1 12 2 13 3 3 2 2 3
2 23 3 21 1 1 3 3 1
3 31 1 32 2 2 1 1 2
d d d d d d
de d e d e d e d e
de d e d e d e d e
de d e d e d e d e
( 4. 7 )
The corresponding “vector”is (note that these components have been defined in the
rotating frame)
3
1
k k
k
d d e
( 4. 8 )
and we can rewrite the infinitesimal changes in the unit vectors as
where we should think of this as the instantaneous angular velocity of the rotating
frame as viewed from the inertial frame. Thus we have for the rotating unit vectors
j
j j
de
d
e e
dt dt
( 4. 12 )
Returning to Eq. ( 4. 3 ) we (finally) obtain
3 3 3
1 1 1
j j j j j j
j j j
r r e r e r e
( 4. 13 )
or
inertial rotating
dr t dr t
r t
dt dt
( 4. 14 )
The velocity we measure with respect to the inertial frame differs from the velocity
we measure with respect to the rotating frame by the time dependent vector
r t
.
(Note that this vector is the cross product of a polar vector and a pseudovector and is
thus a polar vector.)
This last result holds for the difference between the time derivatives measured in the
two frames for any vector. In particular, it holds for the (pseudo) vector
t
itself.
Thus the angular acceleration is the same in both frames
inertial rotating rotating
d t d t d t
t
dt dt dt
( 4. 15 )
Finally we want to consider the form of the (usual) acceleration in the two frames so
thatwecanaddressNewton’slaws.Wehave
2
2
rotating inertial
rotating
2
2
rotating rotating
d r t dr t d
r t
dt dt dt
dr t
r t
dt
d r t dr t
dt dt
d
r t r t
dt
( 4. 16 )
Note again that while the instantaneous angular velocity is measured in the inertial
frame, the angular acceleration is the same in both frames.
Finally let us include the possibility of both a rotation and an accelerating translation
(Eq. 3.4,
0
r r r
) between the two frames. With both sorts of acceleration present
we have
2 2 2
0
2 2 2
inertial inertial inertial
2 2
0
2 2
acceler inertial acceler
d r t d r t d r t
dt dt dt
d r t d r t dr t
dt dt dt
d
r t r t
dt
( 4. 17 )
ApplyingthistoNewton’ssecondlaw,
2
inertial 2
inertial
d r t
m F
dt
( 4. 18 )
we find the corresponding equation in the accelerating frame, the effective force, to
be
2
inertial 2
acceler
2
2
ˆ ˆ 6378 km ˆ,
2 rad
24 60 60 s
E E
E
E E
E E
r t r t R x R x x
z
d r t
m F m r t
dt
mgx m R z z x
xm g R
( 4. 20 )
Thus the fractional decrease in the apparent weight is given by
2
E E
R
w
g
( 4. 21 )
For most of us this is of order one half a pound – not much of a diet plan.
Next we want to consider the motion of particles near the
surface of the earth in the rotating frame, but now at arbitrary
polar angle (colatitude) . As indicated in the figure (
y
is into
the figure) we place the rotating frame at the surface of the
earth (assumed to be spherical for now) with the
z ˆ
unit vector
aligned with the radial direction,
r ˆ , in the inertial system fixed
at the center of the earth. Again there is no angular
acceleration. We start with
E
r R r r
( 4. 22 )
Newton’slaw (in the accelerating frame), Eq. ( 4. 19 ), becomes
2 2
inertial 2 2
acceler acceler inertial
2
acceler
E
E
E
E
d r t d R r dr t
m F m m
dt dt dt
m r t
GM m
r mR r t
R z
dr t
m m r t
dt
( 4. 23 )
We can further simply our analysis by the (usually very reasonable) assumption that
E
r R
. Now we can include the first, second and fourth terms on the right hand
side as an effective, dependent, gravitational acceleration (recall the previous
example), with components in both the
r ˆ (
z ˆ
) and
directions, but with the largest
component in the
z
direction,
2
2
acceler acceler
acceler
eff
eff
d r t dr t
m mg m
dt dt
dr t
mg z m
dt
( 4. 24 )
To the extent that the earth is composed of an elastic material it will assume a shape
so that the local gravitational acceleration is perpendicular to its surface (the surface
is an equipotential). Thus the earth is slightly flattened at the poles. We will address
some of these details in the HW (see F&W exercise 2.7, our problem III.3). Here we
will simply assume that the acceleration of gravity is in the – zdirection in the frame
fixed to the surface of the earth. So now consider the specific problem of a particle
released from rest (in the rotating frame) at height h abovetheearth’ssurface
(
E
h R ). We can proceed by looking for (small!) deviations from the usual, non-
rotating, result due to the Coriolis force. Define
eff
r g tz t
( 4. 25 )
positive
x
direction) with speed
0
v
, instead of just being dropped. So now we start
with
0
eff
r t g tz v x t
( 4. 29 )
and try to solve
0 0
0
0
0
2 sin 2 sin
2 sin cos ˆ.
eff eff
E eff
eff E E
E eff
g tz v x g tz v x
g t z z v z x
g t y v y
g t v y
( 4. 30 )
Thus the motion, as viewed in the rotating frame is (same approximations as above)
0
ˆ sin 2 cos.
eff E E
r g t z t y v x t y
( 4. 31 )
The first term on the right-hand-side is just as above, while the initial horizontal
velocity leads to the second term. This term dependent on the initial velocity now
does distinguish between the hemispheres, i.e ., cos changes sign. As viewed from
above in the rotating frame, this term has clockwise sense in the northern hemisphere
(a velocity first in the
x
direction plus a growing term in the
y
direction) and a
counter-clockwise sense in the southern hemisphere (a velocity first in the
x
direction plus a growing term in the
y
direction). Again the qualitative effect can
be simply understood in terms of the skater problem (angular momentum
conservation) as viewed from the inertial frame. In the northern hemisphere the
horizontal velocity towards the equator is moving the particle to a larger distance
from the axis of rotation and thus produces a reduction in the velocity in the
y
direction. In the southern hemisphere the particle is thrown away from the equator to
a smaller distance from the rotation axis and thus produces an increase in the
y
velocity.
A related question is the impact of the Coriolis force on weather patterns (a very
important issue over the last few months!). In the northern hemisphere air flowing
from a high pressure region to a low pressure region will be pushed to the right ( i.e .,
flowing air tends to have high pressure on its right and low pressure on its left with
respect to the flow direction). Thus the effective flow pattern around a low pressure
region will tend to be in a counterclockwise direction (viewed from above, see
below). The opposite is true in the southern hemisphere. This is meant to explain the
differing flow patterns in the two hemispheres. (This consideration also explains why
hurricanes do not develop near the equator.) Likewise winds near a relatively straight
front between a high pressure region and low pressure region will flow along the
front with the high pressure to the right in the northern hemisphere.
Hurricane in the Northern Hemisphere
