Paleomagnetism and Earth History - Quantitative Plate Tectonics - page 186

Fig. 6.6 Geometrical
relations between a rock
unit at location S , whose
time-averaged
magnetization vector M
has declination D and
inclination I ,andthe
corresponding paleopole,
at location P ,ofa
geocentric paleomagnetic
dipole field with magnetic
moment m . Angle ™ is the
site paleo-colatitude

simply the paleopole , associated with this field.
In fact, for a magnetic dipole field, the inclination
is related to the distance from the field pole by
the dipole equation ( 4.49 ) , while the declination
tells us the direction where to find this pole.
Figure 6.6 illustrates the geometric relationships
between the various parameters.
Let S (™ S ,¥ S ) be the geographic location (co-
latitude and longitude) of a rock unit, for which
a time-averaged remnant magnetization vector M
with declination D and inclination I has been
determined. If this is considered as equivalent
to an NRM that was acquired by exposition to
a geocentric dipole field, then from the dipole
equation ( 4.49 ) we have that the paleocolatitude
™ of the site can be calculated readily by the
following formula:
™ D cot 1 1

Therefore,
“ D arcsin sin D sin ™
sin ™ P

(6.43)

This formula constrains “ to be in the range
[- /2, C /2]. In fact, there is a source of am-
biguity arising from the fact that a longitude
difference “ and a difference - “ give the
same sine. Therefore, using ( 6.43 ) we cannot
distinguish a situation where a paleopole P lies
in the same hemisphere of S from a situation in
which P is in the opposite hemisphere. Applying
again the law of cosines to the spherical triangle
( S , P , N )ofFig. 6.6 , we see that the two situations
give:

cos ™ D cos ™ P cos ™ S C sin ™ P sin ™ S
cos “
cos . “/

2 tan I

S and P in the same hemisphere
S and P in opposite hemispheres
(6.44)

(6.40)

To determine the colatitude of the paleopole
P , we can use the spherical version of the law of
cosines:

Therefore,

cos ™ D cos™ P cos ™ S ˙ sin ™ P sin ™ S cos“
(6.45)

cos™ P D cos ™ S cos™ C sin ™ S sin ™ cosD
(6.41)

A little bit more complicate is to determine the
paleopole longitude ¥ P .Let“ be the longitude
difference between paleopole and site. By the
spherical version of the law of sines we have that:

Now we note that sin™ P sin™ S cos“ 0inany
case, thereby, the two situations can be distin-
guished comparing cos™ with cos™ P cos™ S :

¥ P ¥ S
¥ S ¥ P C

for cos™ cos™ P cos™ S
for cos™<cos™ P cos™ S
(6.46)

“ D

sin ™
sin “ D

sin ™ P
sin D

(6.42)