PART I.5 - Physical Mathematics
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This document discusses matrix operations and properties including: - Addition of matrices is done by adding corresponding elements. - Multiplication of matrices involves multiplying rows of the first by columns of the second. - For multiplication, the number of columns of the first must equal the rows of the second. - The inverse of a matrix M exists if the determinant of M is non-zero. - Differentiation and integration are done on matrix elements individually.
![The purpose of this chapter is to describe the geometry of an idealized situation: a
vacuum with an Earth rotating uniformly with respect to fixed inertial coordinates – in
general, they are labelled X, Y, and Z. We will also throw in some material relevant for
PART III – QUANTUM MECHANICS.
Up to now we have met a few situations where a vector x (or [x1 x2 x3]T,as we shall
see later) can be changed into another vector y by a linear transformation given by:
where aij are numbers (or angles) and the subscripts i and j take on the values 1, 2, 3
which is useful in computations. We can also express yi more as an abstract vector:
A rectangular array M of numbers aij is called a matrix. Matrix algebra is the
expression of algebraic operations on arrays of quantities, such as the
transformations y1, y2, y3 above or in compressed notation such as yi or y.
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Matrix Operations
This set of equations can be abbreviated as:
3
3332321312
3232221212
3132121111
xaxaxay
xaxaxay
xaxaxay
++=
++=
++=
∑=
=
3
1j
jjii xay
xMy =
which in your mind you can visualize using the dyadic y=M•x (e.g., inertia dyadic ϒϒϒϒ).](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-3-320.jpg)
![The word ‘matrix’ was introduced in 1850 by Sylvester (1814-1897) and the matrix M
represents, in general, a rectangular array of quantities:
=≡
mnmm
n
n
ji
aaa
aaa
aaa
M
L
MMM
L
L
21
22221
11211
][M
where aij are called elements (of the i-th row and j-th column): they may be real (or
complex) numbers or functions. The matrix M has m rows and n columns and is called a
matrix of order m×n (i.e., m by n). If m=n, the matrix is called a square matrix and in this
case the main diagonal of the square matrix consists of the elements a11, a22, …, ann.
( )1
10
01
01
10
0
0
01
10
321 −=
−
+
=
−+
−−
=
+
−
=
+
+
= i
i
i REEL/DIAG
PARITY
TICOMPLEX/ANREEL/SYM
and, σσσ
These are the Pauli (1900-1958) ‘spin’ matrices and they have the following properties:
3122121
2
3
2
2
2
1332211 2],[1 σσσσσσσσσσσσσσσσ i=−≡=++=++ and
e.g., the direct products σ1⊗σ3 and σ1⊗σ2 and their addition (σ1⊗σ3) ⊕(σ1⊗σ2) isgivenby:
−
−
−
−
=
−
−
⊕
−
−
=
⋅⋅
⋅⋅
⊕
⋅⋅
⋅⋅
=⊗⊕⊗
001
001
100
100
000
000
000
000
0010
0001
1000
0100
01
10
01
10
)()(
22
22
33
33
2131
i
i
i
i
i
i
i
i
σσ
σσ
σσ
σσ
σσσσ
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As a general example in what follows, consider the following 2×2 complex matrices:
4](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-4-320.jpg)
![Vector Matrix: A vector, or column matrix, x, is a matrix that has only one column:
=≡
n
i
x
x
x
x
M
2
1
][x
Square Matrix: A square matrix, or quadratic matrix, has the same number of rows and
columns. A square null matrix can also be composed of just zeros, 0, or null vectors.
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Orthogonal Matrix: An orthogonal matrix is a square matrix whose calculated
determinant detM≡|M|=|aij | is ±1, and whose inverse is equal to its transpose.
5
Rotation Matrix: A Rotation matrix, Ri (θ) where, in Group Theory notation,θ is a
parameter, is an orthogonal matrix whose determinant is +1 (reflections have a
determinant −1):
−=
100
0cossin
0sincos
)(3 θθ
θθ
θR
θθθ sinsincos01 −=+=+======= jkkjkkjjikkiijjiii rrrrrrrrr and,,,
These rules are consistent with a right-handed coordinate system and positive signs for
counterclockwise rotation, as viewed looking toward the origin from the positive axis. For
example:
Rotation matrices are also called direction cosine matrices.](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-5-320.jpg)
![Transpose (or Inverse) Matrix – MT: The transpose (or inverse M−1) of an arbitrary
matrix M=aij is written MT=M−1 = aji and essentially end up being obtained by
interchanging corresponding rows and columns of M. For example:
−
=≡
−
= −
0
0
0
0 1
222
i
i
i
i
σσσ T
and
If M* = M, the matrix is real.(N.B., This operationjust amounts to replacing i=√(−1) by −i).
Complex Conjugate Matrix – M*: The complex conjugate of an arbitrary matrix M is
formed by taking the complex conjugate of each element. For example:
−
=
−
=
0
0
0
0 *
22
i
i
i
i
σσ and
Hermitian Conjugate Matrix – M†: The Hermitian conjugate of an arbitrary matrix M is
obtained by taking the complex conjugate of the matrix and then the transpose of the
complex conjugate matrix. For example:
If M† = M the matrix is said to be a Hermitian matrix.
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7
If the transpose N of a matrix M is a matrix whose elements bij satisfy the rule:
ijji ab =
This is the reason that the transpose of a matrix M is generally denoted by MT.
+
−
=⊕=⊕
−
+
=⊕
+
−
=⊕
01
10
)(])[(
01
10
)(
01
10 †
21
*
21
*
2121
i
i
i
i
i
i
σσσσσσσσ T
finallyandthen](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-7-320.jpg)
![Operations in matrix algebra are given as follows.
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jijiji bac +=
8
Addition: The sum C of two matrices M and N of equal dimension has elements cij that
satisfy the rule:
such that |l,s; j,mj 〉 is simultaneously an ‘eigenfunction’ of the total quantum angular
momentum component along the z-axis, Jz, or the square of total quantum angular
momentum vector, J2, an invariant. So, technically speaking, [x,y,z] is no more different
than |l,s;ml,ms 〉 for |ΨΨΨΨkmo
( j,mj)〉→U(W)|ΨΨΨΨ(k,mj)〉=Σm′j
Dm′jmj
( j)[W(Λ,p)]|ΨΨΨΨ(k,m′j)〉 but if
you want to be ‘relativistic-lingo’ precise you would use |ΨΨΨΨ(Λp;j,mj)〉=|ΨΨΨΨ(Λp)〉|l,s; j,mj 〉.
The operation of summing is denoted by:
NMC ++++=
in about the same way as learned in vector algebra (e.g., components cij vs sets [x,y,z]).
We will see, in PART IV – QUANTUM FIELDS, how these cij above end up being
quantities like Cj
mlms
=〈l,s;ml,ms |l,s; j,mj 〉 that are numerical coefficients which are known
as Clebsch-Gordan (CG) coefficients that are used for the space-time translation of a
physical ket state vector |l,s;ml,ms 〉 in a complex Hilbert space ends up looking like a
transformation of standard momentum k≡kµ (i.e., contravariant momentum four-vectors):
∑∑
±=
+=
±=
==
)21(
21
,;,,;,,;,,;,,;,
s
mmm
s
j
mm
mm
jssj
sj
s
s
mmsmjsmmsmmsmjs
l
l
l
l
l
ll lllll C](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-8-320.jpg)
![Differentiation: The operations of differentiation and integration of a matrix are applied
to each element separately; that is:
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∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
=
∂
∂
x
a
x
a
x
a
x
a
x
a
x
a
x
a
x
a
x
a
x
nmmm
n
n
L
MOMM
L
L
21
22212
12111
M
and similarly for integration.
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
321
321
],,[
],,[
det
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
xxx
yyy
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
=
∂
∂
=
∂
∂
x
y
Jacobian: A Jacobian – Jacobi (1804-1851) – is a matrix of partial derivatives of the
elements of one vector with respect to those of another, and is given by:
If the elements of M in y=Mx are not functions of the elements of x, then M is the
Jacobian of y with respect to x. If, on the other hand, the determinant of M in y=Mx
is 1, then it is a rotation matrix, and y=Rx expresses a rotation of coordinate axes.](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-16-320.jpg)
![A convenient way to express the above equations is by the following matrix equation:
−
=
−
−
y
x
YY
XX
θθ
θθ
cossin
sincos
o
o
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The square matrix with the direction cosines for elements is called the transfer matrix
which, in this case, transforms the body coordinates to the fixed inertial coordinates.
To obtain the inverse transformations from the fixed coordinate system to the moving
coordinate system, we can start with our equation above for [Xi′++++Yj′++++Zk′] (with Z=z=0)
arranged as follows:
ˆ ˆ ˆ
jiji ˆ)(ˆ)(]ˆˆ[ oo ′′= YYXXyx −−−−++++−−−−++++
and from the dot product with i and j:ˆ ˆ
jjijjiii ˆˆ)(ˆˆ)(ˆˆ)(ˆˆ)( oooo •′•′=•′•′= YYXXyYYXXx −−−−++++−−−−−−−−++++−−−− and
The above equations in matrix notation becomes:
−
−
−
=
−
−
•′•′
•′•′
=
o
o
o
o
cossin
sincos
ˆˆˆˆ
ˆˆˆˆ
YY
XX
YY
XX
y
x
θθ
θθ
jjji
ijii
which is the inverse of the matrix equation above. So, this transfer matrix is the inverse:
1
cossin
sincos
cossin
sincos
−
−
=
− θθ
θθ
θθ
θθ
17](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-17-320.jpg)
![A point on a rigid body can be defined in terms of body-fixed axes x, y, z. To determine
the orientation of the body itself, we now introduce Euler’s angles α, γ, β which are
three independent quantities capable of defining the position of the x, y, z, body axes
relative to the inertial X, Y, Z axes, as shown in the Figure - Left.
Body axes x, y, z defined relative to inertial
axes, X, Y, Z by Euler’s angles α, γ, β.
With the x, y, z axes coinciding with X, Y, Z axes, allow the x, y, z coordinates to rotate
about the Z axis through an angle α so as to take up the position ξ′η′ζ ′. The
relationship between the two coordinates is the given by the transfer matrix (see Figure -
Right):
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Rotation about Z axis through angle α.
[ ]T
ζηξαα
αα
ζ
η
ξ
′′′=
−=
′
′
′
Z
Y
X
100
0cossin
0sincos
Z
Y
X
O
η
Line of nodes
z
xα γ
ξ
β
ζ
y
Z
Y
X
η′
α
ξ ′
ζ ′
19](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-19-320.jpg)
![We next allow a rotation β about the ξ′ axis as shown in the Figure - Left and let the
new position of the ξ′, η′, ζ ′ axes be ξ, η, ζ with transfer matrix
Rotation about node axis ξ′=ξ through angle β.
Finally we allow a spin γ about the axis ζ, as shown in the Figure - Right, to arrive at
the body axes x, y, z. The transfer matrix for this rotation is:
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Rotation about z =ζ axis through spin angle γ.
[ ]T
ζηξ
ζ
η
ξ
ββ
ββ
ζ
η
ξ
=
′
′
′
−
=
cossin0
sincos0
001
η′
Z
Y
X
O
η
α
β
ζ
ξ ′ξ
ζ ′
[ ]T
zyx
z
y
x
=
−=
ζ
η
ξ
γγ
γγ
100
0cossin
0sincos
z
Z
Y
X
O
y
xα γ
ξ
β
ζ
20](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-20-320.jpg)
![In arriving at the final position of the body axes, we have encountered four sets of
orthogonal axes: X, Y, Z; ξ′, η′, ζ ′; ξ, η, ζ; and x, y, z. Some of these axes coincide,
such as the Zζ ′, the ζ z, the ξ′ξ; however both letters will be retained to identify the
coordinate system referred to. Of particular interest is the ξ′ξ axis, called the line of
nodes. It represents the intersection of the transverse body plan xy and the horizontal
inertial plane XY.
Other transformations between these coordinates can be obtained by the multiplication
of the two or more transfer matrices. For instance, by substituting [ξ′ η′ ζ ′]T into [ξ η ζ ]T
we obtain the following transformation between [X Y Z]T into [ξ η ζ ]T axes:
−
−=
−
−
=
Y
Y
X
Y
Y
X
βαβαβ
βαβαβ
αα
αα
αα
ββ
ββ
ζ
η
ξ
coscossinsinsin
sincoscossincos
0sincos
100
0cossin
0sincos
cossin0
sincos0
001
−
+−−−
+−
=
Y
Y
X
z
y
x
βαβαβ
βγαβγαγαβγαγ
βγαβγαγαβγαγ
coscossinsinsin
sincoscoscoscossinsinsincoscoscossin
sinsincoscossinsincossincossincoscos
Substituting this transfer matrix into [x y z]T, we obtain the transformation from the XYZ
axes to the body axes xyz:
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21](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-21-320.jpg)
![The inverse transformation from x, y, z body axes to X, Y, Z inertial axes can be
obtained in a similar manner by writing [ξ′ η′ ζ ′]T and [x y z]T in the reverse order:
&c.
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−+−+
−−−
=
z
y
x
Z
Y
X
βγβγβ
αβαβγαγαβγαγ
αβγβααγαβγαγ
coscossinsinsin
cossincoscoscossinsincoscossinsincos
sinsincoscossincossinsincossincoscos
′
′
′
−
=
ζ
η
ξ
αα
αα
100
0cossin
0sincos
Z
Y
X
Exercise: Derive this matrix.
Rules are also available for the direct inversion of matrices. The inverse [x y z]T above
is:
22](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-22-320.jpg)
![Let Earth-fixed positions be represented by a Cartesian coordinate system x, with
the x axis toward latitude 0°, latitude 0°; the y axis toward latitude 0°, longitude 90°E; and
z axis toward latitude 90°N, the north pole. The connection is through an inertially fixed-
coordinate system X, with the X axis toward the vernal equinox ( ), the point where
the sun’s orbit intersects the equator (i.e., when the Sun is over your head at the equator
on the longest day of the year!) The angle Ω between the equinox and the Greenwich
meridian – 0° longitude – is known as the Greenwich Sidereal Time.
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It is convenient to refer the position of a satellite to Cartesian coordinates q fixed in
an ellipse inclined to the equatorial plane, as shown in the Figures below. Geodesic
students are interested in Earth-fixed coordinates such as x=[x,y,z]; we need to
connect Earth-fixed positions to positions referred to this tilted ellipse!
Perigee
Ellipse
Apogee
l
ea
ψ
rb
a ξ
η
P
W
θ
10 << e
1=e
Circle
Q
××××
12
2
2
2
=+
ba
ηξ
z
O
i
Ω
ω Perigee
NodeVERNAL
EQUINOX
Focus
rp
Prime Meridian
or Greenwich
(0° longitude)
Y
Z
X
North Pole
x
y
F](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-26-320.jpg)
![30
Our interest going forward is in satellite geodesy mainly due to the fact that the Earth’s
gravitational field is non-central as studied in detail in the Gravitational Potential chapter;
that is, the equation r=−GM⊕ r/r3 should be replaced by:
where the potential, V, has a non-central form such as:
where m=ω2a/geq, or the spherical harmonic version:
V∇∇∇∇=r&&
⋅⋅⋅⋅⋅⋅⋅⋅
)1(
3
1
1 20
2eq
40
4
420
2
2 Pr
a
mg
P
r
a
JP
r
a
J
r
GM
U −+
−
−
−= ⊕
L
∑ ∑
∞
= =
+
+=
0 0
1
)]sin()cos()[(sin
1
l
l
llllll
l
lll
m
mmm mSmCP
r
V λλφ
⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅
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However, even for this non-central field the Keplerian ellipse and its orientation can be
regarded as a coordinate system, alternative to Cartesian or polar coordinates, analo-
gous to the use of geodetic latitude and altitude for position in an Earth-fixed system. At
any instant the situation of a satellite in Earth-centered, inertially fixed coordinates can
be described by the Cartesian components of position [X,Y,Z] and velocity [X,Y,Z]. In
place of these six number, the six numbers of the Keplerian ellipse (a,e, M⊕,i,ω,Ω) may
be used. The relationship between the two systems can be expressed by the rotation
from a coordinate system in the orbital plane referred to perigee to the inertially fixed
system, as given by the equations X=RXqq and X=RXqq.
⋅⋅⋅⋅ ⋅⋅⋅⋅](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-30-320.jpg)
![∑=
−
=+==
l
lll l
l
m
s
ssmsmmi
i
s
m
im
0
sincosRe])sin[(cosRe]e[Re)cos( ξξξξξ ξ
In order to convert the spherical harmonic potential V to Keplerian elements, we
require some trigonometric identities such as:
where i=√(−1), Re denotes the real part, is the binomial coefficient:
)!(!
!
sms
m
s
m
−
=
l
ll
∑=
−−
=+−=−=
l
lll l
l
m
s
ssmsmmi
i
s
m
iiim
0
1
sincosRe])sin(cos[Re]e[Re)sin( ξξξξξ ξ
and:
∑∑
∑∑∑∑
= =
+
= =
−−+
+
=
−−
=
−−−−
−−++−−+−
−
=
−
−
=
⋅
−
−
=
+
−−=
a
c
b
d
c
ba
aa
a
c
b
d
dcbaic
ba
aab
d
didbi
b
a
c
cicaic
a
aab
ii
a
iiba
dcbaidcba
i
d
b
c
ai
d
b
c
aiii
0 0
0 0
)22(
0
)(
0
)(
]})22sin[(])22{cos[()1(
2
)1(
e)1(
2
)1(
ee
2
1
ee)1(
2
)1(
)ee(
2
)ee(
2
cossin
ξξ
ξξ ξξξξξξξξξ
Then, we also need:
and the usual:
)sin(
2
1
)sin(
2
1
sincos)sin(
2
1
)sin(
2
1
cossin
)cos(
2
1
)cos(
2
1
sinsin)cos(
2
1
)cos(
2
1
coscos
babababababa
babababababa
−−+=−++=
−++−=−++=
and
,,
s
ml
31
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](sin[)](cos[)](cos[)cos( ϕαϕαλ −ΩΩ−−−ΩΩ−= lllll mmmmm
)]sin()cos()[(sin1
eq
λλφ llllll
l
l llll
mSmCP
r
rGM
V mmmm += +
⊕
where req is the equatorial radius of the Earth (N.B., by applying the factor GMreq
l we
have made Clml
and Slml
non-dimensional). We then substitute [ml(α −Ω)+ml(Ω−ϕ)] for
mlλ, where α is right ascension and ϕ is Greenwich Sidereal Time:
)](sin[)](cos[)](cos[)](sin[)sin( ϕαϕαλ −ΩΩ−+−ΩΩ−= lllll mmmmm
Orbit-equator-meridian triangle.
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In the spherical triangle formed by the orbit, the equator, and
the satellite meridian (see Figure), we have:
Ω−+Ω−=+
2
π
cossin)sin(cos)cos()ωcos( φαφαθ
32
and:
icos)sin()ωsin()cos()ωcos()ω(cos Ω−++Ω−+=+ αθαθθφ
we then get:
φ
θ
α
φ
θ
α
cos
icos)ωsin(
)sin(
cos
)ωcos(
)cos(
+
=Ω−
+
=Ω− and
and:
)ωsin(isinsin θφ +=
Ω
α – Ω
φ
i
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![If we apply cos(mlξ) and sin(mlξ) to the α −Ω functions in cos(mlλ) and sin(mlλ) and
substitute cos(α −Ω) and sin(α −Ω) therein, we get:
and:
∑
∑
=
−+−−
=
−−
+
⊕
++
×
−Ω+++−Ω−×
=
l
ll
llll
l
ll
ll
lllllll
l
ll
l
l
m
s
ssmstms
mmmm
k
t
tm
tm
e
m
i
s
m
mCiSmmSiC
T
r
rGM
V
0
2
0
2
1
icos)ω(cos)ω(sin
)]}(sin[)()]sin()](cos[)Re{(
isin
ϕϕ
ϕλϕ
33
)]}(sin[)({cos[
cos
icos)ω(sin)ω(cos
Re)cos(
0
ϕϕ
φ
ϕϕ
λ −Ω+−Ω⋅
++
= ∑=
−
ll
l
l
l
l
l
mimi
s
m
m
m
s
m
sssm
s
)]}(cos[)({sin[
cos
icos)ω(sin)ω(cos
Re)sin(
0
ϕϕ
φ
ϕϕ
λ −Ω−−Ω⋅
++
= ∑=
−
ll
l
l
l
l
l
mimi
s
m
m
m
s
m
sssm
s
If we substitute sinφ=sinisin(ω+ϕ) for sinφ in Plml
(sinφ)=cosmlφ ΣtTlml t sinl−ml −2tφ, which
is the definition of Plml
, and then substitute both Plml
and the two trigonometric functions
cos(mlλ) and sin(mlλ) in Vlml
above, by cancelling out the cosmlφs, we have:
where k is the integer part of (l−ml)/2.
2016
MRT](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-33-320.jpg)
![On applying the sinaξ cosbξ identity to our last equation for Vlml
, with a=l−ml −2t+s, and
b=ml −s:
)]}ω)(222sin[()]ω)(222{cos[(
2
)1(
2
)(
icos
)]}(sin[)()]sin()](cos[)Re{(
isin
0
2
0 0
2
2
0
2
1
ϕϕ
ϕλϕ
+−−−++−−−×
−
+−−
−
−
×
−Ω+++−Ω−×
=
∑ ∑ ∑
∑
=
+−−
=
−
=
−
+−−
=
−−
+
⊕
dctidct
d
sm
c
stmi
i
s
m
mCiSmmSiC
T
r
rGM
V
m
s
stm
c
sm
d
c
t
stm
ss
mmmm
k
t
tm
tm
e
m
ll
ll l ll
llll
l
ll
l
ll
l
l
l
lllllll
l
ll
l
l
34
2016
MRT](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-34-320.jpg)
![By applying the trigonometric identities cosacosb=½cos(a+b)+½cos(a−b), &c. to the
products of Ω−φ and ω+θ trigonometric function of Vlml
above, and dropping any term
with an odd power of i=√(−1) as a coefficient (since Vlml
is real, such a term has another
term cancelling it out), we obtain:
−Ω++−−−
+
−Ω++−−−
−
×
−
+−−
−×
−=
−
−
−
−
+−−
=
−
=
= =
−
+−−
+
⊕
∑ ∑
∑ ∑
)]()ω)(222sin[(
)]()ω)(222cos[(
2
)1(
2
icos
)1(isin
2
0 0
0 0
2
2
1
eq
ϕϕ
ϕϕ
l
l
l
l
l
l
l
l
l
l
l
ll
l
ll
ll
l
l
l
l
l
l
l
l
l
l
l
l
l
l l
l
l
ll
mdct
C
S
mdct
S
C
d
sm
c
stm
s
m
T
r
rGM
V
m
m
m
m
m
m
m
m
stm
c
sm
d
c
k
t
m
s
t
s
tktm
tmm
even
odd
even
odd
35
2016
MRT
It is desirable to transform this last equation so that the terms with the same argument,
[(l−2p)(ω+θ)+ml(Ω−ϕ)], are collected together.](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-35-320.jpg)
![By substituting p for t+c+d necessitates, in turn, the elimination of one subscript from
the factors. Putting p−t−c in place of the d summation place limits on the possible values
of c, which turn out to be simply those making the binomial coefficient non-zero. In
addition, t≤p. The expression for the spherical harmonic potential Vlml
thus becomes:
−Ω++−
+
−Ω++−
−
=
−
−
=
−
−
+
⊕
∑
)]()ω)(2sin[(
)]()ω)(2cos[()i(
0
1
eq
ϕθ
ϕθ
l
l
l
l
l
l
l
l
l
l
l
ll
l
l
l
l
l
l
l
l
l
l
l
l
ll
mp
C
S
mp
S
C
F
r
rGM
V
m
m
m
m
p
m
m
m
m
pmm
even
odd
even
odd
36
2016
MRT
Substituting from Tlmlt =(−1)t(2l−2t)!/[2lt!(l−t)!(l−ml −2t)!] yields the Kaula expansion:
∑∑
∑
−−
−
+−−
−
×
−−−
−
=
−
=
−−
−
c
kc
m
s
s
t
tm
tpm
ctp
sm
c
stm
s
m
tmtt
t
F
lll
l
l
l
l
l
ll
l
l
l
l
2
)1(icos
isin
2)!2()!(!
)!22(
)i(
0
2
22
Here k is the integer part of (l−ml)/2, t is summed from 0 to the lesser of p or k, and c
is summed over all values making the binomial coefficient non-zero.](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-36-320.jpg)
![R=Ro +r
Consider the case where the position of a particle P in space is defined in terms of the
displacement vector r relative to the moving coordinates x, y, z (see Figure - Top). If the
displacement of the origin of the moving coordinate system is Ro, the displacement of P
relative to the fixed coordinate X, Y, Z is:
2016
MRT
][]ˆˆˆ[]ˆˆˆ[ ++++++++++++++++++++++++++++++++ KJIKJI =⇒= ZYX
Letting unit vectors along a fixed and moving axes be
designated by I, J, K and i, j, k, respectively, the above equation
can be written as:
We can determine the component of the above vector in any
direction by forming the dot product of the above equation with a
unit vector in the desired direction (e.g., the X component is
obtained by the dot product of the above equation with I, &c.)
These three rectangular components along the fixed coordinates
are, then:
where the dot product of the various unit vector represents the di-
rection cosine between the coordinates. For a plane motion with
Z =z=0 (see Figure - Bottom), the direction cosines involved are:
ˆ ˆ ˆ
Rotation Transformations
Transformation between coordinates x, y, z and
X, Y, Z.
ˆ ˆ ˆ
ˆ
kKjKiK
kJjJiJ
kIjIiI
ˆˆˆˆˆˆ
ˆˆˆˆˆˆ
ˆˆˆˆˆˆ
•+•+•+=
•+•+•+=
•+•+•+=
Z
Y
X
iJjIjJiI ˆˆsinˆˆˆˆcosˆˆ •−=−=••==• θθ and
θθθθ cossinsincos oo yxYYyxXX +=−−=− and
O
X
Z
Y
P(x,y,z)
x
z
yr
ˆi
jˆ
kˆ
Kˆ
ˆI Jˆ
Ro
R
so that the equation above reduces to:
Coordinate transformation in a plane.
Y
X
x
y
P(x,y)
θ
Iˆ
Jˆ
iˆ
jˆ
R Ro r
Xo
Yo
Zo
z
z
z
x
x
x
y
y
y
zx yXo Yo Zo
ˆ ˆ ˆki j
38](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-38-320.jpg)
![So, the rotation matrix about the z-axis by a finite angle ϕ is:
Similarly, the rotation matrix about the x-axis and y-axis by a finite angle ϕ are:
In general, a rotation is defined as a linear transformation which leads the scalar product
of two vectors invariant and whose determinant is +1.
2016
MRT
−=
=+−=+=
100
0cossin
0sincos
)(
cossincoscos
3 ϕϕ
ϕϕ
ϕ
ϕϕϕϕ
R
zzyxyyxx and,
−
=
−
=
ϕϕ
ϕϕ
ϕ
ϕϕ
ϕϕϕ
cos0sin
010
sin0cos
)(
cossin0
sincos0
001
)( 21 RR and
Suppose an observer O is located in a coordinate system[x,y,z]. Owoulddescribe, say,
the temperature T which varies from point to point according to some law described by T
= f(x, y,z)≡ f(r). There is nothing special about the coordinate system [x,y,z]; the law
expressing the variation of temperature from point to point could equally well be referred
to by an observer O in the coordinate system O(x,y,z). However, the form of thelaw, i.e.,
the dependence on the coordinates, will generally be different in the two systems. If we
let T=g(x, y,z)≡ g(r) in the [x,y,z] system, then the temperature at any point – say
described by T= f(x,y,z) and T=g(x,y,z) – must be the same. The physics describing
the temperaturefieldwillnot change because of a rotation(assuming it is not too fast).
40](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-40-320.jpg)
![As an example,suppose we have a hydrogen atom surrounded by three unit charges q
located at positions 1, 2, and 3 which form an equilateral triangle of side a; the proton is
at the origin O and the electron at r (see Figure). The Hamiltonian H of the electron is:
in which |r1|, |r2|, and |r3| are the distances O1, O2, and O3, respectively, and
++−
+∇−=++=+= 2
3
2
2
2
1oo
2
2
e
2
o
111
επ4επ42
)(
rrrrrr −−−−−−−−−−−−
qe
r
Ze
m
WVTWHH
h
)(
232
)(
232
)(
3
3
2
22
2
32
2
22
2
21
22
2
2
1 rrrrrrrrr fz
a
y
a
xfz
a
y
a
xfzy
a
x ≡+
++
−=≡+
−+
−=≡++
+= −−−−−−−−−−−− and,
Hydrogen atom surrounded by three charges
located at the corners of an equilateral triangle.
−−−=⇒= −
zyxyxfzyxfDDffD ,
2
1
2
3
,
2
3
2
1
),,()()()()( 1
DD rr
−−
−
=
100
0
2
1
2
3
0
2
3
2
1
D
zzyxyyxx =−−=+−= &,
2
1
2
3
2
3
2
1
)()()(
)()()(
)(
22
1
2
3
322
3
2
1
)()(
21
13
3
2
22
2
rr
rr
rr
ffD
ffD
fz
a
yx
a
yxfD
=
=
=+
−−+
−−−=
D
D
D
and Ho is invariant under all three-dimensional rotations.
so that the function f (r) is indeed invariant since:
)()()()]()()()[( 321321 rrrrrr ffffffD ++=++D
−
−−
=−
100
0
2
1
2
3
0
2
3
2
1
1
D
2016
MRT
O
y
x
z
a
r2
1
3
e
Ze
q
q
q
O
r
1
r1
|r –r1|
O
r
3
r3
|r –r3|
O
r
2
r2
| r – r2 |
a
a
The interaction potential W remains to be investigated.
Let us take, as a practical example, the coordinate transformation
D which, as we just saw before, is a positive (counter clockwise)
rotation of coordinates about the z-axis through 120°. We have:
Therefore:
44](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-44-320.jpg)
![For rockets and space vehicles it is often necessary to consider the general problem of
the spinning body under thrust. The concern here is the body attitude and the motion of
the center of mass. We will consider only problems where the rate of mass variation is
small enough to be negligible.
2016
MRT
To outline the problem at hand, we will consider a rigid body and define a set of body-
fixed axes x, y, z rotating with angular velocity ωωωω, and with the origin coinciding with the
center of mass. Although it is always desirable to let the body axes coincide with the
principal axes, this is often not possible, so that in the general case, the moments and
products of inertia will be defined as:
FIEIDICIBIAI yzxzxyzyx ====== &,and&,
Using the equations Lx =Ix ωx −Ixy ωy − Ixz ωz, &c., the angular momentum becomes:
kjiL ˆ)ωωω(ˆ)ωωω(ˆ)ωωω( yxzxzyzyx FECDFBEDA +−+−−+−−=
and the vector moment equation about the body axes:
LωLM ××××+= ][ &
can be written out in terms of the components given by Euler’s equations (i.e., Mx =Lx +
ωy Lz− ωz Ly, &c. obtained earlier) for body-fixed coordinates, x, y, and z:
yzyxxxzyyxzz
xyxzzzyxxzyy
zxzyyyxzzyxx
EDADFBFECM
FECEDADFBM
DFBFECEDAM
ω)ωωω(ω)ωωω()ωωω(
ω)ωωω(ω)ωωω()ωωω(
ω)ωωω(ω)ωωω()ωωω(
−−−−−++−=
+−−−−+−−=
−−−+−+−−=
&&&
&&&
&&&
45
Space Vehicle Motion
⋅⋅⋅⋅](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-45-320.jpg)
![We next let the velocity of the center of mass be expressed by the equation:
2016
MRT
and the force as:
Since the x, y, z coordinates are rotating with the body, the vector force equation is:
kjiv ˆˆˆ
zyx vvv ++++++++=
and the force components along x, y, z directions are determined from the equation:
vω
v
pωpF m
td
d
m ××××++++××××++++
== ][ &
If the resultant of the above force does not pass through the center of mass
coinciding with the origin of the x, y, z, axes, our equations for Mx, My, Mz, and Fx, Fy,
Fz, above become coupled owing to the moment of the force. Also these equations
define the motion of the body only in terms of the linear and angular velocities
referred to body axes, and their solution and transformation to displacements and
angles relative to inertial coordinates X, Y, Z are problems of considerable difficulty
which can be accomplished only under simplifying assumptions.
)ωω(
)ωω(
)ωω(
yxxyzz
xzzxyy
zyyzxx
vvvmF
vvvmF
vvvmF
−+=
−+=
−+=
&
&
&
kjiF ˆˆˆ
zyx FFF ++++++++=
46](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-46-320.jpg)
![T
1
2
A
C
3
O
A
L1
∫∫ =+=
−+=
−+=
−+=
++==
mdCmdA
ACA
III
LLLM
MMM
2
2
2
3
2
2
321
322311
233211
321
2)(
ωω)(ω
ω)ω(ω
ωω
ˆˆˆ][
lll
&
&
&
&
&
ntmisalignmethrustforso
where
321LLM ××××ωωωω++++ ω3 L2
ω2 L3
ωωωω1
−−−−Pitch
++++Pitch
−−−−Yaw
++++Yaw
++++Roll
y
x
z
++++Spin
z
ωωωω2
ωωωω3
L2
L3
ωωωω1
⋅⋅⋅⋅
l2
Let us now consider thrust misalignment. We will consider a simple problem of a
spinning missile with a misalignment of the thrust line. We will assume that the missile is
symmetric so that the x, y, z, axes coincide with the principal axes 1, 2, 3 with I1 =I2 =A
and I3 =C. With A=B, we can rotate the 1, 2 axes so that one of these axes, say 1, is
perpendicular to the plane containing the thrust and axis 3, as shown in the Figure.
Euler’s moment equations, M=[L]+ωωωω××××L, for the missile (N.B., A=B & C=I3) are then:
The third equation tells us that ω3 =n is a constant and recall that L=r××××mv.
0ωωωωωω
0ωω)(ωωωωωω
0ωω)(ωωωωωω
3211233
312133122
321322311
==−+=
=−−=−+=
≠−+=−+=
&&
&&
&&
CAACM
ACACAAM
ACAACAM
Thrust misalignment resulting in moment M1 on
missile of principal axes 1, 2, 3. Art circa 1960.
Although C is generally less than A for missiles, we let:
−
=
A
AC
nλ
0ωωωω 12
1
21 =−=+ λλ && and
A
M
and rewrite the first two equations as:
From the second equation we find ω1 =(1/λ)ω2 and when we
substitute it into the first equation we get:
⋅⋅⋅⋅
)()(ωω 12
2
2 tMAλλ =+&&
which has the solution:
∫ −++=
t
tdtttM
A
ttt
0
1
2
22 )](sin[)(
1
sin
)0(ω
cos)0(ω)(ω λλ
λ
λ
&
2016
MRT
47
⋅⋅⋅⋅](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-47-320.jpg)
![The solution of the previous thrust misalignment problem is in terms of body-fixed
coordinates which are rotating.
Adding ω1 and ω2 in quadrature:
γ
βαβ i
ii −
+=+= e)sin(ωωω 2112 &&
48
( )0cosω
sincossinω
cossinsinω
33
2
1
==+=
−=
+=
Mn forconstantaγβα
γβγβα
γβγβα
&&
&&
&&
From ω3:
β
γ
α
cos
&
&
−
=
n
which, substituted into equation ω12 =ω1 +iω2 above, yields:
γ
βγβ i
ni −
−+= e]tan)([ω12 &&
2016
MRT
In order to transform from the body-fixed coordinates x, y, and z to the inertial
coordinates X, Y, and Z it is necessary to introduce Euler’s angles α, γ, and β. The
transformation for our principal axes, 1, 2, and 3 is:](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-48-320.jpg)
![Although this last equation relates the angular velocity ω12 about the body-fixed
coordinates in terms of Euler’s angles referenced to inertial axes, further simplification
generally requires a small angle approximation for β. Such an approximation is often
justified when dealing with rockets and missiles whose spin axis must not deviate greatly
from a fixed direction of flight.
When β is small, tanβ can be replaced by β:
At this point, we introduce a complex angle of attack (H. Leon, 1958), which uncouples
this equation into:
γ
βγβ i
ni −
−+= e])([ω12 &&
49
2016
MRT
γ
ββ i−
= e12
Differentiating gives:
γ
βγββ i
i −
−= e)(12 &&
The equation above for ω12 now take the following form:
121212 ω=+ ββ ni&
so that when ω12 is a known function of time we have a first-order ordinary differential
equation in β12 to solve. It must be remembered however that the above procedure is
limited to problem when β is small.](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-49-320.jpg)
![At this point the significance of the term exp(−iγ) appearing in the various equations
should be pointed out. For example, consider the equation for ω12 obtain earlier:
Writing exp(−iγ)=cosγ −isinγ, ω12 becomes:
50
Velocity components in transverse plane tilted
by an angle β about the node axis ξ.
2016
MRT
Thus the multiplication of the components β +iαsinβ along the
node coordinate system by exp(−iγ ) results in ω1 +iω2, the
components along the body-fixed axes rotated through an angle γ
from the node axis, ξ. It follows then that, if we multiply the
components ω1 +iω2 along the body axes by exp(iγ ), we should
obtain the vector ω12 in terms of the node axis components as
follows:
ξη
γγ
βαβ ω)sin(eωe)ωω( 1221 =+==+ && ii ii
We can now attach physical significance to the complex angle
of attack β12=β exp(−iγ). Since β is multiplied by exp(−iγ), β12 is
resolved into components along the body-fixed axes 1 and 2. To
restore β along its node axis we multiply β12 by exp(iγ) (i.e., β =β12
exp(iγ)). Furthermore, if we wish to examine β12 in terms of inertial
components, we need to multiply β by exp(iα), or βexp(iα)=
[β12exp(iγ)]exp(iα)=β12exp[i(γ + α)]≅β12exp(in t) where γ +α ≅nt.
Z
O
η
ξ
α
2
1
ω2
ω1
α⋅⋅⋅⋅
β
⋅⋅⋅⋅
ω12
where all the components in this equation lie in the tilted transverse plane which is
shown in the Figure.
γ
βαβ i
ii −
+=+= e)sin(ωωω 2112 &&
)sincossin()sinsincos(ω12 γβγβαγβαγβ &&&& −++= i
The real and imaginary parts of this equation are, however, equal to the components of
β and αsinβ along axes 1 and 2.
⋅⋅⋅⋅ ⋅⋅⋅⋅
⋅⋅⋅⋅ ⋅⋅⋅⋅](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-50-320.jpg)
![Z
O
ξ
Φ
y
x
z
β
Y
X
1
θ
γα
3
To solve for the angular velocities, we first write down the transformation from the
missile axes to the principal axes, assuming θ to be small:
56
Adding ω1 and iω2 we obtain the complex angular velocity:
−
Φ−ΦΦ−
ΦΦΦ
=
z
y
x
ω
ω
ω
10
coscossin
sinsincos
ω
ω
ω
3
2
1
θ
θ
θ
where ωz ≅ n. By multiplying the last equation by exp(iΦ), this equation may also be
written in the inverse form, ωxy =ω12exp(iΦ)−iθn.
Φ−
Φ−
+≅
++=+=
i
yx
i
zyx
ni
iii
e)ω(
e]ω)ωω[(ωωω 2112
θ
θ
Principal axes 1, 2, 3 referred to missile axes x,
y, z, which in turn are referred to node axis ξ
and inertial axes X, Y, Z.
2016
MRT
The Figure shows the relationship between the missile axes x,
y, z, the principal axes 1, 2, 3, the inertial axes X, Y, Z, and the line
of nodes ξ. The missile axis x is normal to axes 3 and z, whereas
the line of nodes ξ is normal to axes z and Z.
The principal axes 1, 2, 3 are then referenced to the x, y, z axes
by fixed angles θ and Φ.
The position of the missile axes x, y, z is obtained by starting
with the missile spin axis z coinciding with Z and performing three
rotations as follows:
1. Rotation of α about Z ;
2. Rotation of β about ξ;
3. Rotation of γ about the spin axis z.
X
Y
Z
z x
y
3
1
2
I1
I2
I3](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-56-320.jpg)
![2016
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References
68
M. Klein, Calculus, 2-nd Edition, Dover, 1977.
New York University
Believe it or not, my father had the first edition in 2 volumes and when I found them and browsed through them I was amazed at the
Motion in One Dimension in a Resisting Medium part which eventually ended up being the case for this treatment of air resistance in
ballistic problems. Since then, I’ve always enjoyed Klein’s way of presenting calculus using physical concepts. The only problem
though is his use of 32 [ft/sec] everywhere! Most of the gravitation discussion related to hollow or filled up spheres is from this book.
H. Benson, University Physics, Revised Edition, 1996.
Vanier College
Amazing course which is somewhat similar to the one I had with Halliday and Resnick back in 1986-88. It is a great page turner with
special topics everywhere. The problems in themselves are worth solving (e.g., kinematics of a basketball shot!) and many have an
equation to prove that is actually displayed! My edition has 44 chapters spread across 942 pages covering everything from vectors,
kinematics, inertia, particle dynamics, work and energy, conservation of energy, momentum, rotations about a fixed axis, gravitation,
solids and fluids, oscillations, waves and sound, temperature and the ideal gas law, thermodynamics and entropy, electrostatics, the
electric field, pretty much everything ‘electric and magnetic’, Maxwell’s equations, light and optics, special relativity , quantum theory
and wave mechanics, atoms and solids. Finally nuclear physics and elementary particles and a (1997) view of Grand Unified Theory.
C. Harper, Introduction to Mathematical Physics, Prentice Hall, 1976.
California State University, Haywood
This is my favorite go-to reference for mathematical physics. Most of the differential equations presentation and solutions, complex
variable and matrix definitions, and most of his examples and problems, &c. served as the primer for this work. Harper’s book is so
concise that you can pretty much read it in about 2 weeks and the presentation is impeccable for this very readable 300 page
mathematical physics volume.
D.G. Zill, W. S. Wright, Advanced Engineering Mathematics, 4-th Edition, Jones & Bartlett, 2011.
Loyola Mary-mount University
If you are going to go into some scientific field that makes you learn engineering physics you should get this book as a reference.
Besides being set in color, it is easily readable and with just enough conceptual ‘meat’ around the physical ‘bone’ to capture your
attention without being to mathematical about it. I mean, when I browsed through it I wished I had used this book for my course!
F. P. Beer, E. R. Johnson, W. E. Clausen, Vector Mechanics for Engineers, Volume II - Dynamics, 8-Edition, McGraw-Hill, 2007.
Lehigh University, University of Connecticut, Ohio State University
Great reference and a constant page turner with amazing color graphics. It is fairly advanced though but requires some study.
T. Allen Jr, R.L. Ditsworth, Fluid Mechanics, McGraw Hill, 1972.
College of Engineering Sciences at Arizona State University
My father’s book. I first skimmed through this book when I was 14 and fell in love with it instantly. The fluid mechanics here is all theirs.
W.J. Thomson, Introduction to Space Dynamics, Dover, 1986.
University of California at Santa Barbara
Great introduction to vectors, kinematics, dynamics and earth & satellite-related applications. Most of the mathematical treatment for
kinematics, rotation transformations and satellite or ballistic dynamics here is based on this book since it is succinct and to the point.](https://image.slidesharecdn.com/1-170114010522/85/PART-I-5-Physical-Mathematics-68-320.jpg)
PART I.5 - Physical Mathematics
- 1. BAAABAAA XQQPQQ βαβαµ µ βαβα εδσ == },{2},{ and&& From First Principles PART I – PHYSICAL MATHEMATICS January 2017 – R4.2 Maurice R. TREMBLAY BAAABAAA XQQPQQ βαβαµ µ βαβα εδσ == },{2},{ and&& Chapter 5
- 2. Contents PART I – PHYSICAL MATHEMATICS Useful Mathematics and Infinite Series Determinants, Minors and Cofactors Scalars, Vectors, Rules and Products Direction Cosines and Unit Vectors Non-uniform Acceleration Kinematics of a Basketball Shot Newton’s Laws Moment of a Vector Gravitational Attraction Finite Rotations Trajectory of a Projectile with Air Resistance The Simple Pendulum The Linear Harmonic Oscillator The Damped Harmonic Oscillator General Path Rules Vector Calculus Fluid Mechanics Generalized Coordinates 2016 MRT The Line Integral Vector Theorems Calculus of Variations Gravitational Potential Kinematics of Particles Motion Under a Central Force Particle Dynamics and Orbits Space Vehicle Dynamics Complex Functions Derivative of a Complex Function Contour Integrals Cauchy’s Integral Formula Calculus of Residues Fourier Series and Fourier Transforms Transforms of Derivatives Matrix Operations Rotation Transformations Space Vehicle Motion Appendix 2
- 3. The purpose of this chapter is to describe the geometry of an idealized situation: a vacuum with an Earth rotating uniformly with respect to fixed inertial coordinates – in general, they are labelled X, Y, and Z. We will also throw in some material relevant for PART III – QUANTUM MECHANICS. Up to now we have met a few situations where a vector x (or [x1 x2 x3]T,as we shall see later) can be changed into another vector y by a linear transformation given by: where aij are numbers (or angles) and the subscripts i and j take on the values 1, 2, 3 which is useful in computations. We can also express yi more as an abstract vector: A rectangular array M of numbers aij is called a matrix. Matrix algebra is the expression of algebraic operations on arrays of quantities, such as the transformations y1, y2, y3 above or in compressed notation such as yi or y. 2016 MRT Matrix Operations This set of equations can be abbreviated as: 3 3332321312 3232221212 3132121111 xaxaxay xaxaxay xaxaxay ++= ++= ++= ∑= = 3 1j jjii xay xMy = which in your mind you can visualize using the dyadic y=M•x (e.g., inertia dyadic ϒϒϒϒ).
- 4. The word ‘matrix’ was introduced in 1850 by Sylvester (1814-1897) and the matrix M represents, in general, a rectangular array of quantities: =≡ mnmm n n ji aaa aaa aaa M L MMM L L 21 22221 11211 ][M where aij are called elements (of the i-th row and j-th column): they may be real (or complex) numbers or functions. The matrix M has m rows and n columns and is called a matrix of order m×n (i.e., m by n). If m=n, the matrix is called a square matrix and in this case the main diagonal of the square matrix consists of the elements a11, a22, …, ann. ( )1 10 01 01 10 0 0 01 10 321 −= − + = −+ −− = + − = + + = i i i REEL/DIAG PARITY TICOMPLEX/ANREEL/SYM and, σσσ These are the Pauli (1900-1958) ‘spin’ matrices and they have the following properties: 3122121 2 3 2 2 2 1332211 2],[1 σσσσσσσσσσσσσσσσ i=−≡=++=++ and e.g., the direct products σ1⊗σ3 and σ1⊗σ2 and their addition (σ1⊗σ3) ⊕(σ1⊗σ2) isgivenby: − − − − = − − ⊕ − − = ⋅⋅ ⋅⋅ ⊕ ⋅⋅ ⋅⋅ =⊗⊕⊗ 001 001 100 100 000 000 000 000 0010 0001 1000 0100 01 10 01 10 )()( 22 22 33 33 2131 i i i i i i i i σσ σσ σσ σσ σσσσ 2016 MRT As a general example in what follows, consider the following 2×2 complex matrices: 4
- 5. Vector Matrix: A vector, or column matrix, x, is a matrix that has only one column: =≡ n i x x x x M 2 1 ][x Square Matrix: A square matrix, or quadratic matrix, has the same number of rows and columns. A square null matrix can also be composed of just zeros, 0, or null vectors. 2016 MRT Orthogonal Matrix: An orthogonal matrix is a square matrix whose calculated determinant detM≡|M|=|aij | is ±1, and whose inverse is equal to its transpose. 5 Rotation Matrix: A Rotation matrix, Ri (θ) where, in Group Theory notation,θ is a parameter, is an orthogonal matrix whose determinant is +1 (reflections have a determinant −1): −= 100 0cossin 0sincos )(3 θθ θθ θR θθθ sinsincos01 −=+=+======= jkkjkkjjikkiijjiii rrrrrrrrr and,,, These rules are consistent with a right-handed coordinate system and positive signs for counterclockwise rotation, as viewed looking toward the origin from the positive axis. For example: Rotation matrices are also called direction cosine matrices.
- 6. Unit Matrix: The unit (identity) matrix I (or plainly 1 as used in Group Theory) is given by I=δij where IM=MI=I. Recall that the Kronecker delta δij has the following property: ( )iD ji ji iiji allforor when when 1 0 1 = ≠ = =δ 2016 MRT Diagonal Matrix: Here we write: Singular Matrix: If the determinant is null, i.e., when detM≡|M|=|aij |=0, then M is said to be a singular matrix. For example: 0det 00 01 =≡ = AAA then ( )jiDDD jijiji ≠== 0orδ For example: −= 400 010 002 D Since, by definition, the determinant is given by (Exercise): 12 12 11 11 2 1 1 1 2221 1211 AaAaAa aa aa A j j j +=== ∑= where: 1221 2112 2222 1111 )1()1( aaAaaA −=−==−= ++ and On substituting A11 and A12 into the expression for |A|, we obtain: 21122211 aaaaA −= 6
- 7. Transpose (or Inverse) Matrix – MT: The transpose (or inverse M−1) of an arbitrary matrix M=aij is written MT=M−1 = aji and essentially end up being obtained by interchanging corresponding rows and columns of M. For example: − =≡ − = − 0 0 0 0 1 222 i i i i σσσ T and If M* = M, the matrix is real.(N.B., This operationjust amounts to replacing i=√(−1) by −i). Complex Conjugate Matrix – M*: The complex conjugate of an arbitrary matrix M is formed by taking the complex conjugate of each element. For example: − = − = 0 0 0 0 * 22 i i i i σσ and Hermitian Conjugate Matrix – M†: The Hermitian conjugate of an arbitrary matrix M is obtained by taking the complex conjugate of the matrix and then the transpose of the complex conjugate matrix. For example: If M† = M the matrix is said to be a Hermitian matrix. 2016 MRT 7 If the transpose N of a matrix M is a matrix whose elements bij satisfy the rule: ijji ab = This is the reason that the transpose of a matrix M is generally denoted by MT. + − =⊕=⊕ − + =⊕ + − =⊕ 01 10 )(])[( 01 10 )( 01 10 † 21 * 21 * 2121 i i i i i i σσσσσσσσ T finallyandthen
- 8. Operations in matrix algebra are given as follows. 2016 MRT jijiji bac += 8 Addition: The sum C of two matrices M and N of equal dimension has elements cij that satisfy the rule: such that |l,s; j,mj 〉 is simultaneously an ‘eigenfunction’ of the total quantum angular momentum component along the z-axis, Jz, or the square of total quantum angular momentum vector, J2, an invariant. So, technically speaking, [x,y,z] is no more different than |l,s;ml,ms 〉 for |ΨΨΨΨkmo ( j,mj)〉→U(W)|ΨΨΨΨ(k,mj)〉=Σm′j Dm′jmj ( j)[W(Λ,p)]|ΨΨΨΨ(k,m′j)〉 but if you want to be ‘relativistic-lingo’ precise you would use |ΨΨΨΨ(Λp;j,mj)〉=|ΨΨΨΨ(Λp)〉|l,s; j,mj 〉. The operation of summing is denoted by: NMC ++++= in about the same way as learned in vector algebra (e.g., components cij vs sets [x,y,z]). We will see, in PART IV – QUANTUM FIELDS, how these cij above end up being quantities like Cj mlms =〈l,s;ml,ms |l,s; j,mj 〉 that are numerical coefficients which are known as Clebsch-Gordan (CG) coefficients that are used for the space-time translation of a physical ket state vector |l,s;ml,ms 〉 in a complex Hilbert space ends up looking like a transformation of standard momentum k≡kµ (i.e., contravariant momentum four-vectors): ∑∑ ±= += ±= == )21( 21 ,;,,;,,;,,;,,;, s mmm s j mm mm jssj sj s s mmsmjsmmsmmsmjs l l l l l ll lllll C
- 9. Difference: The difference D of M and N is similarly defined by: 2016 MRT jijiji bad −= 9 and denoted by: NMD −−−−= Product: The product E of M and N has elements eik that satisfy the rule: ∑= = n j kjjiji bae 1 Hence the number of columns of M must equal the number of rows of N. In matrix notation, multiplication is denoted by: NME = Matrix multiplication satisfies the associative rule: ΓBΑΓΒΑ )()( = but, in general, does not satisfy the commutative rule: ΑBΒΑ ≠ also written: TTT ΑBΒΑ =)(
- 10. If the determinant detM≡|M|=|aij | of a square matrix M is non-zero, then there exists one and only one matrix, which is called the inverse, or reciprocal, matrix of M and is denoted M−1, for which: 2016 MRT 10 1MMMM == 11 −−−−−−−− The elements of M−1 are given by: ji ji ij a A a =−1 where Aij is the cofactor of the element aij in the determinant |aij |, namely, (−1)i +j times the minor obtained from |aij | by taking away the i-th row and the j-th column. The operations of differentiation and integration of a matrix are applied to each element separately; that is: ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ x a x a x a x a x a x a x a x a x a x nmmm n n L MOMM L L 21 22212 12111 M and similarly for integration.
- 11. Adjoint of a Matrix: The adjoint of a matrix is written as adjM; it is defined as the cofactor transpose, that is: − − = − − = = 13 31 13 21 12 31 cc T hencethenIf AAA e.g., (Exercise), 2016 MRT Tc adj MM = Cofactor Matrix: The cofactor matrix is written as Mc and is defined by: ji MM =c For example: = = 332331 322221 312111 c 333231 232221 131211 AAA AAA AAA M aaa aaa aaa M thenIf where the minors are: .,, ,,, ,,, 2221 12113333 2321 13112332 2322 13121331 3231 12113223 3331 13112222 3332 13121221 3231 22213113 3331 23212112 3332 23221111 )1()1()1( )1()1()1( )1()1()1( aa aa A aa aa A aa aa A aa aa A aa aa A aa aa A aa aa A aa aa A aa aa A +++ +++ +++ −=−=−= −=−=−= −=−=−= 11
- 12. Self-Adjoint Matrix: If adjA = A, A is said to be self-adjoint. For example: 2016 MRT Symmetric and Antisymmetric Matrices: If ST=S, S is said to be a symmetric matrix whereas if AT=−A, A is said to be an antisymmetric (skew) matrix. For example: 222111 0 0 0 0 01 10 01 10 σσσσσσ −= − = − =+= = = i i i i TT then:tricAntisymme;then:Symmetric Hermitian Matrix: If H†=H, H is said to be a Hermitian matrix. For example: 2 † 2222 0 0 )( 0 0 0 0 σσσσσ = − == − = − = ∗∗ i i i i i i T henceandIf In quantum mechanics, all physical observables (e.g., the Hamiltonian or total energy operator) are represented by Hermitian operators (matrices). So, aij =aji vs aij =−aji. Unitary Matrix: If UU†=I (or U†=U−1) U is said to be a unitary matrix. For example: I= = == = = = ∗∗ 10 01 01 10 01 10 01 10 )( 01 10 01 10 † 11 † 1111 σσσσσσ andhencethenIf T Unitary matrices are important in quantum theory (e.g., the scattering matrix) and even more important in quantum field theory in that they ensure the conservation of probability and further ensures that particle ‘ghosts’ do not crop into the equations describing the scattering processes. AAAAA = − − == − − = − − = 10 01 adj 10 01 10 01 cc T hencethenIf 12
- 13. Orthogonality Relation: If OOT=I, where O is an Orthogonal matrix. For example: I= = =+= = = 10 01 01 10 01 10 01 10 01 10 11111 TT thenand σσσσσ Trace of a Matrix: The trace of a matrix A is given by the sum of its diagonal components: ∑= k kkaATr Inverse Matrix: For the inverse matrix, M−1, we require that: IMM =−1 We will have to work out an explicit expression for M−1. The value of a determinant is: 2016 MRT wherek=1,2,3,…,n,sums up the diagonalelements of a square n×n matrix.For example: 972Tr 73 42 2211 2 12221 1211 =+=+== = = ∑= aaaA aa aa A k kkand ∑∑ == == n j jk jiki n j ji ji AaMAaM 11 δor Let bjk =Ak j, that is, N=McT, so that this expression for the determinant becomes: T or c 1 MMNMMIbaM n j kjjiki === ∑= δ where δik =I. 13
- 14. On dividing I|M|=MMcT by |M|, we obtain: The quantity in brackets must be M−1 because of MM−1 =I. For A, we have |A|=1−6=−5 for the determinant, and: − − = = 13 21 2221 1211 c AA AA A and: 2016 MRT = 12 31 A Hence the inverse of A is: = M M MI Tc Let us try out an example and find the inverse of: − − = 12 31cT A − − −==− 12 31 5 1c 1 A A A T Check: QED 10 01 50 05 5 1 12 31 12 31 5 1c 1 I A A AAA = = − − −= − − −= =− T 14
- 15. The matrix method may be used to solve a system of linear equations. Consider the following system of equations to illustrate the method: In matrix form, we write: 2016 MRT CXA = 3223 =+=+ yxyx and 15 where: = = = 3 2 12 31 C y x XA and, The solution of the matrix equation AX=C is: C A A CAX Tc 1 == − where: − − =−= 12 31 5 cT and AA On substituting the value of |A| into the expression for X, we obtain: = − − −= = 3 1 5 7 3 2 12 31 5 1 y x X or x=7/5 and y=1/3.
- 16. Differentiation: The operations of differentiation and integration of a matrix are applied to each element separately; that is: 2016 MRT 16 ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ x a x a x a x a x a x a x a x a x a x nmmm n n L MOMM L L 21 22212 12111 M and similarly for integration. 3 3 2 3 1 3 3 2 2 2 1 2 3 1 2 1 1 1 321 321 ],,[ ],,[ det x y x y x y x y x y x y x y x y x y xxx yyy ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ = ∂ ∂ x y Jacobian: A Jacobian – Jacobi (1804-1851) – is a matrix of partial derivatives of the elements of one vector with respect to those of another, and is given by: If the elements of M in y=Mx are not functions of the elements of x, then M is the Jacobian of y with respect to x. If, on the other hand, the determinant of M in y=Mx is 1, then it is a rotation matrix, and y=Rx expresses a rotation of coordinate axes.
- 17. A convenient way to express the above equations is by the following matrix equation: − = − − y x YY XX θθ θθ cossin sincos o o 2016 MRT The square matrix with the direction cosines for elements is called the transfer matrix which, in this case, transforms the body coordinates to the fixed inertial coordinates. To obtain the inverse transformations from the fixed coordinate system to the moving coordinate system, we can start with our equation above for [Xi′++++Yj′++++Zk′] (with Z=z=0) arranged as follows: ˆ ˆ ˆ jiji ˆ)(ˆ)(]ˆˆ[ oo ′′= YYXXyx −−−−++++−−−−++++ and from the dot product with i and j:ˆ ˆ jjijjiii ˆˆ)(ˆˆ)(ˆˆ)(ˆˆ)( oooo •′•′=•′•′= YYXXyYYXXx −−−−++++−−−−−−−−++++−−−− and The above equations in matrix notation becomes: − − − = − − •′•′ •′•′ = o o o o cossin sincos ˆˆˆˆ ˆˆˆˆ YY XX YY XX y x θθ θθ jjji ijii which is the inverse of the matrix equation above. So, this transfer matrix is the inverse: 1 cossin sincos cossin sincos − − = − θθ θθ θθ θθ 17
- 18. The velocity of an arbitrary fixed point on the moving coordinate system x, y, z with respect to the fixed-axis system is: 2016 MRT This equation indicates that we can start with the displacement equation in terms of the Cartesian components and differentiate, holding x and y as constants. For instance, if we differentiate the equations X−Xo =xcosθ − ysinθ and Y−Yo =xsinθ + ycosθ: Comparing with X − Xo =xcosθ − ysinθ and Y − Yo =xsinθ + ycosθ, these equations can be written as: θθθθθθ &&&&&& )sincos()cossin( oo yxYYyxXX −=−+−=− and rRR ××××ωωωω++++o && = θθ &&&&&& )()( oooo XXYYYYXX −=−−−=− and 18
- 19. A point on a rigid body can be defined in terms of body-fixed axes x, y, z. To determine the orientation of the body itself, we now introduce Euler’s angles α, γ, β which are three independent quantities capable of defining the position of the x, y, z, body axes relative to the inertial X, Y, Z axes, as shown in the Figure - Left. Body axes x, y, z defined relative to inertial axes, X, Y, Z by Euler’s angles α, γ, β. With the x, y, z axes coinciding with X, Y, Z axes, allow the x, y, z coordinates to rotate about the Z axis through an angle α so as to take up the position ξ′η′ζ ′. The relationship between the two coordinates is the given by the transfer matrix (see Figure - Right): 2016 MRT Rotation about Z axis through angle α. [ ]T ζηξαα αα ζ η ξ ′′′= −= ′ ′ ′ Z Y X 100 0cossin 0sincos Z Y X O η Line of nodes z xα γ ξ β ζ y Z Y X η′ α ξ ′ ζ ′ 19
- 20. We next allow a rotation β about the ξ′ axis as shown in the Figure - Left and let the new position of the ξ′, η′, ζ ′ axes be ξ, η, ζ with transfer matrix Rotation about node axis ξ′=ξ through angle β. Finally we allow a spin γ about the axis ζ, as shown in the Figure - Right, to arrive at the body axes x, y, z. The transfer matrix for this rotation is: 2016 MRT Rotation about z =ζ axis through spin angle γ. [ ]T ζηξ ζ η ξ ββ ββ ζ η ξ = ′ ′ ′ − = cossin0 sincos0 001 η′ Z Y X O η α β ζ ξ ′ξ ζ ′ [ ]T zyx z y x = −= ζ η ξ γγ γγ 100 0cossin 0sincos z Z Y X O y xα γ ξ β ζ 20
- 21. In arriving at the final position of the body axes, we have encountered four sets of orthogonal axes: X, Y, Z; ξ′, η′, ζ ′; ξ, η, ζ; and x, y, z. Some of these axes coincide, such as the Zζ ′, the ζ z, the ξ′ξ; however both letters will be retained to identify the coordinate system referred to. Of particular interest is the ξ′ξ axis, called the line of nodes. It represents the intersection of the transverse body plan xy and the horizontal inertial plane XY. Other transformations between these coordinates can be obtained by the multiplication of the two or more transfer matrices. For instance, by substituting [ξ′ η′ ζ ′]T into [ξ η ζ ]T we obtain the following transformation between [X Y Z]T into [ξ η ζ ]T axes: − −= − − = Y Y X Y Y X βαβαβ βαβαβ αα αα αα ββ ββ ζ η ξ coscossinsinsin sincoscossincos 0sincos 100 0cossin 0sincos cossin0 sincos0 001 − +−−− +− = Y Y X z y x βαβαβ βγαβγαγαβγαγ βγαβγαγαβγαγ coscossinsinsin sincoscoscoscossinsinsincoscoscossin sinsincoscossinsincossincossincoscos Substituting this transfer matrix into [x y z]T, we obtain the transformation from the XYZ axes to the body axes xyz: 2016 MRT 21
- 22. The inverse transformation from x, y, z body axes to X, Y, Z inertial axes can be obtained in a similar manner by writing [ξ′ η′ ζ ′]T and [x y z]T in the reverse order: &c. 2016 MRT −+−+ −−− = z y x Z Y X βγβγβ αβαβγαγαβγαγ αβγβααγαβγαγ coscossinsinsin cossincoscoscossinsincoscossinsincos sinsincoscossincossinsincossincoscos ′ ′ ′ − = ζ η ξ αα αα 100 0cossin 0sincos Z Y X Exercise: Derive this matrix. Rules are also available for the direct inversion of matrices. The inverse [x y z]T above is: 22
- 23. We now express angular velocities ωx, ωy, ωz about the body axes x, y, z in terms of Euler angles. Rotation about node axis ξ′=ξ through angle β. Resolve the angular velocity α along ζ and η axes so that the orthogonal components of α, γ, and β are β along ξ, α sinβ along η, and γ +α cosβ along ζ, as shown in the Figure. γβα γβγβα γβγβα && && && += −= += cosω sincossinω cossinsinω z y x 2016 MRT ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ Next resolve the components along the ξ and η axes to the x and y directions, the result being: The inverse for of these equations is: γγβ γγ β β γ γγ β α sinωcosω )cosωsinω( sin cos ω )cosωsinω( sin 1 yx yxz yx −= +−= += & & & In matrix form, these equations become: − −−= −= z y x z y x ω ω ω 0sinsinsincos sincoscoscossin 0cossin 01cos sin0cossin cos0sinsin ω ω ω βγβγ ββγβγ γγ β γ α β γ α β γγβ γγβ & & & & & & & 23 Z Y X O η β ζ ξ ζ ′ α& γ& α β ⋅⋅⋅⋅
- 24. Problem: A convenient coordinate system for surface navigation on Earth is the longitude-latitude system with the origin coinciding with the moving vehicle shown. kjiω ˆsin)(ˆcos)(ˆ λφλφλ Ω++Ω++−= &&& a) Show that the angular velocities along the coordinates are: 2016 MRT where i, j, k vectors are along the x, y, z directions, and x is the Earth’s rotational speed.ˆ ˆ ˆ b) Determine the x, y, z components of the acceleration. The x, y axes lie in the horizontal plane along the latitude and longitude lines. 24 ΩΩΩΩ x λ zy Latitude φ Longitude Equator (Latitude = 0) jˆ iˆ kˆ
- 25. Problem: A satellite s circles the Earth with the orbit plane making an angle α with the Earth equatorial plane. The X axis is oriented so that it passes through the intersection of the orbit and the equatorial planes. The position of the satellite at any time can be given in terms of rs, the distance from the Earth’s center, φs the angle of the meridian plane measured from the X axis, and λs the longitude; the corresponding coordinates of an observer station O are Ro, φo and λo as shown. a) Determine the angle ϕ measured from the X axis to rs in the plane of the orbit, in terms of φs, λs, and α. 2016 MRT y z x Y Z X O λ φ rs α s b) Determine the cosine of the angle between Ro and rs, and the straight-line distance between O and s. Use h for altitude. h c) Determine the direction cosine of the line Os relative to X, Y, Z. d) Determine the direction cosine of the line Os relative to a coordinate system x, y, z with the origin at the observation station (as shown in the above Figure). 25 Ro
- 26. Let Earth-fixed positions be represented by a Cartesian coordinate system x, with the x axis toward latitude 0°, latitude 0°; the y axis toward latitude 0°, longitude 90°E; and z axis toward latitude 90°N, the north pole. The connection is through an inertially fixed- coordinate system X, with the X axis toward the vernal equinox ( ), the point where the sun’s orbit intersects the equator (i.e., when the Sun is over your head at the equator on the longest day of the year!) The angle Ω between the equinox and the Greenwich meridian – 0° longitude – is known as the Greenwich Sidereal Time. 2016 MRT 26 It is convenient to refer the position of a satellite to Cartesian coordinates q fixed in an ellipse inclined to the equatorial plane, as shown in the Figures below. Geodesic students are interested in Earth-fixed coordinates such as x=[x,y,z]; we need to connect Earth-fixed positions to positions referred to this tilted ellipse! Perigee Ellipse Apogee l ea ψ rb a ξ η P W θ 10 << e 1=e Circle Q ×××× 12 2 2 2 =+ ba ηξ z O i Ω ω Perigee NodeVERNAL EQUINOX Focus rp Prime Meridian or Greenwich (0° longitude) Y Z X North Pole x y F
- 27. Performing the multiplication above, we have: 2016 MRT 27 For an Earth rotating counterclockwise uniformly about an axis fixed with respect to inertial space, the shift from the Earth-fixed coordinate system x to the inertially fixed system X will be a simple clockwise rotation about the Z axis through an angle ϕ, the Greenwich Sidereal Time: xxRX − =−= 100 0cossin 0sincos )( ϕϕ ϕϕ ϕZ zZyxYyxX =+=−= and, ϕϕϕϕ cossinsincos Using an alternate notation of subscripts to the rotation matrix, denoting the vectors transformed by the rotation, we get: )( ϕ−= ZRR xX and inversely: )(ϕZRR Xx =
- 28. 2016 MRT 28 For rotation from the X coordinates to the q coordinates – with q1 =P towards the point of the ellipse closest to the origin (called Perigee – see previous Figure), q2 =Q in the orbital plane (as defined by the previous Figure), and q3 =W normal to the orbital plane – we require first a clockwise rotation about the Z-axis from the vernal equinox ( ) to the intersection of the inclined plane with the equator, called the nodes (see previous Figure). This rotation is denoted by: )(ΩZR Next, a counterclockwise rotation about the X-axis, from the equatorial plane to the orbital plane is given by: )()i( ΩZX RR And finally a counterclockwise rotation about the Z-axis from the node to perigee is given by: )()i()ω( Ω= ZXZ RRRR Xq ΩΩΩΩ, i, and ωωωω are identical with the Euler angles relating the q and X coordinate axes. Conversely we have: )ω()i()( −−Ω−= ZXZ RRRR qX
- 29. 2016 MRT 29 Applying rii =1, rij =rji =rik =rki =1, rjj =rkk = +cosϕ, rjk =+sinϕ, and rkj =−sinϕ and multiplying the matrices together (Exercise), we get: which is required for use – sooner or later if you want to tilt things in your head! An alternative notation often used is P, Q, W for the unit vectors along the q axes referred to the X axes: Ω−Ω+Ω−Ω+Ω ΩΩ−Ω−Ω−Ω = icosωcosisinωsinisin isincosωcosicoscosωsinsinωsinicoscosωcossin isinsinωcosicossinωsincosωsinicossinωcoscos qXR We studied in detail earlier in the Particle Dynamics and Orbits chapter how a particle of negligible mass, under the assumption that even at a distance r away, will be attracted by another and larger point mass M in accordance with a=GM/r2. That means that those prior developments apply solely to motion in a purely central field where the origin of the coordinates is at the center of the mass M and everything is pretty much described by: rr 3 r GM −=&& The acceleration vector r Is therefore colinear with the position vector r. If we define the equatorial plane as the plane determined by the position vector and the velocity vector r, the particle will never depart from the plane. ⋅⋅⋅⋅⋅⋅⋅⋅ ⋅⋅⋅⋅
- 30. 30 Our interest going forward is in satellite geodesy mainly due to the fact that the Earth’s gravitational field is non-central as studied in detail in the Gravitational Potential chapter; that is, the equation r=−GM⊕ r/r3 should be replaced by: where the potential, V, has a non-central form such as: where m=ω2a/geq, or the spherical harmonic version: V∇∇∇∇=r&& ⋅⋅⋅⋅⋅⋅⋅⋅ )1( 3 1 1 20 2eq 40 4 420 2 2 Pr a mg P r a JP r a J r GM U −+ − − −= ⊕ L ∑ ∑ ∞ = = + += 0 0 1 )]sin()cos()[(sin 1 l l llllll l lll m mmm mSmCP r V λλφ ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ 2016 MRT However, even for this non-central field the Keplerian ellipse and its orientation can be regarded as a coordinate system, alternative to Cartesian or polar coordinates, analo- gous to the use of geodetic latitude and altitude for position in an Earth-fixed system. At any instant the situation of a satellite in Earth-centered, inertially fixed coordinates can be described by the Cartesian components of position [X,Y,Z] and velocity [X,Y,Z]. In place of these six number, the six numbers of the Keplerian ellipse (a,e, M⊕,i,ω,Ω) may be used. The relationship between the two systems can be expressed by the rotation from a coordinate system in the orbital plane referred to perigee to the inertially fixed system, as given by the equations X=RXqq and X=RXqq. ⋅⋅⋅⋅ ⋅⋅⋅⋅
- 31. ∑= − =+== l lll l l m s ssmsmmi i s m im 0 sincosRe])sin[(cosRe]e[Re)cos( ξξξξξ ξ In order to convert the spherical harmonic potential V to Keplerian elements, we require some trigonometric identities such as: where i=√(−1), Re denotes the real part, is the binomial coefficient: )!(! ! sms m s m − = l ll ∑= −− =+−=−= l lll l l m s ssmsmmi i s m iiim 0 1 sincosRe])sin(cos[Re]e[Re)sin( ξξξξξ ξ and: ∑∑ ∑∑∑∑ = = + = = −−+ + = −− = −−−− −−++−−+− − = − − = ⋅ − − = + −−= a c b d c ba aa a c b d dcbaic ba aab d didbi b a c cicaic a aab ii a iiba dcbaidcba i d b c ai d b c aiii 0 0 0 0 )22( 0 )( 0 )( ]})22sin[(])22{cos[()1( 2 )1( e)1( 2 )1( ee 2 1 ee)1( 2 )1( )ee( 2 )ee( 2 cossin ξξ ξξ ξξξξξξξξξ Then, we also need: and the usual: )sin( 2 1 )sin( 2 1 sincos)sin( 2 1 )sin( 2 1 cossin )cos( 2 1 )cos( 2 1 sinsin)cos( 2 1 )cos( 2 1 coscos babababababa babababababa −−+=−++= −++−=−++= and ,, s ml 31 2016 MRT
- 32. Let a particular term of V be: and: )](sin[)](sin[)](cos[)](cos[)cos( ϕαϕαλ −ΩΩ−−−ΩΩ−= lllll mmmmm )]sin()cos()[(sin1 eq λλφ llllll l l llll mSmCP r rGM V mmmm += + ⊕ where req is the equatorial radius of the Earth (N.B., by applying the factor GMreq l we have made Clml and Slml non-dimensional). We then substitute [ml(α −Ω)+ml(Ω−ϕ)] for mlλ, where α is right ascension and ϕ is Greenwich Sidereal Time: )](sin[)](cos[)](cos[)](sin[)sin( ϕαϕαλ −ΩΩ−+−ΩΩ−= lllll mmmmm Orbit-equator-meridian triangle. 2016 MRT In the spherical triangle formed by the orbit, the equator, and the satellite meridian (see Figure), we have: Ω−+Ω−=+ 2 π cossin)sin(cos)cos()ωcos( φαφαθ 32 and: icos)sin()ωsin()cos()ωcos()ω(cos Ω−++Ω−+=+ αθαθθφ we then get: φ θ α φ θ α cos icos)ωsin( )sin( cos )ωcos( )cos( + =Ω− + =Ω− and and: )ωsin(isinsin θφ += Ω α – Ω φ i 2016 MRT
- 33. If we apply cos(mlξ) and sin(mlξ) to the α −Ω functions in cos(mlλ) and sin(mlλ) and substitute cos(α −Ω) and sin(α −Ω) therein, we get: and: ∑ ∑ = −+−− = −− + ⊕ ++ × −Ω+++−Ω−× = l ll llll l ll ll lllllll l ll l l m s ssmstms mmmm k t tm tm e m i s m mCiSmmSiC T r rGM V 0 2 0 2 1 icos)ω(cos)ω(sin )]}(sin[)()]sin()](cos[)Re{( isin ϕϕ ϕλϕ 33 )]}(sin[)({cos[ cos icos)ω(sin)ω(cos Re)cos( 0 ϕϕ φ ϕϕ λ −Ω+−Ω⋅ ++ = ∑= − ll l l l l l mimi s m m m s m sssm s )]}(cos[)({sin[ cos icos)ω(sin)ω(cos Re)sin( 0 ϕϕ φ ϕϕ λ −Ω−−Ω⋅ ++ = ∑= − ll l l l l l mimi s m m m s m sssm s If we substitute sinφ=sinisin(ω+ϕ) for sinφ in Plml (sinφ)=cosmlφ ΣtTlml t sinl−ml −2tφ, which is the definition of Plml , and then substitute both Plml and the two trigonometric functions cos(mlλ) and sin(mlλ) in Vlml above, by cancelling out the cosmlφs, we have: where k is the integer part of (l−ml)/2. 2016 MRT
- 34. On applying the sinaξ cosbξ identity to our last equation for Vlml , with a=l−ml −2t+s, and b=ml −s: )]}ω)(222sin[()]ω)(222{cos[( 2 )1( 2 )( icos )]}(sin[)()]sin()](cos[)Re{( isin 0 2 0 0 2 2 0 2 1 ϕϕ ϕλϕ +−−−++−−−× − +−− − − × −Ω+++−Ω−× = ∑ ∑ ∑ ∑ = +−− = − = − +−− = −− + ⊕ dctidct d sm c stmi i s m mCiSmmSiC T r rGM V m s stm c sm d c t stm ss mmmm k t tm tm e m ll ll l ll llll l ll l ll l l l lllllll l ll l l 34 2016 MRT
- 35. By applying the trigonometric identities cosacosb=½cos(a+b)+½cos(a−b), &c. to the products of Ω−φ and ω+θ trigonometric function of Vlml above, and dropping any term with an odd power of i=√(−1) as a coefficient (since Vlml is real, such a term has another term cancelling it out), we obtain: −Ω++−−− + −Ω++−−− − × − +−− −× −= − − − − +−− = − = = = − +−− + ⊕ ∑ ∑ ∑ ∑ )]()ω)(222sin[( )]()ω)(222cos[( 2 )1( 2 icos )1(isin 2 0 0 0 0 2 2 1 eq ϕϕ ϕϕ l l l l l l l l l l l ll l ll ll l l l l l l l l l l l l l l l l l ll mdct C S mdct S C d sm c stm s m T r rGM V m m m m m m m m stm c sm d c k t m s t s tktm tmm even odd even odd 35 2016 MRT It is desirable to transform this last equation so that the terms with the same argument, [(l−2p)(ω+θ)+ml(Ω−ϕ)], are collected together.
- 36. By substituting p for t+c+d necessitates, in turn, the elimination of one subscript from the factors. Putting p−t−c in place of the d summation place limits on the possible values of c, which turn out to be simply those making the binomial coefficient non-zero. In addition, t≤p. The expression for the spherical harmonic potential Vlml thus becomes: −Ω++− + −Ω++− − = − − = − − + ⊕ ∑ )]()ω)(2sin[( )]()ω)(2cos[()i( 0 1 eq ϕθ ϕθ l l l l l l l l l l l ll l l l l l l l l l l l l ll mp C S mp S C F r rGM V m m m m p m m m m pmm even odd even odd 36 2016 MRT Substituting from Tlmlt =(−1)t(2l−2t)!/[2lt!(l−t)!(l−ml −2t)!] yields the Kaula expansion: ∑∑ ∑ −− − +−− − × −−− − = − = −− − c kc m s s t tm tpm ctp sm c stm s m tmtt t F lll l l l l l ll l l l l 2 )1(icos isin 2)!2()!(! )!22( )i( 0 2 22 Here k is the integer part of (l−ml)/2, t is summed from 0 to the lesser of p or k, and c is summed over all values making the binomial coefficient non-zero.
- 37. Expressions for Flml p (i) up to lmlp=332 are given in the Table and F333 =15(1−cosi)/8. 37 2016 MRT Flml p (i) Flml p (i) 2 0 0 3 0 3 2 0 1 3 1 0 2 0 2 3 1 1 2 1 0 3 1 2 2 1 1 3 1 3 2 1 2 3 2 0 2 2 0 3 2 1 2 2 1 3 2 2 2 2 2 3 2 3 3 0 0 3 3 0 3 0 1 3 3 1 3 0 2 3 3 2 l ml p isin 8 3 2 200 −=F 2 1 isin 4 3 2 201 −=F isin 8 3 2 202 −=F )icos1(isin 4 3 2 210 +=F icosisin 2 3 211 −=F )icos1(isin 4 3 212 −−=F 2 220 )icos1( 4 3 +=F isin 2 3 2 221 =F 2 222 i)cos1( 4 3 −=F l ml p isin 16 5 3 300 −=F isin 4 3 isin 16 15 3 301 −=F isin 4 3 isin 16 15 3 302 +−=F isin 16 5 3 303 =F i)cos(1isin 16 15 2 310 +−=F )icos1( 4 3 )icos31(isin 16 15 2 311 +−+=F )icos1( 4 3 )icos31(isin 16 15 2 312 −−−=F i)cos(1isin 16 15 2 313 −−=F 2 320 i)cos1(isin 8 15 +=F i)cos3icos1(isin 8 15 2 321 −−=F i)cos3icos21(isin 8 15 2 322 −+−=F 2 323 i)cos31(isin 8 15 −−=F 3 330 i)cos1( 8 15 −=F i)cos1(isin 8 45 2 331 +=F i)cos1(isin 8 45 2 332 −=F
- 38. R=Ro +r Consider the case where the position of a particle P in space is defined in terms of the displacement vector r relative to the moving coordinates x, y, z (see Figure - Top). If the displacement of the origin of the moving coordinate system is Ro, the displacement of P relative to the fixed coordinate X, Y, Z is: 2016 MRT ][]ˆˆˆ[]ˆˆˆ[ ++++++++++++++++++++++++++++++++ KJIKJI =⇒= ZYX Letting unit vectors along a fixed and moving axes be designated by I, J, K and i, j, k, respectively, the above equation can be written as: We can determine the component of the above vector in any direction by forming the dot product of the above equation with a unit vector in the desired direction (e.g., the X component is obtained by the dot product of the above equation with I, &c.) These three rectangular components along the fixed coordinates are, then: where the dot product of the various unit vector represents the di- rection cosine between the coordinates. For a plane motion with Z =z=0 (see Figure - Bottom), the direction cosines involved are: ˆ ˆ ˆ Rotation Transformations Transformation between coordinates x, y, z and X, Y, Z. ˆ ˆ ˆ ˆ kKjKiK kJjJiJ kIjIiI ˆˆˆˆˆˆ ˆˆˆˆˆˆ ˆˆˆˆˆˆ •+•+•+= •+•+•+= •+•+•+= Z Y X iJjIjJiI ˆˆsinˆˆˆˆcosˆˆ •−=−=••==• θθ and θθθθ cossinsincos oo yxYYyxXX +=−−=− and O X Z Y P(x,y,z) x z yr ˆi jˆ kˆ Kˆ ˆI Jˆ Ro R so that the equation above reduces to: Coordinate transformation in a plane. Y X x y P(x,y) θ Iˆ Jˆ iˆ jˆ R Ro r Xo Yo Zo z z z x x x y y y zx yXo Yo Zo ˆ ˆ ˆki j 38
- 39. O r A rotation of the x-y plane about the z-axis by an angle ϕ. In the bottom part of this figure, β = ϕ since ϕ +γ = π/2 and γ +β = π/2 (180°/2 = 90°) because a and b are similar triangles. The relations between the bared and ‘unbared’ axes are: 2016 MRT ϕϕ ϕϕ ϕϕϕ sincos sin)(cos sinsincos 21 12 yxx x xx += ++= ++= ll ll ϕϕ ϕϕ ϕ ϕ sincos coscos cos)( cos 2 2 1 xyy y y y −= −= −= = l l l since y= l1 +l2 and since l1 = y −l2 and l2cosϕ = x sinϕ and since we rotate the x-y plane around the z-axis, it remains unchanged: zz = In matrix form, these equations become: = −= ⋅+⋅+⋅ ⋅++− ⋅++ = z y x R z y x zyx zyx zyx z y x )( 100 0cossin 0sincos 100 0cossin 0sincos 3 ϕϕϕ ϕϕ ϕϕ ϕϕ where R3(ϕ) denotes the 3×3 rotation matrix as a function of the angle ϕ . It is to be noted that infinitesimal rotations commute with one another whereas finite rotations do not. rrr r )(ϕR=→ whereRr(ϕ) isarotationbyanangleϕ andr=xi++++yj++++zk if Cartesian coordinates are used.ˆ ˆ ˆ y x z x y ϕ P(x,y) P(x,y) Typically, the positive +ϕ direction is that of the right- hand screw but to facilitate the math a bit we show here ϕ ≡−ϕ . For the true convention, replace ϕ by −ϕ . Which gives cos(−ϕ) = cosϕ and sin (−ϕ) = −sinϕ . l1 l2 l2sinϕ l1sinϕ β xcosϕ l1cosϕ γ y x x ϕ P(x,y) P(x,y) • b a M M α ϕ ϕϕ αϕαϕαϕ ϕϕ αϕαϕαϕ sinsin )sincoscos(sin)sin( sincos )sinsincos(cos)cos( yx OPOPy yx OPOPx += +=+= −= −=+= y 39 The relation between the labels of the points of three-dimensional space for two observers whose coordinate systems are rotated with respect to one another about a common origin is given by:
- 40. So, the rotation matrix about the z-axis by a finite angle ϕ is: Similarly, the rotation matrix about the x-axis and y-axis by a finite angle ϕ are: In general, a rotation is defined as a linear transformation which leads the scalar product of two vectors invariant and whose determinant is +1. 2016 MRT −= =+−=+= 100 0cossin 0sincos )( cossincoscos 3 ϕϕ ϕϕ ϕ ϕϕϕϕ R zzyxyyxx and, − = − = ϕϕ ϕϕ ϕ ϕϕ ϕϕϕ cos0sin 010 sin0cos )( cossin0 sincos0 001 )( 21 RR and Suppose an observer O is located in a coordinate system[x,y,z]. Owoulddescribe, say, the temperature T which varies from point to point according to some law described by T = f(x, y,z)≡ f(r). There is nothing special about the coordinate system [x,y,z]; the law expressing the variation of temperature from point to point could equally well be referred to by an observer O in the coordinate system O(x,y,z). However, the form of thelaw, i.e., the dependence on the coordinates, will generally be different in the two systems. If we let T=g(x, y,z)≡ g(r) in the [x,y,z] system, then the temperature at any point – say described by T= f(x,y,z) and T=g(x,y,z) – must be the same. The physics describing the temperaturefieldwillnot change because of a rotation(assuming it is not too fast). 40
- 41. We therefore require that (i.e., to preserve coordinate invariance): Thus D(R) acting on a function f(r) produces a new function g(r) such that g(r) is the transformed function under the coordinate transformation r=Rr r. The main point to keep in kind is that Rr transforms coordinates while D(R) transforms functions of coordinates. Now, a function f (r) is said to be an invariant function under a coordinate transforma- tion r=Rrr if: 2016 MRT where D(R) is a rotation operator – an engine that transforms functions of coordinates. )()()()()( 1 rrrr fRffRg === − D )()( rr gf = We now ask the following question: Given a scalar function f(r)= f(x,y,z) and a coordinate transformationr =Rr r, what is the formal method for finding g(r) such that f(r) =g(r)? For this purpose we define an operation D(R) by the relation: )()()()( 1 rrr fRffR == − D == === − 100 010 001 1 EE zzyyxx and, Now, we state a few special rotation transformations (and do notice the symmetries): E – The identity transformation: 41
- 42. 2016 MRT − −== −=−== − 100 010 001 1 AA zzyyxx and, B – A rotation of coordinates about the O2-axis through 180°: − − == −=−=+−= − 100 0 2 1 2 3 0 2 3 2 1 2 1 2 3 2 3 2 1 1 BB zzyxyyxx and, A – A rotation of coordinates about the x-axis through 180°: C – A rotation of coordinates about the O3-axis through 180°: − − −− == −=+−=−−= − 100 0 2 1 2 3 0 2 3 2 1 2 1 2 3 2 3 2 1 1 CC zzyxyyxx and, O y x z 2 1 3 z y O y x z 2 1 3 z O y x z 2 1 3 z y x y 30° x 60° x 42
- 43. D – A positive (counterclockwise) rotation of coordinates about the z-axis through 120°: 2016 MRT FDD zzyxyyxx = − − −− = − −− − = =+−=+−= − 100 0 2 1 2 3 0 2 3 2 1 100 0 2 1 2 3 0 2 3 2 1 2 1 2 3 2 3 2 1 1 and and, F – A negative (clockwise) rotation of coordinates about the z-axis through 120°: E A B C D F E E A B C D F A A E D F B C B B F E D C A C C D F E A B D D C A B F E F F B C A E D DFF zzyxyyxx = −− − = − −− = =−=−−= − 100 0 2 1 2 3 0 2 3 2 1 100 0 2 1 2 3 0 2 3 2 1 2 1 2 3 2 3 2 1 1 and and, Below is a Multiplication Table fortheelementsof D3 (e.g.,AC=F but CA=D and F−1F=E). O y x z 2 1 3 z y x 120° O y x z 2 1 3 x y z 120° Applied Second AppliedFirst This Table establishes that the rotations E, A, B, C, D and F are the elements of a group – the D3 group. 43
- 44. As an example,suppose we have a hydrogen atom surrounded by three unit charges q located at positions 1, 2, and 3 which form an equilateral triangle of side a; the proton is at the origin O and the electron at r (see Figure). The Hamiltonian H of the electron is: in which |r1|, |r2|, and |r3| are the distances O1, O2, and O3, respectively, and ++− +∇−=++=+= 2 3 2 2 2 1oo 2 2 e 2 o 111 επ4επ42 )( rrrrrr −−−−−−−−−−−− qe r Ze m WVTWHH h )( 232 )( 232 )( 3 3 2 22 2 32 2 22 2 21 22 2 2 1 rrrrrrrrr fz a y a xfz a y a xfzy a x ≡+ ++ −=≡+ −+ −=≡++ += −−−−−−−−−−−− and, Hydrogen atom surrounded by three charges located at the corners of an equilateral triangle. −−−=⇒= − zyxyxfzyxfDDffD , 2 1 2 3 , 2 3 2 1 ),,()()()()( 1 DD rr −− − = 100 0 2 1 2 3 0 2 3 2 1 D zzyxyyxx =−−=+−= &, 2 1 2 3 2 3 2 1 )()()( )()()( )( 22 1 2 3 322 3 2 1 )()( 21 13 3 2 22 2 rr rr rr ffD ffD fz a yx a yxfD = = =+ −−+ −−−= D D D and Ho is invariant under all three-dimensional rotations. so that the function f (r) is indeed invariant since: )()()()]()()()[( 321321 rrrrrr ffffffD ++=++D − −− =− 100 0 2 1 2 3 0 2 3 2 1 1 D 2016 MRT O y x z a r2 1 3 e Ze q q q O r 1 r1 |r –r1| O r 3 r3 |r –r3| O r 2 r2 | r – r2 | a a The interaction potential W remains to be investigated. Let us take, as a practical example, the coordinate transformation D which, as we just saw before, is a positive (counter clockwise) rotation of coordinates about the z-axis through 120°. We have: Therefore: 44
- 45. For rockets and space vehicles it is often necessary to consider the general problem of the spinning body under thrust. The concern here is the body attitude and the motion of the center of mass. We will consider only problems where the rate of mass variation is small enough to be negligible. 2016 MRT To outline the problem at hand, we will consider a rigid body and define a set of body- fixed axes x, y, z rotating with angular velocity ωωωω, and with the origin coinciding with the center of mass. Although it is always desirable to let the body axes coincide with the principal axes, this is often not possible, so that in the general case, the moments and products of inertia will be defined as: FIEIDICIBIAI yzxzxyzyx ====== &,and&, Using the equations Lx =Ix ωx −Ixy ωy − Ixz ωz, &c., the angular momentum becomes: kjiL ˆ)ωωω(ˆ)ωωω(ˆ)ωωω( yxzxzyzyx FECDFBEDA +−+−−+−−= and the vector moment equation about the body axes: LωLM ××××+= ][ & can be written out in terms of the components given by Euler’s equations (i.e., Mx =Lx + ωy Lz− ωz Ly, &c. obtained earlier) for body-fixed coordinates, x, y, and z: yzyxxxzyyxzz xyxzzzyxxzyy zxzyyyxzzyxx EDADFBFECM FECEDADFBM DFBFECEDAM ω)ωωω(ω)ωωω()ωωω( ω)ωωω(ω)ωωω()ωωω( ω)ωωω(ω)ωωω()ωωω( −−−−−++−= +−−−−+−−= −−−+−+−−= &&& &&& &&& 45 Space Vehicle Motion ⋅⋅⋅⋅
- 46. We next let the velocity of the center of mass be expressed by the equation: 2016 MRT and the force as: Since the x, y, z coordinates are rotating with the body, the vector force equation is: kjiv ˆˆˆ zyx vvv ++++++++= and the force components along x, y, z directions are determined from the equation: vω v pωpF m td d m ××××++++××××++++ == ][ & If the resultant of the above force does not pass through the center of mass coinciding with the origin of the x, y, z, axes, our equations for Mx, My, Mz, and Fx, Fy, Fz, above become coupled owing to the moment of the force. Also these equations define the motion of the body only in terms of the linear and angular velocities referred to body axes, and their solution and transformation to displacements and angles relative to inertial coordinates X, Y, Z are problems of considerable difficulty which can be accomplished only under simplifying assumptions. )ωω( )ωω( )ωω( yxxyzz xzzxyy zyyzxx vvvmF vvvmF vvvmF −+= −+= −+= & & & kjiF ˆˆˆ zyx FFF ++++++++= 46
- 47. T 1 2 A C 3 O A L1 ∫∫ =+= −+= −+= −+= ++== mdCmdA ACA III LLLM MMM 2 2 2 3 2 2 321 322311 233211 321 2)( ωω)(ω ω)ω(ω ωω ˆˆˆ][ lll & & & & & ntmisalignmethrustforso where 321LLM ××××ωωωω++++ ω3 L2 ω2 L3 ωωωω1 −−−−Pitch ++++Pitch −−−−Yaw ++++Yaw ++++Roll y x z ++++Spin z ωωωω2 ωωωω3 L2 L3 ωωωω1 ⋅⋅⋅⋅ l2 Let us now consider thrust misalignment. We will consider a simple problem of a spinning missile with a misalignment of the thrust line. We will assume that the missile is symmetric so that the x, y, z, axes coincide with the principal axes 1, 2, 3 with I1 =I2 =A and I3 =C. With A=B, we can rotate the 1, 2 axes so that one of these axes, say 1, is perpendicular to the plane containing the thrust and axis 3, as shown in the Figure. Euler’s moment equations, M=[L]+ωωωω××××L, for the missile (N.B., A=B & C=I3) are then: The third equation tells us that ω3 =n is a constant and recall that L=r××××mv. 0ωωωωωω 0ωω)(ωωωωωω 0ωω)(ωωωωωω 3211233 312133122 321322311 ==−+= =−−=−+= ≠−+=−+= && && && CAACM ACACAAM ACAACAM Thrust misalignment resulting in moment M1 on missile of principal axes 1, 2, 3. Art circa 1960. Although C is generally less than A for missiles, we let: − = A AC nλ 0ωωωω 12 1 21 =−=+ λλ && and A M and rewrite the first two equations as: From the second equation we find ω1 =(1/λ)ω2 and when we substitute it into the first equation we get: ⋅⋅⋅⋅ )()(ωω 12 2 2 tMAλλ =+&& which has the solution: ∫ −++= t tdtttM A ttt 0 1 2 22 )](sin[)( 1 sin )0(ω cos)0(ω)(ω λλ λ λ & 2016 MRT 47 ⋅⋅⋅⋅
- 48. The solution of the previous thrust misalignment problem is in terms of body-fixed coordinates which are rotating. Adding ω1 and ω2 in quadrature: γ βαβ i ii − +=+= e)sin(ωωω 2112 && 48 ( )0cosω sincossinω cossinsinω 33 2 1 ==+= −= += Mn forconstantaγβα γβγβα γβγβα && && && From ω3: β γ α cos & & − = n which, substituted into equation ω12 =ω1 +iω2 above, yields: γ βγβ i ni − −+= e]tan)([ω12 && 2016 MRT In order to transform from the body-fixed coordinates x, y, and z to the inertial coordinates X, Y, and Z it is necessary to introduce Euler’s angles α, γ, and β. The transformation for our principal axes, 1, 2, and 3 is:
- 49. Although this last equation relates the angular velocity ω12 about the body-fixed coordinates in terms of Euler’s angles referenced to inertial axes, further simplification generally requires a small angle approximation for β. Such an approximation is often justified when dealing with rockets and missiles whose spin axis must not deviate greatly from a fixed direction of flight. When β is small, tanβ can be replaced by β: At this point, we introduce a complex angle of attack (H. Leon, 1958), which uncouples this equation into: γ βγβ i ni − −+= e])([ω12 && 49 2016 MRT γ ββ i− = e12 Differentiating gives: γ βγββ i i − −= e)(12 && The equation above for ω12 now take the following form: 121212 ω=+ ββ ni& so that when ω12 is a known function of time we have a first-order ordinary differential equation in β12 to solve. It must be remembered however that the above procedure is limited to problem when β is small.
- 50. At this point the significance of the term exp(−iγ) appearing in the various equations should be pointed out. For example, consider the equation for ω12 obtain earlier: Writing exp(−iγ)=cosγ −isinγ, ω12 becomes: 50 Velocity components in transverse plane tilted by an angle β about the node axis ξ. 2016 MRT Thus the multiplication of the components β +iαsinβ along the node coordinate system by exp(−iγ ) results in ω1 +iω2, the components along the body-fixed axes rotated through an angle γ from the node axis, ξ. It follows then that, if we multiply the components ω1 +iω2 along the body axes by exp(iγ ), we should obtain the vector ω12 in terms of the node axis components as follows: ξη γγ βαβ ω)sin(eωe)ωω( 1221 =+==+ && ii ii We can now attach physical significance to the complex angle of attack β12=β exp(−iγ). Since β is multiplied by exp(−iγ), β12 is resolved into components along the body-fixed axes 1 and 2. To restore β along its node axis we multiply β12 by exp(iγ) (i.e., β =β12 exp(iγ)). Furthermore, if we wish to examine β12 in terms of inertial components, we need to multiply β by exp(iα), or βexp(iα)= [β12exp(iγ)]exp(iα)=β12exp[i(γ + α)]≅β12exp(in t) where γ +α ≅nt. Z O η ξ α 2 1 ω2 ω1 α⋅⋅⋅⋅ β ⋅⋅⋅⋅ ω12 where all the components in this equation lie in the tilted transverse plane which is shown in the Figure. γ βαβ i ii − +=+= e)sin(ωωω 2112 && )sincossin()sinsincos(ω12 γβγβαγβαγβ &&&& −++= i The real and imaginary parts of this equation are, however, equal to the components of β and αsinβ along axes 1 and 2. ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅
- 51. When the geometric axes x, y, and z (corresponding to yaw, pitch, and spin) of a missile are not principle axes, the solution in terms of such body coordinates will require the solving of the general equation: 51 Principal axes 1, 2, 3 displayed from missile axes x, y, z. Axis x is normal to plane zO3. 2016 MRT Without loss of generality, the transverse axis x can be chosen normal to the plane zO3, and the other two axes 1 and 2 are defined by the angle Φ between axis 1 and the transverse axis x which intersects the planes 1, 2, and xy. If we assume that the principal inertias I1 ≅I2, then elementary solutions are available in terms of principal axes 1, 2, 3. The motion of the geometric axes x, y, z can then be obtained by a transformation of coordinates with θ and Φ known. yzzxyxyxxxxxyzzyyyyzyxzxzzz xyzyxzxzzzzzxyxyxxxxyzzyyyy zxxyzzyyyyyzyxzxzzzzxyxyxxx IIIIIIIIIM IIIIIIIIIM IIIIIIIIIM ω)ωωω(ω)ωωω()ωωω( ω)ωωω(ω)ωωω()ωωω( ω)ωωω(ω)ωωω()ωωω( −−−−−++−= +−−−−+−−= −−−+−+−−= &&& &&& &&& z x Φ 2 1 3 y We recognize first that every body has a set of principal axes 1, 2, and 3. For the near symmetric body, the principal axis 3 deviates only by a small angle θ from the spin axis z, as shown in the Figure. where Ixx =Ix, Iyy =Ix, and Izz =Iz. These equations do not lead to a simple solution, even for small products of inertia, and it is desirable to take a different approach (N.B., When the origin of the body axes coinciding with the center of mass, we can orient the x, y, z axes to coincide with the principal axes 1, 2, 3 of the body to eliminate the product of inertia terms in the moment of momentum expressions with L1 =Aω1, L2 =Bω2, and L3 =Cω3).
- 52. The transformation we obtained earlier: 52 can now be used to convert things to our application since α →0 (cos0=1 and sin0=0), β →θ, and γ →Φ using x, y, z (see Figure): −+−+ −−− = z y x Z Y X βγβγβ αβαβγαγαβγαγ αβγβααγαβγαγ coscossinsinsin cossincoscoscossinsincoscossinsincos sinsincoscossincossinsincossincoscos The direction cosines used in the following developments are obtained from the matrix above for the transformation between coordinates x, y, z and 1, 2, 3 with lengths along it. Components of r in two coordinate systems. 2016 MRT ΦΦ −ΦΦ Φ−Φ = ••• ••• ••• = = z y x z y x z y x lll lll lll zyx zyx zyx θθθ θθθ coscossinsinsin sincoscoscossin 0sincos ˆˆˆˆˆˆ ˆˆˆˆˆˆ ˆˆˆˆˆˆ 3 2 1 333 222 111 k3j3i3 k2j2i2 k1j1i1 The inverse will be required: − ΦΦΦ− ΦΦΦ = = 3 2 1 10 coscossin sinsincos 3 2 1 333 222 111 θ θ θ zyx zyx zyx lll lll lll z y x and since the angle θ is small, the approximation sinθ ≅θ and cosθ ≅1 is used such that the transformation matrix is: − ΦΦΦ− ΦΦΦ = 3 2 1 cossin0 cossincoscossin sinsincossincos θθ θθ θθ z y x O r x z y P 1 3 2 ˆi kˆ3ˆ 2ˆ jˆ θ ˆ1 Φ
- 53. 53 where I23 =I1, I13 =I2, and I12 =I3 and the lξ 1 , lξ 2 , lξ 3 are the direction cosines of the principal axes ξ =1,2,3 with respect to the x, y, z axes, as follows: 33 2 322 2 211 2 1 3 1 2 )()()()( IlIlIlIlI i iii ξξξξξξ ++== ∑= moments 2016 MRT They are related, in the first part, to the principal moments of inertia I11 =I1, I22 =I2, and I33 =I3, by the moments of inertia equation: We assume that the moments and products of inertia about the missile axes x, y, z are: ,,,and,, FIEIDICIBIAI zyzxyxzyx ====== 3 22 2 22 1 3 2 2 2 1 2 3 2 32 2 21 2 1 3 22 2 2 1 3 2 2 2 1 2 3 2 32 2 21 2 1 2 2 2 1 3 2 2 2 1 2 3 2 32 2 21 2 1 cossin )1()cos()sin( cossin )()(cos)(sin sincos )0()sin()(cos IIIC III IlIlIlII IIIB III IlIlIlII IIA III IlIlIlII zzzzzz yyyyyy xxxxxx +Φ+Φ= +Φ+Φ= ++=≡ +Φ+Φ= −+Φ+Φ= ++=≡ Φ+Φ= +Φ−+Φ= ++=≡ θθ θθ θ θ > > >
- 54. And in the second part, by the products of inertia equation: 54 2016 MRT 3 2 2 2 1 32 2 1 2 321 333222111 21 21 321 333222111 21 21 321 333222111 )cossin( cossin )1)(()cos)((cos)sin)((sin cossin)( sincossincos )1)(0()cos)(sin()sin)((cos cossin)( sincossincos ))(0())(cossin())(sin(cos IIIF III III IllIllIllI IIE II III IllIllIllI IID II III IllIllIllI zyzyzyzy zxzxzxzx yxyxyxyx θθ θθθ θθθ θ θθ θθ θ −Φ+Φ=− −Φ+Φ= −+ΦΦ+ΦΦ= ++=− ΦΦ−=− ΦΦ−ΦΦ= +ΦΦ−+ΦΦ= ++=− ΦΦ−=− ΦΦ−ΦΦ= −+ΦΦ−+ΦΦ= ++=− > > > 333322221111 3 1 IllIllIllIllI i iiii ηξηξηξηξηξ ++==− ∑= products where the lξi and lη j are direction cosines of the principal axes ξ,η =1,2,3 with respect to the x, y, z axes and is given by:
- 55. Collecting all terms we get, in the case of three independent moments I1, I2, and I3: 55 2016 MRT and from this last equation for F, the angle θ becomes: 31 2 3 2 1 1 IICI IIBI IAI z y x +=≡ +=≡ =≡ θ θ 1313 II I II F zy − = − =θ )( 0 0 31 IIFI EI DI zy zx yx −−=≡ =≡ =≡ θ and: If we assume I1 =I2 (N.B., I3 ≠0), which is mostly the case for a missile, our equations for A, B, C, −D, −E, and −F reduce to the following: 3 2 2 2 13 22 2 22 1 123 22 2 2 1 12 2 2 2 1 )cossin(cossin cossin)(cossin cossin)(sincos IIIFIIIICI IIEIIIIBI IIDIIIAI yzz xzy xyx θθθθ θθ +Φ+Φ−=≡+Φ+Φ=≡ ΦΦ−=≡+Φ+Φ=≡ ΦΦ−=≡Φ+Φ=≡ and
- 56. Z O ξ Φ y x z β Y X 1 θ γα 3 To solve for the angular velocities, we first write down the transformation from the missile axes to the principal axes, assuming θ to be small: 56 Adding ω1 and iω2 we obtain the complex angular velocity: − Φ−ΦΦ− ΦΦΦ = z y x ω ω ω 10 coscossin sinsincos ω ω ω 3 2 1 θ θ θ where ωz ≅ n. By multiplying the last equation by exp(iΦ), this equation may also be written in the inverse form, ωxy =ω12exp(iΦ)−iθn. Φ− Φ− +≅ ++=+= i yx i zyx ni iii e)ω( e]ω)ωω[(ωωω 2112 θ θ Principal axes 1, 2, 3 referred to missile axes x, y, z, which in turn are referred to node axis ξ and inertial axes X, Y, Z. 2016 MRT The Figure shows the relationship between the missile axes x, y, z, the principal axes 1, 2, 3, the inertial axes X, Y, Z, and the line of nodes ξ. The missile axis x is normal to axes 3 and z, whereas the line of nodes ξ is normal to axes z and Z. The principal axes 1, 2, 3 are then referenced to the x, y, z axes by fixed angles θ and Φ. The position of the missile axes x, y, z is obtained by starting with the missile spin axis z coinciding with Z and performing three rotations as follows: 1. Rotation of α about Z ; 2. Rotation of β about ξ; 3. Rotation of γ about the spin axis z. X Y Z z x y 3 1 2 I1 I2 I3
- 57. Problem: Body axes x, y, and z initially coincide with the inertial axes X, Y, and Z are given the following sequence of rotations. Rotation β3 about z followed by rotation β2 about the displaced y axis and a rotation β1 about the final position of the x axis. a) Derive the transfer matrix expressing the body axes in terms of the inertial axes, and its inverse. b) Assume angular velocities β3, β2, and β1 about axes z, y, and x in the sequence given above, and write the equations for the angular velocities ω1, ω2, and ω3 about the final position axes x, y, z. c) Referring to the Figure below, assume the missile to be symmetric so that Iy=Iz, and determine the equation for the attitude deviation β2+ iβ3 of the longitudinal axis due to a constant yawing torque Mz. 2016 MRT 57 ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ x β1 β2 β3 β1 ⋅⋅⋅⋅ β2 ⋅⋅⋅⋅ β3 ⋅⋅⋅⋅ X y Y z Z β1 β2 β3
- 58. ωωωωo 1 2 3 Problem: A space vehicle of moment of inertia I1, I2, and I3 is in a circular orbit with constant angular velocity ωωωωo about the axis 2 to maintain the direction of axis 1 always tangent to the orbit as shown. 2016 MRT Assuming small disturbances θ1, θ2, θ3, derive the differential equation of motion for the torques about the body axes 1, 2, and 3. 12213 o 312133 o 22 2 1233 o 1 sin2sin)( 2 3 cos2sin)( 2 3 cos2sin)( 2 3 θθθθθθ II R GM MII R GM MII R GM M −=−=−= and, where Ro is the distance from the center of Earth to the vehicle center of mass. Problem: Assume the body axes 1, 2, and 3 of the space vehicle of the Problem above to deviate from the orbit axes 1′, 2′ and 3′ by angles θ3, θ2, and θ1 in the following sequence: a rotation θ1 about z followed by rotation θ2 about the displaced y axis and a rotation θ1 about the final position of the x axis. Using the procedure used in the calculation of the effect of the Earth’s oblateness and a spherical Earth, show that the gravity force on the space vehicle results in torques about the body axes equal to: 58
- 59. 2016 MRT 59 Appendix PART II – MODERN PHYSICS Charge and Current Densities Electromagnetic Induction Electromagnetic Potentials Gauge Invariance Maxwell’s Equations Foundations of Special Relativity Tensors of Rank One 4D Formulation of Electromagnetism Plane Wave Solutions of the Wave Equation Special Relativity and Electromagnetism The Special Lorentz Transformations Relativistic Kinematics Tensors in General The Metric Tensor The Problem of Radiation in Enclosures Thermodynamic Considerations The Wien Displacement Law The Rayleigh-Jeans Law Planck’s Resolution of the Problem Photons and Electrons Scattering Problems The Rutherford Cross-Section Bohr’s Model Fundamental Properties of Waves The Hypothesis of de Broglie and Einstein Appendix: The General Theory of Relativity References We list here, as a reference, the Contents of the remaining parts of this 10-PART Series which as a whole makes for quite a thorough review of Theoretical Physics (N.B., Since Superstring Theory is still being developed, this content is accurate up to year 1990-ish).
- 60. 60 PART III – QUANTUM MECHANICS Introduction Symmetries and Probabilities Angular Momentum Quantum Behavior Postulates Quantum Angular Momentum Spherical Harmonics Spin Angular Momentum Total Angular Momentum Momentum Coupling General Propagator Free Particle Propagator Wave Packets Non-Relativistic Particle Appendix: Why Quantum? References 2016 MRT
- 61. PART IV – QUANTUM FIELDS Review of Quantum Mechanics Galilean Invariance Lorentz Invariance The Relativity Principle Poincaré Transformations The Poincaré Algebra Lorentz Transformations Lorentz Invariant Scalar Klein-Gordon & Dirac One-Particle States Wigner’s Little Group Normalization Factor Mass Positive-Definite Boosts & Rotations Mass Zero The Klein-Gordon Equation The Dirac Equation References 61 2016 MRT
- 62. 62 PART V – THE HYDROGEN ATOM What happens at 10−−−−10 m? The Hydrogen Atom Spin-Orbit Coupling Other Interactions Magnetic & Electric Fields Hyperfine Interactions Multi-Electron Atoms and Molecules Appendix – Interactions The Harmonic Oscillator Electromagnetic Interactions Quantization of the Radiation Field Transition Probabilities Einstein’s Coefficients Planck’s Law A Note on Line Broadening The Photoelectric Effect Higher Order Electromagnetic Interactions References 2016 MRT
- 63. 63 PART VI – GROUP THEORY Symmetry Groups of Physics Basic Definitions and Abstract Vectors Matrices and Matrix Multiplication Summary of Linear Vector Spaces Linear Transformations Similarity Transformations Dual Vector Spaces Adjoint Operator and Inner Product Norm of a Vector and Orthogonality Projection, Hermiticity and Unitarity Group Representations Rotation Group SO(2) Irreducible Representation of SO(2) Continuous Translational Group Conjugate Basis Vectors Description of the Group SO(3) Euler Angles α, β & γ Generators and the Lie Algebra Irreducible Representation of SO(3) Particle in a Central Field Transformation Law for Wave Functions Transformation Law for Operators Relationship Between SO(3) and SU(2) Single Particle State with Spin Euclidean Groups E2 and E3 Irreducible Representation Method Unitary Irreducible Representation of E3 Lorentz and Poincaré Groups Homogeneous Lorentz Transformations Translations and the Poincaré Group Generators and the Lie Algebra Representation of the Poincaré Group Normalization of Basis States Wave Functions and Field Operators Relativistic Wave Equations General Solution of a Wave Equation Creation and Annihilation Operators References 2016 MRT
- 64. 64 PART VII–QUANTUM ELECTRODYNAMICS Particles and Fields Second Quantization Yukawa Potential Complex Scalar Field Noether’s Theorem Maxwell’s Equations Classical Radiation Field Quantization of Radiation Oscillators Klein-Gordon Scalar Field Charged Scalar Field Propagator Theory Dirac Spinor Field Quantizing the Spinor Field Weyl Neutrinos Relativistic Quantum Mechanics Quantizing the Maxwell Field Cross Sections and the Scattering Matrix Propagator Theory and Rutherford Scattering Time Evolution Operator Feynman’s Rules The Compton Effect Pair Annihilation Møller Scattering Bhabha Scattering Bremsstrahlung Radiative Corrections Anomalous Magnetic Moment Infrared Divergence Lamb Shift Overview of Renormalization in QED Brief Review of Regularization in QED Appendix I: Radiation Gauge Appendix II: Path Integrals Appendix III: Dirac Matrices References 2016 MRT
- 65. 65 Fermion Masses and Couplings Why Go Beyond the Standard Model? Grand Unified Theories General Consequences of Grand Unification Possible Choices of the Grand Unified Group Grand Unified SU(5) Spontaneous Symmetry Breaking in SU(5) Fermion Masses Again Hierarchy Problem Higgs Scalars and the Hierarchy Problem Appendix References PART VIII – THE STANDARD MODEL The Particles The Forces The Hadrons Scattering Field Equations Fermions Particle Propagators Noether’s Theorem and Global Invariance Local Gauge Invariance in QED Yang-Mills Gauge Theories Quantum Chromodynamics (QCD) Renormalization Strong Interactions and Chiral Symmetry Spontaneous Symmetry Breaking (SSB) Weak Interactions The SU(2)⊗U(1) Gauge Theory SSB in the Electroweak Model Gauge Boson Masses Gauge Boson Mixing and Coupling 2016 MRT
- 66. 66 PART IX – SUPERSYMMETRY Motivation Introduction to Supersymmetry The SUSY Algebra Realizations of the SUSY Algebra The Wess-Zumino Model Lagrangian with Mass and Interaction Terms The Superpotential Supersymmetric Gauge Theory Spontaneous Breaking of Supersymmetry F-type SUSY Breaking D-type SUSY Breaking The Scale of SUSY Breaking The SUSY Particle Spectrum Supersymmetric Grand Unification General Relativity The Principle of Equivalence General Coordinates Local Lorentz Frames Local Lorentz Transformations General Coordinate Transformations Covariant Derivative The Einstein Lagrangian The Curvature Tensor The Inclusion of Matter The Newtonian Limit Local Supersymmetry A Pure SUGRA Lagrangian Coupling SUGRA to Matter and Gauge Fields Higher-dimensional Theories Compactification The Kaluza Model of Electromagnetism Non-Abelian Kaluza-Klein Theories Kaluza-Klein Models and the Real World N=1 SUGRA in Eleven Dimensions References 2016 MRT
- 67. 67 2016 MRT PART X – SUPERSTRING THEORY A History of the Origins of String Theories The Classical Bosonic String The Quantum Bosonic String The Interacting String Fermions in String Theories String Quantum Numbers Anomalies The Heterotic String Compactification and N=1 SUSY Compactification and Chiral Fermions Compactification and Symmetry Breaking Epilogue: Quantum Gravity Appendix I: Feynman’s Take on Gravitation Appendix II: Review of Supersymmetry Appendix III: A Brief Review of Groups and Forms Appendix IV: The Gamma Function Appendix V: The Beta Function References
- 68. 2016 MRT References 68 M. Klein, Calculus, 2-nd Edition, Dover, 1977. New York University Believe it or not, my father had the first edition in 2 volumes and when I found them and browsed through them I was amazed at the Motion in One Dimension in a Resisting Medium part which eventually ended up being the case for this treatment of air resistance in ballistic problems. Since then, I’ve always enjoyed Klein’s way of presenting calculus using physical concepts. The only problem though is his use of 32 [ft/sec] everywhere! Most of the gravitation discussion related to hollow or filled up spheres is from this book. H. Benson, University Physics, Revised Edition, 1996. Vanier College Amazing course which is somewhat similar to the one I had with Halliday and Resnick back in 1986-88. It is a great page turner with special topics everywhere. The problems in themselves are worth solving (e.g., kinematics of a basketball shot!) and many have an equation to prove that is actually displayed! My edition has 44 chapters spread across 942 pages covering everything from vectors, kinematics, inertia, particle dynamics, work and energy, conservation of energy, momentum, rotations about a fixed axis, gravitation, solids and fluids, oscillations, waves and sound, temperature and the ideal gas law, thermodynamics and entropy, electrostatics, the electric field, pretty much everything ‘electric and magnetic’, Maxwell’s equations, light and optics, special relativity , quantum theory and wave mechanics, atoms and solids. Finally nuclear physics and elementary particles and a (1997) view of Grand Unified Theory. C. Harper, Introduction to Mathematical Physics, Prentice Hall, 1976. California State University, Haywood This is my favorite go-to reference for mathematical physics. Most of the differential equations presentation and solutions, complex variable and matrix definitions, and most of his examples and problems, &c. served as the primer for this work. Harper’s book is so concise that you can pretty much read it in about 2 weeks and the presentation is impeccable for this very readable 300 page mathematical physics volume. D.G. Zill, W. S. Wright, Advanced Engineering Mathematics, 4-th Edition, Jones & Bartlett, 2011. Loyola Mary-mount University If you are going to go into some scientific field that makes you learn engineering physics you should get this book as a reference. Besides being set in color, it is easily readable and with just enough conceptual ‘meat’ around the physical ‘bone’ to capture your attention without being to mathematical about it. I mean, when I browsed through it I wished I had used this book for my course! F. P. Beer, E. R. Johnson, W. E. Clausen, Vector Mechanics for Engineers, Volume II - Dynamics, 8-Edition, McGraw-Hill, 2007. Lehigh University, University of Connecticut, Ohio State University Great reference and a constant page turner with amazing color graphics. It is fairly advanced though but requires some study. T. Allen Jr, R.L. Ditsworth, Fluid Mechanics, McGraw Hill, 1972. College of Engineering Sciences at Arizona State University My father’s book. I first skimmed through this book when I was 14 and fell in love with it instantly. The fluid mechanics here is all theirs. W.J. Thomson, Introduction to Space Dynamics, Dover, 1986. University of California at Santa Barbara Great introduction to vectors, kinematics, dynamics and earth & satellite-related applications. Most of the mathematical treatment for kinematics, rotation transformations and satellite or ballistic dynamics here is based on this book since it is succinct and to the point.