Gardner. Differential geometry and relativity (lecture notes, web draft, 2004)(198s)_PGr_.pdf
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Differential Geometry (and Relativity) - Summer
2000
Classnotes
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Chapter 1: Introduction.
PDF.
PS.
DVI.
Section 1-1: Curves.
PDF.
PS.
DVI.
Section 1-2: Gauss Curvature.
PDF.
PS.
DVI.
Section 1-3: Surfaces in
E
3
.
PDF.
PS.
DVI.
Section 1-4: First Fundamental Form.
PDF.
PS.
DVI.
Section 1-5: Second Fundamental Form.
PDF.
PS.
DVI.
Section 1-6: The Gauss Curvature in Detail.
PDF.
PS.
DVI.
Section 1-7: Geodesics.
PDF.
PS.
DVI.
Section 1-8: The Curvature Tensor and the
Theorema Egregium
.
PDF.
PS.
DVI.
Section 1-9: Manifolds.
PDF.
PS.
DVI.
Chapter 2: Special Relativity: The Geometry of Flat Spacetime.
PDF.
PS.
DVI.
Section 2-1: Inertial Frames of Reference.
PDF.
PS.
DVI.
Section 2-2: The Michelson Morley Experiment.
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PS.
DVI.
Section 2-3: The Postulates of Relativity.
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Section 2-4: Relativity of Simltaneity.
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DVI.
Section 2-5: Coordinates.
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DVI.
Section 2-6: Invariance of the Interval.
PDF.
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DVI.
Section 2-7: The Lorentz Transformation.
PDF.
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DVI.
Section 2-8: Spacetime Diagrams.
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DVI.
Section 2-9: Lorentz Geometry.
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DVI.
Section 2-10: The Twin Paradox.
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Section 2-11: Temporal order and Causality.
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DVI.
Chapter 3: General Relativity: The Geometry of Curved Spacetime.
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DVI.
Section 3-1: The Principle of Equivalence.
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DVI.
Section 3-2: Gravity as Spacetime Curvature.
PDF.
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DVI.
Section 3-3: The Consequences of Einstein's Theory.
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DVI.
Section 3-6: Geodesics.
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DVI.
Section 3-7: The Field Equations.
PDF.
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DVI.
Section 3-8: The Schwarzschild Solution.
PDF.
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DVI.
Section 3-9: Orbits in General Relativity.
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DVI.
Section 3-10: The Bending of Light.
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DVI.
Black Holes.
PDF.
PS.
DVI.
Chapter 1. Surfaces and the
Concept of Curvature
Notation.
We shall denote the familiar three dimensional Euclidean
space (tradiationally denoted
R
3
)as
E
3
.
Recall.
The Euclidean metric on
E
3
is
x
=
(
x, y, z
)
=
x
2
+
y
2
+
z
2
.
1
1.1 Curves
Definition.
A
curve
in
E
3
is a vector valued function of the parameter
t
:
α
(
t
)=(
x
(
t
)
,y
(
t
)
,z
(
t
))
.
Note.
We assume the functions
x
(
t
),
y
(
t
), and
z
(
t
) have continuous
second derivatives.
Definition.
The
derivative vector
of curve
α
is
α
(
t
)=(
x
(
t
)
,y
(
t
)
,z
(
t
))
.
If
α
(
t
) is the position of a particle at time
t
,then
α
(
t
)isthe
velocity
vector
of the particle and
α
(
t
)isthe
acceleration vector
of the particle.
The
speed
of the particle is the scalar function
α
(
t
)
.
Note.
According to Newton’s Second Law of motion, the force acting
on a particle of mass
m
and position
α
(
t
)is
F
(
t
)=
mα
(
t
)
.
Definition.
The
length
(or
arclength
)ofthecurve
α
(
t
)for
t ∈
[
a, b
]is
S
=
b
a
α
(
t
)
dt.
Note.
If
β
(
t
)isacurvefor
t ∈
[
a, b
], then
β
can be written as a
function of arclength (which we will denote
α
(
s
)) as follows. First,
S
(
t
)=
t
a
β
(
t
)
dt
1
(that is,
S
(
t
) is an antiderivative of speed which satisfies
S
(
a
)=0).
Therefore
S
is a one to one function and
S
−
1
exists.
S
−
1
gives the time
at which the particle has travelled along
β
(
t
) a (gross) distance
s
.So
we denote this as
t
=
S
−
1
(
s
). Second, we make the substitution for
t
:
β
(
t
)=
β
(
S
−
1
(
s
))
≡ α
(
s
)
.
However, it may be algebraically impossible to calculate
t
=
S
−
1
(
s
)
(see page 11, number 5).
Recall.
If
f
is differentiable on an interval
I
and
f
is nonzero on
I
,
then
f
−
1
exists (i.e.
f
is one-to-one on
I
)on
f
(
I
)and
f
−
1
is differen-
tiable of
I
. In addition,
df
−
1
dx
x
=
f
(
a
)
1
=
df
dx
x
=
a
1
or
f
−
1
(
f
(
a
)) =
f
(
a
)
.
Note.
If
β
(
t
) is parameterized as
α
(
s
)asabove,then
β
(
t
)=
β
(
S
−
1
(
s
)) =
α
(
s
)
and
=
dβ
β
(
t
)
β
(
t
)
.
dS
−
1
dS
−
1
dα
ds
1
1
S
(
t
)
=
=
β
(
S
−
1
(
s
))
S
(
S
−
1
(
s
))
=
β
(
t
)
ds
Notice
dα
ds
=
α
(
s
) is a unit vector in the direction of the velocity vector
of
β
(
t
)
.
2
Definition.
If
α
(
s
) is a curve parameterized in terms of arclength
s
,
then the
unit tangent vector
of
α
(
s
)is
α
(
s
)=
T
(
s
). (
α
(
s
) is called a
unit speed curve
since
α
(
s
)
=1.)
Example 3 (page 6).
Consider the circular helix
β
(
t
)=(
a
cos
t, a
sin
t, bt
)
(see Figure I-3, page 6). Parameterize
β
(
t
) in terms of arclength
α
(
s
)
and calculate
T
(
s
)
.
Solution.
We have
β
(
t
)=(
−a
sin
t, a
cos
t, b
)
.
With
S
(
t
) the total
arclength travelled by a particle along the helix at time
t
,wehave
S
(
t
)=
β
(
t
)
=
√
a
2
+
b
2
.
Therefore,
S
(
t
)=
t
√
a
2
+
b
2
(taking
S
(0) = 0). Hence
s
t
=
S
−
1
(
s
)=
√
a
2
+
b
2
and
s
α
(
s
)=
β
(
t
)=
β
(
S
−
1
(
s
)) =
β
√
a
2
+
b
2
s
s
bs
√
a
2
+
b
2
√
a
2
+
b
2
√
a
2
+
b
2
a
cos
,a
sin
,
.
=
Also,
1
s
s
T
(
s
)=
α
(
s
)=
√
a
2
+
b
2
√
a
2
+
b
2
√
a
2
+
b
2
−a
sin
,a
cos
,b
.
Notice that
β
(
t
)
β
(
t
)
β
(
S
−
1
(
s
))
T
=
β
(
S
−
1
(
s
))
.
=
3
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