Mechanics and Relativity 2
February 2016
Appendix B1: Dot Product
The dot product or scalar product between two vector is defined to be:
~a · ~b ≡ ax bx + ay by + az bz
It gives a scalar as result. Here some properties of the Dot Product:
1. ~a · ~b = ~b · ~a
2. (~a + ~b) · ~c = ~a · ~c + ~b · ~c
3. ~a · ~a = a2x + a2y + a2z = |~a|2 = a2 = ~a2
4. ~a · ~b = |~a| · |~b| · cos θ
where θ is the angle between the 2 vectors. If the dot product is equal to 0, it means that
the vectors are perpendicular to each other.
Demonstration of property 4
Consider the dot product of the vector c≡a+b with itself, which is simply the square of length c.
Using the commutative and distributive properties, we have
c2 = (~a + ~b) · (~a + ~b) = a2 + b2 + 2~a · ~b
From the law of cosines applied to the triangle, we have
c2 = a2 + b2 − 2ab cos γ = a2 + b2 + 2~a~b cos θ
because γ = π − θ. Comparing both equations, we get that ~a · ~b = ab cos θ. The angle between two
vectors is therefore given by
~a · ~b
cos θ =
|~a| · |~b|
Appendix B2: Cross Vector Product
The cross product takes 2 vectors and produced another vector and it’s defined like
x̂ ŷ ẑ
a × b ≡ ax ay az = (ay bz − az by )x̂ − (ax bz − az bx )ŷ + (ax by − ay bx )ẑ
bx by bz
Here some properties of the Cross Product:
1
, 1. ~a × ~b = −~b × ~a
2. ~a × (~b + ~c) = ~a × ~b + ~a × ~c
3. ~c = ~a × ~b c is perpendicular to a and b. The direction of c can be determined with the
right-hand rule.
4. |~c| = |~a × ~b| = |~a| · |~b| · sin θ n̂
If ~a × ~b = 0 this means that the two vectors are paralel to each other.
Demonstration of property 3
If we make use of the fact that if the dot product of two vectors is zero, then the vectors are
perpendicular. We have
~a · (~a × ~b) = ax (ay bz − az by ) + ay (az bx − ax bz ) + az (ax by − ay bx ) = 0
Demonstration of property 4
|~a × ~b| = ab sin θ
|~a × ~b|2 = a2 b2 (1 − cos2 θ) = a2 b2 − (a · b)2
Written in terms of the components, we see that they are equal:
(ay bz − az by )2 + (az bx − ax bz )2 + (ax by − ay bx )2 = (a2x + a2y + a2z )(b2x + b2y + b2z ) − (ax bx + ay by + az bz )
Chapter 4.2: Harmonic Oscillator
For a spring
Hooke’s law: F (x) = −kx
Equation of motion:
mẍ = −kx
r
2 k 2π
ẍ + w x = 0 → w= T =
m w
Solution of the differential equation:
x(t) = Acos(wt) + Bsin(wt) = A0 cos(wt + ϕ)
For a pendulum
Equation of motion:
m(lθ̈) = −mg sinθ
r
2 g
θ̈ + w θ = 0 (sinθ ≈ θ) → w=
l
Solution of differential equation:
θ(t) = Acos(wt + ϕ)
Page 2
February 2016
Appendix B1: Dot Product
The dot product or scalar product between two vector is defined to be:
~a · ~b ≡ ax bx + ay by + az bz
It gives a scalar as result. Here some properties of the Dot Product:
1. ~a · ~b = ~b · ~a
2. (~a + ~b) · ~c = ~a · ~c + ~b · ~c
3. ~a · ~a = a2x + a2y + a2z = |~a|2 = a2 = ~a2
4. ~a · ~b = |~a| · |~b| · cos θ
where θ is the angle between the 2 vectors. If the dot product is equal to 0, it means that
the vectors are perpendicular to each other.
Demonstration of property 4
Consider the dot product of the vector c≡a+b with itself, which is simply the square of length c.
Using the commutative and distributive properties, we have
c2 = (~a + ~b) · (~a + ~b) = a2 + b2 + 2~a · ~b
From the law of cosines applied to the triangle, we have
c2 = a2 + b2 − 2ab cos γ = a2 + b2 + 2~a~b cos θ
because γ = π − θ. Comparing both equations, we get that ~a · ~b = ab cos θ. The angle between two
vectors is therefore given by
~a · ~b
cos θ =
|~a| · |~b|
Appendix B2: Cross Vector Product
The cross product takes 2 vectors and produced another vector and it’s defined like
x̂ ŷ ẑ
a × b ≡ ax ay az = (ay bz − az by )x̂ − (ax bz − az bx )ŷ + (ax by − ay bx )ẑ
bx by bz
Here some properties of the Cross Product:
1
, 1. ~a × ~b = −~b × ~a
2. ~a × (~b + ~c) = ~a × ~b + ~a × ~c
3. ~c = ~a × ~b c is perpendicular to a and b. The direction of c can be determined with the
right-hand rule.
4. |~c| = |~a × ~b| = |~a| · |~b| · sin θ n̂
If ~a × ~b = 0 this means that the two vectors are paralel to each other.
Demonstration of property 3
If we make use of the fact that if the dot product of two vectors is zero, then the vectors are
perpendicular. We have
~a · (~a × ~b) = ax (ay bz − az by ) + ay (az bx − ax bz ) + az (ax by − ay bx ) = 0
Demonstration of property 4
|~a × ~b| = ab sin θ
|~a × ~b|2 = a2 b2 (1 − cos2 θ) = a2 b2 − (a · b)2
Written in terms of the components, we see that they are equal:
(ay bz − az by )2 + (az bx − ax bz )2 + (ax by − ay bx )2 = (a2x + a2y + a2z )(b2x + b2y + b2z ) − (ax bx + ay by + az bz )
Chapter 4.2: Harmonic Oscillator
For a spring
Hooke’s law: F (x) = −kx
Equation of motion:
mẍ = −kx
r
2 k 2π
ẍ + w x = 0 → w= T =
m w
Solution of the differential equation:
x(t) = Acos(wt) + Bsin(wt) = A0 cos(wt + ϕ)
For a pendulum
Equation of motion:
m(lθ̈) = −mg sinθ
r
2 g
θ̈ + w θ = 0 (sinθ ≈ θ) → w=
l
Solution of differential equation:
θ(t) = Acos(wt + ϕ)
Page 2
