Making headway on a grand unifying theory of particles

Author: erikdsjostrom

Physics/src/Calculus/Calculus.lhs

@@ -208,6 +208,26 @@ left-associative, and set the precedence.
 > infixl 9 :.
 
 
+HEJ JOHAN! Vet inte vad du vill ha det här så lägger det bara mitt i ◕ ◡ ◕
+fattas några operationer eftersom jag inte visste hur du ville hantera dem.
+> instance Num FunExpr where
+>   (+) = (:+)
+>   (*) = (:*)
+>   (-) = (:-)
+>   fromInteger i = Const (fromInteger i)
+
+> instance Fractional FunExpr where
+>   (/) = (:/)
+>   fromRational r = Const (fromRational r)
+
+> instance Floating FunExpr where
+>   pi     = Const   pi
+>   exp a  = Exp  :. a
+>   log a  = Log  :. a
+>   sin a  = Sin  :. a
+>   cos a  = Cos  :. a
+>   asin a = Asin :. a
+>   acos a = Acos :. a
 
 A structure with class
 ----------------------------------------------------------------------

Physics/src/Calculus/SyntaxTree.hs

@@ -114,7 +114,6 @@ canonify (D e)                = derive e
 -- | Catch all
 canonify (Const x)            = Const x
 canonify Id                   = Id
-canonify Id                   = Id
 canonify e                    = error $ show e
 
 

Physics/src/NewtonianMechanics/SingleParticle.lhs

@@ -7,12 +7,11 @@
 > import           Test.QuickCheck
 > import           Vector.Vector as V
 
-Laws
-1: A body remains at rest or in uniform motion unless acted upon by a force.
-2: A body acted upon by a force moves in such a manner that the time rate of
-change of the momentum equals the force.
-3: If two bodies exert forces on each other, these forces are equal in
-magnitude and opposite in direction.
+Laws:
+
+- A body remains at rest or in uniform motion unless acted upon by a force.
+- A body acted upon by a force moves in such a manner that the time rate of change of the momentum equals the force.
+- If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction.
 
 > type Time    = Double
 > type Mass    = FunExpr
@@ -30,15 +29,26 @@ unit m*s^-1
 > velocity :: Particle -> VectorE
 > velocity = vmap D . pos
 
-Acceleration, derivative of velocity with respect to time
-unit m*s^-2
+Forces & Newton's second law
+------------------------------
 
-> acceleration :: Particle -> VectorE
-> acceleration = vmap D . velocity
+This law expresses the relationship between force and momentum and is
+as follows:
+\begin{equation}
+  \vec{F} = \frac{d \vec{p}}{d t} = \frac{d(m \cdot \vec{v})}{d t}
+\end{equation}
 
-Force: F = m*a
-unit kg * m * s^-1
-this is a bit weird
+The quantity $m \cdot v$ is what we mean when we say momentum. So the law
+states that the net force on a particle is equal to the rate of change of the
+momentum with respect to time. And since the definition of acceleration is $a =
+\frac{d \vec{v}}{d t}$ we can write this law in a more familiar form, namely:
+
+\begin{equation}
+  \vec{F} = m \cdot \vec{a}
+\end{equation}
+
+And thus if the particle is accelerating we can calculate the net force that
+must be acting on it, in code this would be:
 
 > force :: Particle -> VectorE
 > force p = vmap (* m) a
@@ -48,12 +58,29 @@ this is a bit weird
 
 > type Energy = FunExpr
 
+Where the acceleration of particle is found by deriving the velocity of that
+same particle with respect to $t$:
+
+> acceleration :: Particle -> VectorE
+> acceleration = vmap D . velocity
+
 TODO: Write something here
 
 > square :: VectorE -> FunExpr
 > square v = dotProd v v
 
-1/2 m * v^2
+Work and energy
+---------------------
+
+If a constant force $\vec{F}$ is applied to a particle that moves from 
+position $\vec{r_1}$ to $\vec{r_2}$ then the *work* done by the force is defined
+as the dot product of the force and the vector of displacement.
+
+\begin{equation}
+  W = \vec{F} \cdot \Delta \vec{r}
+\end{equation}
+
+where $\Delta \vec{r} = \vec{r_2} - \vec{r_1}$. 
 
 > kineticEnergy :: Particle -> Energy
 > kineticEnergy p = Const 0.5 * m * v2
@@ -67,7 +94,7 @@ total work *W* done on the particle as it moves from position $r_1$ to $r_2$ is
 equal to the change in kinetic energy $E_k$ of the particle:
 
 \begin{equation}
-  W = \delta E_K = E_{k,2} - E_{k,1} = \frac{1}{2} m (v^2_2 - v^2_1)
+  W = \Delta E_K = E_{k,2} - E_{k,1} = \frac{1}{2} m (\vec{v_2}^2 - \vec{v_1}^2)
 \end{equation}
 
 Let's codify this theorem:
@@ -77,7 +104,7 @@ Let's codify this theorem:
 >   where
 >     particle1 = P v1 m -- | Two particles with the same mass
 >     particle2 = P v2 m -- | But different position vector
->     -- | E_k,2 - E_k,1
+>     -- |          E_k,2                     - E_k,1
 >     deltaEnergy = (kineticEnergy particle2) - (kineticEnergy particle1)
 >     displacedParticle = P (v2 - v1) m
 
@@ -90,51 +117,55 @@ Let's codify this theorem:
 > dE = (kineticEnergy p2) - (kineticEnergy p1)
 > p3 = P (v2 - v1) m
 
-If a particles position is defined as a vector representing its displacement
-from some origin O, then its heigh should be x. Or maybe it should be the
-magnitude of the vector, if the gravitational force originates from O. Hmmm
-
+Law of universal gravitation
+-------------------------------------
+Newton's law of universal gravitation states that a particle attracts every
+other particle in the universe with a force that is directly proportional to
+the product of their masses, and is inversely proportional to the square
+of the distance between their centers. 
 
-This seems so weird since I don't know what the frame of reference is...
+This means that every particle with mass attracts every other particle with
+mass by a force pointing along the line intersecting both points. 
 
-> potentialEnergy :: Particle -> Energy
-> potentialEnergy p = undefined
->   where
->     m          = mass p
->     (V3 x _ _) = pos p
+There is an equation for calculating the magnitude of this force which
+states with math what we stated in words above:
 
 \begin{equation}
   F = G \frac{m_1 m_2}{r^2}
 \end{equation}
-Where *F* is the force, *m1* and *m2* are the masses of the objects interacting,
-*r* is the distance between the centers of the masses and *G* is the
-gravitational constant.
+Where *F* is the maginitude of the force, *m1* and *m2* are the masses of the
+objects interacting, *r* is the distance between the centers of the masses and
+*G* is the gravitational constant.
 
-> type Constant = FunExpr
+The gravitational constant has been finely approximated through experiments
+and we can state it out code like this:
 
+> type Constant = FunExpr
+>
 > gravConst :: Constant
 > gravConst = 6.674 * (10 ** (-11))
 
+Now we can codify the law of universal gravitation using our definition
+of particles.
+
 > lawOfUniversalGravitation :: Particle -> Particle -> FunExpr
 > lawOfUniversalGravitation p1 p2 = gravConst * ((m1 * m2) / r2)
 >   where
 >     m1 = mass p1
 >     m2 = mass p2
 >     r2 = square $ (pos p2) - (pos p1)
 
-> instance Num FunExpr where
->   (+) = (:+)
->   (*) = (:*)
->   (-) = (:-)
->   fromInteger i = Const (fromInteger i)
+If a particles position is defined as a vector representing its displacement
+from some origin O, then its heigh should be x. Or maybe it should be the
+magnitude of the vector, if the gravitational force originates from O. Hmmm
 
-> instance Fractional FunExpr where
->   (/) = (:/)
->   fromRational r = Const (fromRational r)
+This seems so weird since I don't know what the frame of reference is...
 
-> instance Floating FunExpr where
->   exp a = Exp :. a
->   log a = Exp :. a
+> potentialEnergy :: Particle -> Energy
+> potentialEnergy p = undefined
+>   where
+>     m          = mass p
+>     (V3 x _ _) = pos p
 
 TODO!!!! Fix prettyCan $ lawOfUniversalGravitation p1 p2