1.1 Elementary principles

Let x(t) a particule trajectory where x is the position vector relative to some orogin and t the "absolute" time, m the constant mass.

The particule has velocity :                     v = dx/dt = D(x,t)
momentum :                 p = m v

Newton second law states that                f = dp/dt  where the vector f is the force acting on the particule. the

m is constant, and we recover familiar expression

f = m a  where a is the acceleration
a = d²x/dt²

The path of a single particule is then determined by a second-order differential equation.

The work done by the force f on a single particule is

W12 = Defint(f . v, t1, t2) = Defint(f . ds, 1, 2)

The final form illustrates that the integral is independant of how the path is parametrized.
From Newton second law, we have

W12 = m Defint(dv/dt . v dt, t1, t2) = m/2 Defint(d/dt(v²) dt, t1, t2)
where v =|v| = sqrt(v²)

It folows that the work done is equal to the change of kinetic energy

T = 1/2 m v²

In the case where the forve is independent of the path from point 1 to point 2 the forve is said to be conservative, and can be write as the gradient of a potential

f = - nabla(V)

for conservative forces the work also evaluates to

W12 = - Defint(ds . nabla(V) = V1 - V2

and the total energy                                   E = T + V is conserved

1.2  Angular momentum

L = x ^ p                 L a bivector
If we defferentiate L we obtain

dL/dt = v ^ (m v) + x ^ (m a) = x ^ f

We fedine the torque N about the origin as        N = x ^ f

so that the torque and angular momentum are related by

dL/dt = N

N is a bivector. Both L and N depend on the origin. Applyng two torques is found by adding the respective bivectors

x
The angular momentum bivewctor can be written in an alternative way by defining first r = |x| and writing

x = r x
We therefore have

dx/dt = d/dt(r x) = dr/dt x  + r dx/dt

So that                                                      L = m x ^ (dr/dt x + r d/dt(x) = m r x  ^ (dr/dt x + r d/dt(x)
L = m r² x  ^ d/dt(x)   (1)
But since x² = 1   we have
0 = d/dt(x²) = é x . d/dt(x)

Wecan therefore eliminate the outer product in equation (1) and write

L = m r² x  d/dt(x) = - m r² d:dt(x) x
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