After playing around for a while with the dimensionless constant of
motion K = z + s*ln(1 + s*r) from Carl's problem (of a vertically
thrown object in a uniform gravitational field and a velocity-squared
resistance) I have figured out a canonical Hamiltonian (& Lagrangian)
formulation of the problem in spite of the dissipation inherent to
the problem. The trick is to use a modified functional form in the
kinetic energy term. The resulting Hamiltonian H is:
H = (1/2)*m*(v_t)^2*F(p/(m*v_t)) + m*g*y
where y is the canonical coordinate (height), p is the canonical
momentum, and the dimensionless function F(x) is defined as:
F(x) == -2*ln(cos(x*sqrt(sgn(x))))/sgn(x)
where the algebraic sign function sgn(x) == x/|x| (when x is
nonzero).
In terms of separate positive and negative values of the argument
x, the function F(x) has the explicit form:
F(x) == -2*ln(cos(x)) when x > 0
F(x) == 2*ln(cosh(x)) when x < 0
F(x) == 0 when x = 0 .
Notice that in the small |x| limit where |x| << 1 the function F(x)
boils down to : F(x) --> x^2. This means that if we take the limit
of an infinite terminal speed v_t the Hamiltonian above, the
kinetic energy term boils down to the usual Newtonian form of the
kinetic energy and the Hamiltonian becomes that for a Newtonian
object in a uniform gravitational field.
[The usual Newtonian form of the kinetic energy (i.e. (1/2)*m*v^2 )
in the Newtonian Lagrangian & Hamiltonian can be related to the
requirement that a free particle must obey Galilean invariance.
IOW the KE must be form-invariant under a Galilean transformation
up to a total time derivative. The modification of the kinetic
energy term in this formulation is manifestly *not* so invariant
under Galilean transformations, but this is to be expected because
it includes the effect of the dissipative interaction with the
background medium through which the object moves. This medium
provides an absolute rest frame for the problem and explicitly
breaks the Galilean invariance in the problem.]
The canonical relationship between our momentum p and the velocity
v = dy/dt for our problem above is given by:
p = m*v_t*arctan(v*sqrt(sgn(v))/v_t)/sqrt(sgn(v)) or equivalently
in terms of explicit real functions for the separate positive and
negative values of v & p we can write the relationship as:
p = m*v_t*arctan(v/v_t) if v > 0
p = m*v_t*arctanh(v/v_t) if v < 0
p = 0 if v = 0 .
Likewise we can write v in terms of p as:
v = (dH/dp)_partial = (1/2)*v_t*F'(p/(m*v_t))
where F'(x) == dF(x)/dx or equivalently
v = v_t*tan(p/(m*v_t)) if P > 0
v = v_t*tanh(p/(m*v_t)) if p < 0
v = 0 if p = 0 .
The equation of motion generated from this Hamiltonian is:
dp/dt = -(dH/dy)_partial = -m*g
or more explicitly,
m*(dv/dt)/(1 + v*|v|/v_t^2) = -m*g
or multiplying through by the denominator gives:
m*(dv/dt) = -m*g*(1 + v*|v|/v_t^2)
which is the usual version in terms of Newton's 2nd law.
If we write the above canonical Hamiltonian H in terms of the
velocity v (instead of the canonical variable p) we get:
H = m*g*y + (1/2)*m*(v_t^2)sgn(v)*ln(1 + v*|v|/v_t^2)
It is left as an exercise for the reader to find the corresponding
canonical form for the Lagrangian L = p*v - H in terms of the
coordinate y and the velocity v.
If we define the dimensionless height z == 2*g*y/(v_t)^2 and define
the dimensionless kinetic energy r == v^2/(v_t)^2, and define the
dimensionless Hamiltonian K == 2*H/(m*(v_t)^2) we get the
dimensionless constant of motion that I had introduced in my most
recent previous post, i.e.
K = z + s*ln(1 + s*r) where s == sgn(v).
If we define the new constant G as the exponential of K, i.e.
G = exp(K) then we have a particularly nice form for seeing the
result that Carl first noted about this problem. In this formulation
we have
G = exp(z)*(1+ r*s)^s
being a constant of motion. If we take r_i as the value of r when
the object is launched upward from the z = 0 height and take r_f as
the final value of r when the object returns to the z = 0 height we
can relate the G-value at launch to its value upon return and get:
1 + r_i = 1/(1 - r_f) .
A bit of algebra immediately yields the result: 1/r_f = 1 + 1/r_i .