Re: Conjugate variables and the universe

Date: Fri, 17 Nov 1995 16:33:50 -0800
From: Paul Easton <paul@brl4.med.nyu.edu>
Reply to: quantum-d@teleport.com
To: quantum-d@teleport.com
Subject: QUANTUM-D: Re: Conjugate variables and the universe

(12 Nov 1995 : QUANTUM-D: Conjugate variables and the universe)

> Hamiltonian dynamics (which as far as I know can express all
> physical laws) is symmetric under the following transformation:
> interchange canonical position and momentum coordinates and
> reverse time. Let us postulate that this symmetry also applies
> to the macroscopic universe.
 
David Finkelstein:
(15 Nov 1995 : QUANTUM-D: Re: Conjugate variables and the universe)

> The transformation you mention is not  a symmetry of the Hamiltonian.
> Therefore there is no reason to expect it to be a symmetry of the universe.
 
Yes I certainly slipped a cog there, but I hope to be able to
fix my mistake.
 
The transformation in question is the interchange of canonical
position with canonical momentum, along with reflection of time
around the midpoint of the duration of the universe. The latter
part not only reverses the direction of time, but equilibrates
the age of the universe in the two coordinate systems.
 
Just to make things clear, there is no question that any physical
law may be reexpressed in the other coordinates, although it
might turn out not to be true, as is certainly the case for
irreversible processes. What I am saying is that any physical law
has a dual form, obtained by applying the transformation to the
equations, which would appear to be true to a hypothetical
observer who lives in the other coordinate system. This dual
form is not necessarily implied by the original, nor is it
necessarily true from our point of view, again as with irreversible
processes.
 
I dont think I agree that one would have to abandon the symmetry
for the macroscopic universe if it did not hold for the laws,
but the latter symmetry makes for a much stronger and more compelling
hypothesis and was intended.
 
The symmetry need not apply to every Hamiltonian, only to those
which do not contain an implicit reference to the direction of time.
That means not only that it must be invariant under time reversal,
but there must not be any constraints that imply position or momentum
space structure, because as I said previously structure is negentropy
and implies a particular direction in time.

Since Newtonian dynamics is my main area of expertise, I will attempt
to symmetrize the Hamiltonian for the gravitational system consisting
of the Earth and the Sun. Let us write
 
   H = sq(p1!)/2*m1 + sq(p2!)/2*m2 + a*sq(r1!)/2 + b*sq(r2!)/2 -
     G*m1*m2/( sqrt( sq(r1!-r2!) + f*sq(p1!-p2!) ) )
 
Here 1 is the earth and 2 is the sun, and "!" indicates a vector.
a, b, and f are unknown constants that are required to keep the
dimensionality consistent. I want to show that one can define
a, b, and f, using known constants, so that the equations
will differ by unobservable amounts from the standard ones.
 
   d(r1!)/dt = (grad(p1!)) H
 
leads to
 
   d(r1!)/dt = p1!/m1 - f*G*m1*m2*(p1!-p2!) / cube(D)
 
where D is the denominator in the interaction term of H.
In the center of mass frame the total momentum is 0, so
p1!-p2! = 2*p1!. The first term is the normal one, and we
need to show that the second divided by the first is small.
 
   2*f*G*m1*m1*m2 / cube(D) << 1
 
>From its position in H we see that f has dimensionality
sq(time/mass). From our limited selection of constants we may
readily choose f = sq( G/cube(c) ). Then we need
 
   2*cube(G)*m1*sq(m2) / cube( D*sq(c) ) << 1
 
Here we have exchanged m1 with m2 to make it harder. Using
 
G = 7 e-8 cube(cm)/gm*sq(sec)
Mearth = 6 e+27 gm
Msun = 2 e+33 gm
D = 1.5 e+13 cm      (The momentum part is negligible.)
c = 3 e+10 cm/sec
 
we find that the ratio is 7 e-30, which is small enough, though
if it were not that small we might try to see it.
 
   d(p1!)/dt = -(grad(r1!)) H
 
leads to
 
   d(p1!)/dt = -a*r1! + G*m1*m2*(r1!-r2!) / cube(D)
 
Doing the same for body 2, and subtracting the second equation from
the first, gives
 
   d(p1!-p2!)/dt = -(a*r1!-b*r2!) + 2*G*m1*m2*(r1!-r2!) / cube(D)
 
The first term on the right represents some small force which is 
proportional to the distance between the bodies. My best guess on
the physical meaning of this is that it has something to do with the
expansion of the universe. This implies the relevance of Hubble's
constant. Let us take a = m1/sq(T), and b  = m2/sq(T), where T is
the inverse of Hubble's constant and is on the order of the age of
universe. Then
 
   d(p1!-p2!)/dt = -(m1*r1!-m2*r2!) / sq(T) + 2*G*m1*m2*(r1!-r2!) / cube(D)
 
This does not mean the force is tied directly to the expansion
of the universe, because that would be first order in 1/T. If there
is a connection it would relate to the slowing of the expansion.
 
By definition m1*r1!+m2*r2! = 0 in center of mass coordinates, so
 
   m2*r2! = - m1*r1!   and   r1!-r2! = -r2!*(1-(m2/m1))
 
Then we can factor out m1*r1! and the absolute ratio between the first
and second terms on the rhs above is
 
   cube(D) / sq(T)*G*(m2-m1)
 
We take 
 
   T = 1 e+10 year = 3 e+17 sec
 
Then the ratio is 3 e-22.
 
The upshot is that for the gravitational force the hypothesis cannot
be disproved, but cannot be confirmed either. It might well be
informative to apply this analysis to other simple forces and
fields. I leave this as a recommended exercise for the reader.
 
I would appreciate if anyone can tell me whether or not the mR/sq(T)
force can be deduced from general relativity.