Nuclear Physics
Logical Inconsistencies in Nuclear Physics
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This presentation was first placed on the Internet in
August 2007.
When a truly rigorous logical analysis is done of the highly
respected NIST data for spectral lines, there is every indication
that at least one of Newton, Coulomb and/or Planck must have made a huge
blunder! One of them has to be quite wrong about some major concepts,
on which all of Nuclear Physics has been built!
The reasoning is presented here as carefully and as strictly as is
possible.
Around 300 years ago, Newton determined the Law of Gravitation,
which is given by the following (simplified) equation:
This equation gives the Force which applies on any object due to
a gravitational attraction of two masses M and m, when they are at
a distance apart of r. G is a constant which gives the Force
in units called newtons (G is called the Gravitational Constant).
μ is commonly used to
represent the product of the central mass M and G.
Work is done, or Energy consumed, when Force is applied over a distance
(by definition). If the Force is of constant amplitude and direction,
and if the distance is considered to be a line segment, IN THE SAME
DIRECTION, then W = E = F * r. This is the traditional
definition of energy.
The quantities F and r are both VECTORS and not simply numbers.
That means that each has both an amplitude (size) and a direction.
Because of this, the above equation is not simply two numbers multiplied
by each other, but what is called a Dot Product. IF they have the exact
same direction, then it is simple multiplication. But if they are
in DIFFERENT directions, where their direction is different by an
angle that we will call θ, then
the multiplication requires the sine of that angle, as:
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If the Force is allowed to constantly change in either or both of amplitude
and direction, and if the motion is not limited to being a straight
line segment, then we have both values which are constantly varying
vector quantities. With that we need the Integral Calculus version of the
same Equation:
The f quantity represents the force which applies at any instant in
any specific location. The fact that f exists everywhere, and for all
times, means that we have a Force Field. If we specify that the
strength of that Force Field is dependent only on position
(for constant masses of both objects), then we have a Conservative
Force Field. For either gravitation or electrostatic
effects, we have the following coordinates (for the simplified case
of being in a single x-y plane, where the three-dimensional form
is similar but more complicated.)
You may have found Newton's Gravitation or Coulomb's electrostatics
presented in a much simpler form! The form above is the technically
correct form, for ANY type of motion in a gravitational or electrostatic
field. IF we assume that the affected object will move in a uniform
circular motion, then the square root of (x2 + y2)
is always the radius of that circular orbit, and this is then
easily simplified into the equation:
We can see that for a target which has unity mass or charge,
this is simply Newton's Gravitation equation with
k = μ, or similarly, it is
Coulomb's Law of electrostatics with k being the central charge.
It will later become important to know the relationship of these
two forces, so we might as well calculate them here. Let us consider
an electron and a proton, the components of a Hydrogen atom, but
for this calculation we are going to place them much farther apart,
only because the standard equations use centimeters as the basic unit
of length. For the Gravitational Force, we have Fg =
G * M * m / r2 (Newton's Law of Gravitation) and for the
Electrostatic Force we have Fe = Q * q / r2 (Coulomb's
Law for electrostatics). The r distance in both cases will be chosen as one
centimeter. F will be calculated in dynes. We know that G is 6.67 *
10-8, the mass of the proton is 1.65 * 10-24, and the
mass of the electron is 0.91 * 10-27 (grams). This calculates to
the Gravitational attraction being 1.0 * 10-58 dyne.
We know that the charge Q and q are identical in strength but opposite,
and that they are each 4.80296 * 10-10 electrostatic units (esu).
This calculates to the Electrostatic attraction being 2.307 * 10-19
dyne. We can see that the Electrostatic Force is immensely more
strong than the Gravitational, by a factor of around 2.3 * 1039.
We will soon use this fact to entirely ignore gravitational effects within
the atom.
Above, we noted that the definition of Work or Energy is the (dot)
product of the external force and the instantaneous radial distance.
For discussion reasons, we will use the simplified form here, and we then
have:
We have chosen to use r = infinity as one limit of the Integral.
This results in the denominator becoming immensely large and the
total quantity therefore disappearing to zero. We are essentially
defining a Work Potential at infinity as being zero.
When we solve this simple Integral for both limits, we get the following:
Notice that ALL Work Potentials are therefore negative.
This situation is identically true for both gravitational and
electrostatic energy fields, as both are inverse-square
dependencies.
These two fields are confirmed as being Conservative Fields.
One consequence of this is that there is Conservation of Energy,
meaning that the total of kinetic energy and potential energy
remains constant unless external energy is either provided or
removed from the system. As we chose to define the potential
energy at infinity as zero, we similarly define the kinetic
energy there, for a total of zero total energy. As an object
falls in toward the source of the force field (whether a gravitational
mass or an electric charge) the potential energy becomes negative
(as given by
) This quantity MUST then equal the (positive) kinetic energy of motion
of the target particle, which is given by the standard kinetic
energy formula of:
If we keep with the simplified view of a circular orbit, then the
velocity of motion in the circular orbit is equal to the length of
a radian times the angular velocity, or:
We therefore have two quantities which must always total zero, in order
to comply with Conservation of Energy:
We can therefore equate the two terms:
We can then cross-multiply to get 2 * k = m *
ω2r3.
If we were considering a Gravitational system, like planets orbiting
around the Sun, we know from above that k would be
This presentation was first placed on the Internet in August 2007.
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C Johnson, Theoretical Physicist, Physics Degree from Univ of Chicago