Nuclear Physics - Logical Inconsistencies

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:


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

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) 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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Carl W. Johnson, Theoretical Physicist, Physics Degree from Univ of Chicago