## Determining the Electron Structure
## The classical electron radiusBefore going further some points need to be made about something called
the ‘classical electron radius’. This is a calculated radius based on an
assumption that the mass-energy potential of an electron is fully contained within a
certain radius [2]. It has a value of 2.82x10
## Falling through the EarthImagine you bored a hole through the Earth – straight down through its centre and out the other side. Next you dropped a stone at the entrance. The stone would fall rapidly at first, gaining in speed until it reached terminal velocity. It would then continue toward the Earth’s centre and past it. Once beyond the mid-point it would decelerate as it neared the other side. The stone would then oscillate several times until it came to rest at the Earth’s centre. Let’s assume instead that there was no air and the stone was able to
move all the way from one side of the Earth to the other and back again, much like a
pendulum. What would be its precise motion of oscillation, and how long would it take for
a full ‘swing’? Where Where
## A simple spring oscillatorTo understand what this means we’ll look at a simple spring oscillator. Above we see an object with mass Where Where And the period of oscillation will be the reciprocal: (a minus sign was inserted to make direction consistent). Here we see that K/m is equivalent to G M/R^{3}Therefore frequency will be: Inserting values for
## Falling through an electronWhat does any of this have to do with electron-positron interactions?
Well, gravity is a good analogy to the electric force because it varies in inverse
proportion to distance. If we think of an electron as spherical and assume that its charge
is uniformly distributed then its force function (exerted on a positron) would be very
similar to Earth’s gravitational field, namely: Where Where In this diagram a positron (red line) falls from a height equal to 10
electron radii. As can be seen the VDCL
(Velocity Dependant Coulombs Law) causes much
dampening during the initial fall, causing the oscillations to lie fully within the
electron (blue lines). Thus we can calculate the motion of oscillation based on the forces
within the electron and ignore the force function outside it. Meaning that frequency of oscillation will be:
## Solving the two body problemAt this point it would be tempting to plug values in and calculate a value
of Determining the force function for two spheres is going to be more complicated. As it happens, the force function will be same as the normal Coulomb force when the spheres are outside each other, i.e. not overlapping. When they are overlapping the function becomes very complex but here’s what it looks like: The x-axis represents the distance from the mid-point of the two spheres because this is the point that the spheres will be mirrored on. As can be seen the force function is ‘rounder’ than the single sphere situation and is not a straight line. Here is a comparison between the two functions: The blue line is the two-sphere force function, red is the single sphere situation (shown for comparison), and the green dotted line represents half the gradient of the red line. As can be seen the force function for when the sphere’s overlap closely approximates half the strength of the single sphere situation. Thus we can relate this back to our simple spring oscillator and find that: Meaning that frequency of oscillation will be: We know that this frequency must correspond to the one calculated from quantum mechanical and relativistic energy equations, namely: Meaning that: Where Isolating the radius term we get: Substituting the standard values for
## DiscussionThis is disappointing. We wanted a radius at least a hundred or so times
smaller than a proton. Instead we calculate one fifty times larger. It’s hard to
believe that this number reflects the true electron radius. ## Is the frequency correct?The first possibility is a question of frequency. The frequency of gamma
radiation arising from electron-positron annihilation is quoted as 1.24x10 When we made the above calculations of electron radius we were assuming that the quoted frequency value was correct. But what if it was higher? For example a frequency of 10 ^{23}
Hz would make the calculated radius 100 times smaller, or 1/20^{th} the radius of
a proton. And a frequency of 10^{26} Hz would set the radius 10000 times smaller,
or 1/2000^{th} that of a proton. Could the frequency be that high?## A radio antenna comparisonSome interesting analogies can be drawn between the high frequencies of
gamma rays and the low frequencies of radio transmission. We know that a radio
receiver’s antenna is most effective when its length matches half the wavelength of
the transmission frequency. For example a Wi-Fi wireless computer network operating on 2.4
GHz uses antennas around 6 cm in length because the wavelength of that frequency is 12.5
cm. ## Energy/frequency conversion formulaWhen determining the gamma ray frequency, we did so from its energy using
the formula
## So what is the answer?Unfortunately this lack of information regarding frequency has left us
unable to determine the size of an electron. However jumping ahead to the information
given in chapters on particle physics
and nucleon structure it should be possible to
estimate an electron’s diameter based on a proton’s. Without giving away
the story in advance, based on this I estimate the radius of an electron to be
approximately 10
## ConclusionsThe fixed-frequency radiation emitted during electron-positron
annihilation indicates that an electron has a spherical structure and its charge is
uniformly distributed within. E=ke.^{2}/r[3] http://www.wbabin.net/physics/yue.pdf There are several sources with different values, but they appear to be around 10 ^{-15}
m. |

Copyright © 2010 Bernard Burchell, all rights reserved.