Hans de Vries
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It was noted that using the \beta_1= 0.855599677 radius (relative to the Compton
radius) as the cut-off radius for the muon we get as the self energy of the
remaining Electric field the value of the electron mass to within 0.038%.
However one would expect that an abrupt cut-off at a certain radius can only
give an approximation since since we do not expect something like that to happen
in Nature. Therefor we apply a more realistic gradual cut off on the 1/r potential
by subtracting a Yukawa potential:
<br /> V \ =\ \frac{q}{4\pi\epsilon r}\ <br /> \left(\frac{1}{r}-\frac{e^{-r/r_o}}{r}\right)<br />
We will see that this leads to exactly the same result as the abrupt cut-off.
The Electric field follows from differentiation:
<br /> E \ =\ \frac{dV}{dr}\ =\ \frac{q}{4\pi\epsilon}\ \left( <br /> -\frac{1}{r^2}+\frac{1}{r^2}e^{-r/r_o} +\frac{1}{r_or}e^{-r/r_o} <br /> \right)\ <br />
The Energy Density is given by:
<br /> {\cal E}_{(r)} \ =\ \epsilon E^2\ =\ \frac{q^2}{(4\pi)^2\epsilon}\ <br /> \left( -\frac{1}{r^2}+\frac{1}{r^2}e^{-r/r_o} <br /> +\frac{1}{r_or}e^{-r/r_o} \right)^2\ <br />
We integrate over r.
<br /> {\cal E} \ =\ \frac{q^2}{(4\pi)^2\epsilon}\ \int_0^\infty 4\pi r^2 <br /> \left( -\frac{1}{r^2}+\frac{1}{r^2}e^{-r/r_o} <br /> +\frac{1}{r_or}e^{-r/r_o} \right)^2 dr<br />
<br /> {\cal E} \ =\ \frac{q^2}{4\pi\epsilon}\ \int_0^\infty \left( <br /> -\frac{1}{r}+\frac{1}{r}e^{-r/r_o} +\frac{1}{r_o}e^{-r/r_o} <br /> \right)^2 dr<br />
To get the result:
<br /> {\cal E} \ =\ -\frac{q^2}{4\pi\epsilon}\ \left| \frac{1}{2r_o}\ <br /> e^{-2r/r_o}\ +\ \frac{1}{r}\left(1-e^{-r/r_o} \right)^2\ <br /> \right|_0^\infty<br />
and with,
<br /> e^x \ =\ 1 + x + \frac{1}{2}x^2 + ... <br />
Using only the linear term for the limit r goes to 0 we obtain the result:
<br /> {\cal E} \ =\ \frac{q^2}{8\pi\epsilon r_o}\ <br />
Which is equal to that of the abrupt cut-off.Regards, Hans
radius) as the cut-off radius for the muon we get as the self energy of the
remaining Electric field the value of the electron mass to within 0.038%.
However one would expect that an abrupt cut-off at a certain radius can only
give an approximation since since we do not expect something like that to happen
in Nature. Therefor we apply a more realistic gradual cut off on the 1/r potential
by subtracting a Yukawa potential:
<br /> V \ =\ \frac{q}{4\pi\epsilon r}\ <br /> \left(\frac{1}{r}-\frac{e^{-r/r_o}}{r}\right)<br />
We will see that this leads to exactly the same result as the abrupt cut-off.
The Electric field follows from differentiation:
<br /> E \ =\ \frac{dV}{dr}\ =\ \frac{q}{4\pi\epsilon}\ \left( <br /> -\frac{1}{r^2}+\frac{1}{r^2}e^{-r/r_o} +\frac{1}{r_or}e^{-r/r_o} <br /> \right)\ <br />
The Energy Density is given by:
<br /> {\cal E}_{(r)} \ =\ \epsilon E^2\ =\ \frac{q^2}{(4\pi)^2\epsilon}\ <br /> \left( -\frac{1}{r^2}+\frac{1}{r^2}e^{-r/r_o} <br /> +\frac{1}{r_or}e^{-r/r_o} \right)^2\ <br />
We integrate over r.
<br /> {\cal E} \ =\ \frac{q^2}{(4\pi)^2\epsilon}\ \int_0^\infty 4\pi r^2 <br /> \left( -\frac{1}{r^2}+\frac{1}{r^2}e^{-r/r_o} <br /> +\frac{1}{r_or}e^{-r/r_o} \right)^2 dr<br />
<br /> {\cal E} \ =\ \frac{q^2}{4\pi\epsilon}\ \int_0^\infty \left( <br /> -\frac{1}{r}+\frac{1}{r}e^{-r/r_o} +\frac{1}{r_o}e^{-r/r_o} <br /> \right)^2 dr<br />
To get the result:
<br /> {\cal E} \ =\ -\frac{q^2}{4\pi\epsilon}\ \left| \frac{1}{2r_o}\ <br /> e^{-2r/r_o}\ +\ \frac{1}{r}\left(1-e^{-r/r_o} \right)^2\ <br /> \right|_0^\infty<br />
and with,
<br /> e^x \ =\ 1 + x + \frac{1}{2}x^2 + ... <br />
Using only the linear term for the limit r goes to 0 we obtain the result:
<br /> {\cal E} \ =\ \frac{q^2}{8\pi\epsilon r_o}\ <br />
Which is equal to that of the abrupt cut-off.Regards, Hans
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