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How is this 137 arrived at ?

Any formula or method to get to this no. ?

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- #1

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How is this 137 arrived at ?

Any formula or method to get to this no. ?

- #2

samalkhaiat

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The classical radius of the electron is given by

How is this 137 arrived at ?

Any formula or method to get to this no. ?

[tex]mc^{2} = \frac{e^{2}}{r}[/tex]

Equating the angular momentum of sphere of radius r and mass m with the spin angular momentum of the electron, leads to

[tex] rmv = \frac{1}{2} \hbar[/tex]

Thus

[tex]v = \frac{1}{2} \left( \frac{c \hbar}{e^{2}}\right) . c = \frac{1}{2} (137) c = 68.5 c[/tex]

regards

sam

- #3

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What is spin?

[Ohanian H.C. – Amer. J. Phys., June 1986, v. 54, № 6, p.500]:

(According to N. Bohr) “After the short period of confusion, caused by the temporary limitation of clarity, there was common the agreement about the replacement of concrete means reached by abstract mathematical symbols. Especially this concerns the picture of rotation, which was substituted with the mathematical characteristics of the representations of a group of rotations in the three-dimensional space”. (In: Theoretical physics in the twentieth century. / ed. M. Fierz, V.F. Weisskopf. - New York: Interscience, 1960, p. 216.]).

Thus, physicists gradually began to consider a spin as the deep quantum property of electron, which has inaccessible physical explanation.

According to the predominant opinion, spin of electron or any other particle is certain mysterious internal torque of the momentum, for which it is not possible to construct the real physical picture and does not exist classical analog. Nevertheless on the basis of calculation, executed by Belinfante in 1939 [Belinfante F.J- Physica, 1939, v.6, p.887], it is possible to show that spin can be considered as the angular momentum, generated by the circulating of the energy flow in the field of electron wave. In exactly the same manner [Gordon W. - Z. Phys., 1928, v.50, p.630] the magnetic moment of electron can be considered as that created by the circulating flow of charge in the field of its wave. This approach gives intuitively attractive picture it denies “internal” nature of spin and magnetic moment, connecting them not with the internal structure of electron, but with the structure of the field of its wave. Furthermore, if we compare the calculations of the angular momentum in the Dirac and electromagnetic fields, then it becomes obvious that electron spin is completely analogous to the angular momentum of the classical circularly polarized wave.

As basic element in the Belinfante calculation is the use of the symmetrized tensor of energy-momentum. From the field theory it is well known that it is possible to construct several tensors of energy - momentum, each of which satisfies the conservation law [itex] \ partial _ \ nu T^ {\ mu \ nu} =0[/itex], and also gives the same resulting energy [itex]\ left ({\ int {T^ {00} d^3x}} \ right)[/itex] and momentum [itex]\ left ({\ int {T^ {k0} d^3x}} \ right)[/itex] as the canonical tensor of energy-momentum. All these tensors are distinguished by the terms of form [itex]\ partial _ \ alpha U^ {\ mu \ nu \ alpha}[/itex], which are antisymmetric on the last two indices [itex]\ left ({U^ {\ mu \ nu \ alpha} =-U^ {\ mu \ of alpha \ nu}} \ right)[/itex], therefore the conservation law [itex]\ partial _ \ nu \ partial _ \ alpha U^ {\ mu \ nu \ alpha} of =0[/itex] is identically carried out. Belinfante showed that with the aid of the suitable selection of term [itex]\ partial _ \ alpha U^ {\ mu \ nu \ alpha}[/itex] always it is possible to construct the symmetrized tensor of energy-momentum [itex]\ left ({T^ {\ mu \ nu} =T^ {\ nu \ mu}} \ right) [/itex]. Its advantage is the fact that the angular momentum, calculated directly from the momentum density of [itex]T^ {k0}[/itex], remains.

- #4

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I did these calculation some years ago, but I don't remember what the results were. They were not very useful calculation of course It could be that it turned out, it would make sense to think electron as two massless points rotating around each others with the speed of light. At least to the extent that if you are merely interested about the relationship of the mass, size and angular momentum.

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Anyone know what the classical circularly polarized wave looks like? Is it a standing wave, or it moves in some direction? If it moves in some direction, does the analogy with the electron spin imply that the spin's direction is locked relative to the direction of motion?electron spin is completely analogous to the angular momentum of the classical circularly polarized wave

- #6

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In other words, if the electron goes round a nucleus, is the spin locked in some fixed direction relative to the rotation around the nucleus, and any other arbitrary motion of the electron?the spin's direction is locked relative to the direction of motion?

- #7

Hans de Vries

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Attempts to link the net charge density with the magnetic moment densityIf one uses the "classical radius of the electron" and the known total angular momentum of the electron, it is easy to calculate that a point on the equator of the electron is moving at about 137 times the speed of light

are bound to fail. This doesn't mean that Maxwell's EM laws are in direct conflict

with QM however. The effective charge of the electron we see can be considered

much smaller as the bare charge. The bare charge is screened by vacuum

fluctuations of virtual particles and anti-particles.

Regards, Hans

- #8

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Your approach gives a speed at the equator half that expected in the OP.The classical radius of the electron is given by

[tex]mc^{2} = \frac{e^{2}}{r}[/tex]

Equating the angular momentum of sphere of radius r and mass m with the spin angular momentum of the electron, leads to

[tex] rmv = \frac{1}{2} \hbar[/tex]

Thus

[tex]v = \frac{1}{2} \left( \frac{c \hbar}{e^{2}}\right) . c = \frac{1}{2} (137) c = 68.5 c[/tex]

But by using classical angular momentum of a sphere as [tex] = rmv [/tex]

Doesn’t that assume all the mass is located at the equator of the sphere?

Wouldn’t assuming a hollow sphere with the mass distributed evenly across the entire surface from pole to pole require a higher speed?

Would a solid sphere with uniform mass density be different than a hollow sphere?

Is there a simple classical formula to relate the surface equator speed to the angular momentum in a solid sphere?

Verses a hollow one or one with all the mass located along the equator?

I also take it that the “137” used here is actually the inverse of the fine structure constant

[tex] alpha = \frac{1}{137.036} [/tex]

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