^{1}

^{*}

^{2}

^{3}

^{4}

In a recent work, we calculated the magnetic field inside a free electron due to its spin, and found it to be about
*B* = 8.3 × 10
^{13} T. In the present study we calculate the spinning speed of a free electron in the current loop model. We show that spinning speed is equal to the speed of light. Therefore it is shown that if electron was not spinning the mass of electron would be zero. But since spinning is an unseparable part of an electron, we say that mass of electron is non-zero and is equal to (
*m* = 9.11 × 10
^{−28} g).

Recently we have calculated the magnetic field inside a free electron due to spinning motion [^{13} T. This field is about 8.3 × 10^{11} times bigger than the highest magnetic field obtained in today’s laboratories [^{3} times bigger than that in neutron stars (magnetars) [_{0} is zero:

m = m 0 1 − v 2 c 2 (1)

where m_{0} is the mass with zero speed. Therefore the Equation (1) can only be non-zero if and only if m_{0} is zero. Since spinning is an unseperable part of electron we may say that mass of electron is non-zero and is equal to 9.11 × 10^{−28}g.

As we said earlier the current loop model [

To calculate the quantum flux for any quantum orbit [_{c}:

Φ = ∫ B ⋅ d a = ∫ 0 T c B 2 ⋅ ( r × d r d t ) d t = ω s ω c ∫ 0 T s B 2 . ( r × d r d t ) d t = ω s ω c π R 2 B (2)

Here we distinguish spin angular frequency ω s from the cyclotron frequency ω c = e B / m c . When an electron is placed in an external magnetic field B, during the cyclotron period T_{c} it completes one turn around the cyclotron orbit, but it spins ( ω s ) times about itself (

Now we want to look at the Equation (2) in detail: To consider the spin dependence in the flux expression [

ŕ = r + R (3)

where r is the vector going from origin to the centre of mass of the electron and R is the vector going from centre of mass to this fictitious point charge ( R ≪ r ). So the vector r in Equation (3) reads:

r = r cos ( ω c t + ϑ c ) x ^ + r sin ( ω c t + ϑ c ) y ^ (4)

here ϑ c is the angle at t = 0.

Depending on the spin orientation, the vectors R ( ↑ ) and R ( ↓ ) namely for spin up and spin down electrons read:

R ( ↑ ) = R cos ( ω s t + ϑ s ) x ^ − R sin ( ω s t + ϑ s ) y ^ (5)

R ( ↓ ) = R cos ( ω s t + ϑ s ) x ^ + R sin ( ω s t + ϑ s ) y ^ (6)

where ϑ s is the angle at t = 0.

Here we distinguish the spin angular frequency ω s from the cyclotron angular frequency ω c = e B / m c . During the cyclotron period, T_{c} electron completes one turn around the cyclotron orbit, but it spins ω s / ω c times about itself and hence the fictitious point charge completes ω s / ω c loops with the area π R 2 . It can be shown that the number of turns, N = ω s / ω c is very large. We will see that ω s = 7.77 × 10 22 rad / sec and ω c = e B / m c can be made as small as possible. But if we take a huge magnetic field let us say for B = 10, T = 10^{5} G then the value of ω c becomes: ω c ≅ 1.76 × 10 12 rad / sec . Therefore we can say ω s ≫ ω c .

If we take the time derivative of (3), we find the corresponding velocities:

v ' = v + V (7)

Here v and v ' are the velocities of the electron itself and fictitious point charge (−e) with respect to the origin and V is the velocity of the fictitious point charge with respect to the centre of mass of the electron.

Next, following Saglamand Boyacioglu [

Φ ′ ( ↑ ) = ∮ B ⋅ r × d r 2 = ∮ 0 T c B 2 ⋅ ( r ′ × v ′ ) d t = ∮ 0 T c B [ ω c r 2 − ω s R 2 + c r o s s t e r m s ] d t (8)

where the cross terms contains the product of different angular frequencies like cos ω c t cos ω s t ⋯ and so on. When we take the integral of cross terms, they vanish and the Equation (8) reduces to

Φ ′ ( ↑ ) = B 2 [ ω c r 2 − ω s R 2 ] 2 π ω c = π r 2 B − ω s ω c π R 2 B (9)

As it is shown in [

π r 2 B = ( n + 1 2 ) Φ 0 = ( n + 1 2 ) h c e (10)

which is the quantum flux without considering electron spin. But the second term in Equation (9) is the contribution of spin to the total flux for spin-up electron. To calculate it we use Equation (20) of the following section:

ω s = ℏ m R 2 ≡ h 2 π m R 2 (11)

Substituting Equation (10) and Equation (11) in Equation (9) and using the relation ω c = e B / m c we find:

Φ ′ ( ↑ ) = ( n + 1 2 ) h c e − h c 2 e = n h c e = n Φ 0 ( n = 0 , 1 , 2 , 3 , ⋯ ) (12)

If we follow a similar procedure for spin-down electron the total flux for spin-down electron takes the form:

Φ ′ ( ↓ ) = π r 2 B + ω s ω c π R 2 B (13)

With a similar procedure we find:

Φ ′ ( ↓ ) = ( n + 1 2 ) h c e + h c 2 e = ( n + 1 ) h c e = ( n + 1 ) Φ 0 ( n = 0 , 1 , 2 , 3 , ⋯ ) (14)

From Equations (12) and (14) it is seen that the spin contribution to the total flux is − Φ 0 / 2 for spin-up electron and Φ 0 / 2 for spin-down electron:

Φ ( ↑ ) = − h c 2 e = − Φ 0 2 (15a)

Φ ( ↓ ) = h c 2 e = Φ 0 2 (15b)

Here the obtained net results show that the current loop model for electron spin is realy a satisfactory model.

To proceed further we write the spin magnetic moment μ for a free electron [

μ = − g μ B S (16)

Here ℏ S the spin angular momentum of the electron.

When we introduce the magnetic field B = B z ^ , the z-component of the magnetic moment for a spin-down electron [

μ z = e ℏ 2 m c (17)

In the current loop model z-component of the magnetic moment for a spin-down electron is:

( μ e ) z = I A c = e R 2 ω s 2 c (18)

where A = π R 2 and R is the radius of the current loop. From Equation (17) and (18) we obtain:

R = ( ℏ m ω s ) 1 / 2 (19)

If we solve ω s , from Equation (19) we find the spinning angular velocity of electron in terms of the radius of the current loop, R:

ω s = ℏ m R 2 (20)

Now with a rough approach we want to calculate spinning speed of electron in the current loop model. If we take the radius of this current loop equal to the electron radius, R ( e l ) = 2.82 × 10 − 13 cm and the mass of electron, m = 9.11 × 10 − 28 g [

ω ′ s = 9.11 × 10 25 rad / sec (21)

and writing v ( e l ) = R ( e l ) ω ′ s we find electron velocity as:

v ( e l ) = 410 × 10 10 cm / sec (22)

which is larger that the speed of light, c. This is imposible! The reason for this is that the radius of the mentioned current loop is not equal to the electron radius. Therefore the right value of the radius of this current loop must be taken in to account. We know that (please see the discussion section of [

v = R ω s = c (23)

When we solve Equations (20) and (23) together we find:

R = 3.86 × 10 − 11 cm (24)

and

ω s = 7.77 × 10 22 rad / sec (25)

So we can say that in the current loop model electron is spinning in a circular ring of radius R = 3.86 × 10 − 11 cm with the speed of light, c and with an angular velocity ω s = 7.77 × 10 22 rad / sec . Furthermore, since v = c , according to the relativity theory [_{0} is zero:

m = m 0 1 − v 2 c 2 (26)

That is to say; If the spinning speed is equal to the speed of light, c, the Equation (26) can only be non-zero if and only if m_{0} is zero. Since spinning is an unseperable part of electron we may say that mass of electron is non-zero and is equal to the mass, m = 9.11 × 10 − 28 g .

We have calculated the spinning speed of a free electron in the current loop model which is a correct one as it produced the magneticflux due to spin of electron as Φ e ( s ) = h c 2 e = Φ 0 / 2 .

By using the Equation (20) and R ω s = c , we were able to calculate the radius of this current loop R and cyclotron frequency, ω s of electron on this current loop. These values are: R = 3.86 × 10 − 11 cm and ω s = 7.77 × 10 22 rad / sec .

More importantly it is shown that if electron was not spinning the mass of electron would be zero. But since spinning is unseparable part of electron we say that mass of electron is non-zero and is equal to m = 9.11 × 10 − 28 g .

The authors declare no conflicts of interest regarding the publication of this paper.

Saglam, M., Bayram, B., Saglam, Z. and Gur, H. (2020) Calculation of the Spinning Speed of a Free Electron. Journal of Modern Physics, 11, 9-15. https://doi.org/10.4236/jmp.2020.111002