An electron has an internal angular momentum, called spin, which contributes to its total angular momentum [16] . The electron spin angular momentum is $S^2=s\left(s+1\right){\hslash }^2$. The allowed values of s are 0, 1/2, 1, 3/2, so on. For an electron $s=1/2$ and the magnitude of its spin angular momentum vector is

$\left|S\right|=\hslash \sqrt{s\left(s+1\right)}=\frac{\sqrt{3}}{2}\hslash$ [11] . The z-component has two possible values $\pm \frac{1}{2}\hslash$ [16] [17] .

The moment of inertia of a thin disk of mass M and radius R about an axis

through the centre and perpendicular to the thin disk is $I=\frac{1}{2}M{R}^2$ [18] . The angular momentum of the thin disc is $Iw=\frac{1}{2}M{R}^{2}w$ ( Figure 1 ), where w is the angular velocity of the thin disc [18] .

We assume that the electron has a thin, disk-like structure, as shown in Figure 1 . Let its intrinsic spin angular momentum $\hslash /2$, is equal to its angular momentum

$I{w}_s$, where $I=\frac{1}{2}{m}_0{r}_s^2$ and $w_s$ is the spin angular velocity of the electron [16] [17] [19] . Thus, the intrinsic angular momentum [20] :

$L=\frac{\hslash }{2}=\frac{1}{2}{m}_0{r}_s^2{w}_s$(6)

where the mass electron at rest is m₀ and its radius is $r_s$. Rewriting the above equation when the time unit is kshana, we obtain,

$\frac{{\hslash }^{\prime }}{2}=\frac{1}{2}{m}_0{r}_s^2{w^{\prime }}_s$(7)

where, ${\hslash }^{\prime }=\hslash /n$ J.kshana and ${w^{\prime }}_s=\pi /2$ rad/kshana and n is the number of kshana in a second. Thus,

$\frac{\hslash }{2n}=\frac{1}{2}{m}_0{r}_s^2\frac{\pi }{2}$(8)

But, from Equation (2), $r_s=2c^{\prime }/\pi$. Therefore,

$\frac{\hslash }{2n}=\frac{1}{2}{m}_0{\left(\frac{2c^{\prime }}{\pi }\right)}^2\left(\frac{\pi }{2}\right)=\frac{1}{2}{m}_0{\left(\frac{2c}{n\pi }\right)}^2\left(\frac{\pi }{2}\right)=\frac{m_0{c}^2\pi }{n^2{\pi }^2}$(9)

Since velocity of light $c^{\prime }=c/n$ m/kshana, and $m_0{c}^2=0.511$ MeV, therefore, we have,

$n=\frac{2m_0{c}^2}{\pi \hslash }=\frac{1.022\text{MeV}}{\pi \hslash }$(10)

$n=\frac{1.022\times {10}^6\times 1.60217662\times {10}^{-19}}{\pi \times 1.054571800\times {10}^{-34}}=4.94237\times {10}^{20}\text{kshana}/\text{sec}$(11)

Thus, the number of kshanas per second is 4.94237 × 10²⁰, and the value of one kshana, is, 2.02332079 × 10⁻²¹ s. The spinning period of electron T_s = 4 × 2.02332079 × 10⁻²¹ = 8.09328316 × 10⁻²¹ sec. The results obtained for the number of kshana in a second n, and the value of a kshana in a second are the same as those obtained from Equation (3) and in an article published by Wanjerkhede S. M. [10] [21] .

Table 1 shows the value of number of kshana “n”, and the value of a kshana in second for various methods adopted to find them. Equation in the first row is derived based on Maharishi Vyasa’s definition of kshana which is very generic in nature. Value of “n” depends on the radius of the circular orbit. May be a radius of the spinning electron or any other spinning fundamental particle radius or Bohr radius of the first orbit of hydrogen atom [9] . The value of “n” in the second row is when we considered the ratio of spin and orbital periods of hydrogen atom.

Again, re-writing Equation (10) using the mass-energy equivalence relation $E=m_0{c}^2=h{\nu }_{\gamma }$, where ${\nu }_{\gamma }$ is the threshold frequency of gamma radiation [20] [22] - [25] , we obtain,

$n=\frac{2h{\nu }_{\gamma }}{\pi \hslash }=4{\nu }_{\gamma },\text{since}\hslash =\frac{h}{2\pi }$(12)

${\nu }_{\gamma }=\frac{n}{4}=1.2355925\times {10}^{20}\text{Hz}$(13)

$T_{\gamma }=\frac{4}{n}=8.093283\times {10}^{-21}=T_s$(14)

where, $T_{\gamma }$ and $T_s$ are the period of gamma radiation and the spinning period of an electron, respectively. Equation (14) shows that when the gamma radiation disappears, an electron is produced which has the same spinning period [10] .

Here, it is worth to note that there is connection between the gamma ray and electron spin. According to Ohanian, H. C., “the spin of the electron may be regarded as an angular momentum generated by a circulating flow of energy in the wave field of the electron [26] ” and for Richard Gauthier “The electron is a charged photon” [27] .

The electron has finite spin angular momentum of $S=\frac{\sqrt{3}}{2}\hslash =\frac{\sqrt{3}h}{4\pi }$ [16] and this spin angular momentum is equal to $I{w}_s$ [17] [28] , where, I is the moment of inertia of the electron and is equal to $\frac{2}{3}{m}_0{r}_s^2$, whose rest mass is $m_0$ and revolve around the centre with radius $r_s$ ( Figure 2 ).

Equating the total spin angular momentum to the angular momentum of the spherical shell yields,

$S=\frac{\sqrt{3}h}{4\pi }=\frac{2}{3}{m}_0{r}_s^2{w}_s$(15)

The above equation is expressed in time unit kshana, as shown in Figure 2 . We get,

$S=\frac{\sqrt{3}{h}^{\prime }}{4\pi }=\frac{2}{3}{m}_0{r}_s^2{w^{\prime }}_s$(16)

Substituting for $h^{\prime }=h/n$, ${w^{\prime }}_s=\pi /2$, and $r_s$ from Equation (5), we have

$\frac{\sqrt{3}h}{4\pi n}=\frac{2}{3}{m}_0{r}_s^2\frac{\pi }{2}=\frac{\pi m_0}{3}{\left(\frac{1.90853806\times {10}^8}{n}\right)}^2$(17)

Now, number of kshana n in a second will be,

$n=\frac{4{\pi }^2{m}_0{\left(1.90853806\times {10}^8\right)}^2}{3\sqrt{3}h}=3.8046303796\times {10}^{20}\text{kshana}/\text{sec}$(18)

Thus, the number of kshanas in a second $n=3.804630379\times {10}^{20}$ kshana and $1\text{kshana}=2.628376\times {10}^{-21}\text{s}$.

The z-component of spin angular momentum of an electron is $\frac{1}{2}\hslash$. The relationship between velocity v and angular momentum for a spherical shell of mass m₀ and radius $r_s$ was $\frac{2}{3}{m}_{0}v{r}_s$ [28] . For a spherical shell, the moment of inertia is $\frac{2}{3}{m}_0{r}_s^2$ [16] [17] . By equating the angular momentum of a spherical shell and the z-component of the spin angular momentum, we obtain:

$S=\frac{h}{4\pi }=\frac{2}{3}{m}_0{r}_s^2{w}_s$(19)

Re-writing Equation (19) when time is in kshana, we have:

$S=\frac{h^{\prime }}{4\pi }=\frac{2}{3}{m}_0{r}_s^2{w^{\prime }}_s$(20)

Substituting for $h^{\prime }=h/n$, ${w^{\prime }}_s=\pi /2$, and $r_s$, we get,

$n=\frac{4{\pi }^2{m}_0{\left(1.90853806\times {10}^8\right)}^2}{3h}=6.589813\times {10}^{20}\text{kshana}/\text{sec}$(21)

Thus, the number of kshanas in the second $n=6.589813\times {10}^{20}$ kshana and $1\text{kshana}=1.5174937\times {10}^{-21}\text{s}$. The average value of Equations (18) and (21) for the number of kshanas in a second is $n=5.1972216895\times {10}^{20}$ kshana which is the same as that shown in Equation (11) and is in good agreement with the values found in a previously published article [10] .

Following Figure 3 shows the solid sphere spinning around the z-axis.

Again, let the spin angular momentum $\frac{\sqrt{3}h}{4\pi }$ be equal to the angular momentum $\frac{2}{5}{m}_0{r}_s^2{w}_s$, where the angular frequency is $w_s$ rad/sec, and the moment of inertia is $I=\frac{2}{5}{m}_0{r}_s^2$. Therefore,

$\frac{\sqrt{3}h}{4\pi }=\frac{2}{5}{m}_0{r}_s^2{w}_s$(22)

Now, substituting the value of $r_s$ from Equation (3) in Equation (22) in which the angular frequency w rad/sec is w' rad/kshana, and the Planck constant h J.sec is h' J.kshana, we have,

$\frac{\sqrt{3}{h}^{\prime }}{4\pi }=\frac{2}{5}{m}_0{r}_s^2\frac{\pi }{2}=\frac{\pi m_0{r}_s^2}{5}$(23)

$\frac{\sqrt{3}h}{4\pi n}=\frac{\pi m_0}{5}{\left(\frac{1.90853806\times {10}^8}{n}\right)}^2$(24)

where, $n=h/h^{\prime }$ sec/kshana. Solving the above for n, we have,

$n=\frac{4{\pi }^2{m}_0{\left(1.90853806\times {10}^8\right)}^2}{5\sqrt{3}h}$(25)

$n=\frac{4{\pi }^2{m}_0{\left(1.90853806\times {10}^8\right)}^2}{5\sqrt{3}h}=2.28277822775\times {10}^{20}\text{kshana}/\text{sec}$(26)

Thus, the number of kshanas in a second $n=2.2827782277\times {10}^{20}$ kshana, and $1\text{kshana}=4.3806270\times {10}^{-21}\text{s}$.

The Z-component of the spin angular moment is $\hslash /2$, and the resulting equation for n is:

$\frac{h^{\prime }}{4\pi }=\frac{2}{5}{m}_0{r}_s^2\frac{\pi }{2}=\frac{\pi m_0{r}_s^2}{5}$(27)

$n=\frac{4{\pi }^2{m}_0{\left(1.90853806\times {10}^8\right)}^2}{5h}=3.953887872\times {10}^{20}\text{kshana}/\text{sec}$(28)

Thus, the number of kshanas in a second $n=3.953887872\times {10}^{20}$ kshana and the value of a kshana is $1\text{kshana}=2.52915619\times {10}^{-21}\sec$. The average value of n is equal to 3.11833304985 × 10²⁰ which is much lower than the results obtained previously, where the number of kshana $n=4.942359849\times {10}^{20}$ kshana, and a kshana is 2.0233249508 × 10⁻²¹ sec (as shown in Equation (3)) [10] .

Validation of kshana in seconds is described in the following subsections.

The spin angular frequency of the electron, in a semi-classical model, is related to its rest energy by $\hslash w_s=m_0{c}^2$ and the relativistic velocity $v=c=w_s{r}_s$, where $w_s$ is the angular frequency (and c is the velocity of light with which it is spinning) of the spinning electron with a mass of $m_0$ [19] [29] . Thus, the radius of the electron is:

$r_s=\frac{\hslash }{m_{0}c}={{\lambda }^{\prime }}_c$(29)

where, ${{\lambda }^{\prime }}_c$ denotes the reduced Compton wavelength. For the simple thin circular plate model of Zhao (2017), the electron radius being $r_s={{\lambda }^{\prime }}_c$, and the spin angular momentum [19] are

$J=\frac{m_0{r}_s^2{w}_s}{2}=\frac{\hslash }{2}$(30)

$m_0{r}_s^2{w}_s=\hslash =\frac{h}{2\pi }$(31)

When time is in kshana, $w_s=\frac{\pi }{2}$ and $h^{\prime }=h/n$, where

$n=4.96240709788\times {10}^{20}$ (see Equations (3) and (11)) is the number of kshana in a second, then we have,

$m_0{r}_s^2{w^{\prime }}_s=\frac{h^{\prime }}{2\pi }$(32)

$m_0{r}_s^2\frac{\pi }{2}=\frac{h}{2n\pi }$(33)

$r_s^2=\frac{h}{n{\pi }^2{m}_0}$(34)

$r_s=\sqrt{\frac{h}{n{\pi }^2{m}_0}}$(35)

Substituting the values of $h$, $n$, and $m_0$ we have,

$r_s=\sqrt{\frac{6.626070040\times {10}^{-34}}{4.96240709788\times {10}^{20}{\pi }^2\times 9.10938356\times {10}^{-31}}}$(36)

$r_s=3.8537847098\times {10}^{-13}\text{m}$(37)

The radius of the electron is equal to the reduced Compton wavelength, which confirms the accuracy of the kshana.

In electron-positron pair generation, the threshold energy of the gamma radiation required is 1.022 MeV and for electron generation the gamma ray energy is 0.511 MeV [30] [31] . The frequency of gamma radiation is,

${\nu }_{\gamma }=\frac{0.511}{h}=1.233914\times {10}^{20}\text{cps}$(38)

where $h$ is Planck constant. The period $T_{\gamma }=\frac{1}{{\nu }_{\gamma }}=8.104293\times {10}^{-21}$ sec and wavelength ${\lambda }_{\gamma }$ are,

${\lambda }_{\gamma }=\frac{c}{{\nu }_{\gamma }}=2.42960597\times {10}^{-12}\text{m}$(39)

If $r_s$ is the radius of the spinning electron, the relativistic spinning velocity of the electron is $v^{\prime }=c^{\prime }$ m per kshana. According to Vyasa’s definition of kshana-, ( Figure 1 ), is given by the following equation:

$v^{\prime }=c^{\prime }=\frac{2\pi r_s}{T_s}=\frac{\pi r_s}{2}\text{m}/\text{kshana}$(40)

where, the electron spinning period is $T_s=4$ kshana. However, $c^{\prime }={{\nu }^{\prime }}_{\gamma }{\lambda }_{\gamma }$, assuming that the period of gamma radiation is equal to the spinning period of the electron that is, $T_{\gamma }=T_s$ (Equation (14)). The attribute of the cause was found to be present in the effect [10] [32] . Therefore, in this study, it is assumed that the period (an attribute) of the spinning electron or positron (effect) is the same as that of the photon (cause) [10] [32] . From the above equation, we obtain the following:

$c^{\prime }=\frac{{\lambda }_{\gamma }}{T_{\gamma }}=\frac{{\lambda }_{\gamma }}{4}=\frac{\pi r_s}{2}\text{or}{\lambda }_{\gamma }=2\pi r_s$(41)

Thus, the circumference $2\pi r_s={\lambda }_{\gamma }$, and the radius of the electron is $r_s={\lambda }_{\gamma }/2\pi$ [10] . Keeping the value of ${\lambda }_{\gamma }$ of Equation (39) in Equation (41), we obtain:

$r_s=\frac{{\lambda }_{\gamma }}{2\pi }=\frac{2.42960597\times {10}^{-12}}{2\pi }=3.86683799\times {10}^{-13}\text{m}$(42)

This indicates that the radius $r_s$ of the spinning electron is consistent with the Compton radius [13] . Now, if n is the number of kshanas in a second, then from the above equation, we obtain:

$n=\frac{c}{c^{\prime }}=\frac{2c}{\pi r_s}=\frac{2c2\pi }{\pi {\lambda }_{\gamma }}=\frac{4c}{{\lambda }_{\gamma }}=4.9356556\times {10}^{20}\text{kshana}$(43)

Therefore, $1\text{kshana}=1/n=2.02607329\times {10}^{-21}\sec$. Again, this shows that the value of the number of kshana n in a second and the value of a kshana in seconds are in good agreement with the values previously determined in the article [10] .

Electron spin is the particles’ intrinsic angular momentum S given by the equation

$S=\hslash \sqrt{s+1}$(44)

where, $s=n/2$, and n is any non-negative integer [11] . In hydrodynamics, the magnitude of the angular momentum of the vortex L is proportional to moment of inertia I and angular speed $\omega$ radians per second. Therefore,

$L=I\omega$(45)

The angular momentum of the vortex L is the angular momentum relative to the centre. For a single particle $I=r_s^2{m}_0$ and $\omega =c/r_s$ for circular motion [11] . Now, angular momentum is:

$L=\frac{r_s^2{m}_{0}c}{r_s}=r_s{m}_{0}c$(46)

where $r_s$ is the radius of the vortex, and m₀ is the electron rest mass. Equating above Equation (46) to $S=\hslash \sqrt{s+1}$, for $s=0$, we have,

$\hslash =r_s{m}_{0}c$(47)

When time unit is kshana, we have,

${\hslash }^{\prime }=r_s{m}_0{c}^{\prime }$(48)

Substituting the relativistic spinning velocity $c^{\prime }=\pi r_s/2$ meter/kshana in the above Equation (48), we get,

${\hslash }^{\prime }=\frac{m_0\pi r_s^2}{2}$(49)

Since ${\hslash }^{\prime }=h^{\prime }/2\pi$

$\frac{h^{\prime }}{2\pi }=\frac{m_0\pi r_s^2}{2}$(50)

$h^{\prime }=m_0{\pi }^2{r}_s^2=1.340669281507\times {10}^{-54}\text{joule}\cdot \text{kshana}$(51)

Alternatively, we find $h^{\prime }$ as shown below: The velocity of the fluid element instantaneously passing through a given point in space in the vortex with radius $r_s$ is constant in time; therefore, the circulation or the vorticity $\tau =2\pi r_{s}c$ is constant, where c is the speed of light. For electron mass $m_0$, conserved momentum is $m_0\tau$. Therefore, $2\pi rc{m}_0$ is constant, which equal to the Planck constant [11] . Thus,

$h=2\pi m_0{r}_{s}c$(52)

Rewriting the above Equation (52) when time unit is kshana, we have,

$h^{\prime }=2\pi m_0{r}_s{c}^{\prime }$(53)

From Equation (2), where c' is the relativistic spinning velocity.

$c^{\prime }=\frac{\pi r_s}{2}$(54)

Substituting this relativistic spinning velocity in the above Equation (54), we get

$h^{\prime }=2\pi m_0{r}_s\frac{\pi r_s}{2}={\pi }^2{m}_0{r}_s^2$(55)

$h^{\prime }={\pi }^2{m}_0{r}_s^2=1.340669281507\times {10}^{-54}$(56)

Now, the value of number of kshana in a second is:

$n=\frac{h}{h^{\prime }}=\frac{6.626070040\times {10}^{-34}}{1.340669281507\times {10}^{-54}}=4.94235985816\times {10}^{20}\text{kshana}$(57)

Thus, the value of n is same as previously found by Wanjerkhede S. M. as shown in Table 1 [10] [21] .