Radii and Volumes of Electromagnetic Wave Quanta

1.1.1 As is well known, an electromagnetic wave in a vacuum is a local variable vortex electric field propagating with speed $c$ (speed of light), which creates (induces) a local variable vortex magnetic field. And, accordingly, vice versa.

Moreover, the energy density of an electromagnetic wave (we will denoted it by $\varphi_\mu$) is described by the following equivalent formulas (in the Gaussian unit system) \begin{align*} \varphi_\mu = \frac{1}{8\pi}(E_{\max}^2 + H_{\max}^2) \end{align*} or \begin{align*} \varphi_\mu = \frac{E_{\max}^2}{4\pi} = \frac{H_{\max}^2}{4\pi} \end{align*} (here: $E_{\max}$, $H_{\max}$ are the maximum values of the electric field and magnetic field strength in the electromagnetic wave).

1.1.2 In the second half of the 19th century, J. Maxwell constructed the theory of electromagnetic waves (later in the text we will also use the abbreviated name: EM waves).

1.2.1 In the process of studying light (light waves), it was established (in the same 19th century) that the light wave is transverse. This leads to the need to raise the question of the transverse size of the electromagnetic wave.

1.2.2 However, it is impossible to find the transverse size of the EM wave based on Maxwell's equations—fundamentally impossible. Therefore, the single geometric characteristic of electromagnetic waves is only the wavelength: from the time of Maxwell (and earlier) to the present day, all experimental and theoretical studies operate exclusively with the wavelength $\lambda$.

1.3.1 Further, in the present work, the cross-section of an electromagnetic wave is established. In order to establish this cross-section, its area and configuration, some initial assumptions about the physical properties of EM waves are adopted.

1.3.2 The solution of the above problem make it possible to calculate the volumes of individual quanta of electromagnetic waves (in particular, quanta of gamma radiation, X-ray radiation and the spectrum of waves of a hydrogen atom). This allows to conduct a general study of quanta of EM waves of various types.

2.1.1 Let us examine the currently accepted graph of an electromagnetic wavelength $\lambda_\gamma$, which is built in a coordinate system $x, y, z$ in accordance with Maxwell's theory (this graph is well-known, standard, and therefore is not provided). Such a graph always shows two identical sinusoids, reflecting synchronous harmonic oscillations (changes) in the values of the vectors of the electric field strength ($\vec{E}$) and the vectors of the magnetic field strength ($\vec{H}$). Synchronous changes in the values of these vectors are described in two mutually perpendicular planes:

• the plane of oscillations (let us adopt that this is the plane corresponding to the axes $x$ and $y$);
• the plane of polarization (this is the plane corresponding to the axes $x$ and $y$).

2.1.2 The study of electromagnetic waves (in particular, light waves) was started by Huygens. He assumed that light is a longitudinal wave. But (as was established later) an electromagnetic wave—in this case, light—is a transverse wave. In the three and a half centuries that have passed since Huygens' work, the physical properties of EM waves have been fully studied, but their transverse geometric dimensions are still unknown.

2.2.1 Let us establish the cross-section of an electromagnetic wave. On the $x$-axis, let us arbitrarily select a point, which we will denote by $k$. To the point $k$ corresponds to time $t_k$ (let us adopt that $t_k=t_\gamma/3$). Let us assume that an infinite plane is constructed that is perpendicular to the $x$-axis (point $k$ lies in this plane). We will call this plane as plane $k$.

2.2.2 As is well known, according to Maxwell's theory, in the plane $k$, the electric field strength (vectors $\vec{E}_{k}$) have the same absolute value and the same direction at any point selected (taking into account the following) on the specified plane. Also, the magnetic field strength (vectors $\vec{H}_{k}$) have the same magnitude and direction at the same points of the plane $k$.

2.2.3 Constructed plane $k$, perpendicular to the $x$-axis, is infinite. However, that part of the plane $\bm{k}$, in which the physical properties of an electromagnetic wave are manifested (that is, the vectors $\bm{\vec{E}_{k}}$ and $\bm{\vec{H}_{k}}$ are manifested), cannot be infinitely large. This, finite, part of the plane $\bm{k}$ let us call it a $\bm{\mu}$-area. The final $\mu$-area must have some boundaries.

2.2.4 Let us assume that the boundaries of the ${\mu}$-area are blurred, that is, they are not defined uniquely. Let us assume that at the boundaries of the $\mu$-area there is a gradual attenuation of the electric and magnetic fields, that is, there is a gradual decrease to zero of the absolute value of the vectors $\bm{\vec{E}_{k}}$ and $\bm{\vec{H}_{k}}$.

2.2.5 However, such an assumption (presumption) is incorrect, since it was stated above—in §2.2.2—that in accordance with Maxwell's theory, at any point chosen on the plane $k$ (selected within $\mu$-areas): $E_k = \text{const}$ and $H_k = \text{const}$ ($E_k$, $H_k$—absolute values of the examined vectors).

2.2.6 Consequently, the boundaries (boundary) of the $\mu$-area can only be some closed line. Inside this line exist (manifest) vectors $\vec{E}_{k}$ and $\vec{H}_{k}$—the same absolute value at any point of the $\mu$-area; outside the line: $\vec{E}_{k} = 0$, $\vec{H}_{k} = 0$. Let us call this closed line a $\bm{\mu}$-area contour line.

2.3.1 Let us indicate the fundamentally possible shapes of the $\mu$-area contour line.

• A. Rectilinear configurations: equilateral triangle (1), square (2), regular (equilateral) polygon (3).
• В. Curvilinear configurations: ellipse (4), circle (5).

Let us assume that the $x$-axis passes through the geometric centers of the listed figures.

2.3.2 Of all the fundamentally possible configurations of the examined closed line, the simplest shape—a circle is emphasized. In any other case, the following additional conditions must exist:

• the requirement that the contour of the $\mu$-area be limited by a broken straight line and not a curve (for figures $1, 2, 3$);
• the requirement that the points lying on the contour line be at different, strictly defined distances from the $x$-axis (for all four figures: $1, 2, 3, 4$);
• the requirement that the mathematical description of the contour line be built on two parameters, and not on one (two different semi-axes of the ellipse, sizes of the angles and sides of the polygon).

2.4.1 In order to completely define any electromagnetic wave, it is necessary and sufficient to specify a single characteristic—wavelength $\bm{\lambda}$ (or wave frequency $c/\lambda$). This means that the wavelength $\lambda$ and the radius of the above circle $R$ are uniquely related.

2.4.2 Let us introduce the following notations: $\lambda_\gamma$ and $R_\gamma$. Let us note that both quantities—$\lambda_\gamma$ and $R_\gamma$—have the same dimension ($\si{\m}$, $\si{\cm}$, etc.). Consequently, we arrive at the following ratio between $\lambda_\gamma$ and $R_\gamma$ \begin{align*}\tag{1} {\lambda}_{\gamma} = \eta{R}_{\gamma} \end{align*} (here $\eta$ is a dimensionless (numerical) coefficient).

2.4.3 Let us adopt the following initial premise: the length of the circle line limiting the $\bm{\mu}$-area must be equal to the wavelength $\bm{\lambda_\gamma}$.

2.4.4 This means \begin{align*}\tag{2} {\lambda}_{\gamma} = 2\pi{R}_{\gamma}\:, \end{align*} \begin{align*}\tag{3} \eta = 2\pi \end{align*} (The premise introduced above is based on the assumption that the dimensionless coefficient $\eta$ cannot be any arbitrary number).

3.1.1 According to Maxwell's theory, the energy of the electric field and the energy of the magnetic field are distributed along the entire wavelength $\bm{\lambda_\gamma}$. The cross-section of the electromagnetic wave was established in Section 2. This section is $\boldsymbol{\mu}$-area. The $\mu$-area contour line is a circle of radius $R_\gamma$ that encloses the cross-section of the wave.

3.1.2 Since \begin{align*} {\lambda}_{\gamma}= 2 \pi {R}_{\gamma}, \end{align*} the volume of one quantum of an electromagnetic wave (we will denote it by ${V}_{\gamma}$) can be represented by the following formula \begin{align*}\tag{4} {V}_{\gamma}= {(\pi {R}_{\gamma}^{2})}{\lambda}_{\gamma} ; \end{align*} then we obtain \begin{align*}\tag{5} {V}_{\gamma}=\pi {\left( \frac{\lambda_\gamma}{\eta} \right)}^{2}{\lambda}_{\gamma} = \frac{\pi}{\eta^2} \lambda^3_\gamma = 2 \pi^2 R^3_\gamma \end{align*}

3.2.1 As is well known, atomic nuclei emit (and absorb) gamma-quanta with energy in the range from ${\sim}0.01\si{\MeV}$ to ${\sim}5\si{\MeV}$. Let us choose for study, such gamma radiation whose quanta have an energy of ${\sim}0.1\si{\MeV}$.

3.2.2 As an example, let us examine the the nucleus of iridium $\ce{_{77}{Ir}^{191}}$. It emits and absorbs quanta with an energy of $0.129\si{\MeV}$ (we adopt in calculations that the energies of the emitted quantum and the absorbed quantum are approximately the same). The wavelength of the specified quantum ($\lambda_\gamma$) will be equal to $9.611 \cdot 10^{-12} \si{\m}$, and the radius of the circle limiting the $\mu$-area will be equal to \begin{align*} R_{\gamma} = 1.5301 \cdot 10^{-12}\si{\metre} \end{align*} Therefore, according to formula (5), the volume of the examined gamma quantum will be \begin{align*}\tag{6} V_\gamma = 2 \pi^2 R^3_\gamma = 7.063 \cdot 10^{-35}\si{{\m}^3} \end{align*}

3.3.1 The radius of sufficiently heavy nuclei can be calculated using the well-known approximate formula [1] \begin{align*}\tag{7} R \approxeq 1.4 \cdot {A}^{\frac{1}{3}} \cdot 10^{-15}\si{\metre}\:, \end{align*} where $A$ is the number of nucleons in the nucleus. The radius of the $\ce{_{77}{Ir}^{191}}$ nucleus—according to the specified formula—is \begin{align*}\tag{8} R \approxeq 1.4 \cdot {191}^{\frac{1}{3}} \cdot 10^{-15} = 8.064 \cdot 10^{-15}\si{\m} \end{align*} Therefore, the volume of the iridium nucleus is equal to \begin{align*}\tag{9} V \approxeq \frac{4}{3}\pi R^3=2.197 \cdot 10^{-42}\si{{\m}^3} \end{align*}

3.3.2 Let us correspond the calculated volume of the gamma quantum examined above with the calculated volume of the iridium nucleus \begin{align*} V_\gamma/V \approx 3.215 \cdot 10^{7} \end{align*} The resulting ratio leads to the question: how can the iridium $\boldsymbol{\ce{_{77}{Ir}^{191}}}$ nucleus absorb such gamma quanta if the volume of each quantum $\boldsymbol{{\sim}10^7}$ times exceeds the volume of the nucleus?

3.4.1 It is well known that the first excited level of the nucleus of the iron $\ce{_{26}{Fe}^{57}}$ isotope has an energy of $0.0144\si{\MeV}$. The transition to the unexcited state is associated with the emission of the corresponding gamma quantum by the iron nucleus (we will examine the possibility of the opposite case—absorption of the gamma quantum by the iron nucleus). The wavelength of the specified quantum ($\lambda^{*}_\gamma$) is $8.610 \cdot 10^{-11}\si{\m}$. Therefore, the radius of the circle limiting the cross-section of the quantum is \begin{align*} R^{*}_\gamma = 1.371 \cdot 10^{-11}\si{\m} \end{align*}

3.4.2 Let us indicate the calculated quantum volume of the studied gamma radiation (let us assume that the emitted quantum and the absorbed quantum must be approximately the same in their energy and size). \begin{align*} V^{*}_\gamma = 5.082 \cdot 10^{-32}{\si{\m}}^{3} \end{align*} Let us establish the radius of the iron $\ce{_{26}{Fe}^{57}}$ isotope nucleus \begin{align*} R^{*} \cong 1.4 \cdot 57 ^{\frac{1}{3}} \cdot 10 ^{-15} = 5.376 \cdot 10^{-15}\si{\m} \end{align*} Let us calculate the volume of the nucleus—in accordance with the given radius \begin{align*} V^{*} \cong 6.509 \cdot 10^{-43}\si{{\m}^3} \end{align*} The ratio between the calculated gamma quantum volume and the calculated nucleus volume is equal to \begin{align*} V^{*}_\gamma / V^{*} \approx 7.808 \cdot 10^{10} \end{align*}

4.1.1 Let us study an atom—a hydrogen atom. Let us analyze the conditions for the absorption of the following quanta by this atom: ultraviolet waves, visible light waves, and infrared waves (this range of electromagnetic waves is the spectrum of a hydrogen atom).

4.1.2 Let us examine the following well-known series of electromagnetic waves emitted and absorbed by a hydrogen atom.

1. Lyman series of EM waves. The longest wave of this series, lying in the ultraviolet part of the spectrum of a hydrogen atom, has wavelength $\lambda_{\gamma.1}$, which is equal to $1.216 \cdot 10^{-7} \si{\m}$ [2]. This wavelength corresponds to the transition of an electron—when absorbing a quantum—from the energy level $n=1$ to the energy level $n=2$.
2. Balmer series of EM waves (in the visible light part of the hydrogen atom spectrum there are four waves of this series). The longest wave is also selected from this series; its $\lambda_{\gamma.2}$ is $6.563 \cdot 10^{-7} \si{\m}$ [2] (this corresponds to the transition of an electron from the level $n=2$ to the level $n=3$).
3. Pashen series of EM waves. Let us examine the longest wave, which is in the infrared part of the above spectrum; its wavelength ${\lambda}_{\gamma.3} = 1.875 \cdot 10^{-6} \si{\m}$ [2] (this corresponds to the transition of an electron from the level $n=3$ to the level $n=4$).
4. Bracket series of EM waves. The longest wave is chosen, which is also in the infrared part of the spectrum of a hydrogen atom; its wavelength ${\lambda}_{\gamma.4} = 4.05 \cdot 10^{-6} \si{\m}$ [2] (this corresponds to the transition of an electron from the level $n=4$ to the level $n=5$).
5. Series of Pfund EM waves. The selected wave of this series, which has a length $\mathrm{{\lambda}_{\gamma.5} = 7.4 \cdot 10^{-6}\si{\m}}$, [3] also belongs to the infrared part of the considered spectrum of electromagnetic waves (this corresponds to the transition of an electron from a level from level $n=5$ to level $n=6$).

4.2.1 Volume of a hydrogen atom ($\boldsymbol{V_H}$). Let us adopt that for our approximate calculations we can use the Bohr atom model. Based on the wavelengths of the studied electromagnetic waves, let us choose the following values of $n: 2, 3, 4, 5, 6$. Let us calculate the volume of a hydrogen atom (nominal volume), corresponding to the indicated values of $n$.

4.2.2 Let us define radii of a hydrogen atom (we will denote them by $R_2, R_3, R_4, R_5, R_6$) by the well-known formula—according to the Bohr atom model \begin{align*} R_n = n^2 R_0 \:, \end{align*} where $R_0 = 5.292 \cdot 10^{-11} \si{\m}$ is the minimum possible radius of the atom. Let us write down calculated values of the radii: \begin{align*} R_2 &= 2.117 \cdot 10^{-10}\si{\m},\quad R_3 = 4.763 \cdot 10^{-10}\si{\m}, \quad R_4 = 8.467 \cdot 10^{-10}\si{\m},\\ R_5 &= 1,323 \cdot 10^{-9}\si{\m},\quad R_6 = 1,905 \cdot 10^{-9}\si{\m}. \end{align*} The calculated volumes of a hydrogen atom we will denote by $\boldsymbol{V_{H.n}}$ . According to the formula $\boldsymbol{V_H = \frac{4}{3}\pi R^3_n}$, we obtain: \begin{align*} V_{H.2} &= 3.975 \cdot 10^{-29} \si{\m^{3}}, \quad V_{H.3} = 4.528 \cdot 10^{-28} \si{\m^{3}},\quad V_{H.4} = 2.543 \cdot 10^{-27} \si{\m^{3}},\\ V_{H.5} &= 9.7 \cdot 10^{-27}\si{\m^{3}},\quad V_{H.6} = 2.896 \cdot 10^{-26} \si{\m^{3}} \:. \end{align*}

4.3.1 Let us calculate the volumes of electromagnetic wave quanta described in §4.1.2

1. Quantum volume at $\lambda_{\gamma.1} = 1.216 \cdot 10^{-7} \si{\m}$
   • a) Radius $R_{\gamma.1}$ \begin{align*} R_{\gamma.1} = \frac{\lambda_{\gamma.1}}{2\pi}=1.936 \cdot 10^{-8} \si{\m} \end{align*} ($R_{\gamma.1}$ is the radius of the circle limiting the cross-section of the quantum of the EM wave of the wavelength $\lambda_{\gamma.1}$).
   • b) Cross-section area of the quantum $S_{\gamma.1}$ \begin{align*} S_{\gamma.1}=\pi R^2_{\gamma .1}=1.177 \cdot 10^{-15} \si{\m^2} \end{align*}
   • c) Quantum volume $\boldsymbol{V_{\gamma.1}}$ \begin{align*} V_{\gamma.1} = S_{\gamma.1} \cdot \lambda_{\gamma.1} = \boldsymbol{1.431 \cdot 10^{-22} \si{{\m}^3}} \end{align*}
2. Quantum volume at $\boldsymbol{\lambda_{\gamma.2} = 6.563 \cdot 10^{-7} \si{\m}}$ \begin{align*} R_{\gamma.2} = 1.045 \cdot 10^{-7} \si{\m}; \quad S_{\gamma.2} = 3.429 \cdot 10^{-14} \si{\m^{2}}; \quad \boldsymbol{V_{\gamma.2} = 2,251 \cdot 10^{-20}\si{\m^{3}}} \end{align*}
3. Quantum volume at $\boldsymbol{\lambda_{\gamma.3} = 1,875 \cdot 10^{-6} \si{\m}}$ \begin{align*} R_{\gamma.3}= 2,986 \cdot 10^{-7} \si{\m}; \quad S_{\gamma.3} = 2,799 \cdot 10^{-13} \si{\m^{2}}; \quad \boldsymbol{V_{\gamma.3} = 5,249 \cdot 10^{-19} \si{\m^{3}}} \end{align*}
4. Quantum volume at $\boldsymbol{\lambda_{\gamma.4} = 4,05 \cdot 10^{-6} \si{\m}}$ \begin{align*} R_{\gamma.4}= 6,449 \cdot 10^{-7} \si{\m}; \quad S_{\gamma.4} = 1,306 \cdot 10^{-12} \si{\m^{2}}; \quad \boldsymbol{V_{\gamma.4} = 5,289 \cdot 10^{-18} \si{\m^{3}}} \end{align*}
5. Quantum volume at $\boldsymbol{\lambda_{\gamma.5} = 7.4 \cdot 10^{-6} \si{\m}}$ \begin{align*} R_{\gamma.5} = 1.178 \cdot 10^{-6} \si{\m}quad ; \quad S_{\gamma.5} = 4.359 \cdot 10^{-12} \si{\m^{2}}; \quad \boldsymbol{V_{\gamma.5} = 3.225 \cdot 10^{-17} \si{\m^{3}}} \end{align*}

4.3.2 We state: the ratios between the volumes of quanta of electromagnetic waves, emitted and absorbed by a hydrogen atom, and the volume of this atom, lead to the question that was formulated in §3.3.2.

5.1 The main task of this work was to establish a correspondence (or discrepancy) between the volumes of nuclei and atoms, and the volumes of those quanta of electromagnetic waves that these nuclei and atoms emit and absorb.

5.2 The iridium $\ce{_{77}{Ir}^{191}}$ isotope nucleus, the iron $\ce{_{26}{Fe}^{57}}$ isotope nucleus, and the hydrogen atom were chosen as examples.

5.3 To calculate the volumes of individual quanta of electromagnetic waves, it was necessary to establish their cross-section—area and configuration.

5.4 The study of the cross-section of the EM wave led to the need to introduce a dimensionless coefficient $\eta$, for which the value $2\pi$ was adopted.

5.5 We also established the base value of the factor $\eta$ (see appendix) \begin{align*} \eta = 2\pi \cdot 137 \end{align*}

5.6 Solving the problem of the cross-section of EM wave quanta made it possible to correspond the volumes of gamma radiation quanta of iridium $\ce{_{77}{Ir}^{191}}$ and iron $\ce{_{26}{Fe}^{57}}$ with the volumes of nuclei that these quanta emit and absorb.

5.7 A similar study was conducted for the line spectrum of emission and absorption of electromagnetic waves by a hydrogen atom.