$E(\gamma) = E \left({\mathbf{H}}_{i} \right) - E \left({\mathbf{H}}_{f} \right) = - \Delta E({\mathbf{H}})$

This approach works very well, to perhaps a few parts in a billion. But experimental science marches on, and now observations of the ${\mathrm{Lyman}} \, \alpha \,$ photons are being reported with a precision of two parts in $10^{15}$. We have to consider ever more subtle possibilities just to stay apace. So for WikiMechanics, we extend analysis to considering what causes a hydrogen atom to change state, and what other effects may be impressed upon the atom in addition to emitting a photon. We account for the fine structure of hydrogen using a more detailed thermodynamic process that explicitly includes an interaction with some kind of field, $\mathscr{F}$. Thus we write

${\mathbf{H}} _{i} + { \mathscr{F} } \to {\mathbf{H}} _{f} + \gamma$

And we also consider that the force due to absorbing this field may do some work, $W$, on the atom by changing its shape. Then for $\Delta \! W \equiv W ( \mathbf{H}_{f} ) - W ( \mathbf{H}_{i} )$ we have

$E(\gamma) = - \Delta E({\mathbf{H}}) - \Delta \! W ({\mathbf{H}})$

## Atomic Transition Forces

Here is a repertoire of little kicks and torques used to get hydrogen to jump from state to state. These field quanta are generically noted by $\mathscr{F}$. There are no electrochemical quarks in these fields. Energies are typically stated in micro electron-volts. The most important characteristic for classifying these fields is their helicity, $\delta_{z} \,$. The atomic quantum numbers $\, \ell$ and $s$ are also relevant. The letter $Ł$ notes a frequently used collection of quarks that is called the Lamb quantum.

Recall that the internal energy of quark $\sf{z}$ is noted by $U ^{\sf{z}}$. Then a conjugate difference $\; \Delta \hspace{-2px} U^{\sf{Z}}$, and a conjugate mean $\tilde{U}^{\sf{Z}}$, describe the relationship between quarks and anti-quarks

$\begin{align} \Delta \hspace{-2px} U^{\sf{Z}} \equiv \frac{U^{\sf{\overline{z}}} - \, U^{\sf{z}}}{2} \end{align}$ | $\begin{align} \tilde{U}^{\sf{Z}} \equiv \frac{U^{\sf{\overline{z}}} + \, U^{\sf{z}}}{2} \end{align}$ |

Usually we assume that the internal-energies of down-quarks are small enough to be completely negligible. Then we write $\Delta \hspace{-2px} U^{\sf{D}} \! =0$ and $\tilde{U}^{\sf{D}} \! =0$. But the conjugate mean for down quarks, as found from hydrogen observations, is $\; \tilde{U}^{\sf{D}}= -27$ (µeV). Also, we generally make an assumption of conjugate symmetry so that $\Delta \hspace{-2px} U^{\sf{Z}} \! =0$ and $\tilde{U}^{\sf{Z}} \! = U^{\sf{\overline{z}}} = U^{\sf{z}}$. But again, these assumptions are not good enough for hydrogen where the differences shown in the accompanying table provide a more accurate description of fine structure in the spectrum. The energy of $\mathscr{F}$ is also a function of its spin angular momentum quantum number, $s \,$. The constant of proportionality is $\; k_{\sf{spin}} = -91$ (µeV), so the dependence is slight. But for field quanta, quarks and anti-quarks are paired, mass and charge are always absent, and small effects are noticeable. Let $N$ note the quark coefficients of $\mathscr{F}$. Then the internal energy is obtained from a sum of these small terms

$\begin{align} U ( \mathscr{F} ) \; \equiv \; s k_{\sf{spin}} \; + \; N^{\sf{D}} \hspace{1px} \tilde{U}^{\sf{D}} \; + \; \sum_{\zeta =1}^{16} N^{\zeta} \Delta \hspace{-2px} U^{\zeta} \end{align}$

## Hydrogen Transitions

WikiMechanics understands the fine structure of the hydrogen spectrum by analyzing the thermodynamic processes that lead to the production of photons. For example, the ${\mathrm{Lyman}} \hspace{2px} {\mathrm{\delta}} \,$ quartet of photons supposedly result from the four different interactions

$\begin{align} 5{\mathrm{S}} \, + \, {\mathscr{F}} ^{\large{⤸}} \, &\to 1{\mathrm{S}} + \gamma \\ \\ 5{\mathrm{P}}_{1/2} \, + \, {\mathscr{F}}^{\large{⇊}} \, &\to 1{\mathrm{S}} + \gamma \\ \\ 5{\mathrm{P}}_{3/2} \, + \, Ł \, &\to 1{\mathrm{S}} + \gamma \\ \\ 5{\mathrm{D}}_{5/2} \, + \, {\mathsf{2}} {\mathscr{F}}^{\large{↑}} \, &\to 1{\mathrm{S}} + \gamma \end{align}$

Each of these processes exhibit a change in the principal quantum number of $\Delta {\mathrm{n}} \equiv {\mathrm{n}}_{f} - {\mathrm{n}}_{i} = -4$. This gross effect characterizes the multiplet, along with a **transition mean** of ${\mathrm{\tilde{n}}} \equiv \left( {\mathrm{n}}_{f} + {\mathrm{n}}_{i} \right) /2 =3 \;$, and a **transition volume** of $\tilde{V} \equiv \left( a_{\mathsf{o}} {\mathrm{ñ}} \right) ^{3}$ where $a_{\mathsf{o}}$ is the Bohr radius. Any spin 'flips' are described by

$\begin{align} \Delta _{\mathsf{Z}} \equiv \frac{ \delta_{z} ( {\mathbf{H}} _{f} ) + \delta_{z} ( {\mathbf{H}} _{i} ) }{2} \end{align}$

And a **signed volume** is specified as $\hat{V} \equiv \left( a_{\mathsf{o}} \Delta _{\mathsf{Z}} \right)^{3}$. Changes in other atomic quantum numbers like $\Delta j$ and $\Delta \ell$ then describe smaller variations in the resulting photon. Different sorts of transitions are associated with different forces. For example ${\mathscr{F}}^{\large{⇊}}$, the 'double down' force mentioned above, also accounts for the following processes which are generically noted by $1{\mathrm{S}} \! - \! X{\mathrm{P}}_{1/2}$

$\begin{align} 3{\mathrm{P}}_{1/2} \! + {\mathscr{F}}^{\large{⇊}} \! \to \! 1{\mathrm{S}} + \gamma \end{align}$ | $\begin{align} 4{\mathrm{P}}_{1/2} \! + {\mathscr{F}}^{\large{⇊}} \! \to \! 1{\mathrm{S}} + \gamma \end{align}$ | $\begin{align} 6{\mathrm{P}}_{1/2} \! + {\mathscr{F}}^{\large{⇊}} \! \to \! 1{\mathrm{S}} + \gamma \end{align}$ | $\begin{align} 7{\mathrm{P}}_{1/2} \! + {\mathscr{F}}^{\large{⇊}} \! \to \! 1{\mathrm{S}} + \gamma \end{align}$ | … |

There are two transitions that can occur without any precipitating force, while still conserving momenta. They are $3{\mathrm{S}} \! - \! 3{\mathrm{D}}_{3/2}$ and $4{\mathrm{S}} \! - \! 4{\mathrm{D}}_{3/2}$. These are the simplest processes, but in general, quark models of photons may be much more complicated. And interactions could possibly involve endlessly more complex loops and wiggles. So to make simple models, we impose a boundary condition: Out of all possible interactions, the only transitions that we actually attend to are described by a few specific values of $\xi$, the **transition type** where

$\xi \equiv \Delta j + \Delta \ell + \Delta N^{Ⓛ}$

Recall that $N^{Ⓛ}$ notes the quantity of levo quarks in an atomic state. For hydrogen $\xi$ takes on some integer values between -8 and +5. For example, consider the following transitions which are generically noted as $1{\mathrm{S}} \! - \! {\mathit{X}}{\mathrm{S}}$

$\begin{align} 2{\mathrm{S}} + {\mathscr{F}} ^{\large{⤸}} \! \to \! 1{\mathrm{S}} + \gamma \end{align}$ | $\begin{align} 7{\mathrm{S}} + {\mathscr{F}} ^{\large{⤸}} \! \to \! 1{\mathrm{S}} + \gamma \end{align}$ | $\begin{align} 9{\mathrm{S}} + {\mathscr{F}} ^{\large{⤸}} \! \to \! 1{\mathrm{S}} + \gamma \end{align}$ | … |

Mathematically, the entire class is identified by $\xi =+4$. And here is another series, generically written as $1{\mathrm{S}} \! - \! {\mathit{X}}{\mathrm{D}}_{5/2}$ and mathematically described by $\xi =-4$

$5{\mathrm{D}}_{5/2} \! + {\mathsf{2}} {\mathscr{F}}^{\large{↑}} \! \to \! 1{\mathrm{S}} + \gamma$ | $7{\mathrm{D}}_{5/2} \! + {\mathsf{2}} {\mathscr{F}}^{\large{↑}} \! \to \! 1{\mathrm{S}} + \gamma$ | $9{\mathrm{D}}_{5/2} \! + {\mathsf{2}} {\mathscr{F}}^{\large{↑}} \! \to \! 1{\mathrm{S}} + \gamma$ | … |

All the foregoing interactions yield a photon and nothing else, no debris. They may also do some work $W$, on the atom by changing its shape. We use $\xi$ to describe $\Delta \! W$ by assessing changes to the norm of the radius vector; $\Delta \rho \equiv \left\| \, \overline{\rho}_{f} \right\| - \left\| \, \overline{\rho}_{i} \right\|$. Field quanta are $\mathsf{q \bar{q} }$ pairs that are usually presumed to have perfect conjugate symmetry. For this ideal case $\Delta \rho =0$. But in the finely-balanced mechanical system of a hydrogen atom, we notice a small change described by

$\begin{align} \Delta \rho = {\mathrm{h}} \hspace{1px} {\mathrm{ñ}}^{\hspace{1px} 3\epsilon} \! + {\mathrm{b}} \end{align}$

where $\epsilon \equiv (-1)^{\xi}$. The distances given by ${\mathrm{h}}(\xi)$ and ${\mathrm{b}}(\xi)$ characterize each transition-type. They are tiny, effects are measured in micro electronvolts as shown in the adjoining table. The change of shape is combined with the change of internal-energy caused by absorbing $\mathscr{F}$ to define the **transition energy density** as

$\begin{align} \tilde{\varrho} \equiv \frac{ U({\mathscr{F}}) + k_{\mathsf{F}} \Delta \rho }{\tilde{V}} \end{align}$

where the constant $k_{\sf{F}}$ was introduced earlier. Then the work done by $\mathscr{F}$ on $\mathbf{H}$ is given by

$\Delta \! W = \tilde{\varrho} \hspace{1px} \hat{V}$

## Photon Models

Quark coefficients of the excited states of hydrogen are known. And coefficients for the transition forces $\mathscr{F}$ are specified above. Then since quarks are indestructible we can make photon models just by adding and subtracting quark coefficients as prescribed by the following processes which are generically written as ${\mathbf{H}} _{i} + { \mathscr{F} } \to {\mathbf{H}} _{f} + \gamma$. This method automatically conserves charge, momentum, etc.

Lyman Multiplets | Balmer Multiplets | Other Multiplets |

## Photon Wavelengths

Measurements of photons report on their wavelength, which is related to their energy by $\lambda = hc/E(\gamma)$. Wavelengths may also depend on a photon's surroundings. Then the symbol $\lambda _{\sf{o}}$ is used to indicate a wavelength where environmental effects are negligible. We assume this to write

$\begin{align} \lambda_{\sf{o}} = \frac{hc}{-\Delta E - \Delta \! W } \end{align}$

Values of $E({\mathbf{H}}_{i})$ and $E({\mathbf{H}}_{f})$ are obtained from the description of atomic hydrogen. And, as shown above, shape changes are described by $\Delta \! W = \tilde{\varrho} \hspace{1px} \hat{V}$. The wavelengths calculated from these formulae are compared with experimental observations^{1}^{,}^{2}^{,}^{3} and shown below. Results marked by X are outside of experimental uncertainty.

Note that some photons contain the same quarks, but nonetheless have different wavelengths. This is because photon energies also depend on how quarks are brought together and arranged, as described by $W$. For more detail about these calculations, please see the spreadsheets in the wiki-files stored here.

Related WikiMechanics articles.