The formula is ready. Let's proceed to the comparisons. I will bring back the last graph from the previous section with some enhancements.
The 0 position represents the point where the first ether appears in the ground state. Since the reference for the horizontal shift is the first ether, setting the difference in values to 0 is acceptable. Additionally, a violet 1 position for the d orbital has been added. Blue represents the s orbital, green represents the p orbital. The blue values of the s orbital are all changes corresponding to the increase in circular ether. The first value in green, representing the p orbital, is the change from [X] to [-], and the rest are changes corresponding to the increase in circular ether. Finally, the violet values for the d orbital also show an increase in linear ether for the first value, and the rest indicate an increase in circular ether.
The increase in circular ether in the s and p orbitals seems to exhibit a pattern of initially peaking, then dropping, and rising again before converging to a focal point. However, the pattern for d is different. There could be some error. Is it not converging well to a focal point? It's too early to give up hope.
The added red line represents the change in the number of linear ether: [X], [-], [--], [---]. Clearly, the other values correspond to changes in the number of circular ether, but the d orbital aligns with the changes in linear ether. If the accumulation of circular and linear ether determines the emitted energy, circular should follow a circular pattern, and linear should follow a linear pattern. That's how addition works. It failed. The hypothesis is broken. Farewell, everyone.
Which comes first, the circle or the line?
Let's raise a brief question here. Between [-O] and [O-], which one seems correct to you? Some might wonder what difference it makes, and others might think as I do that the first one is correct. Those who think the first one is correct may base their orbital classification on the number of linear ether. For example, as the p orbital changes in [-], [-O], [-OO], they habitually place the linear part at the beginning. However, if you consider it as [O-], the meaning completely changes.
The ether in the first position is very large, while the one in the second position is relatively small. The values for circular and linear ether are different. When these three are combined, the values for [-O] and [O-] must be different if it's an addition. If[O] is 10, [-] is 11, and [O-] is 12, for instance. If linear comes first, then in [-O], circular should be 1, and if circular comes first, in the second position of [O-], linear should be 2. So, it can be said that what accumulates first is very important.
[X] => [O] => [OO] => [OOO] => [OOOO] represents the s orbital and circular ether. [X] => [-] => [--] => [---] => [----] represents the linear ether, changing as 1s => 2p => 3d => 4f. [O] => [O-] => [O--] => [O---] represents the stacking of linear ether on one circular ether, changing as 2s => 3p => 4d => 5f.
Let's rearrange the values.
The blue line remains the s orbital, showing the pattern of increasing circular ether, just like before. The red line also represents the change in the number of linear ether, as in the previous plot. The green line represents the changes in values where one circular ether is fixed and linear ether increases. In other words, the blue line shows changes in circular ether, while the others show changes in linear ether. Can you see it, humans? Behold! How beautiful is this pattern! Placing circular first seems to exhibit a better pattern. The orbital classification based on the number of linear ether, like s, p, d, may be incorrect.
A hint was caught in the specific case of helium. But can this be applied to all atoms? Let's make it. After all, I am a computer programmer.
So, I created it.
I will distinguish the classification that prioritizes linear (as in traditional orbital classification) from the classification that prioritizes circular ether and denote it as “Ether Classification.”
Hydrogen
Unfortunately, hydrogen lacks distinctiveness in this context. Firstly, hydrogen already has a very low error with the existing Rydberg formula. The second reason is that the r value I used (the height of the graph in hydrogen) is derived from the ionization energy values obtained here, which may have a slight margin of error. This is true for all atoms, but hydrogen has such a small error that even a change of 0.0001 significantly alters the graph.
However, we cannot simply overlook this. Hydrogen's r value is a crucial factor used as the ratio of the graph in many atoms. Let's make a correction. The obtained value is 1312 J, but after some trial and error, I found it to be 13.5984355 eV. I applied this value and redraw the graph.
It still lacks distinctiveness. The Rydberg formula resembles a logarithmic function that decreases in variation as you go further back. Naturally, as the graph converges backward, it becomes challenging to observe changes. To slightly widen that gap, I'll make a very small modification to the equation.
Now that it's a bit easier to see, let's adopt this formula consistently.
Helium
Helium's graph, as shown earlier in this conversation, is identical to the one using the weighted function D(x), with the added emphasis on regularities. Still, there are no clear patterns in the orbitals, while patterns can be observed in the ethers.
Lithium
The change in lithium orbitals may not have been as noticeable messed up, but the weighting with D(x) emphasizes the differences in the higher positions. Even so, a similar pattern can be observed in lithium. However, as a non-expert, I find it challenging to discern which configurations might be grouped together as the atomic number increases. Unfortunately, this limits my ability to create a more definitive resource on the topic.
Analysis of the Ground-State of Atoms beyond Lithium
The curve traced by hydrogen fits well into the Rydberg formula. Helium exhibits an upward trend in the graph, with values rising along with the peak. However, starting from lithium, the trend shifts downward, and thereafter, it becomes erratic. What could be the reason for this? Let's first examine the values of circular ether for hydrogen-like atoms and helium-like atoms. While eV is suitable for calculating energy, it lacks intuitiveness, so the values are expressed in Rydberg units
| H | He | Li | Be | B | |
|---|---|---|---|---|---|
| I | 0.7496 | 1.5152 | |||
| II | 2.9997 | 4.4777 | |||
| III | 6.7501 | 8.9412 | |||
| IV | 12.0013 | 18.7540 |
Surprisingly, the approximate values fall quite neatly: 0.75, 1.5, 3, 4.5,6.75, 9, 12, 18.75. They are all multiples of 0.75. Notably, 0.75 is also the value when x is 1 in the Rydberg formula. So, how many times is each of these values a multiple of 0.75?
Let's focus on the red values first. The values along the first diagonal are all 1, 4, 9, 16, matching Z squared, indicating the Rydberg formula expressing the atomic number's increase as Z squared. Looking at the diagonal for horizontal changes, it follows a consistent pattern of +1, +2, +3, +4. This could be understood as the shift in the vertical direction, which corresponds to the s value. The vertical changes also follow a consistent pattern of +2, +3, +4. Predicting changes in values as integer multiples of 0.75 seems plausible. This suggests that Li I could be 3, and Be I could be 4. Assuming this, looking at the vertical changes, we can observe that lithium becomes , and beryllium becomes . Another possibility is assuming the horizontal axis changes as . In this case, the vertical changes would increase by 1, like . Between the two possibilities, the first one seems more convincing. This is because the impact of [X] on the vertical shift becomes constant at 0.75. Hydrogen has no [X], making its peak 0.75. For helium, 0.75 plus the impact of one [X] results in 1.5. For lithium, it becomes 2.25 with the impact of two [X]. Let's follow Occam's razor, which suggests that the simpler explanation is more likely to be true. If illustrated, it would look like the following.
The ground state of lithium is [X]2[O]. The last [O] cannot be [X] due to the Pauli exclusion principle in the filling of multi-electron atoms. Therefore, the ground state of lithium can be considered as if the x-axis itself has shifted upward to the ground state. This demonstrates that the Rydberg formula can still be applicable even in this case.
Beryllium
Beryllium is also beautiful. Let's tell a slightly different story than lithium. Can we also shift the graph for beryllium ‘as it originally was’ like lithium? The answer is ‘possible, but challenging.’
The ground state of beryllium is [X]2[O]2. Two circular orbitals are involved. Let's assume that the vertical shift due to [X] is the same as lithium. However, what impact do the two [O] have? To infer this, we can look at the case where there are two ethers in helium and lithium.
For helium, the eV value of [X][O] is 20.6157751334, and the peak is at 24.58732518. However, [O][-] is as high as 58.311. The same pattern exists for lithium. The value for the next state after the ground state, [X]2[OO], is 3.373129, but[X][O][-] is 57.469. When two ethers are present, the values skyrocket. Therefore, understanding the vertical shift caused by two[X] and the vertical shift caused by two [O] is necessary to determine the ground state of beryllium.
Energy Distribution of Lithium To decipher beryllium, one must understand the rules of the constellations above
States beyond beryllium would require further exploration to unravel. However, what I want to emphasize here is the prediction that it will also be based on the Rydberg formula. Why? Because the graph of beryllium is also beautiful.
Sodium
Finally, let's look at sodium. As mentioned earlier, it is very challenging for non-experts to accurately classify values up to neon due to their limitations. However, looking at the approximate shape of oxygen, I expect that everything else will also show a pattern as long as the values are correctly arranged.
Interestingly, after beryllium, the atom where the pattern becomes clear is sodium. It is a friend in the same group as hydrogen on the periodic table. It is a friend with a similar structure where orbitals are initially arranged in a circular shape in the shell.
