I have read this question:
This question is not about the energy levels, negative and kinetic, why it is lower (more negative) close to the nucleus. I understand those. I understand electrons have kinetic and potential energy. I understand as you go closer to the nucleus, the electron will have more kinetic and less potential (more negative) energy. I understand that electrons do not orbit in a classical way, I understand they are existing around the nucleus at a certain energy level as per QM.
This question is about the energy shift from s to d because of relativistic effects (and because of that gold absorbs blue color).
Where Johannes's answer says:
For gold (with atomic number 79 and hence a highly charged nucleus) this classical picture translates into relativistic speeds for electrons in s orbitals. As a result, a relativistic contraction applies to the s orbitals of gold, which causes their energy levels to shift closer to those of the d orbitals (which are localized away from the nucleus and classically speaking have lower speeds and therefore less affected by relativity).
I understand that relativistic effects come into play, because of the Heisenberg uncertainty principle, and not because at s orbitals electrons just move fast like classical planets. On an s energy level, electrons do have a higher probability of being closer to the nucleus, so they are constrained into a smaller space, thus we can tell their position with more precision, thus their momenta will be more uncertain, they will have more kinetic energy.
Electrons on the d orbitals (away from the nucleus) have higher energy (potential), which means (since electrons have negative EM energy) smaller kinetic energy in absolute term (less negative).
Electrons on the s orbital (closer to the nucleus) have lower (potential) energy, which means (since electrons have negative EM energy) higher kinetic energy in absolute term (more negative).
What this answer does not say, is how the s orbital's energy level will shift closer to the d orbital's energy level because of relativistic effects.
I do not understand if s orbitals have higher kinetic energy, and d orbitals lower kinetic energy, how can relativistic effects (Heisenberg uncertainty principle) shift the energy level of s orbital electrons closer to the d orbital electrons, so s orbital electron's energy level will be shifted to lower kinetic energy?
It should be the opposite, closer to the nucleus, the Heisenberg uncertainty principle should give (because of less uncertainty in the position, more uncertainty in the momentum) more kinetic energy to the electron.
I do not understand how can an s orbital electron act like a d orbital electron?
how can relativistic effects (Heisenberg uncertainty principle) shift the energy level to lower kinetic energy (from s to d) without the electron actually moving to the higher energy (d orbital) level?
How can an s orbital electron's energy level shift closer to a d orbital electron's energy level without moving to that higher energy level as per QM and change the absorption frequency?