Why does an electron's orbital contract as its relativistic speed increases? And why does this contraction make the orbital more stable?
 A: I'm assuming you are talking about the average radius of the orbital decreasing as the kinetic energy increases, and not the Lorentz contraction effect.
Lower orbitals have more kinetic energy $T$ by the virial theorem. Specifically for a $1/r$ potential V:
$$2\langle T\rangle=-\langle V\rangle.$$
(See the proof on the Wikipedia page I linked).
So if an electron is closer to the nucleus is has a more negative potential $V$ and thus a higher kinetic energy. At the same time it is more stable because $$\langle E\rangle=\langle T\rangle+\langle V\rangle=-\langle T\rangle,$$
so it is in a lower overall energy state the more kinetic energy it has.
A: The original answer by user octonium I think misses the point... I'm assuming that the question is referring to the fact that when we take relativity into account, orbitals with fast-moving electrons are smaller than we would otherwise expect from a "classical" quantum treatment, due to the relativistic increase in the electron mass:
$$m = {m_0 \over \sqrt{1 - (v/c)^2}}$$
This effect comes from the fact that as a particle's speed nears the speed of light, the more we "push" it the less its speed increases. In other words for a given force we have less acceleration. Since we want $F=ma$, this resistance we encounter is equivalent to an increased inertial mass $m$, but I don't know how to derive the exact correction factor.
There is an article on Wikipedia about applying this effect to atomic orbitals, but it is lacking both clarity and citations.
I think the story is that to account for relativity, we need to replace the
Schrödinger equation with the
Dirac equation, and that we get a better picture from the Klein-Gordon equation or Quantum Field Theory, but the Klein-Gordon equation is good enough and is a jumping-off point for computational approaches, see Desclaux 1972. I don't understand any of this and I don't know where to get ahold of the computer programs used by Desclaux and others, which should be trivial to run on modern machines.
However, solutions of the spherical coordinate version of the classical Schrödinger approximation produce electron wavefunctions in which the radial parameter $r$ is divided by the Bohr radius $$a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{m_e e^2}$$ which as we see is an inverse function of $m_e$. Thus we can consider that when the effective mass is relativistically increased by some factor, the orbital geometry shrinks by the same factor. This is similar to the reason why the nucleus of an atom is much smaller than the electron cloud, because it is much more massive. It is of course just an approximation, because as the orbital shrinks, the electron spends more time near the nucleus, and perhaps we have to account for changes in kinetic and potential engergy and so on; but we can hope that this line of reasoning gives the correct intuition. Another way to think about it is that angular momentum is quantized and classically $L=mvr$, so an increase in $m$ must be compensated by a corresponding decrease in $r$.
According to Wikipedia, the energy of an electron in the classical Schrodinger approximation is negative and inversely proportional to the square of the principal quantum number $n$, while being (roughly) proportional to $m_e$. Also, for the valence electrons of heavy metals, there is an additional effect which is that as an orbital shrinks from relativistic effects, it may become smaller than shells with lower energy. This causes it to experience less screening of nuclear charge, so it "sees" more charge in the nucleus, giving the orbital a decreased potential energy (and increased kinetic energy). These two considerations may help to explain why valence electrons become "more stable" in our atomic models when we take relativity into account.
