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Einstein's famous mass-energy equivalence equation is still used for calculations, but still often considered less physically meaningful since the atoms that comprise a material don't actually gain more...atoms as the object increases its velocity. Or do they?

Despite the revolution in quantum physics that shows the motion of electrons around atoms isn't, in any way, actually like a planet orbiting a star, for some reason this doesn't seem to stop actual graduate scientists from continuing to conform to this inaccurate planet concept.

Multiple chemistry professors I have encountered, as well as apparently people like this fellow continue to state that this mass-energy equivalence is responsible for effects in heavy atoms under classical descriptions of velocity, and it turns out that these assumptions are somehow accurate.

So, if it is already established that electrons don't actually orbit a nucleus in a classical manner, how exactly is it that their "speed" around a stationary nucleus (with respect to the lab frame) can exhibit these localized special relativistic effects? What exactly is velocity supposed to mean in this context?

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  • $\begingroup$ As someone who pondered this very question, I believe there is no answer at the moment: we will need to have a formulation of spacetime fully compatible with QM (aka quantum gravity) in order to get a conceptually clear picture of the situation here. $\endgroup$
    – Stéphane Rollandin
    Apr 10, 2018 at 18:38
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    $\begingroup$ @Stephane that is not true, QM is perfectly compatible with special relativity. $\endgroup$
    – Javier
    Apr 11, 2018 at 11:05
  • $\begingroup$ @Javier. I can't wait to read your answer to the OP question, notably the last part: "what exactly is velocity supposed to mean in this context". $\endgroup$
    – Stéphane Rollandin
    Apr 11, 2018 at 11:22
  • $\begingroup$ @John Joe I am guessing that your question is motivated by the common explanation for $L\cdot S$ coupling, which picture motion in classical terms. You might get better answers by editing you question. $\endgroup$
    – Bert Barrois
    Apr 13, 2018 at 12:08

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This, basically isn't going to answer the questions you asked. Instead it is going to trace the way that relativistic expressions for energy find their way into the definition of the orbitals for heavy atoms.

The thing to understand is the quantum mechanics is a Hamiltonian theory, by which we mean that it is structured like classical Hamiltonian mechanics. Now, without getting into the harry details, many classical systems are described by a Hamiltonian of the form1 $$ H = T + U \;, $$ where $H$ is the Hamiltonian, $T$ is the kinetic energy of the system and $U$ is the potential energy of the system. A second critical feature is that the Hamiltonian is regarded as a function the generalized coordinates of the system $\vec{q}$ and the generalized momenta of the system $\vec{p}$ so that: $$ H = H(\vec{q},\vec{p}) \;.$$ This means that we have to write the kinetic energy of the system in terms of it's momenta and not in terms of velocities.

Just as an example, for point masses in translational motion this is \begin{align} T_\text{classical} &= \frac{p^2}{2m} \\ T_\text{relativistic} &= \sqrt{(mc^2)^2 - (cp)^2} - mc^2 \;. \end{align} So the relativistic Hamiltonian is different from the classical one.

Now, in passing to quantum mechanics we regard the Hamiltonian as on operator and find the stationary states with respect to that operator (and the energies of those states are the eigenvalues). These stationary states are the orbitals.

Finally if you start with a different operator, you will get different stationary states. As with classical mechanics in the limit of low relative velocities the two formalisms give the same results (that is the same orbitals and energies).

However, because nature is relativistic with there is a difference the actual states are those that you get from the relativistic Hamiltonian.

1 In a relativistic framework we'd include the mass energy in there in a manner like \begin{align} H &= T + mc^2 + U \\ &= E_\text{free} + U \;, \end{align} but that doesn't affect the hand-waving that follows.

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  • $\begingroup$ This tells me about the energy but not much about how an electron has speed in a direction, just like how elastic collisions don't tell you about the angle of deflection without a force law. $\endgroup$
    – user182023
    Apr 11, 2018 at 17:43
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    $\begingroup$ It doesn't tell you how an electron has a speed because it doesn't assume or require that such a thing is well defined. But it does tell you that quantum mechanics cares about how energy changes with momentum. Like I said at the beginning, it doesn't answer the question you asked, it answers the question that makes sense in the context of quantum mechanics. $\endgroup$
    – dmckee --- ex-moderator kitten
    Apr 11, 2018 at 19:05
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For atoms and molecules and solid state in general the solutions of the non relativistic Schrodinger equation are adequate. You are correct that in the Schrodinger solution the electrons are in orbitals, and not orbits.

Nevertheless there is an approximation where the average radius for the orbitals give approximately the Bohr solutions . This justifies in using for particular models the semiclassical approximation of the Bohr model where a single momentum can be assigned to the orbital and add relativistic corrections to fit the data, as the wikipedia link you give states.

The higher the Z the higher possible energy of the the orbitals for a given $n$ occupied and the corresponding "velocity" of the electron,(see formula 3.13 and 3.22 ) so the corrections may be measurable .

BTW the confusions introduced by using the relativistic mass is reflected in your

since the atoms that comprise a material don't actually gain more...atoms as the object increases its velocity.

Particle physics is no longer using the relativistic mass formula, but uses the invariant mass concept, and the Energy Momentum vector, covariant under Lorenz trasformations.

$E^2 -p^2c^2=m_0^2c^4$

Where $E$ is the energy, $p$ is the momentum of the particle ( or system).

I was interested to see that the use of relativistic mass survives in atomic studies.

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  • $\begingroup$ The use of relativistic mass also survives in relativistic plasmas, where the plasma frequency $\omega = \sqrt{\frac{n e^2}{\epsilon_0 m}}$ becomes $\omega = \sqrt{\frac{ne^2}{\epsilon_0 \gamma m}}$, leading to things like the relativistic self-focusing of high intensity lasers. $\endgroup$
    – J. Murray
    Apr 10, 2018 at 21:37
  • $\begingroup$ Okay, so these electrons have a kind of angular momentum and potential/kinetic energy, but what does the velocity behind those terms translate too physically? It still seems like this classical mass-equivalence is still pervasive in this interpretation. $\endgroup$
    – user182023
    Apr 11, 2018 at 17:44
  • $\begingroup$ Yes, it is , and it is just an approximate modeling that seems to be successful. Now it means if you managed to measure the energy of an electron for many samples with the same boundary conditions, the average velocity would be the value given by the calculations of the Bohr type model. $\endgroup$
    – anna v
    Apr 11, 2018 at 17:51

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