I've never been satisfied with the explanation that electrons have a “speed” and move close to the speed of light, thus incurring relativistic effects that make gold pretty and mercury wet. I learned that the electron is not moving, but is pinned to the location of the nucleus. A recent answer here explained different meanings of “moving”, and I'm thinking that is what's going on here, too.

An electron isn't really zipping around like a planet at nearly the speed of light. At its simplest, it takes energy to put it in different orbitals. I suppose that energy becomes added mass. But why is the relativistic massive object function shown, as if the electron had some orbit radius and resulting velocity? That would mean some energy goes into increased mass and some energy goes into velocity, so how does that give the same useful answer?

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    $\begingroup$ All very good questions and they were asked by the brightest minds 100 years ago and the only thing that has described (not explained) the electron behavior is quantum mechanics. $\endgroup$ Nov 17, 2014 at 3:53
  • $\begingroup$ I know the real math gets things right. I'm just wondering about the value of imagining classical movement, which is done even by people who really know the stuff (see recent issue of Scientific American). $\endgroup$
    – JDługosz
    Nov 17, 2014 at 4:20
  • $\begingroup$ I think you have to give a link or copy the paragraph that bothers you. $\endgroup$
    – anna v
    Nov 17, 2014 at 4:39
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    $\begingroup$ To be fair, there are plenty of useful ways to envision things, or more precisely, tools to aid in predicting results, that are analogies. Like the Hooke spring law. Chemist even talk about atoms swapping protons because for them everything is about charge in one way or another and it is convenient. The electrons are not really pined to the location of the nucleus either. Those orbital diagrams are probability distributions where the orbital shape is a cut-off at some reasonable probability of finding the electron within - the cloud you get if you look at millions of atoms in like conditions. $\endgroup$ Nov 17, 2014 at 7:02
  • $\begingroup$ Yes, I'll find the passage in SA later, and ask for a blow-by-blow interpretation. $\endgroup$
    – JDługosz
    Nov 18, 2014 at 0:15

2 Answers 2


Here is (a bit) of the rigourous math to talk about this phenomena. You learn this in a first upper division course in QM, so I'm going to abridge it and not prove several things factoids. I will then use this to provide you with a quantitative way to relate this to the Heisenberg Uncertainty Principle, and then apply it to your question. First I have to introduce to you the idea of the expectation value, which can be thought of as the average of the state your interested in, and it tells you the average of a large group of samples; this may also be called the ensemble average. It is represented by:


In quantum mechanics we have a probability density over $x$ which we normalize to one since the notion of probability doesn't make sense without all the probabilities summing up to one (or 100% chance of occuring).

$<1>=\int_{-\inf} ^\inf \Psi^* \Psi dx = 1$

To look at the expectation value of a state of interest

$<f(x)>=\int_{-\inf} ^\inf \Psi^* f(x) \Psi dx $.

To calculate standard deviation of this function we would need to also determine

$<f(x)^2>=\int_{-\inf} ^\inf \Psi^* f(x)^2 \Psi dx $.

And the standard deviation of f(x) can be calculated by:

$\sigma_{f(x)} = \sqrt{<f(x)^2>-<f(x)>^2}$

We can now formally state the uncertainty principle in a mathematically meaningful language. The most commonly encountered version of the uncertainty principle informally is saying that you can't know a particles position and momentum

$\sigma_x \sigma_p \geq {\hbar\over2}$

From this you can tell immediately that if the momentum was always zero then $\sigma_p = 0$, so it follows that there must be "movement" for the electon. This movmenet doesn't need to correspond to the classical idea of movement, and actually if you examine the de Broglie wavelength of an electron you will learn that your everyday electron is ineligible to have a classical counterpart.

One thing that is very convenient about this notation is kinetic energy is easy to represent, that is the kinetic energy $T$ is determined by


  • $\begingroup$ was in the process of fixing that as you commented on it $\endgroup$
    – Skyler
    Nov 18, 2014 at 6:22

First of all, the electron is not a particle that moves one way or another, it is a WAVE. Let's take the simplest case, an atom of hydrogen with the electron on the ground state. This state is bound, the electron won't leave it (unless we supply the ionization energy by bombarding the atom e.g. with an incident photon). Now, let's take a look at the Schrodinger equation. Since the ground level is spherically symmetrical the Schrodinger equation will contain only the radial part, ψ(r). We can develop ψ(r) in a superposition of functions of the type sin(kr) and cos(kr).

Do you know what sin(kr) and cos(kr) mean? We can write them as

sin(kr) = (1/2i)[exp(ikr) - exp(-ikr)],

cos(kr) = (1/2)[exp(ikr) + exp(-ikr)].

Each expression contains two spherical, travelling waves, one in the direction of the radius, the other one returning. Thus, we can't say that the electron doesn't move. But, in quantum domain the things work in a way unconceivable for a classical mind. The electron wave expands simultaneously in two directions, k and -k.

  • $\begingroup$ OK, k is the radial direction, but where's t? These fill space and don't get smaller far away from the r=0 point. $\endgroup$
    – JDługosz
    Nov 18, 2014 at 0:14

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