I understand that special relativity says if I hold a ruler in my own inertial-reference frame, then if a ruler moving relative to me passes me, that I will see the moving ruler as shorter than the one I am holding. Applying this principle to an atom that is moving relative to me, even to near the speed of light, it seems that the expectation value distance between the electron(s) and the nucleus decrease because of Lorentz contraction of the atom.

My question is whether the ionization energy of the electrons increase because of the contracted distance between electron and nucleus. I understand that in a stationary reference frame, the potential energy of the system becomes increasingly negative as the electron gets closer to the nucleus. However, I have also read that Coulomb's law doesn't strictly apply in moving reference frames, which is important for this question because the "motion" of the electron around the nucleus seems to be the quantum version of the classic moving reference frame where Coulomb's law is explained not to apply without modification.

Taking this idea further, if there is a lasing medium with atoms moving relative to one another but emitting from the same emission level, does the relative motion of the atoms mean that at least in theory the emitted photons have a range of frequencies due to Lorentz contraction? I've read about Doppler shift in a laser, but I'm curious if Lorentz contraction of the energy levels is a separate source of frequency spread.


3 Answers 3


This question (the 1st part), is fraught with difficulties.

It's true the electric field of a moving proton is not a $1/r^2$ Coulomb field. For a velocity $\vec v = v\hat x$, one finds (https://www.feynmanlectures.caltech.edu/II_26.html):

$$ E_x = \frac{+e}{4\pi\epsilon_0\sqrt{1-v^2}} \frac{-v^2(x-vt)/(1-v^2)}{\big[\frac{(x-vt)^2}{1-v^2}-y^2+z^2 \big]^{\frac 3 2}}$$

while the field in any transverse direction follows:

$$E_y = \frac{+e}{4\pi\epsilon_0\sqrt{1-v^2}} \frac{y}{\big[\frac{(x-vt)^2}{1-v^2}-y^2+z^2 \big]^{\frac 3 2}} $$

Moreover, there is a magnetic field associated with the current, given by:

$$ \vec B = \vec v \times \vec E$$

The electric field looks like this:

enter image description here

Given that the proton is much more massive then the electron, one could assume this field, and try to find analytic bound solutions for an fiducial charge $-e$ with mass $m_e$, using Schrödinger's equation, with a time dependent potential (is that even a thing?).

That's not exactly fair, as the proton is a degree of freedom in its own right, so you might write down the two-particle hamiltonian:

$$ H =\frac{\hat p_p^2}{2M_p} + \frac{\hat p_e^2}{2m_e} -\frac{e^2}{4\pi\epsilon_0|\vec r_p - \vec r_e|} $$

and look for bound solutions with:

$$ \frac{\langle\psi|\hat p_p|\psi\rangle}{M_p} = \frac{\langle\psi|\hat p_e|\psi\rangle}{m_e} = \vec v$$

which sounds difficult.

You might factor out the center-of-mass motion:

$$ \vec R = \frac{M_p\vec r_p + m_e \vec r_e}{M_p+m_e}$$

setting $$\frac{d\vec R}{dt} = \vec v$$

and solve in terms of:

$$ \vec r \equiv \vec r_e-\vec r_p$$

using the reduced mass:

$$ \mu = \frac 1{\frac 1 {M_p}+\frac 1 {m_e}}$$

which is equivalent to solving it in the rest frame. (Note the Schrödinger's equation is non-relativistic, so this doesn't really work, and you'd just be Lorentz boosting normal solutions anyway).

Since none of those are both satisfactory and tractable, one must adopt a simpler approach:

The mass of a hydrogen atom is:

$$ M_H = M_p + M_e - hc\frac{R_{\infty}}{1 +\frac{m_e}{M_P}} \approx M_p + M_e + R_E $$

($R_E = \frac 1 2 m_ec^2\alpha^2 \approx 13.6\,$eV is the Bohr energy).

Thus, if a hydrogen atom is moving relativistically at velocity $v$, its total energy is:

$$ E_T = \gamma M_H = \gamma M_P + \gamma m_e + \gamma R_E $$

with $\gamma = (1-v^2/c^2)^{-\frac 1 2}$.

The energy of the unbound constituents is:

$$ E_C = \gamma M_P + \gamma m_e $$

The difference is the binding energy:

$$ B.E. = E_T-E_C = \gamma R_E$$

which is greater than the "at-rest" binding energy of $R_E$.

Note, though, that it is not the lowest energy of the system. If we consider just the electron degree of freedom, that occurs for an unbound state in which the electron has velocity $-\vec v$. The total energy is then:

$$ E_{min} = \gamma M_P + m_e $$

If we consider adding the proton DoF, then it too would be moving at $-\vec v$, perhaps bound to the electron...which would be a hydrogen atom at rest.

So: yes, the binding energy is larger, but we expected that, as energy is not a Lorentz scalar. It really doesn't mean a thing.

We can't solve the moving hydrogen atom analytically, but it looks exactly like a Lorentz boosted spherically symmetric ground state hydrogen atom, which undergoes Lorentz contraction into an oblate spheroid.

There is one caveat though, if we could solve the moving hydrogen atom analytically, it would not be Lorentz contracted per se: it would just be squished flat, as that is the solution in our untransformed frame. If we then transformed that into the rest frame of the proton, we would find that it is Lorentz dilated (a lá Bell's Spaceship Paradox) into a spherically symmetric wave function.


My question is whether the ionization energy of the electrons increase because of the contracted distance between electron and nucleus.

Actually the ionization energy of the electrons decrease because of the contracted distance between electron and nucleus. As there is a shorter distance to climb out of the potential well.

Now, you may ask what if the electron climbs to the direction perpendicular to the motion? There is no length contraction in that direction. Well, then there is a repulsive magnetic force that helps. So a smaller force is needed to climb.

Now how can we check that this is correct? Well, we can't. If we send some fuel and some oxidizer to another frame, there is a loss of energy of fusion of the fuels, so the sent fuel can ionize just the normal number of atoms.


You are asking this the wrong way around. The ionization energy is an internal property of an atom and holds by definition in its rest frame. It does not make sense to ask for the ionization energy in a different reference frame. If the atom is moving with regard to let's say a light source, then it will obviously see a different frequency of the radiation due to the Doppler effect, which may affect the ability to ionize the atom, or at any rate lead to a different ionization cross section and different energy of the emitted photo-electron. A similar argument would apply for collisional ionization as the approaching particle would have a different kinetic energy. So the ionization energy never changes by definition, but only the energy of the ionizing agent in the atom's reference frame.


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