Continuing on with my questions regarding multiverses, I have observed in most theories that at least, the physical constants between universes would have changed. So parallel universes could have different strength in their four forces and all... or their particles could either be heavier or smaller.

Thus, I ask, just what would happen if leptons are heavier or lighter than they should. I am not talking about under gravity and all, but their mass, plain and simple.

More precisely, I want to know how molecules and macroscopic structures would be altered because of the different constituent lepton masses.

PS: Let me be frank. Assume the mass of all particles have changed in such a universe. It makes sense this way. It would not if only one particle was heavier. Also, I think that the inhabitants of that universe would not feel anything out of the ordinary, so I instead ask about effects of such particles in comparison to our universe.

And also, AlexP gave the reminder that Higgs Field changing only really affects leptons, so thanks. I will think about the mass change of baryons in another question.

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    $\begingroup$ Do you mean first, second and third generations of matter? Don't worry they usually required more energy to exist otherwise current physics still applicable ;D $\endgroup$ – user6760 Jun 13 '20 at 11:02
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    $\begingroup$ The question is probably unanswerable in its current form. Read en.wikipedia.org/wiki/Fine-tuned_universe, and think about those ideas. $\endgroup$ – John Dallman Jun 13 '20 at 12:41
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    $\begingroup$ Assuming that electrons and nucleons are heavier. while the elementary charge $e$ and the quantum of action $h$ stay the same, and the rest of the physics works in the same way, then atoms will be smaller and electrons will be more strongly bound to the nuclei. This guarantees that the chemistry in that universe, if any, will be profoundly different from the chemistry in ours. But then, the assumption that nucleons and electrons are heavier while the rest of physics is unchanged is not reasonable. Unless you tell us how come that particles are heavier, the question is not answerable. $\endgroup$ – AlexP Jun 13 '20 at 13:02
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    $\begingroup$ Nobody is questioning that if the fundamental constants will be different then everything will be different. What we are saying is that the question does not even begin to provide the minimum data required for an answer. "Electrons are heavier" is not enough, not even remotely. It is important to specify why they are heavier, with specifics. (And, yes, electrons being heavier will wreak havoc on chemistry, because atomic radii will change and thus electron binding energies will change. Chemistry is entirely about atoms taking electrons from other atoms or sharing electrons with other atoms.) $\endgroup$ – AlexP Jun 13 '20 at 14:41
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    $\begingroup$ What are your expectations for an answer? "What happens?" is by definition too broad and a reason to close the question. Are you asking if the heavier leptons will affect a couple's romance next Tuesday? or if life can exist? or if an Earth-like moon's orbital radius changes? or if falling cereal from a box will break the bowl? You can't expect us to even scratch the surface of what could be different and the instance of all-answers-are-equally-correct is disallowed in the help center. What, exactly and narrowly is the scope of your expected answer? What justifies a "best answer?" $\endgroup$ – JBH Jun 14 '20 at 19:55

The key changes in molecular structure arise primarily from changes in electron orbitals and electronic and molecular energy levels.


By solving the Schrödinger equation for an electron moving under a Coulomb potential, we can show that the energy levels an electron can be in for an atom of $Z$ and nuclear mass $m_N$ satisfy $$E_n=-\frac{c^2Z^2\alpha^2m_e}{2n^2}$$ with $\alpha$ the fine structure constant and $m_e$ the mass of the electron. Electrons are quite light, so regardless of the mass of the exact element, $m_e\ll m_N$; here, we have made the approximation that the reduced mass $\mu$ of the atom is equal to the electron mass. The true dependence of $E_n$ on $m_e$ is approximately linear - close enough, at the least. Therefore, doubling $m_e$ will effectively double the energy of each energy level; halving it will cut each energy level in half.

The mass of the electron also shows up when calculating the Bohr radius $a_0$, which gives the most likely separation between the nucleus and the electron in the ground state of hydrogen. The expression is $$a_0=\frac{\hbar}{m_e c\alpha}$$ again assuming $m_e\ll m_N$. The Bohr radius sets the scale of the electron orbitals; it shows up the radial portions of the electron's wavefunctions. For example, for the wavefunction of the $n=3$ state with quantum numbers $l=0$ and $n_r=2$, we get $$R_{3,0}(r)=2\left(\frac{Z}{3a_0}\right)^{3/2}\left[1-\frac{2Zr}{3a_0}+\frac{2(Zr)^2}{27a_0^2}\right]e^{-Zr/3a_0}$$ Notice how this is really a function of the dimensionless variable $x\equiv r/a_0$, and so the Bohr radius really does characterize the size of the atom.


Let's say we have a diatomic molecule of some sort. This molecule has additional (non-electronic) energy levels, arising from vibrational and rotational states. If we argue that, to second order, the molecule vibrates like a harmonic oscillator, we can use a dimensional analysis argument to say that it should have vibrational energy levels scaling like $$E_{n_{\nu}}\sim\left(n_{\nu}+\frac{1}{2}\right)\left(\frac{m_e}{m_N}\right)^{1/2}\left(\frac{m_ee^4}{\hbar^2}\right)$$ or a dependence of $m_e^{3/2}$. Why is this the case? The size of the molecule is set largely by the size of the electron orbitals, so therefore the strength of the vibrations depends on the electron mass.

We can actually make a similar argument for the rotational energy levels. We see that they are characterized by the angular quantum number $l$ and are given by $$E_l=\frac{l(l+1)\hbar^2}{2I}$$ with $I$ the appropriate moment of inertia of the molecule, $I=\mu r_0^2$, with $\mu$ now the reduced mass of the molecule, not the atom, and $r_0$ the equilibrium separation. $\mu$ has no dependence on electron mass, but $r_0$ does, as it is on the same scale as the Bohr radius. Therefore, the electron manages to creep into formulas for transitions that at first glance appear to have nothing to do with electrons!

The van der Waals gas

The ideal gas model is effective in a number of situations, but it's not perfect - after all, it's only valid for a truly ideal gas. More realistic equations of state are required for cases outside the domain of the ideal gas law. One example is the van der Waals gas, which has the equation of states $$\left(p+\frac{a}{V^2}\right)(V-b)=k_BT$$ where $p$ is pressure, $V$ is a scaled volume, $T$ is temperature and $a$ and $b$ are the van der Waals constants (we recover the ideal gas law when we set $a=b=0$). $a$ characterizes intermolecular interactions, while $b$ accounts for the fact that gases aren't point particles - it's a volume of sorts. We can argue informally that as the Bohr radius depends on $m_e$, so too does $b$. A smaller electron mass means a larger Bohr radius and thus a larger $b$ - and therefore a greater departure from the ideal gas law.

More quantitatively, we can rearrange the van der Waals equation as a sum of an attractive interaction and a repulsive interaction: $$p=p_{\text{repulsive}}+p_{\text{attractive}}=\frac{k_BT}{V-b}-\frac{a}{V^2}$$ The same recipe shoes up again and again in other real gas models, including the Dieterici equation (which accounts for the attractive force by simply multiplying the repulsive pressure by an exponential) and the virial expansion, which writes the ratio $pV/k_BT$ as a power series in $1/V$; the coefficients depend on $b$ (and $a$).


We have the following key scaling dependences:

  • For electronic energy levels in atoms, $E_n\propto m_e$.
  • For the Bohr radius characterizing atomic scales, $a_0\propto m_e^{-1}$.
  • For the vibrational energy levels of a molecule, $E_{n_{\nu}}\propto m_e^{3/2}$.
  • For the rotational energy levels of a molecule, $E_l\propto m_e^2$ (through the $I^{-1}\propto r_0^{-2}\sim a_0^{-2}\propto m_e^2$ dependence on the Bohr radius).

Addendum: Increasing the mass of leptons

Given the Higgs field $\phi$, the mass of a particular fermion $f$ can be calculated by $$m_f=\frac{g_fv}{\sqrt{2}}$$ where $g_f$ is the Yukawa coupling constant for $f$ and $v=\langle\phi\rangle$ is known as the non-zero vacuum expectation value of the Higgs. If we want to uniformly increase the mass of all leptons, we could either increase $v$ (which is the same for all leptons) or increase each $g_f$ individually. It seems like you may want to change $v$ for simplicity's sake, although bear in mind that this will have an impact on the mass of the Higgs and the gauge bosons it gives mass to, the W and Z bosons.


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