What happens to the electrons when metal gets cut or broken?

Question

For example, if I take an iron rod (or thread) and cut it in half, could it happen that one side remains with an extra electron or would it balance out too fast? If possible, do the parts simply remain charged or does something else happen?

Answer 1

The electrons in a metal can be described surprisingly well as a gas of free electrons. So let me rephrase your question as:

If I take a container with a gas in and rapidly partition it into two, will the pressure be the same on both side?

If we look at a gas on the atomic scale it's a mass of atoms/molecules whizzing around at random. So on the large scale the density will be the same everywhere but on the atomic scale we get random pressure variations. So when you partition the container of gas it's very unlikely that the number density of gas molecules will be exactly the same on both sides, and we would expect a random pressure difference. However it's extremely unlikely that the pressure difference will be big enough to be measurable.

The same applies to your metal. The electron motion within it is random, just like molecules in a gas, so when you break the metal there will be a tiny difference in the density of electrons on the two sides of the break. However the difference will be far too small to measure.

Answer 2

Maybe a beginning of answer to your question could be found in Lindhard theory. Considere a fermionic field $$ \Psi(\textbf{x},t)=\frac{1}{\Omega}\sum_{\textbf{k}_1,\textbf{k}_2}e^{\mathrm{i}(\textbf{k}_\alpha-\textbf{k}_\beta)\cdot\textbf{x}+\mathrm{i}(E_{\textbf{k}_\alpha}-E_{\textbf{k}_\beta})t/\hbar}a_{\textbf{k}_\alpha}^\dagger a_{\textbf{k}_\beta} $$ which describe spin-less electrons with indiviual states set as plane waves, i.e. $E_{\textbf{k}_{\alpha,\beta}}=\frac{\hbar^2\textbf{k}_{\alpha,\beta}^2}{2m}$.

The goal of Lindhard theory is to study how does $\Psi(\textbf{x},t)$, or more precisely the particle density $\rho(\textbf{x},t)=\Psi^\dagger(\textbf{x},t)\Psi(\textbf{x},t)$, react to a perturbation using linear response theory.

Let say that for $t<t'$ the electrons system is at thermal equilibrium. Such equilibrium state is characterized by the temperature $T$ and the chemical potential $\mu$, since we are consedering an open system. At $t=t'$ we introduce a local chemical potential perturbation $\delta\mu(\textbf{x},t)$ in the system. It can be anything : adding a particle, induce a defect e.g. by starting to cut your piece of metal, etc... The only condition on $\delta\mu(\textbf{x},t)$ is that it should be small enough so that the associated interaction hamiltonian : $$ \mathcal{H}_i(t)=\int\mathrm{d}\textbf{x}\,\rho(\textbf{x})\delta\mu(\textbf{x},t) $$ can be treated perturbatively compared to the non-perturbed system hamiltonian $\mathcal{H}_0$. Then the linear respose theory simply says that such perturbation induces a fluctuation on $\rho$. The associated response function is often given as : $$ \chi(\textbf{x},\textbf{x}',t,t')=\frac{\mathrm{i}}{\hbar}\langle\left[\rho(\textbf{x},t),\rho(\textbf{x}',t')\right]\rangle_0\;\Theta(t-t') $$ where $\langle\cdot\rangle_0\equiv \mathrm{Tr}(f_0\;\cdot\;)$ with $f_0=\left(e^{\mathcal{H}_0/k_BT}+1\right)^{-1}$ the Fermi-Dirac distribution (particle distribution at thermal equilibrium). $\Theta$ stands for the Heaviside step function and is here to ensure the causality of $\chi$.

By using the expression of $\Psi$, one can compute : $$ \chi(\textbf{x},\textbf{x}',t,t')=\frac{\mathrm{i}}{\Omega^2\hbar}\sum_{\textbf{k}_1,\textbf{k}_2}e^{\mathrm{i}(\textbf{k}_\alpha-\textbf{k}_\beta)\cdot(\textbf{x}-\textbf{x}')+\mathrm{i}(E_\alpha-E_\beta)(t-t')/\hbar}(f_0(E_\alpha)-f_0(E_\beta)) $$ More interesting features can be extracted if you perform space-time Fourier transform : $$ \chi(\textbf{q},\omega)=\frac{\mathrm{i}}{\Omega^2\hbar}\sum_{\textbf{k}}\frac{f_0(E_{\textbf{k}+\textbf{q}})-f_0(E_{\textbf{k}})}{\omega+E_{\textbf{k}+\textbf{q}}-E_{\textbf{k}}+\mathrm{i}0^+} $$ with $\textbf{q}=\textbf{k}_\alpha-\textbf{k}_\beta$. Then, if you are interested in the local response of the electronic system (spatialy local $\textbf{q}\rightarrow 0$ and temporally local $\omega\rightarrow 0$), one can show that : $$ \chi(\textbf{q},\omega\sim 0)\underset{\textbf{q}\rightarrow 0}{\sim}\sum_{\textbf{k}}\frac{\partial f_0}{\partial E_{\textbf{k}}} $$ which is independant from $\textbf{q}$ and $\omega$. That is to say, when coming back in real space and time notations : $$ \chi(\textbf{x},\textbf{x}',t,t')\sim\delta(\textbf{x}-\textbf{x}')\delta(t-t') $$ What we get shows that the response of an electronic system to an external perturbation is very fast (the $\delta(t-t')$ part) and very local (the $\delta(\textbf{x}-\textbf{x}')$ part) so that there is not really such a thing as typical time to re-distribute density fluctuation in a electronic system. This comes from the fact that Thomas-Fermi screening is very efficient.

Answer 3

In presence of external electric fields, If you cut it fast enough (at a speed faster than the conduction speed, you can have the two alved ending up with different charges. This is because the external field redistributes the charges on the conductor' surface to annulate the field there. so the charges will not be uniformly distributed. In absence of external fields, random noise fluctuations contribute to a non uniform distributions, although much smaller, and the charge difference between the two cuts will be very small instead of macroscopic.