# Do electrons contribute to the overall specific heat capacity?

Do electrons contribute to the overall specific heat capacity? Please explain why and why not. I researched for a bit and most of them said mostly phonons contribute to the specific heat but never mention electrons. So please help thank you.

As has been mentioned by @annav it is difficult to answer your question within a fair degree of background knowledge of quantum mechanics and mathematics but I will try.

One of the mysteries of physics before the advent of quantum mechanics was why the valence (free) electrons in metals do not contribute to the heat capacity.

There are electrons in metals which behave like a gas in that they are free to move within a metal and are not "attached" to any metal nucleus.
Heat capacity is all to do with adding energy to a system and with this multitude of free electrons within a metal they should contribute to the heat capacity of a metal as they all can pick up energy at will.
Well they do pick up energy but only to the extent of about $$1\%$$ at room temperature.

Quantum mechanics predicts that the free electrons can have definite amounts of energy (energy levels) and because they are fermions which cannot all exist in the same state.

If the metal was at absolute zero, then one would find that all the free electrons would be in the lowest possible allowed energy levels up to the maximum energy which is called the Fermi energy.
Think of it as filling up energy levels in a ground state atom up to some maximum energy.
This immediately shows you that the predictions of Quantum Mechanics differ radically from those of classical physics as here you have the electrons have kinetic energy even though the metal is at absolute zero.
The amount of energy these electrons have is enormous even at this very low temperature with an average kinetic energy for copper being equivalent to a gas molecule at $$33\,000\,\rm K$$ with the highest energy free electrons having an energy equivalent to a temperature $$55\, 000\,\rm K$$!

So why is it that these highly energetic free electrons contribute so little to the heat capacity of a metal and indeed do not melt the metal - copper has a melting point of $$1\,356 \rm \, K$$?
The reason is purely quantum mechanical.

For one of these free electrons to gain energy it must move from the energy state that it is occupying to an unoccupied energy state but all state below the Fermi energy are occupied.
This means that the free electron must be given an enormous amount of energy to jump up to a state above the Fermi energy or its energy must one close to the Fermi energy so that it can jump up into an unoccupied energy level above the Fermi energy.

It the second process which happens in a metal as its temperature rises above absolute zero with electrons being promoted (given some extra energy) to occupy energy levels above the Fermi energy and leaving "vacancies" below the Fermi energy.
However note that as the extra energy these free electron are given at room temperature $$(300\,\rm K)$$ is so much smaller than the temperature equivalent of the Fermi energy, $$55\, 000\,\rm K$$, comparatively few free electrons are disturbed by the increase in temperature of the metal with most of the free electrons having no possibility of gaining energy because there are no accessible empty energy levels.

As I stated above "heat capacity is all to do with adding energy to a system" but it is only those free electrons with energies near the Fermi energy level which can gain and lose energy and there are not very many of them.
So very few of the free electrons contribute to heat capacity of a metal.

Update as a result of a comment from @FredWeasley

The answer to your comment is given in the reference from @annav Electron heat capacity which I will try and explain without the Mathematics.

The fee electrons are allowed to exist in a number of allowed states and a graph of density of states against energy is shown on the left.

The area under the graph, $$\int_0^E (\text {density of states})\, d(\text{energy})$$, gives the total number of states in the range $$\rm energy =0$$ to $${\rm energy} = E$$ and I have shown that all states up to the Fermi energy $$E_{\rm F}$$ are occupied by $$N$$ electrons at $$0\,\rm K$$.

At a higher temperature $$T\,\rm K$$ electrons near Fermi level can be exited into higher energy states as shown in the right hand diagram.
Note that in that region the density of states does no vary very much so the number of electrons in that region ie those that can lose and gain energy, is proportional to the temperature,$$T\,\rm K$$. This is the answer to your question about the number of electrons "available" to take up energy and so contribute to the heat capacity.

Note that my sketch graph is not to scale as $$E_{\rm F}$$ for a metal is equivalent to a temperature of $$\rm 55\,000 K$$ whereas the thermal energy of the electrons may have an equivalent temperature of the order of tens or hundreds of kelvin.

Each of these electrons have a thermal energy of the order $$k_{\rm B} T$$ and so the thermal energy $$U$$ associated will all such electrons is proportional to $$T^2$$.

The heat capacity is $$\frac {dU}{dT}$$ and so the electron contribution to the heat capacity is proportion to $$T$$.
Using hard sums and quite a number of approximations the contribution of electrons to the heat capacity is found to be $$\dfrac{\pi^2}{3}\dfrac{k_{\rm B}T}{E_{\rm F}} = \dfrac{\pi^2}{2}k_{\rm B} \left (\dfrac{T}{T_{\rm F}} \right)$$ as compared with the expectation from classical Physics of $$\dfrac 32 k_{\rm B}$$.
In this expression $$T_{\rm F}$$ is the temperature equivalent of the Fermi energy which is very much greater than the thermal temperature of the electrons and this illustrates the smaller than expected contribution of electrons to the heat capacity of a metal.

• so if there are more free electrons available, the specific heat capacity should increase since there are more electrons near the Fermi energy level? If so, is there a way to calculate the number of freed electrons, let say in one kilogram of copper? Also, thank you for simplifying the concept to me! – Fred Weasley Apr 21 at 12:02