Mica is a good conductor of heat but an electrical insulator.

According to free electron theory (which applies only to metals) free electrons carry heat and electricity. Therefore, thermal conductivity is directly proportional electrical conductivity.

What about dielectric materials? In the case of mica, it conducts heat but not electricity, so what are the carriers responsible for this behavior. If the are electrons as they are in metals then why they don't carry electricity too?

  • 2
    $\begingroup$ Atomic lattice vibrations (aka phonons) can also conduct heat. In metals, the electronic contribution is (by far) the greatest, but not the only one. $\endgroup$
    – Bosoneando
    Jun 28, 2015 at 21:30
  • 2
    $\begingroup$ I edited this question significantly. Please try, as much as possible, to use normal punctuation and grammar. For example, questions should end with a question mark (?). $\endgroup$
    – DanielSank
    Jun 28, 2015 at 23:06
  • $\begingroup$ @Bosoneando why not post that as an answer? $\endgroup$
    – DanielSank
    Jun 28, 2015 at 23:06
  • 1
    $\begingroup$ The logical fallacy here is that free electrons carrying heat does not mean heat is carried by free electrons. $\endgroup$
    – Aron
    Jun 29, 2015 at 5:53

1 Answer 1


Electricity needs charged particles (or quasi-particles) to conduct it. Heat can be conducted with almost any quasi-particle. Diamond is one of the best conductors of heat in existence, and it's because of phonons, ie quasi-particles of lattice vibrations, which are strong because the diamond lattice is strong.

  • 3
    $\begingroup$ To add, at room temperature almost all thermal conductivity is through phonons (lattice conductivity). The electronic component is actually pretty small unless you drop the temperature... $\endgroup$
    – Jon Custer
    Jun 29, 2015 at 1:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.