Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

In some papers, I can see the drift velocity of electrons equaling thermal velocity. Can anyone tell me when both almost equal each other?

share|cite|improve this question

3 Answers 3

The drift velocity is the net velocity of electrons in a certain direction under an applied field. The thermal velocity is has no net direction because it is randomly distributed and occurs in any metal at finite temperatures. Since the two velocities are different, it does not make any sense to say they are qualitatively equal, even though you may equate them quantitatively as in the previous answer.

share|cite|improve this answer
Thermal velocity is due to thermal agitation, and drift velocity is what you get if you superimpose this with the voltage driven motion. It is possible that the two velocities can be of the same value. The question is not asking about qualitative identity, is asking when they can be numerically equal. My simple analysis answers just that. I don't know the significance of this or any effects this might have on the conductivity of a metal wire, but it is certainly a good question. –  JKL Mar 19 '13 at 22:40
It provides no meaningful insight whatsoever! –  mcodesmart Mar 26 '13 at 22:42

The drift velocity of electrons in a metal is given by the equation


where $I$ is the electric current in the metal wire, $n$ is the number of electron density, $A$ is the cross sectional area of the metal wire and $v_D$ is the drift velocity.

From this we get

$v_D= \frac{I}{enA}$

The thermal velocity is given by


where $T$ is the temperature of the metal, $k_B$ is Boltzmann’s constant, $m_e$ is electron mass and $v_T$ is the thermal velocity. From the last equation we get

$v_T= \sqrt{\frac{3k_BT}{m_e}}$

Equating $v_D$ and $v_T$ gives the following

Condition for the two speeds to be equal:

$\frac{I}{enA }=\sqrt{\frac{3k_BT}{m_e}}$.

share|cite|improve this answer

The theory of the above two answers is correct, but the equations provided are not quite correct. The equation $${{1}\over{2}}m_e(v_T)^2 = {{3}\over{2}}k_BT$$ is based on the assumption that the electrons behave as an ideal gas (and is true for semi-conductors). However, in metals, electron velocity actually has a very small dependence on temperature, and a much larger dependence on its Fermi energy. This comes from the application of density of states theory and Fermi-Dirac statistics.

$$E(T) = {{3}\over{5}} \times E_{F0}\times \left[1+\left({{5}\over{12}}\times\pi^2\right)\times(k\times T/(E_{F0})^2\right]$$

Where $E_{F0}$ is the Fermi energy at $T = 0\text{ K}$. Since $E_{F0} \gg kT$, the average energy can usually be taken to be ${{3}\over{5}}E_{F0}$. Depending on the metal you are working with, this is usually readily available in tables on-ne. For example, copper has an $E_{F0} = 7.0\text{ eV}$, meaning that $E_{\text{ave}} = 4.2\text{ eV}$, and $V_{\text{thermal}} = 1.2*10^6 \text{ m/s}$ at $t = 300\text{ K}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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