When there is a temperature change $\Delta T$, the change of resistivity is

(1) proportional to $\Delta T$

(2) proportional to the original resistivity $\rho_0$

Hence we can define the temperature coefficient of resistivity $\alpha$ so that

$$\Delta \rho = \rho_0 \alpha\Delta T$$

I searched on the internet about (2) but it is usually simply stated as a fact or "experiments show that", without explaining why.

Length expansion has similar property but I can understand why intuitively. For the same temperature change, doubling the length will double the change in length as well, because every part of the length expands.

But I don't understand why the change in resistivity should be proportional to the original resistivity.

  • 5
    $\begingroup$ Only in the sense you you can approximate almost any function by a linear expansion in the neighborhood of some point. And if you're approximating everything as linear, then everything is proportional to everything else. $\endgroup$ Mar 7, 2018 at 18:45
  • $\begingroup$ Why the downvote!? I don't think you even get the point of my question. $\endgroup$
    – velut luna
    Mar 8, 2018 at 2:48
  • $\begingroup$ @MichaelSeifert I still don't see why your comment can answer why the change in resistivity should be proportional to the original resistivity. $\endgroup$
    – velut luna
    Mar 8, 2018 at 12:50
  • 1
    $\begingroup$ How do you think the change in resistance should relate to the original resistivity? Whatever physical process that causes resistance is being perturbed slightly. As @MichaelSeifert noted, for small changes you use the first term in the expansion. Note further that over any appreciable temperature change the change in length (or resistance of something) is not linear. $\endgroup$
    – Jon Custer
    Mar 8, 2018 at 14:14
  • $\begingroup$ @JonCuster I KNOW it's a linear approximation for small temperature change. No need to point out again and again. I am asking why it is proportional to the original resistivity. If you know the answer, answer it. Don't ask me the same question I am asking!! $\endgroup$
    – velut luna
    Mar 8, 2018 at 14:19

4 Answers 4


The resistivity of a material actually depends in a complicated way on temperature. For example, the resistivity of a metal is well-modeled by the Bloch–Grüneisen formula:

$$\rho(T)=\rho(0)+A\left(\frac{T}{\Theta_R}\right)^n\int_0^{\frac{\Theta_R}{T}}\frac{x^n}{(e^x-1)(1-e^{-x})}dx\ \ ,$$

where $\rho(0)$ is residual resistivity due to defect scattering, the constant A depends on things like electron velocity, Debye radius and number density of electrons, $\Theta_R$ is the Debye temperature, and $n$ is an integer that depends on whether the primary interaction producing the resistance is electron-electron interaction, s-d electron scattering, or phonon scattering.

As another example, the resistivity of an undoped semiconductor can be modeled as being of the form

$$\rho(T)= \rho_0 e^{-aT}\ \ ,$$

but the relationship between resistivity and temperature can be given more accurately in implicit form by the Steinhart–Hart equation:

$$\frac{1}{T} = A + B \ln(\rho) + C (\ln(\rho))^3\ \ ,$$

where $A$, $B$ and $C$ are the Steinhart–Hart coefficients.

However, in many cases, you mainly just care about the behavior of $\rho(T)$ in the vicinity of some temperature $T_0$. Common values used for $T_0$ in tables are 20°C (roughly room temperature) or 0°C, but the important point about $T_0$ here is that it's just an arbitrary choice used for a table, not a physically significant temperature such as a temperature at which $\rho(T)$ is at a local minimum or maximum for a material. In the vicinity of $T_0$, $\rho(T)$ can be approximated by a Taylor expansion

$$\rho(T) \approx \rho(T_0) + \rho'(T_0) (T - T_0)\ \ .$$

Because $T_0$ has no physical significance for the material, the linear term is in general going to be more important than higher order terms for $T$ sufficiently close to $T_0$.

It would be possible to define a $$\bar{\alpha}=\rho'(T_0)\ \ ,$$ call $\bar{\alpha}$ the "alternative coefficient of resistivity" (ACR) or something, and create a table of the ACR for various materials for some given $T_0$. And tables listing the ACR would be very convenient for calculating

$$\Delta \rho=\rho(T)-\rho(T_0)=\rho(T)-\rho_0$$


$$\Delta \rho=\bar{\alpha}\Delta T\ \ ,$$

where we've defined

$$\Delta T=T-T_0$$


$$\rho_0=\rho(T_0)\ \ .$$

However, in practical calculations, it isn't usually $\Delta \rho$ that's important for the calculation, but rather $\Delta \rho/\rho_0$. For example, if you're designing a resistor for use in electronic devices that will be operated in the temperature range $20±40 °C$, and the resistance of the resistor needs to change from its nominal value by no more than 5% within that temperature range, you need to choose a material to create the resistor from such that, as listed in a table that uses $T_0=20°C$,

$$\left| \frac{\Delta \rho}{\rho_0}\right |=\left| \frac{\bar{\alpha}\Delta T}{\rho_0}\right |=40\left| \frac{\bar{\alpha}}{\rho_0}\right |<0.05\ \ ,$$


$$\left| \frac{\bar{\alpha}}{\rho_0}\right |<0.00125\ \ .$$

Or if you're designing a thermistor, and you want the thermistor's resistance to change by a given percentage for a given change in temperature, you'd wind up needing to choose a material such that $\frac{\bar{\alpha}}{\rho_0}$ has some target value, instead of a maximum value as in the case of a resistor.

Because the expression $\frac{\bar{\alpha}}{\rho_0}$ would keep showing up in practical calculations like that, it's more convenient to just define

$$\alpha=\frac{\bar{\alpha}}{\rho_0}\ \ ,$$

and create tables that list $\alpha$ instead of $\bar{\alpha}$. And since $\bar{\alpha}=\rho_0\alpha$, we have

$$\Delta \rho=\bar{\alpha}\Delta T=\rho_0\alpha\Delta T\ \ .$$

  • $\begingroup$ It is a nice answer but could you make it nicer further by explaining theoretically and intuitively? $\endgroup$
    – user326901
    Aug 5, 2022 at 9:22

It's just a matter of definition. If we expand $R(T)$ in a Taylor series about $T_0$, the first two terms are $$ \rho(T) \approx \rho(T_0) + \rho'(T_0) (T - T_0). $$ The coefficient $\alpha$ is then defined so that $$ \alpha = \frac{\rho'(T_0)}{\rho(T_0)}, $$ which yields $\Delta \rho \approx \rho_0 \alpha \Delta T$ (where $\rho_0 \equiv \rho(T_0)$.)

We could equally well define $\tilde{\alpha}$ to be $\rho'(T_0)$, and then we would have $\Delta \rho \approx \tilde{\alpha} \Delta T$. But the physics would be unaffected by this change of definition. The important part is that the change in resistance is roughly proportional to the change in temperature; how we decide to define the constant of proportionality in this relationship is just a matter of convention.

  • 1
    $\begingroup$ I don't think it's a matter of convention. A table of $\alpha(T_0)$ needs only specify the value of $\alpha$ at what $T_0$ but needs not specify the value of $\rho(T_0)$. This means $\alpha$ is independent of $\rho(T_0)$. If you produce a table of your $\rho^\prime(T_0)$, you need to specify both. So my question is why your $\rho^\prime(T_0)$ is proportional to $\rho(T_0)$. $\endgroup$
    – velut luna
    Mar 8, 2018 at 15:05
  • $\begingroup$ I disagree that the value of $\rho_0$ is irrelevant. If you had a table of $\alpha$ values and you didn't know $\rho_0$, how would you find $\Delta \rho$? $\endgroup$ Mar 8, 2018 at 15:42
  • $\begingroup$ What I said is $\alpha$ is independent of $\rho_0$. I didn't say to find $\rho$ you don't need $\rho_0$. That's the advantage why people define $\alpha$ in this way, instead of your $\tilde{\alpha}$. $\endgroup$
    – velut luna
    Mar 8, 2018 at 15:50

Okay, so if I understand your question properly you're asking for the physical causes behind the change in resistivity being affected by the temperature and why such changes are linearly proportional to the original resistivity.

Well, being just a physics student, I went and, uh, looked at the Wikipedia page on conductivity and resistivity which said this:

"A metal consists of a lattice of atoms, each with an outer shell of electrons that freely dissociate from their parent atoms and travel through the lattice. This is also known as a positive ionic lattice.

Most metals have resistance. In simpler models (non quantum mechanical models) this can be explained by replacing electrons and the crystal lattice by a wave-like structure each. When the electron wave travels through the lattice the waves interfere, which causes resistance. The more regular the lattice is the less disturbance happens and thus resistance lowers. The amount of resistance is thus caused by mainly two factors. Firstly it is caused by the temperature and thus speed of vibration of the crystal lattice. The temperature causes irregularities in the lattice."

Which makes sense to me.


Define a quantity temperature coefficient of resistivity $\alpha$ as $$\alpha=\frac{1}{\rho}\frac{d\rho}{dT}$$

One can clearly see that $\alpha$ is a function of $\rho$ and T. But Resistivity is itself a quantity dependent on Temperature. So essentially for a particular material $\alpha$ is just a function of time.
Now imagine for a solid at room temperature, the molecules just jiggle back and forth. If we increase the temperature by a small amount the atoms will start to giggle faster. This doesn't make the properties upside down. This will change the resistivity by a small amount.
To check how the parameter $\alpha$ is varying with temperature experiments were conducted and found that $\alpha$ doesn't appreciably change near room temperature.
So one can essentially treat $\alpha$ as a constant.

Now integrate the equation, you will get this equation. $$\rho(T)= \rho_0 e^{\alpha (T-T_0)}\ \ $$

I will stop here. I suppose this argument was convincing. Just remember that physics is not directly answering reality. It is about creating a world of analytical models which will be used to explain reality.

PS: If you're wondering why I didn't use the classical $\rho=\rho_0(1+\alpha \Delta T) $ in my answer is because the exponential equation is more precise. In case you are working with numericals you may want to find the resistivity in decimal so you will probably find the exponential of a constant. In that case you may use this linear approximation to yield good numerically good answer. The reason why this is equation doesn't seem convincing is because approximating these things linearly with "variables" may seem vague.Try using numerical values, you wont see much difference.

  • $\begingroup$ The integration you performed in this equation is only accurate to the extent that $\alpha$ is constant over the range of temperature you're interested in. If $\alpha$ varies with temperature, then the exponential equation you've written down won't be any more precise than the linear approximation. $\endgroup$ Mar 11, 2018 at 23:17
  • $\begingroup$ Yes, I agree. I believe I already mentioned this in my answer $\endgroup$ Mar 12, 2018 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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