Why doesn't the variation of resistivity with temperature go both ways?

Question

I've learnt that the variation of resistivity with temperature for a conductor is:

$\rho=\rho_0(1+\alpha (T−T_0))$

Let's consider resistivity at 0℃ and 100℃.

When heating the conductor from 0℃ to 100℃,

$ρ₁₀₀=\rho_0(1+\alpha (100-0))$

α=$\displaystyle \frac{ρ₁₀₀-ρ₀}{100ρ₀}\, $

Now, when cooling the conductor from 100℃ to 0℃,

$\rho_0=ρ₁₀₀(1+\alpha (0-100))$

α=$\displaystyle \frac{ρ₀-ρ₁₀₀}{-100ρ₁₀₀}\, $= $\displaystyle \frac{ρ₁₀₀-ρ₀}{100ρ₁₀₀}\, $=$\displaystyle \frac{ρ₁₀₀-ρ₀}{100ρ₀(1+α(100-0))}\, $=$\displaystyle \frac{α}{1+100α}\, $

Why does this discrepancy exist? Even if the relation only holds for smaller temperature differences, the discrepancy seems to hold, as the new value of α only seems to depend on the old one, as $\displaystyle \frac{α}{1+T'α}\, $.

Answer 1

I've learnt that the variation of resistivity with temperature for a conductor is:

$\rho=\rho_0(1+\alpha (T−T_0))$

This is not a real physical relationship. It is just a convenient first-order approximation. Suppose we have some arbitrary resistivity $\rho(T)$ as a function of temperature. Then, at any $T=T_0$ we can do a series expansion to get: $$\rho(T) = \rho(T_0) + \rho'(T_0) (T-T_0) + O(T-T_0)^2$$ $$ \rho(T) \approx \rho(T_0) \left( 1+\frac{\rho'(T_0)}{\rho(T_0)} (T-T_0) \right) $$ which is the same as your formula with $\rho_0=\rho(T_0)$ and $\alpha = \rho'(T_0)/\rho(T_0)$.

The big issue is that $\alpha$ is not some sort of actual constant of the material itself. It is just the ratio of $\rho'$ to $\rho$ at a specific temperature $T_0$. So you cannot assume that $\alpha$ at $T_0$ is the same as $\alpha$ at any other temperature. You can only use this formula with $T_0$ as the temperature at which $\alpha$ was measured.

Answer 2

More broadly, it's convenient to postulate that the resistivity $\rho$ changes with temperature $T$ in differential form as

$$d\rho=\alpha(T)\rho\,dT,$$

where $\alpha$ is the (temperature-dependent) thermal coefficient of resistivity.

For the purposes of this question, though, we can idealize $\alpha(T)$ as constant. In other words, the temperature variation in $\alpha$ is not the cause of the specific problem you observed that $\rho_0\neq\rho_0(1+\alpha\Delta T)(1-\alpha\Delta T)$.

Integrating, we obtain

$$\ln\left(1+\frac{\Delta\rho}{\rho_0}\right)=\alpha\Delta T,$$

which is symmetric regarding heating and subsequent cooling. That is, $\rho_1=\rho_0e^{\alpha\Delta T}$ is always consistent for successive temperature excursions and returns (in which $\rho_1=\rho_0e^{\alpha\Delta T}e^{-\alpha\Delta T}=\rho_0$).

However, if we choose to simplify the logarithmic relation for small $\Delta\rho$ using a Taylor series expansion as

$$\ln\left(1+\frac{\Delta\rho}{\rho_0}\right)\approx\frac{\Delta\rho}{\rho_0}\left(\approx\alpha\Delta T\right)$$

to obtain the equation you start with, then we lose the advantage of symmetry in exchange for the advantage of a simpler expression.

(Incidentally, the same thing happens with engineering strain $\varepsilon=\frac{\Delta L}{L_0}$; if we apply a certain engineering strain and then its exact opposite, we don't recover the original length. To ensure proper symmetry, we need to use the true strain.)

Answer 3

Maybe we can see this as a purely mathematical misunderstanding, and disregard the discussion about whether such a formula is an approximation (there could in principle exist a material for which the linear relationship was exact, at least in some temperature interval).

So more abstractly, the relation: $$ y = y_0(1 + \alpha x) $$ (with $\alpha\ne 0$) represents a straight line in the $(x,y)$ coordinate system going through the fixed point $(0,y_0)$ and having a slope $y_0\alpha$. However, the equation: $$ y = y_{100}(1 + \alpha(x - 100)) $$ describes a straight line in the $(x,y)$ plane going through $(100,y_{100})$ with slope $y_{100}\alpha$.

So the two equations do not describe the same line. The slopes differ. (And if $y_0=y_{100}$ the equations differ at the 0th-degree coefficient.)

Answer 4

The temperature variation in resistivity $\rho$ is obtained by first principles from the temperature variation in conductivity $\sigma = 1/\rho$. In a (free-electron) conductor, $\sigma$ depends on the number density of free electrons $\rho_{N,e-}$ and their mobility $\mu_{e-}$. The former is a weakly increasing function of temperature and the latter decreases approximately as inverse temperature. To first order, we approximate conductivity as being inversely dependent on temperature. In doing so, we ignore the affects from an increasing carrier density with increasing temperature and mildly non-linear variation ($\approx T^{3/2}$) for a decrease in mobility from phonon scattering as temperature increases.

This leaves us with one typical expression giving a linear dependence for resistivity versus temperature.

$$ \rho = \rho_o(1 - \alpha(T - T_o)) $$

When we want a more accurate expression for $\rho(T)$, we must return to how conductivity varies with temperature and remove the approximations. We do not expand resistivity in higher order powers of $T$.

The parameters $\rho_o$ and $\alpha$ in the above expression are dependent on two things. First, the material. Secondly, the reference temperature $T_o$.

With the above in mind, the mistake in the conjecture is to propose that $\rho_o$ and $\alpha$ should be changed when we change $T$. Alternatively put, we do not report resistivity as though we have made a calculation for a change going from one temperature to another. We report resistivity as being AT a certain temperature relative to a reference temperature.

Consider the case that $\rho_o$ and $\alpha$ are defined at 0$^o$C.

For 100$^o$C

$$\rho_{100} = \rho_o(1 - \alpha(100 - 0))$$

For 0$^o$C

$$\rho_{0} = \rho_o(1 - \alpha(0 - 0))$$

Consider the case that $\rho_o$ and $\alpha$ are defined at 100$^o$C. Use different symbols to avoid confusion.

For 100$^o$C

$$\rho_{100} = \rho_o'(1 - \alpha'(100 - 100))$$

For 0$^o$C

$$\rho_{0} = \rho_o'(1 - \alpha'(0 - 100))$$

For the above, $\rho_o \neq \rho_o'$ and $\alpha \neq \alpha'$.

Answer 5

That $(1 + \alpha \cdot \Delta T)$ term gives a relative change, right? (It multiplies $\rho_0$, after all.)

Forget about the physics, regardless of where this was posted. Even with an unwavering linear dependence assumed, the change in resistivity per degree (or kelvin) is the same everywhere in absolute terms, but relative to the current value, it's different. $\alpha$ depends on the reference temperature since it's specified in relative terms.

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As a purely physics-unrelated and idealized example, say we have $y(x) = 100 + 2x$. Now, at $x_0 = 100$, $y_0 = y(x_0) = 300$, and we can write $$ y(x) = y_0 \cdot (1 + \alpha(x - x_0)), $$ where $\alpha$ is the relative amount of change in $y$ per unit change of $x$ at ($x_0, y_0)$, or $$ \alpha = \frac{\Delta y / \Delta x}{y_0} = 2 / 300. $$

Which, if you work it out for e.g. $x=200$, gives you $y(200) = 500$, the same as the initial equation would.

But, $\alpha$ would be different with another choice of $(x_0, y_0)$, so we can't swap $x_0$ or $y_0$ for other points without adjusting $\alpha$ accordingly. Maybe we should have called them $x_{100}, y_{100}$ and $\alpha_{100}$ or so to make that clearer.

With this clean linear function it'd be much easier to just specify the absolute change, and use it everywhere, but with actual physical properties where the function isn't even linear over all temperatures, I suppose giving the relative change feels better. At least one can see from $\Delta T$ how far they are from the reference point.

Answer 6

You are using the equation incorrectly. $\rho_0$ is the resistivity measured at the reference temperature $T_0$. That reference temperature does NOT change when you are calculating $\rho$ at temperatures other than the reference temperature. For example, if the reference temperature is 20 deg C and you heat the material to 100 deg C, $(T - T_0)$ equals 80 deg C. If you cool the material to 5 deg C, $(T - T_0)$ equals -15 deg C.