I've been wondering: why is the electrical conductivity of a given material defined as the inverse of its electrical resistivity? In other words, why is $$ \sigma \equiv \frac{1}{\rho}~?$$ It indeed makes sense to define a number called conductivity such that, when the resistivity of the material decreases, the conductivity increases. However, there are a bunch of functions for which this property holds. So why aren't the following as convenient as the definition given above? $$ \sigma = \frac{1}{\rho^2} $$ $$ \sigma = - \rho $$ In fact, every decreasing function on $\rho$ could be used here. What is it that makes $\frac{1}{\rho}$ so special and unique?

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    $\begingroup$ By that logic, there would be no problem if we decided to measure time in seconds squared. Go ahead and and try to calculate what time it is one minute squared after noontime (half of a day squared). $\endgroup$ Commented Oct 30, 2020 at 13:07
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    $\begingroup$ I think defining it as the reciprocal is much more "special" than defining it any other way. When we think about inverse relations, the first thing comes into mind is reciprocal. Had it been defined as say 1/(resistivity)^2 , then the same question would have been much more striking as to why it was not defined as simple reciprocal. -- just an alternative way to look at it. $\endgroup$
    – Tony Stark
    Commented Oct 30, 2020 at 15:15
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    $\begingroup$ Does this not fall under the same class of questions as "why is momentum $mv$" or "why is work $\int F.dx$" - the point is that the mathematical expressions arose before we gave them names, we don't make up names then guess expressions for them. $\endgroup$
    – jacob1729
    Commented Oct 30, 2020 at 15:53
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    $\begingroup$ Saw this question on another site as "recommended" and thought to myself, surely someone cannot possibly be asking this question. There must be more to the question than this. Unfortunatly there wasn't. You're essentially asking why is the inverse of "a half" "two". Because it is, and it enables useful things to be done building on that basis. Or why is the inverse of "second", "Hz". No one would talk about "Hz" being the inverse of "second" if it wasn't useful for something. $\endgroup$ Commented Nov 1, 2020 at 1:24
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    $\begingroup$ Why does the word "up" mean the opposite of "down"? Couldn't it mean "left" instead? $\endgroup$
    – knzhou
    Commented Nov 1, 2020 at 18:49

8 Answers 8


In my experience this comes from resistance and conductance in electrical engineering and circuit theory. If you use the loop current analysis method on a circuit of resistors and sources then you get a matrix of linear equations whose coefficients are resistances. If you use the node voltage method on the same circuit you get a matrix whose coefficients are inverse resistances.

So the inverse of resistance shows up very often quite naturally in circuit equations, rather than the negative of resistance or the inverse of resistance squared. Because it shows up naturally it makes sense to give the inverse of resistance a name.

Usually when you run into some quantity that is defined and you are unsure why, that quantity first simply showed up in some important formula. So people needed a way to discuss that part of that formula, and so they gave it a name. But the quantity showed up on its own in the math first and was given a name later.


The usual definition matches up with calculations with parallel resistors. For example, the total resistance of two parallel resistors is $$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}.$$ If the resistors have the same size and shape, then this can be written in terms of their resistivity like so: $$\frac{1}{R} = \frac{A}{L}\left(\frac{1}{\rho_1} + \frac{1}{\rho_2}\right),$$ where $A$ is the cross-sectional area and $L$ is the length of the resistor. In terms of conductance and conductivity, we can write this equation as $$G = \frac{A}{L}\left(\sigma_1 + \sigma_2\right).$$ In fact, just looking at conductance, the equation for a set of parallel resistors is much more intuitive: $$G = \sum_i G_i$$ as opposed to the usual $$\frac{1}{R} = \sum_i \frac{1}{R_i}.$$ The total conductance of a set of parallel resistors is equal to the sum of the conductance of all the resistors. This nicely parallels the case of resistors in series: $$R = \sum_i R_i$$ where the total resistance is the sum of the resistances of each resistor.

Using conductance and conductivity can be useful when trying to calculate the total resistance of a material whose resistivity varies across its geometry. See this question and this answer for an example.

  • $\begingroup$ Perhaps also worth saying something about dimensions, like what units ρ^-2 would have? This does already touch on the fact that pretty much any time you ever used it, you'd be using the square root of it, if defined as ρ^-2, which is necessary to come up with ohms instead of ohms-squared. $\endgroup$ Commented Oct 31, 2020 at 13:17
  • $\begingroup$ @PeterCordes Sure, $1/\rho^2$ has the wrong units, but it's always possible to add a factor with the right unit to make it work: $\sigma = k/\rho^2$, where $k$ has units of ohms. $\endgroup$
    – Mark H
    Commented Nov 1, 2020 at 19:46

The Ohm's law for a conductive material can be expressed locally as a linear relationship between the current density $\boldsymbol{J}$ and the electric field $\boldsymbol{E}$. For an isotropic material, this relationship can take either of the two equivalent forms: $\boldsymbol{J} = \sigma \boldsymbol{E}$ or $\boldsymbol{E} = \rho \boldsymbol{J}$. For these two forms to be really equivalent for a specific material, the relation $\sigma = 1/\rho$ should hold.

In the case of an anisotropic material, current density and electric field are no longer parallel and the above relationships take the forms $\boldsymbol{J} = \boldsymbol{\sigma} \boldsymbol{E}$ or $\boldsymbol{E} = \boldsymbol{\rho} \boldsymbol{J}$, where now $\boldsymbol{\sigma}$ and $\boldsymbol{\rho}$ are matrices related by $\boldsymbol{\sigma} = \boldsymbol{\rho}^{-1}$.

Therefore, you cannot choose an arbitrary relationship between $\sigma$ and $\rho$, for otherwise you would lose one of the two equivalent relationships between the fields.

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    $\begingroup$ This is a good answer, but it actually doesn't answer the question. Is simple transmutes the question into "why is J proportional to E", etc... (or any other arbitary question about definitions). Yes electric fields superpose linearly, but that still doesn't answer the question, because it is a nonsensical question. $\endgroup$ Commented Nov 1, 2020 at 1:27
  • $\begingroup$ @user3728501 You're confusing two different things: the fact that electric fields superpose linearly is independent from the linear relationship between $\boldsymbol{J}$ and $\boldsymbol{E}$. The latter is a property of matter, which is essentially a consequence of the smoothness of response functions. It's not a nonsensical question if one just looks at the qualitative relationship between $\sigma$ and $\rho$, as the OP probably did. $\endgroup$ Commented Nov 1, 2020 at 8:38

Resistivity relates to resistance in the same way that conductivity relates to conductance.

$$R = \rho\frac{l}{A} ~~~~~~~~ G = \sigma\frac{A}{l} $$


$$ \sigma = \frac{1}{\rho} ~~~~~~~~ G=\frac{1}{R} $$

Now, if we consider...

$$ V = IR $$

$$ \frac{V}{R} = I $$

$$ I = VG $$

It's handy because it relates current to voltage in the same way that the canonical form of Ohms law relates voltage to current.

Voltage is current times resistance (R).

Current is voltage times conductance (G).

The other quantities you might imagine to express (like negative resistivity, the square of resistivity, etc) do not have such useful properties, so we do not use them and do not give them names.


In terms of circuits/situations where Ohm's law is obeyed, this is a definition and is physically consistent. As conductivity increases, resistivity decreases. Similarly, as conductivity decreases, resistivity increases. The product of resistivity and conductivity is 1.

It indeed makes sense to define a number called conductivity such that, when the resistivity of the material decreases, the conductivity increases.

That is exactly what the equation

$\sigma = \frac{1}{\rho}$


However, there are a bunch of functions for which this property holds.

No, there is not.

$\sigma = \frac{1}{\rho^2}$

This would imply that incresing the resistivity by two would decrease the conductivity by $\frac{1}{4}$ but increasing the conductivity by two would decrease the resistivity by $\frac{1}{\sqrt{2}}$. Why would nature have such an asymmetry?

$\sigma = - \rho$

This would imply that any resistivity would produce a negative conductivity. What would be the physical meaning of negative conductivity?

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    $\begingroup$ I don't get the symmetry argument - there are certainly reasons to have resistivity and conductivity be "symmetric", but I don't see an a priori reason why they ought to be. Taking the electrostatic force as an example, you can double the distance between charged particles and decrease the force between them to 1/4. But doubling the force only decreases the distance to 1/sqrt(2). I feel like this is a tautological argument, that conductivity and resistivity ought to be inverses, because they are, in fact, defined to be inverses. Why must 2 different physical quantities be symmetric.? $\endgroup$ Commented Oct 30, 2020 at 16:03
  • $\begingroup$ If they are reciprocals of each other they would definitely be symmetric.How would you double the electrostatic force without moving/changing the charges? $\endgroup$
    – joseph h
    Commented Oct 30, 2020 at 22:24
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    $\begingroup$ I'm confused by what you're trying to say in this answer. The asker has stated, completely correctly, that there are a bunch of functions which satisfy the property that "when the resistivity of the material decreases, [that function] increases." You seem to be stating, completely falsely, that the only function which satisfies that property is the reciprocal function. $\endgroup$ Commented Oct 31, 2020 at 19:14
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    $\begingroup$ I think that the asker is talking about the concept of a monotonically decreasing function, whereas you're talking about the concept of a monotonically decreasing function satisfying $f(xy) = f(x)f(y)$, and you're glossing over the fact that these are two different concepts. $\endgroup$ Commented Oct 31, 2020 at 19:14
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    $\begingroup$ Why would nature have such an asymmetry? Both conductivity and resistance are part of a physical model of nature, not part of nature itself. $\endgroup$ Commented Nov 1, 2020 at 19:09

You cannot simply pick any decreasing function; it has to be one that works both for the number and the units. For example $-\log{ρ}$ is for sure a decreasing function, but $-\log{(1\text{Ωm})}$ makes no sense. There are quantities defined using logarithms on ratios, but for that you have to pick a base level, and each is as good as any other, so such a function isn't so simple and independent anymore.

$-ρ$ won't work for another similar reason – it has the same dimension (and hence unit) as $ρ$. This means that an expression like $ρ+σ$ would be well defined, and in this case equal to 0, which again makes no sense. This is what people tried to avoid with units.

The way units work leaves us with nothing but powers, and $ρ^{-1}$ is the simplest there. Plus it is an inverse, so it works both ways.


Ohm's law can be written as $I = V*\frac 1 R$. Given that we're multiplying $V$ by $\frac 1 R$, what is mysterious about the fact that people find it useful to define a quantity equal to this expression $\frac 1 R$ that can be multiplied with voltage to get current?

If there's anything mysterious, it's why conductivity is defined in terms of resistance, rather than resistance in terms of conductivity. Current is driven by voltage drop. Conductivity is a measure of the extent to which this occurs.

The reason that conductivity is the multiplicative inverse of resistance is because current, voltage, and resistance are related through multiplication.

It indeed makes sense to define a number called conductivity such that, when the resistivity of the material decreases, the conductivity increases.


Does it?

It makes sense to define conductivity, and as a result of defining conductivity, we end up with something that has the property of decreasing as resistance increases, but I don't see any reason to have something that decreases as resistance increases, as a goal in and of itself.


What's the resistance of a 10 ohm and a 20 ohm resistor in series? Easy: 10+20 = 30 ohms.

What if the same resistors are in parallel? That's erm...let me get calculator...

$$ \left({1\over 10} + {1 \over 20}\right)^{-1} = 6.67\:\Omega $$

Now what's the conductance of 0.05 siemens in parallel with 0.1 siemens? No calculator required: 0.05+0.1 = 0.15 siemens.

Thus, conductance is convenient whenever dealing with parallel circuits, among other things.

This isn't the only example: you'll find many electrical terms come in duals for the same reasons. For example when applying Thévenin's theorem you'll find it most convenient to think about voltage sources and resistance, but for Norton's theorem you'll want current sources and conductance. When analyzing nontrivial circuits you'll probably convert between the two several times, as that's easier than wrangling an equation that disappears into microscopic text as in:

$$ 1\over...{1\over...{1\over...{1\over ...}}}$$

Another consideration: an ohm is dimensionally equivalent to volt per ampere, whereas siemens is ampere per volt. The physical utility of these units should be pretty obvious when you have some material that obeys Ohm's law. I'm not sure the same could be said of the other possible definitions you propose.


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