Why do we even need the term 'conventional current'?

[etotheipi]

If we take electric current to be the rate of flow of (signed) charge past a certain point in a given reference direction, this unambiguously tells us all the information that we need to know. If we label a current arrow with $-6A$, then in $1$ second we either think of a charge of $-6C$ moving in the direction of the arrow or $+6C$ moving in the opposite direction. The idea of electric current is independent of the type of charge carriers.

My guess is that when discussing conventional current, we can convert all our charge flows into equivalent flows of positive charges. So whenever we refer to a conventional current of $6A$, the direction of the current is implied and we can label it as $6C$ per second in the direction of decreasing potential.

The problem may well be a semantic one; is conventional current just a name we give to abstracting away the charge flows to equivalent positive flows? And if so, why don't we just work with signed electric currents?

 

[Dale]

I dislike the term "conventional current". There is just "current", and there is no need whatsoever to use the term "conventional". The current density of moving charges is given by $\vec J = \rho \vec v$. So if $\rho$ is positive then $\vec J$ is in the same direction of $\vec v$ and if $\rho$ is negative then $\vec J$ is in the opposite direction as $\vec v$. There is no reason whatsoever to assume that $\vec J$ and $\vec v$ need to point in the same direction, so there is no need to use the label "conventional".

Frankly, physics instructors and curricula shouldn't even mention electrons until the first class on quantum mechanics. So now, instead of just not mentioning them at all, we have to waste all sorts of time with concepts like this.

 

[etotheipi]

I dislike the term "conventional current". There is just "current", and there is no need whatsoever to use the term "conventional". The current density of moving charges is given by $\vec J = \rho \vec v$. So if $\rho$ is positive then $\vec J$ is in the same direction of $\vec v$ and if $\rho$ is negative then $\vec J$ is in the opposite direction as $\vec v$. There is no reason whatsoever to assume that $\vec J$ and $\vec v$ need to point in the same direction, so there is no need to use the label "conventional".

You put it much better than I did, I'm glad I'm not alone!

 

[kuruman]

Current is not a vector, current density is. An element of current $dI$ is related to the current density $\vec J$ by $dI=\vec J \cdot \hat n~dA$. Here $\hat n$ is a unit vector perpendicular to area element $dA$. You can see that when $dI$ is positive, this means that the current is flowing through the area in a direction (mostly) parallel to $\hat n$. The opposite is true if $dI$ is negative.

I see that @Dale posted about the usefulness of the current density before me but did not discuss $I$.

 

[kuruman]

Frankly, physics instructors and curricula shouldn't even mention electrons until the first class on quantum mechanics. So now, instead of just not mentioning them at all, we have to waste all sorts of time with concepts like this.

Amen. See here for a related point of view.

 

[vanhees71]

I dislike the term "conventional current". There is just "current", and there is no need whatsoever to use the term "conventional". The current density of moving charges is given by $\vec J = \rho \vec v$. So if $\rho$ is positive then $\vec J$ is in the same direction of $\vec v$ and if $\rho$ is negative then $\vec J$ is in the opposite direction as $\vec v$. There is no reason whatsoever to assume that $\vec J$ and $\vec v$ need to point in the same direction, so there is no need to use the label "conventional".

Frankly, physics instructors and curricula shouldn't even mention electrons until the first class on quantum mechanics. So now, instead of just not mentioning them at all, we have to waste all sorts of time with concepts like this.

Finally you have to understand the connection between current density (a vector field) and a current (a scalar). The current density $\vec{J}(t,\vec{x})$ gives the amount of charge per unit time and unit area flowing through a little area element at $\vec{x}$ at time $t$.

The current through an area element is thus quantitatively defined by defining a surface-unit vector $\mathrm{d}^2 \vec{f}$ which is a vector perpendicular to the infinitesimal surface. Which of the two possible directions of this normal vector is arbitrary. Thus in addition to the polarity of the charges making up the current density the sign of the surface-normal vector also determines the sign of the current, which thus is somewhat arbitrary. Note that this arbitrariness doesn't harm you in any way, because the physical laws where you need it are usually related to a surface and its boundary. E.g., Ampere's law for magnetostatics reads
$$\vec{\nabla} \times \vec{H}=\vec{J},$$
and there you get a relation between the magnetic excitation field $\vec{H}$ and the current by integrating this equation over a surface $A$ with some arbitrarily chosen direction of the surface-normal vectors. This gives you the current on the right-hand side of Ampere's Law
$$I=\int_{A} \mathrm{d}^2 \vec{f} \cdot \vec{J}.$$
On the left-hand side of the equation you use Stokes's theorem, and the orientation of the boundary curve of $\partial A$ is related to the direction of the surface-normal vectors by the right-hand rule. By this definition of the relative orientation between the surface-normal vectors and the orientation of the closed boundary curve it doesn't matter which direction of the surface-normal vector you've chosen in the beginning. You always get Ampere's circuital Law in integral form
$$I=\int_{A} \mathrm{d}^2 \vec{f} \cdot \vec{J} = \int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{H}.$$
If you change the sign of the surface-normal vectors also the orientation of the boundary curve changes, and the above formula stays correct, independent of the choice of this arbitrary orientation of the surface-normal vectors.