electric circuits - Why does current density have a direction and not current?

• Current is a scalar $I$ with units of $\mathrm{[J/s]}$. It is defined as $I=\frac{\mathrm{d}Q}{\mathrm{d}t}$.


• Current density is a vector $\vec{J}$ (with magnitude $J$) with units of $\mathrm{[J/s/m^2]}$. It is current per unit cross-sectional area, and is defined as $\vec{J}=nq\vec{v_d}$ (where $n$ is the number of moving $q$-charges with drift velocity $v_d$).

Why is $I$ defined to not have a direction? Current density $\vec{J}$ is defined as a vector, so why is current $I$ not?

There are many questions about current vs. current density

... like this, this and this, but none answers my question about one being vector and the other being scalar. Is it simply just a definition? It just seems so obvious to define current as a vector too.

Another but equivalent definition of current density is $I = \int \vec{J} \cdot d\vec{A}$. Mathematically, the dot product gives a scalar. But, for me this doesn't give much explanation still, as we could just as well have mathematically defined current as a vector and then used the area in scalar form in an equivalent formula like: $\vec I = \int \vec{J} \cdot dA$.

Is it just a definition without further reason, or is there a point in keeping $I$ in scalar form?

Answer

According to my understanding, indeed you could define a physical quantity like $$\vec{I} = I\;\vec{n_d}$$ where $\vec{n_d}$ is the unitary drift direction. There is no problem with that. But what is the most important is to understand the harmony between different quantities. I mean that there is some little subtleties between $I$ and $\vec{j}$.

The current is defined according to a surface $A_t$ (and is a local quantity: the position of the surface). Since the surface may be tilted (not perpendicular to $\vec{n_d}$), then in general, we should write $$I=n\;q \; \vec{A_t}.\vec{v_d}=n\;q\;A_t \; \vec{n_{A_t}}.\vec{v_d}=n\;q\;A_t \; cos(\theta).v_d$$ where $\theta$ is the angle between $\vec{A_t}$ and $\vec{v_d}$ (If the charge $q$ is negative, $\vec{I}$ and $\vec{n_d}$ would have opposite orientations, which fits with the usual convention of the electric current). Of course, if we have considered a surface $A$ which is perpendicular to the drift, the current would be the same but we would write $$I=n\;q \; \vec{A}.\vec{v_d}=n\;q\;A \; v_d$$

That is, let's talk about the densities. We have $$\mathrm{d}I=n\;q\;v_d\;\mathrm{d}A=n\;q\;v_d\;cos(\theta)\;\mathrm{d}A_t$$ The scalar current density is given by $$j=\frac{\mathrm{d}I}{\mathrm{d}A}=n\;q\;v_d=\frac{1}{cos(\theta)}\frac{\mathrm{d}I}{\mathrm{d}A_t}$$ Yet, you see that you have to be careful about the differential area that you put in the denominator. The current could then be calculated as $$I=\int_A j\;\mathrm{d}A=\int_{A_t} j\;cos(\theta)\;\mathrm{d}A_t$$ Here also, we see a possible source of confusion. This can be fixed if we define a vector current density as $$\vec{j}=j\;\vec{n_d}=n\;q\;v_d\;\vec{n_d}$$ The direction of $\vec{j}$ is resolved too: it's that of the local drift in the considered material position. That direction may be different than that of the average drift of the whole current $\vec{I}$. That is, the expression of the current can be written simply as $$I=\int_A \vec{j}\;\mathrm{d}\vec{A}=\int_{A_t} \vec{j}\;\mathrm{d}\vec{A_t}$$ Or, you could use your own convention and write $$\vec{I}=\bigg(\int_A \vec{j}\;\mathrm{d}\vec{A}\bigg)\;\vec{n_d}=\bigg(\int_{A_t} \vec{j}\;\mathrm{d}\vec{A_t}\bigg)\;\vec{n_d}$$