B (Electric) current is not a vector quantity

[brotherbobby]

TL;DR Summary

Electric current is famous for not being a vector quantity, reason being that two currents $i_1$ and $i_2$ flowing in different directions don't add vectorially to give another current $i_3$ in some other direction. Fair enough.
How about the current density vector $\vec J$? Surely that's a vector. Or am I wrong?

Can we write for two different current density vectors : $$\vec J=\vec J_1+\vec J_2?$$
If we can, which is what I suspect, will it also not mean that the respective currents add up vectorially?

Integrating the current densities above $\displaystyle{\left(i=\iint_S\vec J\cdot d\vec a\right)}$, can we write $\vec i = \vec i_1+\vec i_2$? But that will make electric current a vector quantity, which we believe it isn't.

What is going on?


P.S. : On a different note, the closed surface integral isn't getting compiled by the $\rm{LaTeX}$ editor, so I made it simple above. Let me put it here : $$\displaystyle{\left(i=\oiint_S\vec J\cdot d\vec a\right)}$$
Can someone, for instance @berkeman, look into it?

The $\rm{LaTeX}$ code is correct. Codecogs is compiling it fine :

 

[Dale]

Yes. Current density is a vector.

 

[brotherbobby]

Yes. Current density is a vector.

Does it make electric current a vector too?

 

[nasu]

No. It doesn't. The magnetic field is a vector. Does this make the magnetic flux a vector?

 

[brotherbobby]

No. It doesn't. The magnetic field is a vector. Does this make the magnetic flux a vector?

Given flux $\displaystyle{\Phi_B=\iint_S\vec B\cdot d\vec a}$, of course flux is a scalar like electric current $\displaystyle{\left(i=\iint_S\vec J\cdot d\vec a\right)}$. I get that.

But unlike magnetic flux, electric current has a direction. Where does the direction go in the integral above?

 

[renormalize]

Where does the direction go in the integral above?

The dot-product $\vec{J}\cdot d\,\vec{a}$ you're integrating over doesn't have a "direction".

 

[brotherbobby]

The dot-product $\vec{J}\cdot d\,\vec{a}$ you're integrating over doesn't have a "direction".

I agree, but current does have a direction based on where it flows.

 

[nasu]

The integral of the current density does not have a direction in general. When we talk about a current in a wire usually we have the case of uniform current density and we call the direction of the current is the common direction of the current density, same at all points of the cross section. For a general, non uniform current density, you don't have a defined direction of current.

 

[kuruman]

Force $\vec F$ is a vector but pressure $p$ is a scalar associated with that vector through $\cancel{p=\vec F\cdot \hat n~dA}$ $p=\vec F\cdot \hat n/dA$.

Likewise, Similarly,
Current density $\vec J$ is a vector but current $I$ is a scalar associated with that vector through $I=\vec J\cdot \hat n~dA$.

Neither $p$ nor $I$ have a direction. The vector associated with each does.

Edited to correct typo. Pressure has dimensions of force per unit area.

 

[brotherbobby]

Force $\vec F$ is a vector but pressure $p$ is a scalar associated with that vector through $p=\vec F\cdot \hat n~dA$.

Is this the definition of pressure though? I'd say $p=\dfrac{F}{A}$. I don't think a dot product can be introduced here.
Yes, ${\rm{d}}\,I=\vec J\cdot\hat n\rm{da}$, and clearly $I$ is a scalar. The trouble I am having is why is it that $\vec J$ allows vector addition to be performed on itself whereas $I$, though a scalar, cannot be made into a vector along with the direction it moves, say $\vec I$. Let's say we had some new current defined as $\vec I_{\text{new}}=\iint \vec J da$. Will this new current be a vector?

I think I have resolved my doubt. I will describe my thoughts on it presently.

 

[kuruman]

Is this the definition of pressure though? I'd say $p=\dfrac{F}{A}$. I don't think a dot product can be introduced here.

You may not think that a dot product can be introduced, but the symbol $F$ in your definition is the normal component of the force to area $A$. Otherwise, a vector force $\vec F$ that is in the plane of the area would exert pressure on that surface. So when you write $p=\dfrac{F}{A}$, you really mean $p=\dfrac{\vec F\cdot \hat n}{A}$.

 

[berkeman]

This thread is making me dizzy.

Every current I have ever measured has had a direction.

 

[A.T.]

Is this the definition of pressure though? I'd say $p=\dfrac{F}{A}$.

That's the simplified scalar formula. But in general force is a vector.

I don't think a dot product can be introduced here.

It has to be introduced to get the normal component of a general force vector.

 

[kuruman]

This thread is making me dizzy.

I hope this will undizzy you and perhaps get the OP on the right path.

Consider a closed, for simplicity spherical, surface and three conductors of different cross sectional areas $A_1$, $A_2$ and $A_3$ respectively carrying current densities $\vec J_1$, $\vec J_2$ and $\vec J_3$ as shown on the right. I have deliberately not drawn the directions of the current densities but they are assumed to be uniform and along the conductors, either "into" or "out of" the closed surface.

Start with the charge continuity equation in differential form $$\vec\nabla\cdot\vec J+\frac{\partial \rho}{\partial t}=0$$ Move the second term to the right, multiply both sides $dV$ and integrate over the volume of the sphere. $$\int_V \vec\nabla\cdot\vec J=-\frac{\partial}{\partial t}\int_V\rho dV.$$ Use the divergence theorem to convert the left-hand volume integral into a surface integral. $$\int_S \vec J\cdot \hat n~dA=-\frac{\partial}{\partial t}\int_V\rho dV.$$ Note that

• the symbol $\hat n$ is the outward normal as specified by the divergence theorem;
• on the surface the current density is zero except in the regions of overlap with the conductors. This allows us to write the left hand side as the sum of (in this case) three integrals;
• the right-hand side is the time rate of change of the enclosed charge.

Thus, $$\int_{A_1} \vec J_1\cdot \hat n~dA+\int_{A_2} \vec J_2\cdot \hat n~dA+\int_{A_3} \vec J_3\cdot \hat n~dA=-\frac{\partial q_{enc.}}{\partial t}.$$ Using the definition of the $k$th current $I_k=\int_{A_k}\vec J_k\cdot \hat n~dA$ , we have $$I_1+I_2+I_3=-\frac{\partial q_{enc.}}{\partial t}.$$ In the case of a steady-state circuit, the right-hand side is zero and the expression above says that the charge enclosed by the surface does not change. Whatever charge goes in the spherical surface must come out and nothng stays within. This is also known as Kirchhoff's current law.

So what about the direction (or sign) of the current? That's built in the dot products. The normal $\hat n$ is always outward. If the current density is in the same direction as the normal, the current is positive and positive charge carriers come out of the closed surface. If the current density is opposite to the normal, the current is negative and positive charge carriers go into the closed surface. Also note that this convention is consistent with charge conservation. If the left hand side is positive, i.e. net charge flows out of the surface, the rate of change $~\dfrac{\partial q_{enc.}}{\partial t}~$ of the enclosed charge is negative which means that the net charge crossing the surface must come from within. The opposite will be true if the left hand side is negative.

 

[weirdoguy]

associated with that vector through p=F→⋅n^ dA.

Ok, but still, is this definition of a pressure? Force times area?

 

[kuruman]

Ok, but still, is this definition of a pressure? Force times area?

No it's not. It was a typo in post #9 which I corrected. As others have already indicated, what's in the numerator is not just force, but the component of the force normal to the area. The definition is $$p=\frac{\vec F\cdot \hat n}{dA}.$$

 

[Dale]

I agree, but current does have a direction based on where it flows.

Current does not have a direction (left vs up) but it does have a sense (in vs out).

Consider the most important law for current: Kirchoff’s current law. If you have a node then KCL says that the sum of the currents going in is equal to the sum of the currents going out, that is the sense. It does not say that the current going left is equal to the current going right, that is the direction.

If current were a vector then KVL would treat an inward-left-pointing current the same as an outward-left-pointing current. And it would treat an inward-left-pointing current different from an inward-up-pointing current.

Current is not a vector because it doesn’t behave like a vector in the important physics equations that use current.

Sometimes the direction matters and the sense does not. In those cases we use current density. Other times the sense matters and the direction does not. In those cases we use current. It wouldn’t be useful to only have quantities with direction without sense when sometimes we need quantities without direction and with sense instead.

 

[DaveE]

The direction of a current is a property of the definition of that particular current. It's not the current that has a direction, it's the predefined path (e.g. the wire) that has a direction.

 

[TensorCalculus]

Also, vector addition and resolving vectors has no meaning in the case of current... i.e. I can't just add 2 currents from different parts of the circuit vectorially and then define a "resultant current" - that makes no physical sense. Also if I'm not mistaken (which to be fair I might be), current is defined as being in the direction of flow of positive charge, so you don't specify the direction in the definition of the vector.

 

[brotherbobby]

I am the OP of this thread. Since it's been a while, let me describe the issue I was having in brief.

The issue : Is current $i$ a vector? No, is what people say. But why not? It has both magnitude and direction.

The resolution : A vector is one whose direction is a given. A quantity of charge $q$ can be made to flow along any direction. This makes the direction irrelevant and therefore reduce the quantity to a scalar. For the same reason, pressure $P=\dfrac{F}{A}$ is not a vector, though one might assume it is given that is a vector divided by a scalar. Pressure at a point in a liquid doesn't depend on orientation - it has the same value no matter what the direction of the surface having area $A$ does. For the same depth $d$, the pressure $P$ has the same value in all directions $(P=\rho_L gd)$, making its direction irrelevant. How about the pressure that I exert on the floor on which I stand? Surely that pressure appears to act downwards - does that make it a vector? No. My weight is, since weight is a force that is amenable to vector addition. But the pressure I exert on the floor is the same at all points on the (spherical) globe, long as my weight remains the same. This again makes the direction of pressure irrelevant.

Thank you for your interest.

 

[kuruman]

For the same reason, pressure $P=\dfrac{F}{A}$ is not a vector, though one might assume it is given that is a vector divided by a scalar.

In that case, one would be wrong. Pressure is the scalar component of a vector divided by a scalar. This makes the ratio a scalar.

You are correct in saying that pressure does not depend on orientation. It may depend on position which makes it a scalar field. Temperature is another example of a scalar field in that with every point in space one may associate a temperature value but not a direction. However the gradient of the pressure field and the temperature field are vector quantities and have both magnitude and direction.

You have to understand that what makes a mathematical entity a vector is more than having "magnitude and direction". A vector must transform correctly under a coordinate transformation, e.g. a rotation of the coordinate axes. Current $I$ and pressure $p$ don't do that, current density $\vec J$ and force $\vec F$ do. Perhaps you might profit by studying some linear algebra.

 

[Dale]

For the same reason, pressure P=FA is not a vector, though one might assume it is given that is a vector divided by a scalar. Pressure at a point in a liquid doesn't depend on orientation - it has the same value no matter what the direction of the surface having area A does. For the same depth d, the pressure P has the same value in all directions (P=ρLgd), making its direction irrelevant. How about the pressure that I exert on the floor on which I stand? Surely that pressure appears to act downwards - does that make it a vector? No. My weight is, since weight is a force that is amenable to vector addition. But the pressure I exert on the floor is the same at all points on the (spherical) globe, long as my weight remains the same. This again makes the direction of pressure irrelevant.

I know you didn’t bring it up, but I don’t think the pressure stuff is relevant. Current is not a vector because of how it behaves itself, regardless of whether or not pressure is a vector.

Just look at the equations that use current. They have a sense (in or out) but no direction.

 

[Gavran]

Electric current is the rate of change of electric charge and it is a scalar quantity.
A rate of change of a scalar quantity is a scalar quantity (for example speed is the rate of change of distance) while a rate of change of a vector quantity is a vector quantity (for example velocity is the rate of change of position).

 

[brotherbobby]

Electric current is the rate of change of electric charge and it is a scalar quantity.
A rate of change of a scalar quantity is a scalar quantity (for example speed is the rate of change of distance) while a rate of change of a vector quantity is a vector quantity (for example velocity is the rate of change of position).

Of course, but the trouble is, current has a direction. Distance and speed don't.
That's the confusion I have been having. How can a quantity like current flow in a given direction with a magnitude and yet not be a vector?
Several answers are given and I am sure they are correct. One is current $i$ cannot be added the way the vectors do. Then current (components) don't behave (transform) like vectors do under coordinate transformation. The answer that appeals to me the most is that the direction of current is not a given, but dependent on the conductor. You can change it, unlike a vector like force or displacement.

 

[Dale]

current has a direction

As I have said several times, current does not have a direction (left or right, up or down). It only has a sense (in or out).

Current density has direction (but not sense), and current has sense (but not direction). Look at the equations where they are used.

 

[weirdoguy]

rate of change of electric charge

I have a semantical problem with this, because it kind of implies that in this context charge is something that circuit has and it changes. If, say, charged ball is discharging then it makes sense, but in the context of this thread I think it's iffy. Rate of flow is better, in my opinion.

 

[brotherbobby]

As I have said several times, current does not have a direction (left or right, up or down). It only has a sense (in or out).

Current density has direction (but not sense), and current has sense (but not direction). Look at the equations where they are used.

Yes, I agree with you. Good point.

 

[Gavran]

I have a semantical problem with this, because it kind of implies that in this context charge is something that circuit has and it changes. If, say, charged ball is discharging then it makes sense, but in the context of this thread I think it's iffy. Rate of flow is better, in my opinion.

The definition in the post #23 describes the equation $$ i=\frac{dq}{dt} $$ and whenever the equation is applicable the definition should be applicable too. Officially, the change of charge is not the same as the flow of charge, but these two processes include each other, and exclude each other too. There can only be a difference between a flow of charge with accumulation of charge and a flow of charge with no accumulation of charge. Or, there can only be a difference between a change of charge with accumulation of charge and a change of charge with no accumulation of charge (only charged particles which exit a circuit element or enter a circuit element are included into the change of charge).

 

[Gavran]

I have a semantical problem with this, because it kind of implies that in this context charge is something that circuit has and it changes. If, say, charged ball is discharging then it makes sense, but in the context of this thread I think it's iffy. Rate of flow is better, in my opinion.

You are right. $dq$ means flow of electric charge in the equation $i=dq/dt$.
A rate of flow of a scalar quantity is a scalar quantity, while a rate of flow of a vector quantity is a vector quantity.
Electric charge is a scalar quantity.