Current: Scalar or vector?

Is current a scalar or a vector?

NOPE i change my mind its definitely a scalar

Its a vector, but its not a vector because we said so. Is the general idea behind current being a vector

(edited 7 years ago)

Reply 2

The-Spartan

Original post by Mowerharvey

Is current a scalar or a vector?

This is a good question. Technically, it is a Tensor!
With regards to vector/scalar, current is most suited to be a scalar quantity, even though it can have direction. This is because it does not obey the vector addition rules that vectors must obey.
As well as this, current can be expressed as the dot product of two vectors. Due to this, current is a scalar.

Reply 3

lerjj

Good question - it's a scalar but there are some reasons you might think it's a vector.

Firstly, by definition

I = \frac{dq}{dt}

which is the time rate of change of a scalar, which is another scalar. so current is a scalar.

However, we often use rules such as

naq \vec{v}

which seems to suggest that the current should be a vector. More specifically, the current has the same direction as the drift velocity, surely?

No. What is going on is that current density

\vec{j}

is a vector and

j=nq\vec{v}

. Current is a scalar (dot) product of area and current density, so

I = \vec{A}\cdot\vec{j}=naq\vec{v}

is in fact a scalar. Exactly how treating area as a vector works here is somewhat fastidious and involves defining a normal vector to the surface with magnitude equal to the area of the surface element.

Similarly, there are other rules such as the Biot-Savart law that treat current as a vector (in this case, you have something that looks like

\vec{I} \times \vec{r}

. This again orginates with current density. Current is a scalar.

Reply 4

lerjj

Original post by The-Spartan

I've never seen current treated as a tensor (besides the trivial fact that scalars are zeroth order tensors, but I assume you mean a second order tensor as that's the way the word is usuall used.)

Surely we define

I = \dfrac{dq}{dt} = -\int_s \vec{j}\cdot\vec{dA}

which makes current look like a simple scalar, even when giving a full 3-D treatement?

Reply 5

sengupta

I can not understand why a simple question is being made so complicated. If we take the definition of current, it is charge flowing per unit time. The charge which are basically free electron moves in a fixed direction which is from lower to higher potential ( in other words towards the positive of the battery/ power supply). This makes the flow of charge follow only a particular direction. On the other hand a vector can be in any direction. So the flow of current do not have a vector character. A similar example is that of Pressure. Though it is defined as force/ area and force and area are both vector quantities but pressure is scalar, simply because while defining it we consider only the force which is perpendicular to the surface.

Reply 6

Kyx

Current is a vector. You can have positive current and negative current. The current depends upon the motion of the electrons,

Reply 7

uberteknik

A vector must have both magnitude and direction.

Current is the movement of charge. It's strict definition is therefore a vector.

However, most calculations involving current, impose a convention where electron current flows from the +ve terminal to the -ve terminal.

Since conventional current is entirely opposite to actual electron flow, we can see that there is no need in most cases to invoke current as a true vector.

In that respect, current, although a vector, can be pragmatically treated as a scalar dependent on the context and application in rather the same way as say speed and velocity.

(edited 7 years ago)

Reply 8

djmans

its a vector, google is never wrong

Reply 9

atsruser

Original post by lerjj

No. What is going on is that current density

\vec{j}

is a vector and

j=nq\vec{v}

. Current is a scalar (dot) product of area and current density, so

I = \vec{A}\cdot\vec{j}=naq\vec{v}

is in fact a scalar.

I think that a more interesting question is: why is current density a vector? (apart from being defined as the product of a scalar and a vector).

Offhand, I can't think of a nice way of showing that current density behaves particularly vectorially e.g. if you add a pair of current densities, then we should get a third current density i.e.

\bold{j}_1 + \bold{j}_2 = \bold{j}_3

, but I can think of physical scenarios where that doesn't seem to hold.

Reply 10

Implication

this thread :eyeball:

Reply 11

Implication

Original post by atsruser

\bold{j}_1 + \bold{j}_2 = \bold{j}_3

, but I can think of physical scenarios where that doesn't seem to hold.

i suspect that might just be an artefact of how we use the mathematical formalism to represent the 'real', physical concepts

of course, current (densities) in general need not be vectors - it depends how they are defined. they appear a lot in scalar field theory, for example.

i'm certain we could define current to be a vector if we liked as well, we just don't normally need to because we only tend to represent wires in one dimension- and the difference between a scalar and a one-dimensional vector is probably vanishing to a physicist. choosing a sign convention for the current is effectively equivalent to choosing a basis vector

(edited 7 years ago)

Reply 12

atsruser

Original post by Implication

The idea of current-density-as-a-vector has me a little confused now, in fact. I'll have to think about it. I can't see how current densities add nicely.

and the difference between a scalar and a one-dimensional vector is probably vanishing to a physicist. choosing a sign convention for the current is effectively equivalent to choosing a basis vector

Well, the difference between convergent and non-convergent series is vanishing to a physicist :smile:

But I'm not sure that I agree: for example, the components of 1D velocity flip sign under a rotation by 180 degrees, but the sign of electric charge doesn't, so no physicist should confuse electric charge with a vector.

Reply 13

langlitz

Original post by djmans

its a vector, google is never wrong

Citation? Because they're wrong

Reply 14

lerjj

Original post by atsruser

\bold{j}_1 + \bold{j}_2 = \bold{j}_3

, but I can think of physical scenarios where that doesn't seem to hold.

Yep, that is the more interesting question. I think to answer it it's probably best to go back to a straightforward equation to define

\vec{j}

\displaystyle \frac{q}{t} = - \int_S \vec{j}\cdot\vec{n} dA

Now, in the context of this formula, the first reason we want

\vec{j}

to be a vector is apparent: we need to be able to take the normal projection of it. With regards to adding vectorially, I personally think that's intuitive but I'm sure one could construct a reasonable argument based on defining current density as a sort of charge flux density as above.

For instance, if we have two currents in the same direction, the physically, we expect the left hand side of that equation to double, and so the right hand side must, but

\vec{j}

Reply 15

atsruser

Original post by lerjj

For instance, if we have two currents in the same direction, the physically, we expect the left hand side of that equation to double, and so the right hand side must, but \vec{j}

is the only free variable to double. If we add current densities not parallel to each other, we see that we do not simply add their magnitudes. However, it's hard to see the exact rule for adding them.

Right. In simple scenarios, the vector property seems to make sense, but it also seems to break down easily. For example, I was thinking about two beams of high energy electrons in empty space that meet at some point. In each beam, we can define a current density, but I would expect all kinds of complex scattering effects when the beams cross, and on the other side of the point of interaction, we wouldn't get any kind of nice vector addition going on.

Even if we restrict ourselves to wires, it seems to fail: we can have two wires joining at a point, with the exit wire at a right angle to both. Then again the densities don't add vectorially, since we end up with a exit current density vector that isn't in the span of the two entry vectors. So I'm not at all sure about the limitations of the vector current density idea.

Reply 16

Implication

Original post by atsruser

The idea of current-density-as-a-vector has me a little confused now, in fact. I'll have to think about it. I can't see how current densities add nicely.

Don't worry, I think I said something silly. The currents in all scalar field theories I've met are vectors :tongue:

I don't see why we couldn't represent current density as a scalar in some contexts though.

With regard to addition, can't we just think about what the current density means? It's the flow of charge per unit area through a specified surface/plane. If we have charge flowing in one direction in n-dimensional space, and another charge flowing in another, the net charge flow in each dimension can be found by adding the components. Surely this follows straight-forwardly from modelling the charges' positions (and hence their time derivatives) as vectors?

Well, the difference between convergent and non-convergent series is vanishing to a physicist :smile:

well that depends on how you define a rotation. certainly charge can do :tongue:

more importantly, in a one-dimensional space I'm not sure the concept of rotation is meaningful. You can't really 'rotate' a 1D velocity because doing so implies moving it through other dimensions. A one-dimensional vector space is spanned by a single basis vector, and so a one-dimensional vector field is characterised by a single variable. Flipping the sign is equivalent to flipping the 'direction' by the axioms of a vector space, isn't it?

Reply 17

atsruser

Original post by Implication

With regard to addition, can't we just think about what the current density means? It's the flow of charge per unit area through a specified surface/plane. If we have charge flowing in one direction in n-dimensional space, and another charge flowing in another, the net charge flow in each dimension can be found by adding the components. Surely this follows straight-forwardly from modelling the charges' positions (and hence their time derivatives) as vectors?

See my examples above for reasons as to why I'm confused about the vector addition of current densities.

By a rotation, I mean a rotation of the axes, so that for example, you consider the +ve direction being to the right, and I stand facing you and consider +ve to the left. Then a particle moving with +ve velocity to you has -ve velocity to me. However, we would both agree that the object has -ve charge if it is an electron that is doing the moving.

So the sign of the vector changes under a rotation of axes, but the sign of the scalar charge doesn't.

Reply 18

Implication

Original post by atsruser

Right. In simple scenarios, the vector property seems to make sense, but it also seems to break down easily. For example, I was thinking about two beams of high energy electrons in empty space that meet at some point. In each beam, we can define a current density, but I would expect all kinds of complex scattering effects when the beams cross, and on the other side of the point of interaction, we wouldn't get any kind of nice vector addition going on.

I think you have the same problem with anything you model as a vector. You can define a current density for each beam if you like, but when the beams cease to be two individual beams (as they would in a scattering process), they can no longer be represented by only two current densities. The model you've set up doesn't work anymore. The mathematical formalism doesn't describe the physics. It's the scattering effect that breaks down the addition; not the vectorial properties (or lack thereof) of the current density.

You could make precisely the same argument if you decided to characterise each beam by the position (or velocity) vector of the 'front' of the beam rather than by a current density. The positions wouldn't satisfy 'nice vector addition' after/during a scattering process either, because two position vectors is no longer enough to fully describe the dynamics of the situation.

Even if we restrict ourselves to wires, it seems to fail: we can have two wires joining at a point, with the exit wire at a right angle to both. Then again the densities don't add vectorially, since we end up with a exit current density vector that isn't in the span of the two entry vectors. So I'm not at all sure about the limitations of the vector current density idea.

I'm not sure (this is quite confusing me!), but I think you're making the mistake here of assuming that the basis vectors of the current density vector must correspond to the Cartesian axes/basis vectors we're more familiar with.

The current densities only fail to satisfy vectorial addition here to the extent that the velocity vectors of the charge flows do. The velocity vector before the junction points right (say), and after the junction the two velocity vectors point up and down.

Original post by atsruser