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feynman1

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- Thread starter feynman1
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- #1

feynman1

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It isn’t always the case that the integral is useful, e.g. the flux of the gravitational field has no use that I know of. But mathematically you can always take that integral. So the fact that current and current density both exist is not surprising, one automatically implies the other.

- #3

feynman1

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Thanks. Why is electric current not defined as a vector while electric current density is defined as a vector?

It isn’t always the case that the integral is useful, e.g. the flux of the gravitational field has no use that I know of. But mathematically you can always take that integral. So the fact that current and current density both exist is not surprising, one automatically implies the other.

- #4

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If the direction of current is important then a vector is provided by including as a vector the medium thru which the current flows.Thanks. Why is electric current not defined as a vector while electric current density is defined as a vector?

Example: The ## \bf F = \bf B i \bf L ## law giving force F on a wire length L with current i.## \bf B ## and ## \bf L ## are vectors while i is a scalar but gets its direction from ## \bf L ##.

Thus, the direction of current i is defined by the direction of the medium ## L, ## e.g. a wire. In vector notation ## \bf F = ## i ## ~\bf L x \bf B ##. i doesn't need to be a vector because i flows in ## \bf L ## and ## L~ \bf {is} ## a vector.

Current density is associated with an electric field. Ohm's law for fields is ## \bf j = \sigma \bf E ##. ## \sigma ## is obviously a scalar so since ## \bf E ## is a vector, so must ## \bf j ## be.

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- #5

Ibix

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Because current is the charge passing through some surface in unit time (usually phrased as "charge passing a point in unit time" in circuits). That's not a vector quantity - fundamentally, it's just a count of electrons (or other charge carriers). Current density, on the other hand, is a measure of the flow of charge at a point, which has direction and a magnitude. And mathematically, ##dI=\vec\rho\cdot d\vec\sigma## - that is, the elementary current is the dot product of the current density and some elementary vector area. That's obviously right (you want the current to depend on the projection of the area perpendicular to the flow of charge), and forces current to be a scalar.Thanks. Why is electric current not defined as a vector while electric current density is defined as a vector?

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The difference is that the current density is a local field quantity, while the flux is an integral quantity. For the local description you want to know in detail, how the quantity in question (here the electric charge) flows at any point and at any time. It's more intuitive to think in terms of "electron theory", i.e., the current is made up by a lot of particles, described by a charge density ##\rho(t,\vec{r})##, which gives the amount of charge per unit volume at the place ##\vec{r}## at time ##t##. Now if ##\vec{v}(t,\vec{r})## is the velocity field of the particles, i.e., it gives the velocity of the particles whiach are at at the time ##t## at the point ##\vec{r}##, you can characterize the "flow" of charge by taking some very small surface and ask, how much charge flows through it per unit time. For that we need a surface-normal vector ##\vec{n}## (which you can choose arbitrarily in the one or the other direction perpendicular to the surface). Now in an infinitesimal time intervall ##\mathrm{d}t## the particles along the surface move by an infinitesimal vector ##\mathrm{d} \vec{r}=\mathrm{d}t \vec{v}(t,\vec{r})##. Now these infinitesimal vectors together with the surface sweep out a volume ##\mathrm{d} V## containing the particles flowing through the surface within the infinitesimal time, ##t##. So the amount of charge flowing through is

$$\mathrm{d} Q=\mathrm{d} V \rho(t,\vec{r}).$$

The volume (including a sign!) is given by

$$\mathrm{d} V = \mathrm{d} \vec{r} \cdot \mathrm{d} \vec{f}=\mathrm{d} t \vec{v}(t,\vec{r}) \cdot \mathrm{d}^2 \vec{f},$$

where ##\mathrm{d}^2 \vec{f}## is a vector with a length given by the area of the infinitesimal surface and the direction by the surface normal vector. So the amount of charge running through the surface is

$$\mathrm{d} Q = \mathrm{d} t \rho(t,\vec{r}) \vec{v}(t,\vec{r}) \cdot \mathrm{d}^2 \vec{f}.$$

The sign of this charge gives you the direction of the net flow, i.e., whether the charge moves more in direction than against the direction of the arbitrarily chosen surface-normal vector.

We have not in anyway specified this infintesimal surface element though. So all you need to get the amount of charge per unit time flowing through a surface with a chosen diretion of the surface normal vector ##\mathrm{d}^2 \vec{f}## is the current density (a vector)

$$\vec{j}(t,\vec{r})=\rho(t,\vec{r}) \vec{v}(t,\vec{r}),$$

and then the charge per unit time flowing through this surface is

$$\mathrm{d} Q/\mathrm{d} t=\mathrm{d}^2 \vec{f} \cdot \vec{j}(t,\vec{r}),$$

which is a scalar. For a finite surface of course you have to integrate over this surface, i.e., the net current (flux) is given by

$$I=\int_{F} \mathrm{d}^2 \vec{f} \cdot \vec{j}.$$

- #7

feynman1

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Many thanks. I like your point of 'curent being scalar current flux', then current is more like quantity of a fluid passing thru while current density is more like velocity of a fluid.

The difference is that the current density is a local field quantity, while the flux is an integral quantity. For the local description you want to know in detail, how the quantity in question (here the electric charge) flows at any point and at any time. It's more intuitive to think in terms of "electron theory", i.e., the current is made up by a lot of particles, described by a charge density ##\rho(t,\vec{r})##, which gives the amount of charge per unit volume at the place ##\vec{r}## at time ##t##. Now if ##\vec{v}(t,\vec{r})## is the velocity field of the particles, i.e., it gives the velocity of the particles whiach are at at the time ##t## at the point ##\vec{r}##, you can characterize the "flow" of charge by taking some very small surface and ask, how much charge flows through it per unit time. For that we need a surface-normal vector ##\vec{n}## (which you can choose arbitrarily in the one or the other direction perpendicular to the surface). Now in an infinitesimal time intervall ##\mathrm{d}t## the particles along the surface move by an infinitesimal vector ##\mathrm{d} \vec{r}=\mathrm{d}t \vec{v}(t,\vec{r})##. Now these infinitesimal vectors together with the surface sweep out a volume ##\mathrm{d} V## containing the particles flowing through the surface within the infinitesimal time, ##t##. So the amount of charge flowing through is

$$\mathrm{d} Q=\mathrm{d} V \rho(t,\vec{r}).$$

The volume (including a sign!) is given by

$$\mathrm{d} V = \mathrm{d} \vec{r} \cdot \mathrm{d} \vec{f}=\mathrm{d} t \vec{v}(t,\vec{r}) \cdot \mathrm{d}^2 \vec{f},$$

where ##\mathrm{d}^2 \vec{f}## is a vector with a length given by the area of the infinitesimal surface and the direction by the surface normal vector. So the amount of charge running through the surface is

$$\mathrm{d} Q = \mathrm{d} t \rho(t,\vec{r}) \vec{v}(t,\vec{r}) \cdot \mathrm{d}^2 \vec{f}.$$

The sign of this charge gives you the direction of the net flow, i.e., whether the charge moves more in direction than against the direction of the arbitrarily chosen surface-normal vector.

We have not in anyway specified this infintesimal surface element though. So all you need to get the amount of charge per unit time flowing through a surface with a chosen diretion of the surface normal vector ##\mathrm{d}^2 \vec{f}## is the current density (a vector)

$$\vec{j}(t,\vec{r})=\rho(t,\vec{r}) \vec{v}(t,\vec{r}),$$

and then the charge per unit time flowing through this surface is

$$\mathrm{d} Q/\mathrm{d} t=\mathrm{d}^2 \vec{f} \cdot \vec{j}(t,\vec{r}),$$

which is a scalar. For a finite surface of course you have to integrate over this surface, i.e., the net current (flux) is given by

$$I=\int_{F} \mathrm{d}^2 \vec{f} \cdot \vec{j}.$$

On the other hand, if a current branches off into 2 branches, currents aren't additive vectorially, and current densities in this case aren't vectorially additive either. But a vector needs to be vectorially additive. How to explain this?

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In your gedankenexperiment think about a wire branching into two other wires (e.g., when having a parallel circuit of two resistors). Then one part of the current (charge flux) goes through the one wire and the other part through the other wire. The total amount of charge going through both wires is of course the sum of the charge going through wire 1 and the charge going through wire 2.

- #9

Now in an infinitesimal time intervall dtdt the particles along the surface move by an infinitesimal vector ##\mathrm{d} \vec{r}=\mathrm{d}t \vec{v}(t,\vec{r})##

The volume (including a sign!) is given by $$\mathrm{d} V = \mathrm{d} \vec{r} \cdot \mathrm{d} \vec{f}=\mathrm{d} t \vec{v}(t,\vec{r}) \cdot \mathrm{d}^2 \vec{f},$$ where ##\mathrm{d}^2 \vec{f}## is a vector with a length given by the area of the infinitesimal surface and the direction by the surface normal vector.

Hey @vanhees71, apologies if I'm being daft here, but how does the ##d^2 \vec{f}## show up? Why isn't it just $$\mathrm{d} V = \mathrm{d} \vec{r} \cdot \mathrm{d} \vec{f}=\mathrm{d} t \vec{v}(t,\vec{r}) \cdot \mathrm{d} \vec{f}$$ with ##\mathrm{d} \vec{f} \equiv \mathrm{d}\vec{A}## as the area vector?

- #10

feynman1

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How can one apply ##\vec{j}(t,\vec{r})=\vec{j}_1(t,\vec{r})+\vec{j}_2(t,\vec{r})## for a wire branching into two other wires (e.g., when having a parallel circuit of two resistors)?

In your gedankenexperiment think about a wire branching into two other wires (e.g., when having a parallel circuit of two resistors). Then one part of the current (charge flux) goes through the one wire and the other part through the other wire. The total amount of charge going through both wires is of course the sum of the charge going through wire 1 and the charge going through wire 2.

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We simply cannot apply it. This equation doesn't necessarily hold, though the integral equation holds, that is $$\oint_S \vec{J}\cdot d\vec{S}=\oint_S \vec{J_1}\cdot d\vec{S}+\oint_S\vec{J_2}\cdot d\vec{S}$$ it holds for any closed smooth surface ##S## that encloses the junction point where ##\vec{J}## splits to ##\vec{J_1}## and ##\vec{J_2}##. I suspect the reason for this is that the current densities ##J##,##J_1##,##J_2## are not being continuous (with respect to the spatial variables) in the point (or surface) of junction ##\vec{r_0}##.How can one apply ##\vec{j}(t,\vec{r})=\vec{j}_1(t,\vec{r})+\vec{j}_2(t,\vec{r})## for a wire branching into two other wires (e.g., when having a parallel circuit of two resistors)?

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- #13

feynman1

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I agree. But textbooks usually take current densities being vectorially addable as an argument for them to be vectors as opposed to currents being not addable. Here current densities aren't addable either, thus not supporting the argument that current densities are vectors while currents aren't.We simply cannot apply it. This equation doesn't necessarily hold, though the integral equation holds, that is $$\oint_S \vec{J}\cdot d\vec{S}=\oint_S \vec{J_1}\cdot d\vec{S}+\oint_S\vec{J_2}\cdot d\vec{S}$$ it holds for any closed smooth surface ##S## that encloses the junction point where ##\vec{J}## splits to ##\vec{J_1}## and ##\vec{J_2}##. I suspect the reason for this is that the current densities ##J##,##J_1##,##J_2## are not being continuous (with respect to the spatial variables) in the point of junction ##\vec{r_0}##.

- #14

But textbooks usually take current densities being vectorially addable as an argument for them to be vectors as opposed to currents being not addable.

I'm not too sure I understand this argument. Current density is a vector by definition ##\vec{j} = \rho \vec{v}## so if you ever need to add them it will be vectorially. I would have thought currents are generally the addable ones, i.e. take a surface containing the junction and then the net increase in charge contained by that surface per unit time must be zero, so you get out KCL.

Is it ever necessary to add together two current density vectors? I don't know, perhaps someone more knowledgeable can inform, but I can't think of any physical scenario in which that would be the case.

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That is just history. We certainly could have called the current density “current” and then we would probably have called current “current flux”. I would have preferred that naming. Regardless of what you call the vector field you can always construct an associated scalar by integrating over an area.Thanks. Why is electric current not defined as a vector while electric current density is defined as a vector?

- #16

feynman1

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Textbooks say the currents aren't vectors because they aren't vectorially addable. ButI'm not too sure I understand this argument. Current density is a vector by definition ##\vec{j} = \rho \vec{v}## so if you ever need to add them it will be vectorially. I would have thought currents are generally the addable ones, i.e. take a surface containing the junction and then the net increase in charge contained by that surface per unit time must be zero, so you get out KCL.

Is it ever necessary to add together two current density vectors? I don't know, perhaps someone more knowledgeable can inform, but I can't think of any physical scenario in which that would be the case.

OK thanks a lot. I can accept this history argument.That is just history. We certainly could have called the current density “current” and then we would probably have called current “current flux”. I would have preferred that naming. Regardless of what you call the vector field you can always construct an associated scalar by integrating over an area.

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Of course the equation DOES hold. By definition ##\vec{j}## is a vector field!We simply cannot apply it. This equation doesn't necessarily hold, though the integral equation holds, that is $$\oint_S \vec{J}\cdot d\vec{S}=\oint_S \vec{J_1}\cdot d\vec{S}+\oint_S\vec{J_2}\cdot d\vec{S}$$ it holds for any closed smooth surface ##S## that encloses the junction point where ##\vec{J}## splits to ##\vec{J_1}## and ##\vec{J_2}##. I suspect the reason for this is that the current densities ##J##,##J_1##,##J_2## are not being continuous (with respect to the spatial variables) in the point (or surface) of junction ##\vec{r_0}##.

- #18

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The current density is a proper vector field so that equation always holds. It is not useful for the branching because the ##\vec r## is different before and after the branch. But make no mistake ##\vec j## is a completely valid vector field.How can one apply ##\vec{j}(t,\vec{r})=\vec{j}_1(t,\vec{r})+\vec{j}_2(t,\vec{r})## for a wire branching into two other wires (e.g., when having a parallel circuit of two resistors)?

- #19

feynman1

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We were talking about the addition of j alone, not of its integral.Of course the equation DOES hold. By definition ##\vec{j}## is a vector field!

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I don't think it necessarily holds in the case of a junction in a typical circuit. It doesn't necessarily holds at the junction point.Of course the equation DOES hold. By definition →jj→ is a vector field!

In other cases in electromagnetism it might hold but not in this case not necessarily.

- #21

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No! ##\vec{r}## is the position vector at which you look how large the current density is at this place. There's nothing changing only because there's a flow of electrons branching at a junction in a circuit but ##\vec{j}(t,\vec{r})## precisely describes how much particles flow through each surface anywhere in space.That is just history. We certainly could have called the current density “current” and then we would probably have called current “current flux”. I would have preferred that naming. Regardless of what you call the vector field you can always construct an associated scalar by integrating over an area.

The current density is a proper vector field so that equation always holds. It is not useful for the branching because the ##\vec r## is different before and after the branch. But make no mistake ##\vec j## is a completely valid vector field.

- #22

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I am not sure what the “No!” is for. Nothing I said disagrees with anything you said.No! ##\vec{r}## is the position vector at which you look how large the current density is at this place. There's nothing changing only because there's a flow of electrons branching at a junction in a circuit but ##\vec{j}(t,\vec{r})## precisely describes how much particles flow through each surface anywhere in space.

- #23

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That equation always holds, but in the case of branching that equation doesn’t doesn’t tell you what you want.I don't think it necessarily holds in the case of a junction in a typical circuit. It doesn't necessarily holds at the junction point.

In other cases in electromagnetism it might hold but not in this case not necessarily.

- #24

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The reason it doesn't hold necessarily is that the geometry of the circuit around of junction can be anything:

Counter example on why it doesn't necessarily hold:

Suppose we have a constant current density ##\vec{J}## with magnitude ##|\vec{J}|=\alpha## that is along the x-axis (it is like a dirac delta function ##\alpha\delta (\sqrt {y^2+z^2})\hat x##) and comes from x<0. At the origin it branches with two branches , one ##\vec{J_1}## along the line y=x (x>0) with magnitude ##\frac{\alpha}{3}## and the other ##\vec{J_2}## along the line y=-x (x>0) with magnitude ##\frac{2\alpha}{3}##. I believe if we do the math at the junction point (the origin) it will be $$\vec{J}(\vec{0})\neq \vec{J_1}(\vec{0})+\vec{J_2}(\vec{0})$$

Furthermore if we choose a point ##\vec{r_0}## along the line y=x, other than the origin, ##J(\vec{r_0})=\vec{0}=J_2(\vec{r_0})## but ##J_1(\vec{r_0})\neq \vec{0}##.

Counter example on why it doesn't necessarily hold:

Suppose we have a constant current density ##\vec{J}## with magnitude ##|\vec{J}|=\alpha## that is along the x-axis (it is like a dirac delta function ##\alpha\delta (\sqrt {y^2+z^2})\hat x##) and comes from x<0. At the origin it branches with two branches , one ##\vec{J_1}## along the line y=x (x>0) with magnitude ##\frac{\alpha}{3}## and the other ##\vec{J_2}## along the line y=-x (x>0) with magnitude ##\frac{2\alpha}{3}##. I believe if we do the math at the junction point (the origin) it will be $$\vec{J}(\vec{0})\neq \vec{J_1}(\vec{0})+\vec{J_2}(\vec{0})$$

Furthermore if we choose a point ##\vec{r_0}## along the line y=x, other than the origin, ##J(\vec{r_0})=\vec{0}=J_2(\vec{r_0})## but ##J_1(\vec{r_0})\neq \vec{0}##.

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- #25

jtbell

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In order to add them, ##\vec {j_1}## and ##\vec {j_2}## must be evaluated at the same point. How can you do that when they’re in different wires or resistors?How can one apply ##\vec{j}(t,\vec{r})=\vec{j}_1(t,\vec{r})+\vec{j}_2(t,\vec{r})## for a wire branching into two other wires (e.g., when having a parallel circuit of two resistors)?

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You can add them if you set them to be zero "outside of their domain". But then it is obvious that the equation doesn't hold in all points e.g in a point ##r_2## of the domain of ##J_2## where ##\vec{J_2}(\vec{r_2})\neq\vec{0}## but ##\vec{J}(\vec{r_2})=\vec{J_1}(\vec{r_2})=\vec{0}##In order to add them, ##\vec {j_1}## and ##\vec {j_2}## must be evaluated at the same point. How can you do that when they’re in different wires or resistors?

I believe that my example at post #24 shows that even in the junction point that is common to the domain of all three, the equation doesn't necessarily hold.

- #27

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It holds absolutely 100% regardless of the geometry. You are confusing “holds” with “applies”. “Holds” means that the equation is true, and that equation is always true in classical electrodynamics. “Applies” means that it is useful for answering a particular question.The reason it doesn't hold necessarily is that the geometry of the circuit around of junction can be anything:

Notice very carefully how the equation we are discussing was originally defined:

In contrast your "counter example" is as follows:if you have two sources of flowing charges at each given point ##\vec{r}## at time ##t## the total current density is ##\vec{j}(t,\vec{r})=\vec{j}_1(t,\vec{r})+\vec{j}_2(t,\vec{r})##.

So in your counter example ##\vec J_1## and ##\vec J_2## are not the current densities due to sources but simply unsourced current densities in the individual wires.Counter example on why it doesn't necessarily hold:

Suppose we have a constant current density ##\vec{J}## with magnitude ##|\vec{J}|=\alpha## that is along the x-axis (it is like a dirac delta function ##\alpha\delta (\sqrt {y^2+z^2})\hat x##) and comes from x<0. At the origin it branches with two branches , one ##\vec{J_1}## along the line y=x (x>0) with magnitude ##\frac{\alpha}{3}## and the other ##\vec{J_2}## along the line y=-x (x>0) with magnitude ##\frac{2\alpha}{3}##. I believe if we do the math at the junction point (the origin) it will be $$\vec{J}(\vec{0})\neq \vec{J_1}(\vec{0})+\vec{J_2}(\vec{0})$$

So a proper definition would be ##\vec J_1## is the current density for a source on branch 1, so it includes an amount of current ##\frac{\alpha}{3}## in the x-axis wire, and ##\vec J_2## is the current density for a source on branch 2, so it includes an amount of current ##\frac{2\alpha}{3}## in the x-axis wire. Then indeed $$\vec{J} = \vec{J_1}+\vec{J_2}$$ everywhere including the origin. So it holds but doesn't apply because it already assumes the current bends at the wire which is presumably what you wanted to find out without assuming it.

You cannot change the meaning of the terms and then claim that an equation doesn't hold. If you need to change the meaning of the terms then the equation doesn't apply.

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- #28

jtbell

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$$\int_{S_1} {\vec j \cdot d \vec S} + \int_{S_2} {\vec j \cdot d \vec S} = \int_{S_3} {\vec j \cdot d \vec S} \\ I_1 + I_2 = I_3$$

- #29

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This is confusing ,in my opinion it doesn't hold neither it applies. However you are right that I had in my mind the equation as it was presented in the context of post #10 and not that of @vanhees71 post.It holds absolutely 100% regardless of the geometry. You are confusing “holds” with “applies”. “Holds” means that the equation is true, and that equation is always true in classical electrodynamics. “Applies” means that it is useful for answering a particular question.

Notice very carefully how the equation we are discussing was originally defined:

Ok I see now what you meant, we can agree on that.You cannot change the meaning of the terms and then claim that an equation doesn't hold. If you need to change the meaning of the terms then the equation doesn't apply.

- #30

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$$\vec{j}(t,\vec{x})=\vec{j}_1(t,\vec{x}) + \vec{j}_2(t,\vec{x}).$$

This describes the situation that a current density at one place comes from two sources I labelled with 1 and 2. Of course there's only one total current density in the entire game.

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