Electric current is not a vector while electric current density is a vector

  • Thread starter feynman1
  • Start date
  • #26
Delta2
Homework Helper
Insights Author
Gold Member
4,569
1,857
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?
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}##

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
Dale
Mentor
Insights Author
2020 Award
31,970
8,884
The reason it doesn't hold necessarily is that the geometry of the circuit around of junction can be anything:
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:
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})##.
In contrast your "counter example" is as follows:
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 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.

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.
 
Last edited:
  • Skeptical
  • Like
Likes weirdoguy and Delta2
  • #28
jtbell
Mentor
15,817
4,124
The way I picture this is that there is a single ##\vec j## field that is distributed through the interior of the wires and their junction region. We get the different currents in each wire by integrating over different surfaces that span each wire separately.

current.gif


$$\int_{S_1} {\vec j \cdot d \vec S} + \int_{S_2} {\vec j \cdot d \vec S} = \int_{S_3} {\vec j \cdot d \vec S} \\ I_1 + I_2 = I_3$$
 
  • Like
Likes vanhees71, etotheipi and Delta2
  • #29
Delta2
Homework Helper
Insights Author
Gold Member
4,569
1,857
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:
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.

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.
Ok I see now what you meant, we can agree on that.
 
  • #30
vanhees71
Science Advisor
Insights Author
Gold Member
18,869
9,722
Again: The current density is a vector FIELD, i.e., a function of space and time, and of course vector fields add always taken at the same time and place, i.e., the correct formula written out with all arguments is
$$\vec{j}(t,\vec{x})=\vec{j}_1(t,\vec{x}) + \vec{j}_2(t,\vec{x}).$$
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.
 

Related Threads on Electric current is not a vector while electric current density is a vector

  • Last Post
Replies
21
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
21
Views
13K
Replies
4
Views
4K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
48K
  • Last Post
Replies
4
Views
811
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
2
Replies
26
Views
3K
Top