# Current density in an elliptical loop

1. Aug 18, 2015

### QuasiParticle

Suppose we have an elliptical loop of wire in the x-y plane with constant cross-section. And let's also assume that the cross-section is very small, so we have a thin wire.

$$\vec{r}=\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} a\cos \theta \\ b \sin \theta \end{bmatrix}$$
A tangent unit vector to the loop is

$$\hat{v}= \frac{d\vec{r}/d\theta}{|d\vec{r}/d\theta|}=\frac{1}{\sqrt{(a/b)^2 y^2 + (b/a)^2 x^2}} \begin{bmatrix} -(a/b) y \\ \phantom{-}(b/a) x \end{bmatrix}$$
When current flows in the loop, it sounds at first like a reasonable assumption that it is of the form

$$\vec{J} = J_0 \hat{v},$$
that is, constant and directed along the tangent. According to the continuity equation for a system with time-independent charge density, the divergence of the current density should vanish everywhere.

$$\nabla \cdot \vec{J} = 0$$
If we compute the divergence of the above current density, it is of the form

$$\nabla \cdot \vec{J} = \frac{ab(b^2-a^2)xy}{(\textrm{something} > 0)}$$
This is identically zero if and only if $a=b$, i.e. if the current loop is circular. So it seems that we cannot have an elliptical loop with a constant current density. Instead the "constant" part of the current density should have an angular dependence $J_0 = J_0(\theta)$. I was somewhat surprised to find this out after having (naively) made the assumption $|\vec{J}|=\textrm{const.}$ and I still don't quite understand why the constant current density is not possible. It must have to do with the varying curvature of the loop, so that the current cannot flow perpendicular to the cross-section at all points? I suppose thinking in terms of current density can be misleading. Does anyone have any comments or insight into this issue?

2. Aug 18, 2015

### Staff: Mentor

This is a continuous current density field. You will need at least one, and probably two delta functions to make it represent a wire.

3. Aug 18, 2015

### Philip Wood

I'm worried about your computation of the divergence. I'd do it by considering a right slice of the wire (that is a slice normal to the tangent). In which case the divergence is clearly zero in the circumstances you have described. If you are trying to compute the divergence inside the wire by applying the usual sum of second derivatives in your chosen cartesian co-ordinates the algebra is very messy; are you sure you haven't made any mistakes? You are, imo, being a bit of a masochist because divergence stays the same when you rotate co-ordinates, so you're allowed to re-orientate and do it by the method I described originally.

4. Aug 18, 2015

### marcusl

You butchered your coordinates. Theta is a parametric variable that is not equal to polar angle $\phi=tan^{-1}\left(\frac{y}{x}\right)$, so differentiating with respect to theta is not correct. They are related by $$\phi=tan^{-1}\left(\frac{b}{a}tan\theta\right)$$
The expressions get very complicated so you must be very careful as you solve this problem.

5. Aug 18, 2015

### Philip Wood

I did reach the op's final (cartesian) expression for the tangent unit vector (by a different route).

6. Aug 18, 2015

### marcusl

Try a polar plot of the OP's expressions in polar coordinates $\theta$ and $\rho=\sqrt{x^2(\theta)+y^2(\theta)}$ for, say, a=3 and b=1. It isn't an ellipse because $\theta$ isn't physical angle in the x-y plane.

Last edited: Aug 18, 2015
7. Aug 18, 2015

### Philip Wood

So you say the op's final cartesian expression for the unit tangent vector is wrong?

8. Aug 18, 2015

### marcusl

Yes. Substituting for x and y gives $$\hat{v}=\frac{-\hat{x}a\sin\theta+\hat{y}b\cos\theta}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}$$ where the OP is taking $\theta$ to be polar angle in the x-y plane. This is wrong. I get $$\frac{d\vec{l}}{d\theta}=ab\ \frac{-\hat{x}a^2\sin\theta+\hat{y}b^2\cos\theta}{(b^2\cos^2\theta+a^2\sin^2\theta)^{3/2}}$$

9. Aug 18, 2015

### Philip Wood

I get $\frac{dy}{dx}= -\frac{b^2 x}{a^2 y}$.
This is tan $\psi$ in which $\psi$ is the angle of the tangent to the x axis.
Using trig identities I found cos $\psi$ and sin $\psi$ which are, surely, the x and y components of the unit tangent vector. And lo! They were the same as the op gave, in terms of x and y.

10. Aug 18, 2015

### QuasiParticle

Yes, you are right. In a wire with infinitesimal cross-section there should be deltas and the current must follow the tangent. However, the result in my original post is independent of the size of the wire. So we may consider very thin, but finite size wire.

Maybe this whole thing is something very trivial, but for some reason it is fooling my intuition. Assuming the current density to be along the tangent and constant just seems "natural". But clearly this is wrong.

I don't quite follow. I see no problem in calculating the divergence in Cartesian coordinates. I just plugged everything into Mathematica. If I did the computation, for example, in polar coordinates, I would first have to determine the angular and radial components of the vector field.

You are correct that $\theta$ was not the polar angle. But I did not treat it as such until I was talking about $J_0$, at which time $\theta$ really meant the polar coordinate. Originally I was using another symbol for the parametric variable, but changed it at the last minute to $\theta$. I apologize for the confusion.

But the computation of the tangent vector should nevertheless be correct. The parameter does not need to equal the polar angle.

11. Aug 18, 2015

### marcusl

That's ok then.

12. Aug 18, 2015

### Staff: Mentor

On the contrary, I think that you have shown that you cannot do that. In other words, this is not a valid current distribution for a finite sized wire.

13. Aug 18, 2015

### QuasiParticle

Exactly. I meant in post #10 that the contradiction in the first post exists regardless of the (finite) size of the wire.

I guess I just have to accept that it is so. For some reason my gut feeling is fighting back. It would be nice to have some intuitive picture of why the assumed current distribution is impossible.

14. Aug 18, 2015

### Philip Wood

Don't understand. I'm imagining a small cubic box inside the wire. Divergence seems unproblematically zero if the cube is orientated sensibly, i.e. with normals to two opposite faces parallel to the tangent vector. Shouldn't change if we orientate the cube parallel to axes.

No, sorry, it's all because of the changing direction of the tangent. I'd like to know what the denominator of your expression for the divergence is. Are you sure that the divergence doesn't go to zero as the ellipse becomes very large (so its thickness becomes negligible)?

Last edited: Aug 18, 2015
15. Aug 19, 2015

### Staff: Mentor

The problem is that you started with the assumption of a thin wire and your reasoning was based on that, but then when you actually wrote your distribution you violated that assumption and thereby brought in a bunch of problems that you had neglected previously. You cannot reasonably start a derivation with a thin wire assumption and then come up with a thick wire result.

Let's look at some of the problems that brings.

First, the "stream lines" for your proposed current distribution do not maintain a constant distance from each other, so either your wire cannot be uniform thickness or you must have charge entering and leaving the sides of the wire. If you have a non uniform thickness then the current density will decrease proportionally.

Second, the unit tangent vector is not well defined for a finite thickness elliptical wire. A vector parallel to the inner surface points in a slightly different direction than one parallel to the outer surface.

Third, the outer streamlines are longer than the inner ones. I am not sure if this causes a problem, but I am not sure that it doesn't.

Finally, the angular coordinate is not as simple for an ellipse as for a circle. The coordinate I believe that you are using is not perpendicular to the ellipse except at the principal axes. That probably doesn't matter for a thin wire, but for a thick one it means that angular cross sections will have a different thickness than the wire thickness.

All of these go away if you actually use a mathematical expression for your current distribution which is consistent with your stated assumptions and reasoning.

16. Aug 19, 2015

### Philip Wood

I find that, putting $c=\frac{a}{b}$,
$$\textrm{div} j = j_0 \frac{(1-c^2)c^{2} xy}{(c^{4} y ^{2} + x^{2})^{3/2}}$$

The bottom line of the fraction is of dimensions length cubed, the top, length squared. If we scale up the ellipse uniformly, keeping x and y in the same proportion, but leaving the shape-setter, c, alone, the divergence gets smaller and smaller. This means that your constant cross-section and other assumptions become more and more valid. Hasn't that nailed it?

17. Aug 19, 2015

### QuasiParticle

DaleSpam, thank you for your response.

I think I should restate my assumptions. Let's assume we have a finite size circular cross-section. The center of this cross-section moves along the elliptical path so that the normal of the circular disc always points in the direction of the ellipse tangent and in this manner "draws" the wire. The inner edge tangent of this wire should match the outer edge tangent. What is difficult for me to understand is that current cannot flow so that at every point it is flowing in the direction of the normal of the (circular) cross-section and in such a way that the current density (current through this circular cross-section) is constant, without violating the continuity equation. I think by "constructing" the current density this way avoids problems one, two, and four of your post #15. And I don't see why, in principle, some EM-fields could not accelerate the charge carriers so that problem three would also be solved.

The contradiction with the continuity equation suggests that it is fundamentally impossible to construct the current density in the manner described above without adding and taking off charge along the wire.

You are correct about the angular coordinate not being perpendicular to the ellipse. But I don't think any of my rationale relies on it being perpendicular.

That matches my expression. And indeed this does help somewhat with the mental problem I had with the limiting case of a larger loop. Thank you. I still don't understand why it seems fundamentally impossible to construct a "tangent current density" in a finite wire. Also in your expression, the wire diameter does not appear, so the results are independent of the diameter, and I'm wondering about the scaling law violation.

18. Sep 12, 2015

### Staff: Mentor

I agree. But the current density constructed in this manner will not be the one in the OP.
I am not sure what you mean here. By "the manner described above" are you referring to the current density of the OP or of your more recent post? For the manner of the OP I agree, but you haven't shown any contradiction with the continuity equation for the more recent manner.

19. Sep 13, 2015

### QuasiParticle

To me they seem to be the same. How are they not?

20. Sep 13, 2015

### Staff: Mentor

As far as I can tell, The stream lines for the current density in the OP do not maintain a constant distance from each other. The stream lines for the current density in the recent post do.

Last edited: Sep 13, 2015
21. Sep 15, 2015

### vanhees71

Let's try elliptic cylinder coordinates,
$$(x,y,z)=(a \cosh \eta \cos \varphi,b \sinh \eta \sin \varphi,z),$$
where
$$\eta \in \mathbb{R}_{>0}, \quad \varphi \in [0,2 \pi[, \quad z \in \mathbb{R}.$$
These are orthogonal curvilinear coordinates, because
$$\vec{b}_{\eta}=(a \sinh \eta \cos \varphi, b \cosh \eta \sin \varphi,0),$$
$$\vec{b}_{\varphi}=(-a \cosh \eta \sin \varphi,b \sinh \eta \cos \varphi,0),$$
$$\vec{b}_z=(0,0,1).$$
For the rest of the math, see
https://en.wikipedia.org/wiki/Elliptic_coordinate_system

Now $\eta=\eta_0=\text{const}$, $z=0$ is obviously an ellipse with the semi-axes $a \cosh \eta_0$, $a \sinh \eta_0$. The current density can be described with help of the surface element in the planes perpendicular to the ellipse (i.e., in the planes spanned by $\vec{b}_{\eta}$ and $\vec{b}_{z}$
$$\mathrm{d} F=|\vec{b}_{\eta} \times \vec{b}_{z}| \mathrm{d} \eta \mathrm{d} z=J \mathrm{d} \eta \mathrm{d} z,$$
with
$$J=\frac{a}{\sqrt{2}}\sqrt{\cosh^2(2 \eta)-\cos(2 \varphi)}.$$
The current density thus is given by
$$\vec{j}(\eta,\varphi,z)=\frac{I}{J} \vec{e}_{\varphi}(\eta,\varphi) \delta(\eta-\eta_0) \delta(z-z_0).$$