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Curl and Divergence · Tutorial 913 of 999

Scale Factors in Orthogonal Coordinates

Use orthogonal scale factors to express divergence and curl, and see how infinitesimal flux and circulation lead to the formulas.

Advanced 10 min read

What You'll Learn

  • Define scale factors and unit vectors for orthogonal coordinates.
  • Find the infinitesimal lengths, face areas, and volume from the scale factors.
  • Derive divergence from the net flux through a small coordinate box.
  • Derive curl components from circulation around small coordinate faces.
  • Identify which vector components and scale factors each formula requires.
  • Recognize why differentiating components alone can give an incorrect result.

Why Scale Factors Matter

In Cartesian coordinates, the coordinate directions are perpendicular and a unit change in any coordinate corresponds to a unit distance. In other orthogonal coordinate systems, the coordinate directions remain perpendicular, but the physical distance represented by a coordinate change can vary from point to point.

Divergence and curl measure local behavior in physical space: divergence describes local net outflow, and curl describes local circulation. Their formulas in orthogonal coordinates must therefore account for how coordinate changes translate into physical lengths, areas, and volumes. The scale factors provide exactly that information.

Orthogonal Coordinates and Scale Factors

Let a smooth coordinate map describe position by coordinates \(q_1,q_2,q_3\): \(\mathbf{r}=\mathbf{r}(q_1,q_2,q_3)\). Assume the coordinate directions are nonzero and mutually perpendicular in the region under consideration. Define the scale factors and corresponding unit vectors by

$$ h_i=\left|\frac{\partial \mathbf{r}}{\partial q_i}\right|, \qquad \mathbf{e}_i=\frac{1}{h_i}\frac{\partial \mathbf{r}}{\partial q_i}, \qquad i=1,2,3. $$

Thus a small change \(dq_i\), with the other coordinates fixed, has physical length \(h_i\,dq_i\). We take \(h_i>0\) and choose the order of the coordinates so that \(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\) form a right-handed orthonormal frame. A vector field is written in physical components as \(\mathbf{A}=A_1\mathbf{e}_1+A_2\mathbf{e}_2+A_3\mathbf{e}_3\).

Definition: The scale factor \(h_i\) converts a coordinate increment \(dq_i\) into physical distance along the \(q_i\)-coordinate curve: \(ds_i=h_i\,dq_i\). The components \(A_i\) are the vector’s components along the unit directions \(\mathbf{e}_i\), not rates of change with respect to the coordinates.

The perpendicularity of the coordinate directions makes the small coordinate box’s edge lengths \(h_1dq_1,h_2dq_2,h_3dq_3\). Consequently, its volume and the areas of its coordinate faces are

$$ dV=h_1h_2h_3\,dq_1dq_2dq_3, \qquad dS_i=h_jh_k\,dq_jdq_k \quad (i,j,k\text{ distinct}). $$

Here \(dS_i\) is the area of a face perpendicular to \(\mathbf{e}_i\). These expressions are local: the scale factors are evaluated at the position of the element, and can vary across it.

Worked Example: Scale Factors for Elliptic Coordinates

Worked Example: Scale Factors for Elliptic Coordinates

Consider the coordinates defined by \(\mathbf{r}(\mu,\nu,w)=(a\cosh\mu\cos\nu,\ a\sinh\mu\sin\nu,\ w)\), where \(a>0\) and \(\mu>0\). Find the scale factors and the volume element.

Differentiate the position vector with respect to each coordinate:

$$ \mathbf{r}_{\mu}=(a\sinh\mu\cos\nu,\ a\cosh\mu\sin\nu,\ 0), \qquad \mathbf{r}_{\nu}=(-a\cosh\mu\sin\nu,\ a\sinh\mu\cos\nu,\ 0), \qquad \mathbf{r}_{w}=(0,0,1). $$

The first two directions are perpendicular because

$$ \mathbf{r}_{\mu}\cdot\mathbf{r}_{\nu} =-a^2\sinh\mu\cosh\mu\cos\nu\sin\nu +a^2\sinh\mu\cosh\mu\sin\nu\cos\nu =0. $$

They are also perpendicular to \(\mathbf{r}_w\). Their squared lengths are

$$ |\mathbf{r}_{\mu}|^2 =a^2\left(\sinh^2\mu\cos^2\nu+\cosh^2\mu\sin^2\nu\right) =a^2(\sinh^2\mu+\sin^2\nu), $$

where \(\cosh^2\mu=1+\sinh^2\mu\) and \(\cos^2\nu=1-\sin^2\nu\) give the last equality. The same expansion gives \(|\mathbf{r}_{\nu}|^2=a^2(\sinh^2\mu+\sin^2\nu)\), while \(|\mathbf{r}_w|=1\). Set \(H=a\sqrt{\sinh^2\mu+\sin^2\nu}\). Then \(h_\mu=h_\nu=H\) and \(h_w=1\). In particular, the coordinate-box volume is

$$ dV=H^2\,d\mu\,d\nu\,dw =a^2(\sinh^2\mu+\sin^2\nu)\,d\mu\,d\nu\,dw. $$

As an orientation check, the \(z\)-component of \(\mathbf{r}_\mu\times\mathbf{r}_\nu\) is \(a^2(\sinh^2\mu+\sin^2\nu)>0\). Thus the ordered unit directions \(\mathbf{e}_\mu,\mathbf{e}_\nu,\mathbf{e}_w\) are right-handed where the coordinates are regular, as required.

Divergence from Flux Through a Coordinate Box

Divergence at a point is the limiting net outward flux per unit volume as a region around that point shrinks to zero. To see how the scale factors enter, consider the two faces perpendicular to \(\mathbf{e}_1\). Their leading-order outward fluxes have opposite signs. Their net contribution is the change, across a \(q_1\)-increment, in \(A_1h_2h_3\), multiplied by \(dq_2dq_3\). Dividing that contribution by the box volume gives

$$ \frac{\partial_1(A_1h_2h_3)\,dq_1dq_2dq_3} {h_1h_2h_3\,dq_1dq_2dq_3} = \frac{1}{h_1h_2h_3}\frac{\partial}{\partial q_1}(h_2h_3A_1). $$

The other two pairs of faces give the corresponding expressions for \(A_2\) and \(A_3\). Adding them gives the divergence formula. The argument uses the first-order change in flux across each pair of faces; terms of higher order vanish after division by volume and passage to the limit.

Formula: For a differentiable vector field in a regular, right-handed orthogonal coordinate system,
$$ \nabla\cdot\mathbf{A} = \frac{1}{h_1h_2h_3} \left[ \frac{\partial}{\partial q_1}(h_2h_3A_1) +\frac{\partial}{\partial q_2}(h_3h_1A_2) +\frac{\partial}{\partial q_3}(h_1h_2A_3) \right]. $$

Each term uses one vector component and the area factor for a face perpendicular to that component’s direction. The derivatives act on the whole product, including its scale factors. This is essential: the area of a coordinate face may change with position even when the vector component does not.

Worked Example: Divergence in Elliptic Coordinates

Worked Example: Divergence in Elliptic Coordinates

Use the elliptic coordinates from the previous example, with scale factors \(h_\mu=h_\nu=H\), \(h_w=1\), to find the divergence of a field whose physical components are \(A_\mu=\mu/H\), \(A_\nu=\sin\nu/H\), and \(A_w=w/H^2\). Evaluate it at \(\mu=\operatorname{asinh}(1)\), \(\nu=\pi/2\), with any \(w\).

The scale-factor products in the divergence formula are \(h_\nu h_w=H\), \(h_w h_\mu=H\), and \(h_\mu h_\nu=H^2\). Multiplying each by its component gives

$$ h_\nu h_w A_\mu=\mu,\qquad h_w h_\mu A_\nu=\sin\nu,\qquad h_\mu h_\nu A_w=w. $$

The required derivatives are \(\partial_\mu\mu=1\), \(\partial_\nu\sin\nu=\cos\nu\), and \(\partial_w w=1\). Therefore

$$ \nabla\cdot\mathbf{A} =\frac{1}{H^2}(1+\cos\nu+1) =\frac{2+\cos\nu}{H^2}. $$

At the specified point, \(\sinh\mu=1\), so \(H^2=a^2(1+\sin^2(\pi/2))=2a^2\), and \(\cos(\pi/2)=0\). Hence the value is \(2/(2a^2)=1/a^2\). Substituting these values into the unsimplified numerator also gives \(1+0+1=2\); thus the evaluated result agrees with the three individual contributions.

Curl from Circulation Around a Coordinate Face

The component of curl along a unit normal is the limiting circulation per unit area around a small surface perpendicular to that normal, with the boundary orientation chosen by the right-hand rule. For the component along \(\mathbf{e}_1\), use a small face in the \(q_2,q_3\) directions. Its area is \(h_2h_3\,dq_2dq_3\).

Traverse the face in the positive orientation: first along \(+q_2\), then \(+q_3\), then back along \(-q_2\), and finally along \(-q_3\). Along a \(q_i\)-edge, the line element is \(h_i\,dq_i\), so the tangential contribution to the line integral is \(A_i h_i\,dq_i\). The opposing \(q_2\)-edges therefore contribute a difference in \(h_2A_2\) across \(q_3\); the opposing \(q_3\)-edges contribute a difference in \(h_3A_3\) across \(q_2\). To leading order, the total circulation is

$$ \left[ \frac{\partial}{\partial q_2}(h_3A_3) -\frac{\partial}{\partial q_3}(h_2A_2) \right]dq_2dq_3. $$

Dividing by the face area gives the \(\mathbf{e}_1\)-component of curl. Taking the corresponding positively oriented faces for the other two directions gives the complete formula.

Formula: In a regular, right-handed orthogonal coordinate system, the curl is
$$ \nabla\times\mathbf{A} = \frac{1}{h_2h_3} \left[ \frac{\partial}{\partial q_2}(h_3A_3) -\frac{\partial}{\partial q_3}(h_2A_2) \right]\mathbf{e}_1 + \frac{1}{h_3h_1} \left[ \frac{\partial}{\partial q_3}(h_1A_1) -\frac{\partial}{\partial q_1}(h_3A_3) \right]\mathbf{e}_2 + \frac{1}{h_1h_2} \left[ \frac{\partial}{\partial q_1}(h_2A_2) -\frac{\partial}{\partial q_2}(h_1A_1) \right]\mathbf{e}_3. $$

The component along \(\mathbf{e}_1\) uses \(A_2\) and \(A_3\), not \(A_1\); it measures circulation in the plane perpendicular to \(\mathbf{e}_1\). The other components follow the same cyclic pattern. As in the divergence formula, derivatives act on the entire scale-factor-weighted component.

Worked Example: Curl in Elliptic Coordinates

Worked Example: Curl in Elliptic Coordinates

In the same coordinates, let \(A_\mu=0\), \(A_\nu=0\), and \(A_w=\mu\). Find the curl and evaluate it at \(\mu=\operatorname{asinh}(1)\), \(\nu=\pi/2\).

The \(\mathbf{e}_\mu\)-component uses \(A_\nu\) and \(A_w\), giving \((1/H)\partial_\nu(\mu)=0\). The \(\mathbf{e}_\nu\)-component uses \(A_w\) and \(A_\mu\), giving \((1/H)[\partial_w(0)-\partial_\mu(\mu)]=-1/H\). The \(\mathbf{e}_w\)-component vanishes because both \(A_\mu\) and \(A_\nu\) are zero. Therefore

$$ \nabla\times\mathbf{A}=-\frac{1}{H}\mathbf{e}_\nu. $$

At the specified point \(H=a\sqrt{2}\), so the curl is \(-\mathbf{e}_\nu/(a\sqrt{2})\). Check the sign using circulation on a small \((\mu,w)\)-face oriented with normal \(+\mathbf{e}_\nu\). Its positive boundary traverses \(+w\) at the lower \(\mu\), then \(-w\) at the higher \(\mu\). Since \(A_w=\mu\), the two contributions sum to \(-d\mu\,dw\); the \(\mu\)-edges contribute zero. The face area is \(H\,d\mu\,dw\), so circulation divided by area tends to \(-1/H\), confirming both the sign and the computed component.

A Common Mistake: Differentiating Components Alone

A tempting but generally incorrect shortcut is to add \(\partial A_1/\partial q_1\), \(\partial A_2/\partial q_2\), and \(\partial A_3/\partial q_3\) to obtain divergence. That ignores changes in the areas of the coordinate faces. For a concrete check, take the field \(\mathbf{A}=\mathbf{e}_\mu\) in the elliptic coordinates above. Its components are \(A_\mu=1\), \(A_\nu=A_w=0\). The component-only shortcut gives zero. The correct formula gives

$$ \nabla\cdot\mathbf{A} =\frac{1}{H^2}\frac{\partial H}{\partial\mu}. $$

At \(\mu=\operatorname{asinh}(1)\), \(\nu=\pi/2\), the scale factor has \(H=a\sqrt{2}\) and \(\partial H/\partial\mu=a\). Thus the actual divergence there is \(a/(2a^2)=1/(2a)\), not zero. The missing contribution comes from the changing area \(H\,d\nu\,dw\) of the faces perpendicular to \(\mathbf{e}_\mu\).

How to Use the Formulas Reliably

1
Find the scale factors.
Differentiate the position vector with respect to each coordinate and take the lengths of those tangent vectors.
2
Confirm orthogonality and orientation.
Check that the coordinate directions are perpendicular and that the ordered unit vectors are right-handed.
3
Use physical components.
Write the field as \(A_1\mathbf{e}_1+A_2\mathbf{e}_2+A_3\mathbf{e}_3\), then insert those components in the appropriate formula.
4
Keep the scale factors inside each derivative.
Only after differentiating the full weighted product should you simplify and evaluate.

The formulas require a regular coordinate map: the scale factors must be nonzero in the region where the derivatives are being computed. If a scale factor vanishes or the coordinates cease to describe a regular local coordinate system, the formulas cannot be applied at that point without additional analysis.

Check Your Understanding

Use the definitions and formulas to answer each question.

  1. What physical length corresponds to a small coordinate change \(dq_2\)?
  2. Which vector components enter the curl component along \(\mathbf{e}_3\)?
  3. Why must the divergence formula differentiate a product of a component and scale factors?
  4. For a coordinate box, what is the area of a face perpendicular to \(\mathbf{e}_2\)?
  5. What regularity condition is needed to use these formulas at a point?

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