Vector calculus
Diperkenalkan| English | Makna |
|---|---|
| gradient/ˈɡreɪdɪənt/ | gradien |
| divergence/daɪˈvɜːdʒəns/ | divergensi |
| curl/kɜːl/ | putaran |
A number at every point
- A voltage probe reports one number at each point in space: a whole landscape of values. Later, Maxwell's equations will be statements about that landscape's shape.
- Three operators describe the shape: which way is uphill, where the field flows out, and where it swirls.
Fields and their three operators
- A scalar field gives a number at each point (potential $V$); a vector field gives an arrow (electric field $\mathbf{E}$).
- Gradient 梯度 turns a scalar field into a vector field; divergence 散度 turns a vector field into a scalar; curl 旋度 turns a vector field into another vector field.
Match each operator to what it does.
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Knowing what goes in and what comes out catches most operator mistakes before any arithmetic.
Gradient points uphill
- $\nabla V = \left(\dfrac{\partial V}{\partial x}, \dfrac{\partial V}{\partial y}, \dfrac{\partial V}{\partial z}\right)$ points up the steepest slope of $V$. The electric field points downhill: $\mathbf{E} = -\nabla V$.
- $V = 3x^{2}y$ volts gives $\nabla V = (6xy,\ 3x^{2})$. At $(1, 2)$: $\nabla V = (12, 3)$, so $\mathbf{E} = -(12, 3)\,\dfrac{\text{V}}{\text{m}}$.
V = 3x²y. What is the electric field E at (1, 2)?
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$\nabla V = (6xy, 3x^{2}) = (12, 3)$ at the point, and $\mathbf{E} = -\nabla V = -(12, 3)$.
V = 5x² volts. What is the x-component of E at x = 2 m, in V/m?
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$\dfrac{\partial V}{\partial x} = 10x = 20$ at $x = 2$, so $E_{x} = -20$ V/m: the field points towards lower potential.
Check the gradient with a small step
- Take a small step $\boldsymbol{\delta}$ from the point. The potential changes by about $\nabla V \cdot \boldsymbol{\delta}$, which is $-\mathbf{E} \cdot \boldsymbol{\delta}$.
- From $(1, 2)$, step $\boldsymbol{\delta} = (0.01, 0)$: $\Delta V \approx 12 \times 0.01 = 0.12\,\text{V}$, and directly $3(1.01)^{2}(2) - 6 = 0.1206\,\text{V}$. The field predicts the drop.
V = 3x²y. Using the gradient, estimate the change in V for a step of 0.02 in y from (1, 2), in volts.
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$\dfrac{\partial V}{\partial y} = 3x^{2} = 3$ at the point, so $\Delta V \approx 3 \times 0.02 = 0.06$ V. Directly: $3(1)(2.02) - 6 = 0.06$ exactly, since $V$ is linear in $y$.
Divergence: net outflow
- $\nabla \cdot \mathbf{F} = \dfrac{\partial F_{x}}{\partial x} + \dfrac{\partial F_{y}}{\partial y} + \dfrac{\partial F_{z}}{\partial z}$ is the net outflow per unit volume. For an electric field it measures the charge density there.
- $\mathbf{F} = (x^{2},\ xy,\ 3x^{2}z)$: $\nabla \cdot \mathbf{F} = 2x + x + 3x^{2}$, which is 6 at $(1, 1, 1)$: a net source.
For F = (x², xy, 3x²z), what is the divergence at (1, 1, 1)?
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$\nabla \cdot \mathbf{F} = 2x + x + 3x^{2}$, which is $2 + 1 + 3 = 6$ at the point: a net source there.
Curl: circulation
- The $z$-component of curl is $\dfrac{\partial F_{y}}{\partial x} - \dfrac{\partial F_{x}}{\partial y}$: the circulation per unit area. Electrostatic fields have zero curl; induction is where it is not zero.
- $\mathbf{G} = (-y,\ x,\ 0)$ gives $1 - (-1) = 2$ everywhere. This is a rigid body spinning at 1 rad/s: curl is twice the angular velocity.
What is the z-component of the curl of F = (−3y, 3x, 0)?
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$\dfrac{\partial (3x)}{\partial x} - \dfrac{\partial (-3y)}{\partial y} = 3 + 3 = 6$: rigid rotation at 3 rad/s.
Three common slips
- Operator on the wrong kind of field. Gradient takes a scalar; divergence and curl take a vector.
- Reversing the curl terms. $\dfrac{\partial F_{y}}{\partial x} - \dfrac{\partial F_{x}}{\partial y}$; swapping them flips the rotation.
- "Uniform field, so something flows out." A uniform field has zero divergence and zero curl.
A uniform field F = (1, 0) everywhere has
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Every derivative of a constant is zero, so both operators give zero. No arrows spread apart and none turn.
Recap
- Gradient: steepest uphill (scalar → vector); $\mathbf{E} = -\nabla V$.
- Divergence: net outflow (vector → scalar). Curl: circulation (vector → vector).
- A small step checks a gradient: $\Delta V \approx \nabla V \cdot \boldsymbol{\delta}$.