Scalar and Vector Fields

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2016)
• Priority tier: T4
• Marks (count): 10 (1)
• Average solve time: ~15 min
• Difficulty mix: medium 1
• Section: A | Dominant type: proof

Why This Chapter Matters

This atom appeared once (2016) in Section A. UPSC uses it to test the core vocabulary and verification skills of vector analysis: proving that a given vector field is conservative (irrotational + find a potential), solenoidal (zero divergence), or neither. The question typically gives an explicit $\mathbf{F}$ and asks you to compute $\nabla\times\mathbf{F}$ and/or $\nabla\cdot\mathbf{F}$, then find the scalar potential if the field is conservative.

Minimum Theory

Scalar field

A scalar field is a function $\phi: \mathbb{R}^3 \to \mathbb{R}$, assigning a real number to each point, e.g., temperature, pressure.

Vector field

A vector field is a function $\mathbf{F}: \mathbb{R}^3 \to \mathbb{R}^3$, assigning a vector to each point:

$\mathbf{F}(x,y,z) = F_1\hat{i} + F_2\hat{j} + F_3\hat{k}$

Gradient

For a scalar field $\phi$:

$\nabla\phi = \frac{\partial\phi}{\partial x}\hat{i} + \frac{\partial\phi}{\partial y}\hat{j} + \frac{\partial\phi}{\partial z}\hat{k}$

$\nabla\phi$ is a vector field; it points in the direction of steepest increase of $\phi$.

Divergence

For a vector field $\mathbf{F} = (F_1, F_2, F_3)$:

$\nabla\cdot\mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}$

$\mathbf{F}$ is solenoidal (or divergence-free) if $\nabla\cdot\mathbf{F} = 0$.

Curl

$\nabla\times\mathbf{F} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ F_1 & F_2 & F_3\end{vmatrix} = \left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)\hat{i} - \left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right)\hat{j} + \left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\hat{k}$

$\mathbf{F}$ is irrotational if $\nabla\times\mathbf{F} = \mathbf{0}$.

Conservative field and scalar potential

$\mathbf{F}$ is conservative if $\mathbf{F} = \nabla\phi$ for some scalar field $\phi$ (the scalar potential or potential function).

• In a simply connected domain: $\mathbf{F}$ is conservative $\iff$ $\nabla\times\mathbf{F} = \mathbf{0}$.
• $\mathbf{F}$ conservative $\implies$ $\oint_C \mathbf{F}\cdot d\mathbf{r} = 0$ for every closed curve $C$.

Finding the scalar potential

Given $\mathbf{F} = \nabla\phi$, i.e., $\partial\phi/\partial x = F_1$, $\partial\phi/\partial y = F_2$, $\partial\phi/\partial z = F_3$:

1. Integrate $F_1$ with respect to $x$: $\phi = \int F_1\,dx + g(y,z)$.
2. Differentiate with respect to $y$; match with $F_2$ to find $\partial g/\partial y$; integrate for $g$.
3. Differentiate with respect to $z$; match with $F_3$ to determine any remaining function of $z$.

Key identities

• $\nabla\cdot(\nabla\times\mathbf{F}) = 0$ for any $\mathbf{F}$ (curl is always solenoidal).
• $\nabla\times(\nabla\phi) = \mathbf{0}$ for any $\phi$ (gradient is always irrotational).

Question Archetypes

conservative-solenoidal-check (1 question; 2016)

Recognition Cues

• An explicit vector field $\mathbf{F}(x,y,z)$ is given.
• You are asked to “show”, “prove”, or “verify” that $\mathbf{F}$ is irrotational and/or solenoidal.
• A follow-on part asks you to “find $\phi$ such that $\mathbf{F} = \nabla\phi$”.
• The components of $\mathbf{F}$ are polynomials or simple functions of $x, y, z$.

Solution Template

1. Compute $\nabla\times\mathbf{F}$; show each component is zero to prove irrotational.
2. Compute $\nabla\cdot\mathbf{F}$; show it is zero to prove solenoidal (if asked).
3. To find $\phi$: integrate $F_1$ w.r.t. $x$, introducing $g(y,z)$; differentiate and match $F_2$; differentiate and match $F_3$; write the final $\phi$.
4. Verify: $\nabla\phi = \mathbf{F}$.

Worked Example

2016 Paper 1, 2016-P1-Q3a (10 marks)

Show that the vector field $\mathbf{F} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}$ is conservative, and find a scalar potential $\phi$ such that $\mathbf{F} = \nabla\phi$.

Step 1. Compute $\nabla\times\mathbf{F}$.

$F_1 = 2xy+z^3,\quad F_2 = x^2,\quad F_3 = 3xz^2$

$\hat{i}$ component: $\dfrac{\partial F_3}{\partial y} - \dfrac{\partial F_2}{\partial z} = 0 - 0 = 0$

$\hat{j}$ component: $-\!\left(\dfrac{\partial F_3}{\partial x} - \dfrac{\partial F_1}{\partial z}\right) = -(3z^2 - 3z^2) = 0$

$\hat{k}$ component: $\dfrac{\partial F_2}{\partial x} - \dfrac{\partial F_1}{\partial y} = 2x - 2x = 0$

Since $\nabla\times\mathbf{F} = \mathbf{0}$, the field is irrotational, hence conservative (domain is all of $\mathbb{R}^3$, which is simply connected).

Step 2. Find the scalar potential $\phi$.

We need $\nabla\phi = \mathbf{F}$:

$\frac{\partial\phi}{\partial x} = 2xy + z^3, \qquad \frac{\partial\phi}{\partial y} = x^2, \qquad \frac{\partial\phi}{\partial z} = 3xz^2.$

Integrate (1) w.r.t. $x$:

$\phi = x^2 y + xz^3 + g(y,z)$

Differentiate w.r.t. $y$ and match (2):

$\frac{\partial\phi}{\partial y} = x^2 + \frac{\partial g}{\partial y} = x^2 \implies \frac{\partial g}{\partial y} = 0 \implies g = h(z)$

Differentiate w.r.t. $z$ and match (3):

$\frac{\partial\phi}{\partial z} = 3xz^2 + h'(z) = 3xz^2 \implies h'(z) = 0 \implies h = C$

Step 3. Write the potential.

$\boxed{\phi = x^2 y + xz^3 + C}$

Verification: $\nabla\phi = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k} = \mathbf{F}$. ✓

Common Traps

• Computing the $\hat{j}$ component of curl with the wrong sign: the formula has a minus sign — $-(\partial F_3/\partial x - \partial F_1/\partial z)$ — which is frequently dropped.
• Introducing $g(y,z)$ after integrating in $x$ but then treating $g$ as a function of $y$ only, missing possible $z$-dependence.
• Forgetting to verify the potential at the end; a sign or algebra error in finding $\phi$ is caught only at the verification step.
• Claiming a field is conservative based on its form without actually checking $\nabla\times\mathbf{F} = \mathbf{0}$.

Marks-Aware Writing

This is a 10-mark proof + computation. Examiners expect:

1. Curl calculation — all three components computed and shown to be zero (4 marks).
2. Statement that $\nabla\times\mathbf{F} = \mathbf{0}$ implies conservative (simply connected domain) (1 mark).
3. Integration step — integrate $F_1$ w.r.t. $x$ with $g(y,z)$ introduced (2 marks).
4. Determination of $g$ and $h$ — matching $F_2$ and $F_3$ (2 marks).
5. Final $\phi$ written out and verified (1 mark).

Write every partial derivative explicitly; do not skip to the answer for the curl components.

Practice Set

Only one historical question on this atom (shown above).