VIPSolutionsConsider a vector field given by © = CrP, where C and p are real constants. (a) For what values of C' and p can 7 be expressed as a gradient, i.e. #(7) = Vi)(7). (b) Find ¢(7) for arbitrary C' and p. Treat the case p = —1 separately.
Question:
Consider a vector field given by © = CrP, where C and p are real constants.
(a) For what values of C' and p can 7 be expressed as a gradient, i.e. #(7) = Vi)(7).
(b) Find ¢(7) for arbitrary C' and p. Treat the case p = —1 separately.
Asked by: Adapaka Jagadeesh
Created at: 2025-09-11 19:12:30
Final Answer:
Question:
Good — here’s a fully detailed, step-by-step solution with the reasoning made explicit. I keep parts **(a)** and **(b)** separated and include checks and edge-case discussion. # (a) When is $\mathbf v(\mathbf r)=C\,r^{p}\,\hat r$ a gradient field? **Step 1 — rewrite the field in terms of the position vector.** Let $\mathbf r=(x,y,z)$ and $r=|\mathbf r|=\sqrt{x^2+y^2+z^2}$. The given field can be written $$ \mathbf v(\mathbf r)=C\,r^{p}\,\hat r = C\,r^{p-1}\,(x,y,z) = \varphi(r)\,\mathbf r, $$ where $\varphi(r)=C r^{p-1}$ is a scalar function depending only on $r$. **Step 2 — show the curl vanishes for $r>0$.** Use the vector identity $$ \nabla\times(\phi(r)\,\mathbf r)=\nabla\phi(r)\times\mathbf r + \phi(r)\,\nabla\times\mathbf r. $$ But $\nabla\times\mathbf r=\mathbf0$ (since $\mathbf r=x\hat\imath+y\hat\jmath+z\hat k$ has zero curl) and $\nabla\phi(r)=\phi'(r)\nabla r=\phi'(r)\,\hat r$. Hence $$ \nabla\times(\phi(r)\,\mathbf r)=\phi'(r)\,\hat r\times\mathbf r. $$ Because $\mathbf r=r\hat r$, $\hat r\times\mathbf r=\hat r\times(r\hat r)=r(\hat r\times\hat r)=\mathbf0$. Therefore $$ \nabla\times\mathbf v=\mathbf0\qquad\text{for all }r>0. $$ **Step 3 — use topology (Poincaré lemma).** In three dimensions the region $\mathbb R^3\setminus\{0\}$ (i.e. $r>0$) is simply connected, so a curl-free field on that region is conservative: there exists a scalar potential $\psi$ with $\mathbf v=\nabla\psi$ on the whole region $r>0$. **Conclusion for (a):** * On any domain that excludes the origin (for example the whole set $r>0$ or any spherical shell $a0). $$ So the potentials (valid for $r>0$) are $$ \boxed{\displaystyle \psi(r)= \begin{cases} \dfrac{C}{p+1}\,r^{\,p+1}+K, & p\neq -1,\\[6pt] C\ln r + K, & p=-1, \end{cases} \qquad (K\in\mathbb R).} $$ **Step 3 — verify by differentiation (chain rule).** Differentiate $\psi(r)$ and convert to vector form: * For $p\neq-1$, $\psi'(r)=\dfrac{C}{p+1}(p+1)r^{p}=C r^{p}$, so $\nabla\psi=\psi'(r)\hat r=C r^{p}\hat r=\mathbf v.$ * For $p=-1$, $\psi'(r)=C/r$, so $\nabla\psi = (C/r)\hat r=\mathbf v.$ Equivalently, using Cartesian components and the chain rule: $$ \frac{\partial\psi}{\partial x}=\psi'(r)\frac{\partial r}{\partial x} =\psi'(r)\frac{x}{r} = C r^{p}\frac{x}{r}=C r^{p-1}x, $$ so $\nabla\psi=(C r^{p-1}x,\,C r^{p-1}y,\,C r^{p-1}z)=C r^{p}\hat r=\mathbf v.$ **Step 4 — behavior at the origin (extendability & smoothness).** * The potential $\psi(r)=\dfrac{C}{p+1}r^{p+1}+K$ (when $p\neq -1$) is **finite** at $r=0$ iff $p+1>0$ (i.e. $p>-1$). * If $p>-1$ you can define $\psi(0)=K$ (or the limit value) and $\psi$ is continuous at the origin. * If $p\le -1$ the potential diverges as $r\to0$ (for $p=-1$ it diverges like $\ln r$; for $p0$ (then $\mathbf v\to\mathbf0$). * If $p=0$, $\mathbf v=C\hat r$ has fixed magnitude but direction depends on the approach — no limit at the origin. * If $p0$. **Step 5 — physical examples (quick checks).** * $p=-2$, $C=-GM$: $\mathbf v=-GM r^{-2}\hat r$ (Newtonian gravity). Potential: $\psi(r)=\dfrac{-GM}{-1}r^{-1} = -\dfrac{GM}{r}+K$. (As expected.) * $p=-1$, $C=Q$: $\mathbf v=Q r^{-1}\hat r$. Potential: $Q\ln r +K$ (logarithmic). --- ## Final compact answers **(a)** $\mathbf v$ is curl-free for all $r>0$. On any simply-connected domain that excludes the origin (for example $r>0$), $\mathbf v$ is a gradient field for **all real** $C$ and $p$. The origin is the only singular point and must be treated separately. **(b)** A potential is $$ \boxed{\displaystyle \psi(r)= \begin{cases} \dfrac{C}{p+1}\,r^{p+1}+K, & p\neq -1,\\[6pt] C\ln r + K, & p=-1, \end{cases}\quad (r>0,\;K\in\mathbb R).} $$ **Extendability at $r=0$:** $\psi(0)$ is finite iff $p>-1$; $\psi$ is differentiable at $0$ (so $\nabla\psi$ extends continuously to $0$) iff $p>0$. --- If you want, I can now: * explicitly compute $\nabla\times\mathbf v$ in spherical coordinates term-by-term to show every component is zero, * show the Cartesian computation of the curl to confirm the identity, or * make plots of $\psi(r)$ and $|\mathbf v(r)|$ for several choices of $p$ so you can see the different behaviors near the origin. Which would help most?
Asked by: Adapaka Jagadeesh
Created at: 2025-09-11 19:25:10
Final Answer:
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