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I know of course that If the curl of a vector function is equal to zero, then the vector function is the gradient of some other scalar function, but is this a must?

if so, please give mathematical proof.

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Aug 10 '15 at 9:24
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    $\begingroup$ en.wikipedia.org/wiki/Helmholtz_decomposition $\endgroup$ Aug 10 '15 at 10:36
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    $\begingroup$ This is true on any 3-manifold with trivial first de Rahm cohomology group, as this would then imply that $\text d\omega = 0$ only if $\omega = \text d\alpha$ for some 0-form $\alpha$. Any vector space falls into this class, and hence the Euclidean 3-space. $\endgroup$
    – Phoenix87
    Aug 10 '15 at 13:36
  • $\begingroup$ Wikipedia gives the proof also here. $\endgroup$ Aug 10 '15 at 14:01
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First I would like to note the difference between the following two statements:

If a vector field is the gradient of a scalar function then the curl of that vector field is zero.

If the curl of some vector field is zero then that vector field is a the gradient of some scalar field.

I have seen some trying to prove the first where I think you are asking for the second

I apologize for not giving full details on math here because I'm doing this on my tablet.

Anyway, if the curl of a vector field $F$ is zero $(\nabla\times F=0)$ then the surface integral of the resulting vector field over any arbitrary surface $S$ is also zero.

Stokes theorem (read the Wikipedia article on Kelvin-Stokes theorem) the surface integral of the curl of any vector field is equal to the closed line integral over the boundary curve.

Then since $\nabla\times F=0$ which implies that the surface integral of that vector field is zero then (BY STOKES theorem) the closed line integral of the boundary curve of that (arbitrary) selected surface is also zero.

Since the selection of the surface is arbitrary the. We can say the closed line integral of $F$ over any arbitrary closed curve is zero.

This implies that the line integral of the vector field $F$ is path independent which means the line integral over any curve only depending the initial and final position (not necessarily a closed curve)

To prove this just divide your closed path into two paths from point $P_{1}$ two point $P_{2}$, call those paths $A$ and $B$, the line integral over a closed path $C$ is equal to the summation of the line integral over paths $A$ and $B$ so:

$$ \oint\limits_C F \mathrm{d}\ell = \int\limits_{A_{P_1 \to P_2}} F \mathrm{d}\ell + \int\limits_{B_{P_2 \to P_1}} F \mathrm{d}\ell =0 $$ Then $$ \begin{split} \int\limits_{A_{P_1 \to P_2}}\! F \mathrm{d}\ell &=- \int\limits_{B_{P_2 \to P_1}} F \mathrm{d}\ell\\ &\Updownarrow\\ \int\limits_{A_{P_1 \to P_2}}\! F \mathrm{d}\ell &= \int\limits_{B_{P_1 \to P_2}} F \mathrm{d}\ell \end{split} $$ ( please note that those are alkaline integral and that there should be a for product sign between the two f and dl)

This latter equality implies that it doesn't matter your choice of the path $A$ or $B$ or any path because the result will be the same and it will only depend on the vector field $F$ and the two end points.

And since it only depends on the two points $P_{1}$ and $P_{2}$, then we can DEFINE a scalar field $\Phi(P)$ (note that the points $P_{1}$ and $P_2$ are position vectors) such that $$\Phi(P_{2}) - \Phi(P_{1}) = \int\limits_{P_{1}}^{P_{2}}F \mathrm{d}\ell$$

(Note that the integral doesn't depend on the path and that is the only reason we can write it this way).

Now from the gradient theorem ( look for the Wikipedia article on gradient theorem ) $F=\nabla \Phi$.

Also look for the Wikipedia article on conservative fields.

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  • $\begingroup$ The second statement is not true since there are curl free vector fields $\vec{F}$ which are not gradients $\endgroup$ Sep 7 '20 at 19:14
  • $\begingroup$ This proof is erroneous. It is not enough just to observe that your definition of $\Phi$ is path independent, because it does not follow from this that $F=\nabla \Phi$. Nor does this follow from the gradient theorem. Nor is the proof found on the cited wikipedia article (at the time of writing). $\endgroup$ Jul 7 '21 at 16:28
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If $X$ is a smooth vector field of the form $X=\nabla \phi$ then it follows immediately from the symmetry of partial derivatives that $\mathrm{curl}(X)=0$. The converse statement, however, that for any smooth vector field $X$ satisfying $\mathrm{curl}(X)=0$ there exists a $\phi$ such that $X=\nabla\phi$, which is called the Poincare Lemma, is trickier to establish, and depends in general on the topology of the underlying space.

Rather than get bogged down in a complete answer to this question, which would require advanced topology, I will give a proof of the Poincare Lemma for vector fields on $\mathbb{R}^3$, which extends easily to vector fields on $\mathbb{R}^n$ and to star-shaped subdomains of $\mathbb{R}^n$. I will then give a well-known example which shows that the Poincare Lemma does not hold on the domain $\mathbb{R}^2\setminus(0,0)$, to illustrate that the problem on general domains with "holes" is more complex.

Let X be a smooth vector field defined everywhere on $\mathbb{R}^3$. Note firstly that if there were to exist a $\phi$ such that $X=\nabla \phi$, then without loss of generality we could redefine $\phi$ by adding a constant so that $\phi(0,0,0)=0$, and we could then express $\phi$ in terms of $X$ as

\begin{align} \phi(x,y,z) &= \int_0^1 \frac{d}{ds} \phi(sx, sy, sz) ds \\ &= \int_0^1 \big( x \partial_1 \phi(sx, sy, sz) + y \partial_2 \phi(sx, sy, sz) + z \partial_3\phi(sx, sy, sz) \big) ds \\ &= \int_0^1 \big(x X_1(sx, sy, sz) + y X_2(sx, sy, sz) + zX_3(sx, sy, sz) \big) ds \end{align}

(think of this as evaluating the line integral $\int X \cdot dl$ along the ray from the origin to the point $(x,y.z)$). Motivated by this, we proceed by defining a function $\phi$ by $$ \phi(x,y,z) := \int_0^1 \big(x X_1(sx, sy, sz) + y X_2(sx, sy, sz) + zX_3(sx, sy, sz) \big) ds$$ and we claim that it satisfies $X=\nabla \phi$ as desired.

Indeed, we compute

\begin{align} \partial_1\phi (x,y,z) &= \int_0^1 \big(X_1(sx, sy, sz) + sx\partial_1 X_1(sx, sy, sz) + sy \partial_1X_2(sx, sy, sz) + sz \partial_1X_3(sx, sy, sz)\big) ds \\ &= \int_0^1 \frac{d}{ds} \big( s X_1(sx, sy, sz) \big) ds + \int_0^1 \Big( sy \big( \partial_1X_2(sx, sy, sz) - \partial_2X_1(sx, sy, sz)\big) \\ & \hskip150pt + sz \big( \partial_1 X_3(sx,sy,sz) - \partial_3X_1(sx, sy, sz) \big)\Big)ds \\ &= X_1(x,y,z) \end{align}

where the second integral on the second line vanishes because $\mathrm{curl}(X)=0$, and similarly one can compute $\partial_2 \phi=X_2$ and $\partial_3 \phi = X_3$, showing that $X=\nabla \phi$ as claimed.

Finally, to illustrate the importance of the underlying topology, consider the vector field $X(x,y) = \frac{1}{x^2 + y^2} (-y, x)$ on $\mathbb{R}^2\setminus(0,0)$. This satisfies $\mathrm{curl}(X)=0$ but there cannot exist any $\phi$ such that $X=\nabla\phi$. To see this, let $l(s) = (\cos s, \sin s)$ parameterise the unit circle and observe that $\oint X(l(s)) \cdot dl(s) = 2\pi$. But if we had $X=\nabla \phi$ for some $\phi$, then we would have $\oint X(l(s))\cdot dl(s) = \oint \nabla \phi(l(s)) \cdot dl(s) = \int_0^{2\pi} \frac{d}{ds} \big( \phi(l(s)) \big) ds = 0$, which is a contradiction.

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It is rather sufficient to prove that the curl of a vector function $\mathbf{F}$ which is the gradient of a scalar-function $\phi$ is 0.

Let $\phi(x,y,z)$ be a scalar-function. Then its gradient will be $$\nabla\phi(x,y,z) = \frac{\partial \phi(x,y,z)}{\partial x}\hat{\mathbf{x}} + \frac{\partial \phi(x,y,z)}{\partial y}\hat{\mathbf{y}} + \frac{\partial \phi(x,y,z)}{\partial z}\hat{\mathbf{z}} \quad.$$ By assumption, $$\mathbf{F} = \nabla \phi$$. We have to show that $$\nabla \times \nabla \phi(x,y,z) = \mathbf{0}$$.

Since $\nabla \times \mathbf{h} = \text{vector}$, we can write the components of this vector by the rule of cross-product i.e. $$(\nabla \times \mathbf{h})_x = \nabla_y h_z - \nabla_z h_y$$. So, we can write $$\text{curl}\mathbf{F} = \left(\frac{\partial F_z}{\partial y} - \frac{ F_y}{\partial z}\right)\hat{\mathbf{x}} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right)\hat{\mathbf{y}} +\left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\hat{\mathbf{z}} $$. But we know $$F_x = \nabla\phi_x = \frac{\partial \phi(x,y,z)}{\partial x} \\\\\ F_y = \nabla\phi_y =\frac{\partial \phi(x,y,z)}{\partial y} \\\\\\\ F_z = \nabla \phi_z = \frac{\partial \phi(x,y,z)}{\partial z}$$. Substituting these in the definition of $\text{curl} \mathbf{F}$, we get $$\text{curl} \mathbf{F} = \nabla \times \nabla \phi = \left(\frac{\partial^2 \phi}{\partial y \partial z} - \frac{\partial^2 \phi}{\partial z \partial y}\right)\hat{\mathbf{x}} + \left(\frac{\partial^2 \phi}{\partial z \partial x} - \frac{\partial^2 \phi}{\partial x \partial z}\right)\hat{\mathbf{y}} + \left(\frac{\partial^2 \phi}{\partial x \partial y} - \frac{\partial^2 \phi}{\partial y \partial x}\right)\hat{\mathbf{z}} = \mathbf{0} \qquad .$$

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  • $\begingroup$ @Mohamed Ayman: I'm happy that it helped:) $\endgroup$
    – user142971
    Aug 15 '15 at 17:28
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    $\begingroup$ This proves only the converse -- if the function is a gradient then the curl is zero. $\endgroup$ Dec 26 '19 at 7:22

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