# Why a vector field isn't determined solely by its divergence

To my understanding, Helmholtz theorem tells us that a vector function can be decomposed into the sum of a gradient of a scalar and curl of a vector function. Thus, if you know the divergence of a vector function is non-zero and also that the curl of the same function is zero, you can say that the vector function is wholly determined by the non-zero quantity. This occurs in electrostatics with the vector function for electric field. I understand this decomposition, but I would like to gain a better intuition as to why both quantities (divergence and curl) are needed to determine the vector field.

Suppose we have a vector function $${\textbf{D}}$$ with divergence given by $$\nabla \cdot \textbf{D} = \rho_{f}$$. Let's say its curl is non-zero: $$\nabla \times \textbf{D} = \nabla \times \textbf{P} \not= \textbf{0}$$. The symbols aren't important, but they represent electric displacement ($$\textbf{D}$$), polarization ($$\textbf{P}$$), and free charge density ($$\rho_{f}$$) from 4ed Griffith's Introduction to Electrodynamics.

Is it accurate to draw parallels between the relationship of the derivative and integral of a scalar function and the divergence and anti-divergence (if such thing exists) of a vector function?

For example, let $$y = 1$$ Then, $$\frac{dy}{dx} = 0$$ And, $$\int{\frac{dy}{dx}} = C$$ where C is a constant.

Obviously, in the scalar case the anti-derivative does not have/provide enough information for us to go back to the original function $$y$$. Is this what is happening in the vector function case as well?

It seems I might be conflating boundary conditions (which if provided in the scalar case would give us a unique answer) and something else. Any help would be much appreciated!

• Yes, the divergence of a vector field is a DERIVATIVE. Just as with the derivative of a number valued function, you lose a constant (vector field). May 29 at 0:33
• @user1046533 Let me know if this is not clarifying. We have the derivative and anti-derivative operators. I am supposing that there is a divergence (true) and an anti-divergence operator analogous in function to the anti-derivative operator. Given such a pair of operators exists, does the analysis I presented in the question on the scalar function 1 carry over to an analysis of some vector function. May 29 at 0:45
• If a vector field were determined purely by its divergence, the Maxwell equation $\nabla\cdot B = 0$ would be saying the magnetic field doesn't exist. Jun 1 at 13:12
• multivariable-calculus may be a more appropriate tag to use than calculus. Jun 1 at 16:07
• Fun fact: the divergence and curl of $F = (y+z,x+z,x+y)$ are 0. Jun 1 at 16:59

$$\def\rbf{\mathbf{R}}$$For simplicity consider a smooth function $$f:\rbf\to\rbf$$. Suppose $$f'(x)=g(x)$$ for all $$x$$. Then for any constant $$C$$, one also has $$(f+C)'=g$$. So the derivative $$f'$$ does not determine the function $$f$$ completely. On the other hand, the derivative determines the function "partially". If one knows that $$f'=g'$$, then one must have $$f=g+C$$ by the mean value theorem. One can replace the domain $$\rbf$$ of $$f$$ with any open interval.

$$\def\bg{\mathbf{g}}$$Now let $$f:\rbf^3\to\rbf$$ be a scalar function and suppose $$\nabla f(x)=\bg(x)$$ for all $$x$$. Then $$\nabla(f+C)=\bg$$ for any constant $$C$$. So the gradient $$\nabla f$$ does not determine the function $$f$$ completely. On the other hand, if $$\nabla p = \nabla q$$ for two scalar functions $$p$$ and $$q$$, then similarly, $$p=q+C$$ for some constant $$C$$. The result holds if one replaces $$\rbf^3$$ with any nonempty connected open subset of $$\rbf^3$$.
In the scalar case, by the fundamental theorem of calculus, an antiderivative of the function $$g$$ is given by $$f(x)=\int_0^xg(t)\;dt\;.$$ In the vector-valued function case, an "anti-gradient" of the vector field $$\bg$$, assuming that $$\textrm{curl} (\bg)=0$$, can be found as the line integral $$f(x,y,z)=\int_{(0,0,0)\rightsquigarrow(x,y,z)} \bg\cdot dr$$ where $$(0,0,0)\rightsquigarrow(x,y,z)$$ denotes any path from $$(0,0,0)$$ to $$(x,y,z)$$.

Let $$\lambda:\rbf^3\to\rbf$$ be a scalar field on $$\rbf^3$$ and $$F:\rbf^3\to\rbf^3$$ a vector field with $$\nabla F=\lambda\;.\tag{1}$$ Then every vector field $$F_1:=F+C$$ for some constant $$C$$ satisfies (1). So divergence alone does not determine the vector field. This answers your question in the title.

A theorem says that a vector field can be constructed with both a specified divergence and a specified curl: $$\nabla\cdot F=\lambda,\quad \nabla\times F=q\tag{2}$$ where $$q$$ is a given divergence-free vector field. But since (2) is still invariant up to a constant, $$F$$ is not determined uniquely even by its divergence and curl. One needs an additional "decay property" of $$F$$ for uniqueness. That property can be thought of as some "boundary condition" to the system in (2).

• typo on "∇p=∇q for two scalar functions p and p" May 29 at 8:54
• Hm. I'm not sure that this quite answers the question. I think it's my fault for not presenting the question clearly. To my understanding, the second part of your answer refers to a conservative vector field which has such path integrals. Maybe my question can be simplified to: Suppose you know the divergence of a vector function: $\nabla \cdot \textbf{D} = \lambda$ where $\lambda$ is some arbitrary quantity. Can the curl of this vector function $\nabla \times \textbf{D}$ be considered a boundary condition for solving for the vector function $\textbf{D}$? Jun 1 at 5:02
• If so, could you give a hint (not the answer) as to where the need for two boundary conditions comes from? I haven't taken differential equations yet, but I assume an interpretation can be drawn from there? Jun 1 at 5:04
• @SillyGoose: I somehow misread "divergence" as "gradient". Now the answer is edited. Jun 1 at 13:10
• If you know $\nabla\cdot D=\lambda$ alone, you would need two more "boundary conditions", counting the degrees of freedom informally. The curl can be regarded as one condition; but for uniqueness, you would need yet another one. Jun 1 at 13:12