# Why do we need both Divergence and Curl to define a vector field?

I was reading Classical Electrodynamics by J.D.Jacskon (section 1.5) where he said:

Perhaps some readers know that a vector field can be specified almost completely if its divergence and curl are given everywhere in space

with a comment

(almost)-Up to the gradient of a scalar function that satisfies the Laplace equation.

why is this so

if we want to know $$\vec{E}$$ in 3-D space we really just want to find out is $$E_x,E_y$$ and $$E_z$$ (in cartesian coordinates) which are just 3 scalar functions. If we know the curl is (let's say) zero, that is $$\nabla \times \vec{E}=0$$

in cartesian coordinates that gives us three equations $$\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}=0$$ $$\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}=0$$ $$\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}=0$$

these are three equations for 3 unknown quantities, that should give us enough* information about these three functions, it obviously isn't so hence we need to know the divergence as well, my question is why? is there some theorem that proves this (I suspect there is $$:)$$ )

this is obviously a mathematical question although it is physics motivated so I hope it is ok that I post it here.

(*up to maybe a constant because all the equations involve first derivatives, I can live with that)

• Not up to constants. Up to adding any conservative vector field (i.e., the gradient of a scalar function). – Ted Shifrin Jan 3 at 19:35
• do you by any chance know some reference (book, site..) where I can look up a proof of this ? :) – Alexandar Solženjicin Jan 3 at 19:36
• Don't you know that the curl of a gradient is zero (assuming the function has continuous second-order partial derivatives)? The converse — that on all of $\Bbb R^3$ a vector field with zero curl must be a gradient — is a special case of the Poincaré lemma. You write down the function as a line integral from a fixed point to a variable point; Stokes's Theorem tells you that this gives a well-defined function, and then you check that its gradient is the vector field you started with. – Ted Shifrin Jan 3 at 19:42

If

$$\nabla \times \vec E = 0, \tag 1$$

we know from a standard result that

$$\vec E = \nabla \phi \tag 2$$

for some scalar function $$\phi$$.

If the divergence of $$\vec E$$ is also a known function $$\rho$$,

$$\nabla \cdot \vec E = \rho, \tag 3$$

then combining (2) and (3) we obtain

$$\nabla^2 \phi = \nabla \cdot \nabla \phi = \nabla \cdot \vec E = \rho, \tag 4$$

which can in principle be solved for $$\phi$$ (assuming we admit appropiate boundary conditions on $$\phi$$; $$\phi(x) \to 0$$ sufficiently fast as $$\vert x \vert \to \infty$$ is often used; Jackson's book covers this, mos' likely); thus, we may discover $$\vec E$$ from (2).

We observe that such a solution is not unique; indeed, let $$\psi$$ be any harmonic function,

$$\nabla \cdot \nabla \psi = \nabla^2 \psi = 0; \tag 5$$

then

$$\nabla^2(\phi + \psi) = \nabla^2 \phi + \nabla^2 \psi = \rho + 0 = \rho, \tag 6$$

and if

$$\vec E = \nabla (\phi + \psi), \tag 7$$

we still obtain

$$\nabla \times \vec E = \nabla \times \nabla (\phi + \psi) = \nabla \times \nabla \phi + \nabla \times \nabla \times \psi = 0, \tag 8$$

since the curl of any gradient vanishes. Also,

$$\nabla \cdot \vec E = \nabla \cdot (\nabla \phi + \nabla \psi) = \nabla^2 \phi + \nabla ^2 \psi = \nabla^2 \phi = \rho; \tag 9$$

these last two equations show that we may transform any solution according to

$$\phi \to \phi + \psi, \tag 9$$

$$\vec E = \nabla \phi \to \nabla \phi + \nabla \psi, \tag{10}$$

$$\psi$$ as in (5), and preserve the divergence and curl of $$\vec E$$; so any solution is not unique; uniqueness may be attained by specifying appropriate boundary conditions on $$\phi$$ and $$\psi$$ which can then become unambiguously determined.

The above discussion addresses the relatively simple case (1), (3); we can, in fact, also address the significant generalization

$$\nabla \times \vec E = \vec F, \; \nabla \cdot \vec E = \rho, \tag{11}$$

where $$\vec F$$ is a pre-specified vector field; the situation is more complicated since we may no longer assume $$\vec E$$ is a gradient as in (1)-(2). In this case we instead invoke the vector calculus identity

$$\nabla \times (\nabla \times \vec A) = \nabla (\nabla \cdot \vec A) - \nabla^2 \vec A, \tag{12}$$

where the Laplacian operator $$\nabla^2$$ occurring on the right-hand side is understood to act component-wise on $$\vec A$$; thus we have

$$\nabla \times \vec E = \vec F \tag{13}$$

$$\nabla \times (\nabla \times \vec E) = \nabla \times \vec F, \tag{14}$$

which we trasform according to (12):

$$\nabla (\nabla \cdot \vec E) - \nabla^2(\vec E) = \nabla \times \vec F; \tag{15}$$

by virtue of (3) we write this as

$$\nabla \rho - \nabla^2 \vec E = \nabla \times \vec F, \tag{16}$$

whence we find

$$\nabla^2 \vec E = \nabla \rho - \nabla \times \vec F, \tag{17}$$

which we may in principle solve component-wise for $$\vec E$$; for example,

$$\nabla^2 E_x = \dfrac{\partial \rho}{\partial x} - \left ( \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z} \right) = \dfrac{\partial \rho}{\partial x} - \nabla^2 E_x = \dfrac{\partial \rho}{\partial x} - \dfrac{\partial F_z}{\partial y} + \dfrac{\partial F_y}{\partial z}; \tag{18}$$

of course, we need boundary conditions and perhaps certain restrictions on $$\rho$$ and $$\vec F$$; but I'll leave it to folks like David Jackson to explain such matters.

Apparently solutions to (11), (18) are still not unique; it is still true that adjusting $$\vec E$$ by the gradient of a harmonic function $$\psi$$ gives rise to another solution; of course, we may determine such $$\psi$$ uniquely via appropriate boundary conditions, and then a certain uniqueness is attained for $$\vec E$$.