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The following equations are equations of electrostatics: $$\nabla \times \vec E=0$$ $$\nabla\cdot\vec E=\dfrac{\rho}{\epsilon_0}.$$

These are 4 independent equations, while $\vec E$ has only 3 independent components. Yet these equations do not completely specify the field, as after adding the gradient of a scalar $\nabla \lambda$ that satisfies Laplace equation ($\nabla^2 \lambda$=0) to $\vec E$ leaves the equations unchanged: $$\cases{\nabla \times \vec E=0\\\nabla\cdot\vec E=\dfrac{\rho}{\epsilon_0}}\xrightarrow[\nabla^2\lambda=0]{\vec E'=\vec E+\nabla \lambda}\cases{\nabla \times \vec {E'}=0\\\nabla\cdot\vec {E'}=\dfrac{\rho}{\epsilon_0}}$$ (note the primes on the RightHandSide $\vec E$s)

The system should be overdetermined (4 equations, 3 unknowns) but apparently it is underdetermined.

  • Is the system overdetermined or underdetermined?
  • How do we usually choose the arbitrary $\lambda$ in a problem with $\rho$ given and (say) Neumann boundary condition?
  • Why the first equation (curl) is not enough?
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marked as duplicate by Brandon Enright, Emilio Pisanty, John Rennie, Chris White, Kyle Kanos Feb 5 '14 at 15:26

This question was marked as an exact duplicate of an existing question.

Related: – Qmechanic Feb 4 '14 at 20:07
@Qmechanic's link then points to – dmckee Feb 4 '14 at 20:46
@Qmechanic The static case is completely different from that question. – user215721 Feb 4 '14 at 21:06
The equations in question are partial differential equations, NOT algebraic equations. It is therefore not surprising that the system is underdetermined. For example, consider a function $f(x,y)$ satisfying $\partial_x f = \partial_y f = 0$. We have one function, but two equations. However, the two equations only specify $f(x,y) = c$ (constant). The system is underdetermined upto a constant shift of $f(x,y)$. A similar argument holds here. – Prahar Feb 4 '14 at 21:16
@Prahar So the statement for the each system of (P or O)DEs the total number of equations must be equal to the unknown variables is not generally correct? – user215721 Feb 4 '14 at 21:23

We normally choose $\lambda$ in a way that makes our calculations easiest purely because nothing says we can't do that.

The first equation is not enough to determine the system for the same reason that $\frac{\partial y(x)}{\partial x}=f(x)$ can never uniquely determine $y(x)$. We can always add a constant offset that changes $y(x)$ but not $f(x)$.

As for why we need the second equation, simply stating that an electrostatic field is curl-less is not sufficient to describe the physics. Both equations act as constraints on each other to allow us to determine an electrostatic field given some boundary conditions as well as determine the physics associated with it.

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So the system is underdetermined? – user215721 Feb 4 '14 at 20:31
Yes, it must be undetermined. Otherwise we would have to know the value of the field due to all charges everywhere at a given point – Jim Feb 4 '14 at 20:44

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