In electrostatics, we have Maxwell's equations:

$\nabla \cdot E = \rho$

$\nabla \times E = 0$

These two equations can also be written in terms of the electrostatic potential:

$ -\nabla^2 V=\rho $

$ E = -\nabla V $

Now if we know the positions of every charge in our system, we can find the electrostatic field (completely and entirely, with no additional information required) using Coulomb's law:

$ E(x) = -\nabla \int \frac{\rho(x')}{4\pi|x-x'|} \mathrm{d}^3 x' $

My question is: to what extent can we do the same with Maxwell's Equations? For instance, whenever Coulomb's Law is derived from Maxwell's equations, an appeal needs to be made to spherical symmetry. Must we do this? Can we not use the vanishing curl somehow to reach the same conclusion? Similarly, if we derive Coulomb's Law from Poisson's equation, we must specify boundary conditions. We must specify that the potential is some constant, say zero, at infinity.

It appears that the information content is less.

I have read a little bit about the Helmholtz decomposition, and it appears that the curl and divergence of a vector field (E, in this case) do completely determine the vector field, providing certain restrictions are placed on the smoothness and decay of the field at infinity. In other words, it appears that in fact Maxwell's Equations (in the context of electrostatics) do have less information content than Coulomb's Law.

In electrostatics, we have Maxwell's equations:

$\nabla \cdot E = \rho$

$\nabla \times E = 0$

These four equations (the second line standing for three equations) can also be written in terms of the electrostatic potential:

$ -\nabla^2 V=\rho $

$ E = -\nabla V $

Now if we know the positions of every charge in our system, we can find the electrostatic field (completely and entirely, with no additional information required) using Coulomb's law:

$ E(x) = -\nabla \int \frac{\rho(x')}{4\pi|x-x'|} \mathrm{d}^3 x' $

My question is: to what extent can we do the same with Maxwell's Equations? For instance, whenever Coulomb's Law is derived from Maxwell's equations, an appeal needs to be made to spherical symmetry. Must we do this? Can we not use the vanishing curl somehow to reach the same conclusion? Similarly, if we derive Coulomb's Law from Poisson's equation, we must specify boundary conditions. We must specify that the potential is some constant, say zero, at infinity.

It appears that the information content is less.

I have read a little bit about the Helmholtz decomposition, and it appears that the curl and divergence of a vector field (E, in this case) do completely determine the vector field, providing certain restrictions are placed on the smoothness and decay of the field at infinity. In other words, it appears that in fact Maxwell's Equations (in the context of electrostatics) do have less information content than Coulomb's Law.

In electrostatics, we have Maxwell's equations:

$\nabla \cdot E = \rho$

$\nabla \times E = 0$

These two equations can also be written in terms of the electrostatic potential:

$ -\nabla^2 V=\rho $

$ E = -\nabla V $

Now if we know the positions of every charge in our system, we can find the electrostatic field (completely and entirely, with no additional information required) using Coulomb's law:

$ E(x) = -\nabla \int \frac{\rho(x')}{4\pi\epsilon_0|x-x'|} \mathrm{d}^3 x' $

My question is: to what extent can we do the same with Maxwell's Equations? For instance, whenever Coulomb's Law is derived from Maxwell's equations, an appeal needs to be made to spherical symmetry. Must we do this? Can we not use the vanishing curl somehow to reach the same conclusion? Similarly, if we derive Coulomb's Law from Poisson's equation, we must specify boundary conditions. We must specify that the potential is some constant, say zero, at infinity.

It appears that the information content is less.

Another point of confusion is that the information content of Poisson's equation seems different to that of the Maxwell Equations. Using the divergence theorem and the first Maxwell Equation, surely we can deduce that the electric field must vanish at infinity, and hence the potential be some constant there, from the assumption that the total charge in our system is finite --- if this is the case, then the flux of E out of a sphere at infinity must be finite, and so the field must tend to zero at infinity. It appears that Maxwell's Equations tell us the field must tend to zero at infinity, but this must be added as additional information for Poisson's equation.

I have read a little bit about the Helmholtz decomposition, and it appears that the curl and divergence of a vector field (E, in this case) do completely determine the vector field, providing certain restrictions are placed on the smoothness and decay of the field at infinity. In other words, it appears that in fact Maxwell's Equations (in the context of electrostatics) do have less information content than Coulomb's Law.

In electrostatics, we have Maxwell's equations:

$\nabla \cdot E = \rho$

$\nabla \times E = 0$

These two equations can also be written in terms of the electrostatic potential:

$ -\nabla^2 V=\rho $

$ E = -\nabla V $

Now if we know the positions of every charge in our system, we can find the electrostatic field (completely and entirely, with no additional information required) using Coulomb's law:

$ E(x) = -\nabla \int \frac{\rho(x')}{4\pi|x-x'|} \mathrm{d}^3 x' $

My question is: to what extent can we do the same with Maxwell's Equations? For instance, whenever Coulomb's Law is derived from Maxwell's equations, an appeal needs to be made to spherical symmetry. Must we do this? Can we not use the vanishing curl somehow to reach the same conclusion? Similarly, if we derive Coulomb's Law from Poisson's equation, we must specify boundary conditions. We must specify that the potential is some constant, say zero, at infinity.

It appears that the information content is less.

I have read a little bit about the Helmholtz decomposition, and it appears that the curl and divergence of a vector field (E, in this case) do completely determine the vector field, providing certain restrictions are placed on the smoothness and decay of the field at infinity. In other words, it appears that in fact Maxwell's Equations (in the context of electrostatics) do have less information content than Coulomb's Law.

In electrostatics, we have Maxwell's equations:

$\nabla \cdot E = \rho$

$\nabla \times E = 0$

These four equations can also be written in terms of the electrostatic potential:

$ -\nabla^2 \phi = \rho $

$ E = -\nabla \phi $

Now if we know the positions of every charge in our system, we can find the electrostatic field (completely and entirely, with no additional information required) using Coulomb's law:

$ E(x) = -\nabla \int \frac{\rho(x')}{4\pi|x-x'|} \mathrm{d}^3 x' $

My question is: to what extent can we do the same with Maxwell's Equations? For instance, whenever Coulomb's Law is derived from Maxwell's equations, an appeal needs to be made to spherical symmetry. Must we do this? Can we not use the vanishing curl somehow to reach the same conclusion? Similarly, if we derive Coulomb's Law from Poisson's equation, we must specify boundary conditions. We must specify that the potential is some constant, say zero, at infinity.

It appears that the information content is less.

Another point of confusion is that the information content of Poisson's equation seems different to that of the Maxwell Equations. Using the divergence theorem and the first Maxwell Equation, surely we can deduce that the electric field must vanish at infinity, and hence the potential be some constant there, from the assumption that the total charge in our system is finite --- if this is the case, then the flux of E out of a sphere at infinity must be finite, and so the field must tend to zero at infinity. It appears that Maxwell's Equations tell us the field must tend to zero at infinity, but this must be added as additional information for Poisson's equation.

I have read a little bit about the Helmholtz decomposition, and it appears that the curl and divergence of a vector field (E, in this case) do completely determine the vector field, providing certain restrictions are placed on the smoothness and decay of the field at infinity. In other words, it appears that in fact Maxwell's Equations (in the context of electrostatics) do have less information content than Coulomb's Law.

In electrostatics, we have Maxwell's equations:

$\nabla \cdot E = \rho$

$\nabla \times E = 0$

These two equations can also be written in terms of the electrostatic potential:

$ -\nabla^2 V=\rho $

$ E = -\nabla V $

Now if we know the positions of every charge in our system, we can find the electrostatic field (completely and entirely, with no additional information required) using Coulomb's law:

$ E(x) = -\nabla \int \frac{\rho(x')}{4\pi\epsilon_0|x-x'|} \mathrm{d}^3 x' $

My question is: to what extent can we do the same with Maxwell's Equations? For instance, whenever Coulomb's Law is derived from Maxwell's equations, an appeal needs to be made to spherical symmetry. Must we do this? Can we not use the vanishing curl somehow to reach the same conclusion? Similarly, if we derive Coulomb's Law from Poisson's equation, we must specify boundary conditions. We must specify that the potential is some constant, say zero, at infinity.

It appears that the information content is less.

Another point of confusion is that the information content of Poisson's equation seems different to that of the Maxwell Equations. Using the divergence theorem and the first Maxwell Equation, surely we can deduce that the electric field must vanish at infinity, and hence the potential be some constant there, from the assumption that the total charge in our system is finite --- if this is the case, then the flux of E out of a sphere at infinity must be finite, and so the field must tend to zero at infinity. It appears that Maxwell's Equations tell us the field must tend to zero at infinity, but this must be added as additional information for Poisson's equation.

I have read a little bit about the Helmholtz decomposition, and it appears that the curl and divergence of a vector field (E, in this case) do completely determine the vector field, providing certain restrictions are placed on the smoothness and decay of the field at infinity. In other words, it appears that in fact Maxwell's Equations (in the context of electrostatics) do have less information content than Coulomb's Law.