1

Doing electrostatics, I've found that everyone makes assumptions that nobody proves, and it's related to symmetries.

If we have a system that we want to calculate the E-field, one starts with the general form: $$\mathbf{E}(x,y,z)=E_x(x,y,z)\mathbf{a}_x+E_y(x,y,z)\mathbf{a}_y+E_z(x,y,z)\mathbf{a}_z$$ Let's say the system is a charged infinite plane, people will say that the field points outwards and only depends on z, which is correct but, how can you prove that?

I know its proved with symmetries, I guess that if you have a translation symmetry in the x and y axis, $$\mathbf{E}(z)=E_x(z)\mathbf{a}_x+E_y(z)\mathbf{a}_y+E_z(z)\mathbf{a}_z$$ can't depend on x or y, but how can you prove that $$\mathbf{E}(z)=E_z(z)\mathbf{a}_z$$ The example is with electrostatics but this concerns every vector field, how can I apply symmetries, how those symmetries, individually, affect the vector field??

Tito Diego
  • 13

1 Answers1

0

This really should be a comment, but It’s a little too long for it with the links. People do indeed make quick arguments without proper justification. The ‘true’ reason is that you’re solving some PDE together with suitable boundary conditions. Furthermore, when the system is well-posed (essentially meaning there exists a unique solution) it follows that every symmetry of the problem induces a symmetry of the solution. So, long story short, it’s because of symmetry of the problem + uniqueness of solutions. In your unbounded domains (like yours) it’s a little more mathematically dicey to give fully general results, but still it’s salvageable on a case-by-case basis. But, the bottom line is that you really should not lose sight of the fact that in electrostatics, you’re solving a PDE together with some initial/boundary conditions (or in more general electrodynamics, you’re solving a system of evolution equations with certain initial conditions). So, the properties of the solutions really ought to be consequences of nice properties of the equations being studied.

A simpler example to illustrate this idea is this answer where I show how the familiar properties of the trigonometric functions can be proved starting only from the fundamental differential equation solved by them $f’’=-f$.

Back to the physics, I actually wrote a much more detailed answer on this topic on PhySE: Formalizing the arguments of symmetry. There I focused on the spherically symmetric case, but I think you can see how the argument plays out more generally. The point is you start with a solution, then do some transformations to it (like translations or reflection across one of the planes) and show that the transformed vector field satisfies the same equation (i.e $\nabla\cdot E=\rho/\epsilon_0$) with the same boundary condition (a suitable asymptotic at infinity). Then, by uniqueness of solutions, the thing you started off with is equal to the transformed thing, and hence your solution actually inherits special properties.

peek-a-boo
  • 65,833