4
$\begingroup$

This made a lot of sense to me before, but now that I think about it, I find it rather superficial.

Suppose we are deriving the $\mathbf{B}$ field due to an infinite wire. One usual procedure is to exploit cylindrical symmetry and use Ampere's law $$\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I_{enc}$$

the conclusion with this is that the only component the field will have is in the $\hat{\mathbf{\phi}}$ direction. Similar arguments hold for other fields, like deriving the $E$ field of a spherical (and uniformly charged) conductor.

But I want to know how we can mathematically justify this. For example, when solving Laplace's equation one can sometimes discard certain coordinate dependencies according to translational symmetry (I.e. the environment might look the same irrespective of your position in $z$, therefore the potential has no dependence on $z$).

But how does this go about with vector fields? By invoking some sort of uniqueness theorem? I highly doubt it since we need information of both curls and divergences, which isn't always obvious in the case of $D$ and $H$ fields.

In case I haven't made myself clear, I want to justify the usual use of symmetries in charge or current distributions to then find the vector field associated with it. I want to justify arguments of the type "The field has to be purely radial since it has spherical symmetry". How do you think about this? How do you justify this to yourself?

I'm looking for a mathematical answer to this question. I've asked this before elsewhere and I keep hearing the usual "intuitive arguments", I think I'm safe in saying I know all of these. What I want is a mathematical justification.

$\endgroup$
  • 4
    $\begingroup$ I don't see right now what's non-rigorous about the argument that e.g the situation has spherical symmetry, and thus the fields must be invariant under that symmetry - and fields which are dependent on the angular directions are not invariant. $\endgroup$ – ACuriousMind Nov 14 '15 at 19:54
  • 1
    $\begingroup$ Have you looked at John D. Jackson's book on E&M? I think he has several parts devoted to mathematically rigorous explanations for everything E&M (and if it's not in his book, he has a great set of references within that point to where the answer can be found). $\endgroup$ – honeste_vivere Nov 14 '15 at 21:14
  • 1
    $\begingroup$ A spherically symmetric charge distribution does not mathematically imply a radial electric field. You'd have to add statics or something to get that. $\endgroup$ – Timaeus Nov 18 '15 at 2:02
1
$\begingroup$

In electrostatic, we know the generic solution. The field $\renewcommand{\vec}[1]{\mathbf{#1}}\vec{E}(\vec{x})$ created at point $\vec{x}$ by the distribution of charge $\rho(\vec{y})$ on the domain $D$ is given by

$$\vec{E}(\vec{x}) = \frac{1}{4\pi\epsilon_0}\iiint_D \frac{\vec{x}-\vec{y}}{\|\vec{x}-\vec{y}\|^3}\rho(\vec{y})d^3y.$$

Now let's consider an isometry $R$ (a rotation or a mirror in practice). We have

$$\vec{E}(R\vec{x}) = \frac{1}{4\pi\epsilon_0}\iiint_D R\frac{\vec{x}-R^{-1}\vec{y}}{\|\vec{x}-R^{-1}\vec{y}\|^3}\rho(\vec{y})d^3y,$$

since $R$ preserves the norm of vectors. Now we can do a change of variable $\vec{y}'=R^{-1}\vec{y}$ in the integral. Since we have an isometry, the Jacobian will be equal to 1, and we have

$$\vec{E}(R\vec{x}) = \frac{1}{4\pi\epsilon_0}\iiint_{R(D)} R\frac{\vec{x}-\vec{y}}{\|\vec{x}-\vec{y}\|^3}\rho(R\vec{y})d^3y.$$

Thus if $R$ leaves the domain $D$ invariant, and if it leaves $\rho$ invariant as well, then

$$\vec{E}(R\vec{x})=R\vec{E}(\vec{x}).$$

This is all you need. The same kind of theorem holds for magneto-static but it is more complicated because of the vector product. I can develop if you wish.

$\endgroup$
  • $\begingroup$ rotation should affect the field orientation as well. E.g.) if scalar field transforms as $\phi(x) \to \phi(R^{-1}x)$ then vector field transforms as $E(x) \to RE(R^{-1}x)$ $\endgroup$ – wcc Mar 7 at 19:40
  • $\begingroup$ Sorry typo! Correcting... $\endgroup$ – frapadingue Mar 7 at 20:17
0
$\begingroup$

I want to justify the usual use of symmetries in charge or current distributions to then find the vector field associated with it. I want to justify arguments of the type "The field has to be purely radial since it has spherical symmetry".

For example you want to prove why the electric field of a distribution of elementary charges situated on a circle of radius $R$ has a radial symmetry in the plane of the circle.

You simply apply the Coulomb formula (accepting that an elementary charge has by law a radially symmetric field) and you calculate the field intensity if 3 elementary charges are on that circle (a case illustrated in the attached picture), then you add three more points and approximate the circle by a hexagon of charges. You will see that the field becomes more radially symmetric than in the case of three charges. For a large number of charges the field will reach radial symmetry.

enter image description here

Source

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.