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I have, for instance, a problem with a spherically symmetric charge distribution. I deduce here, in order to solve the problem easily, that the corresponding electric field must be symmetric. How is this type of deduction rigorously justified?

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Proof by contradiction: assume a non-symmetric field configuration, perform a rotation then use the field equations to show that the result is inconsistent. –  Michael Brown Feb 12 '13 at 15:29
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Well, this is just be wrong in general: what's true is that symmetry+uniqueness of the solution imply that the solution is symmetric... –  Yvan Velenik Feb 12 '13 at 16:00
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@Yvan: what you mean by uniqueness ? @ Hayeder: this can be justified by Group theory, there you will learn that such symmetries of the problem will be reflected on differential/algebraical equations that describes this problem/system, and those symmetries will simplifies the solutions too much. –  TMS Feb 12 '13 at 19:20
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@TMS: the solutions to a problem with symmetry are not themselves symmetric in general. The only case in which this is guaranteed is when the solution is unique. What's always symmetric is the set of solutions, not the solutions themselves. –  Yvan Velenik Feb 13 '13 at 8:56
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Of course, in simple situations, like here, uniqueness is not difficult to establish. But, in any case, this is a prerequisite. –  Yvan Velenik Feb 13 '13 at 8:58
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1 Answer

up vote 3 down vote accepted

Let me first answer this question in a particular class of electrostatics problems.

In the case of a localized charge distribution in electrostatics (one in which the charge density vanishes outside of some ball around the origin), the general solution for a potential that vanishes at $\infty$ is $$ \mathbf \Phi(\mathbf x) = k\int_{\mathbb R^3} d^3x' \frac{\rho(\mathbf x')}{|\mathbf x - \mathbf x'|} $$ In this case, we often want to show that if $\rho$ has certain symmetries, then $\Phi$ shares those symmetries. Let's first derive a little result:


Suppose that we make an invertible transformation $T$ on the charge distribution (say a translation for a rotation for example), then the new (transformed) charge distribution $\rho_T$ will be $$ \rho_T(\mathbf x) = \rho(T^{-1}\mathbf x). $$ What will the resulting potential $\Phi$ be? Well, let's compute $$ \Phi_T(\mathbf x) = k\int_{\mathbb R^3} d^3 x' \frac{\rho_T(\mathbf x')}{|\mathbf x - \mathbf x'|} = k\int_{\mathbb R^3} d^3 x' \frac{\rho(T^{-1}(\mathbf x'))}{|\mathbf x - \mathbf x'|} $$ we can perform the resulting integral via a change of variables $$ \mathbf u = T^{-1}(\mathbf x') \implies \mathbf x' = T(\mathbf u) $$ and the formula for changing the variable of integration for volume integrals gives $$ d^3 x' = J_T(\mathbf u) d^3 u $$ where $J_T$ is the jacobian of the transformation, so that the transformed potential becomes $$ \Phi_T(\mathbf x) = k\int_{\mathbb R^3} d^3u \,J_T(\mathbf u)\frac{\rho(\mathbf u)}{|\mathbf x - T(\mathbf u)|} $$ Now back to the show.


To see how this formula for the transformed potential is used to answer your question about symmetries, let's consider a translationally symmetric charge density; $$ \rho(\mathbf x - \mathbf x_0) = \rho(\mathbf x), \qquad \text{for all $\mathbf x_0$} $$ In this case, the transformation $T$ is $T(\mathbf x) = \mathbf x + \mathbf x_0$. The Jacobian is just 1, and our the formula we derived above for the transformed potential gives $$ \Phi_T(\mathbf x) = k\int_{\mathbb R^3} d^3u \frac{\rho(\mathbf u)}{|\mathbf x - (\mathbf u-\mathbf x_0)|} = k\int_{\mathbb R^3} d^3u \frac{\rho(\mathbf u)}{|(\mathbf x - \mathbf x_0) - \mathbf u|} = \Phi(\mathbf x - \mathbf x_0) $$ On the other hand, the translational invariance of the charge density tells us that $$ \Phi_T(\mathbf x) = k\int_{\mathbb R^3} d^3 x' \frac{\rho_T(\mathbf x')}{|\mathbf x - \mathbf x'|} = k\int_{\mathbb R^3} d^3 x' \frac{\rho(\mathbf x')}{|\mathbf x - \mathbf x'|} = \Phi(\mathbf x) $$ so combining these results gives $$ \Phi(\mathbf x - \mathbf x_0) = \Phi(\mathbf x) $$ Namely, the potential is also translationally symmetric. A similar procedure can be used for other symmetries. Try rotational invariance for example on your own!

Hope that helps!

Physics Rocks.

Addendum. I think you can show similar things for generic Neumann or Dirichlet boundary value problems in electrostatics in which, for example, you don't just want to solve Poisson's equation for a localized charge distribution with vanishing potential at infinity, but I haven't worked out the details.

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brilliant. Thank you and +1. –  user27182 Feb 15 '13 at 22:29
    
Great! Glad it helped. –  joshphysics Feb 15 '13 at 22:57
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