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Homework Help: Using symmetry in electrostatic problem

  1. Dec 10, 2011 #1
    here is what I am doing. I am trying to argue that the electric field is zero at the
    center of the sheet using only symmetry arguments.

    consider an insulator in the shape of infinite sheet of thickness 2a and with uniform charge density [itex]\rho[/itex]. Now let me specify the coordinate system. The sheet is from
    z=-a to z=a. Now we need to give convincing arguments for the statement that
    the electric field at z=0 is zero. Before I answer that, lets notice the
    symmtries present in the charge distribution. There is reflection symmetry
    at z=0. There is rotational symmetry about the z axis. And there is translational
    symmetry if we move z axis parallel it itself.

    So lets assume that [itex]\mathbf{E}\neq 0[/itex] at z=0. So it will , in general case, have
    x,y,z components. But if it has z component then it violates relfection symmetry
    about z=0. So lets remove that component. So we are left with x,y components. The net
    [itex]\mathbf{E}[/itex] will have certain direction in the x-y plane. But since charge has
    rotational symmetry around z axis, even this particular direction of [itex] \mathbf{E}[/itex]
    is not possible. There is one last possiblity left. We can let [itex] \mathbf{E}[/itex] emerge
    from (0,0,0) and then it will radially go outwards in the x-y plane. In this way, the
    [itex] \mathbf{E}[/itex] is symmetric about z axis. But since the charge is symmetric about
    any axis parallel to the z axis, this choice of [itex] \mathbf{E}[/itex] is not possible.
    So the only logical option left is the case where [itex] \mathbf{E}=0[/itex] at z=0.
    I hope my argument is right. I have not solved any differential equation so far
    to arrive at this conclusion. I have assumed that the electric field obeys the
    same symmetry as the charge distribution. If we look at the Gauss's law in
    differntial form,

    [tex]\vec{\nabla}\cdot \mathbf{E} =\frac{\rho}{\epsilon_o}[/tex]

    I think its possible to see that the symmetry in charge distribution should result
    in symmetry in electric field. But I don't know how to prove this mathematically.

    I think we might be able to prove this using group theory, I am not sure.
    What do you think about this ?

    It should be possible to apply this line of reasoning to other Maxwell equations
    as well.

    I have not seen the above reasoning in introductory physics books. Please comment and
    let me know any flaws.
  2. jcsd
  3. Dec 10, 2011 #2
  4. Dec 10, 2011 #3
    hi liquid, but is my intuition correct ? that symmetry in charge distribution goes to symmetry
    in the electric field produced by these charges ?
  5. Dec 10, 2011 #4


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    That is right. Read the article:


  6. Dec 11, 2011 #5
    echild, thanks. I have read that article but this particular information I didn't find there.
    Can you point out to the specific information please ?
  7. Dec 11, 2011 #6


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    Read "Invariance in force". Very similar to your argument: The symmetry of charge distribution is mirrored in the symmetry of the electric field. But the whole article is useful; remember it.

  8. Dec 11, 2011 #7
    Yes , I read that. Its related. I was wondering how do we prove this mathematically. There must be some way of doing that. I have not seen my approach in David Griffiths's book even though its so simple and elegant. Now I am wondering if I can extend this approach to solve problems in other areas of EM , like currents and relativistic stuff. But I am looking for more mathematical justification for what I am thinking. Some physicists have a nasty habit of using sloppy arguments (Griffiths does that often). I am trying to think like Herman Weyl, who made important contributions to the study of symmetry and he was a mathematician. And using symmetry arguments is so elegant and beautiful.
    I have not seen US physics books(introductory, e.g. Young and Freedman, Halliday and Resnick , Griffiths's EM) using the kind of argument I gave here to solve physics problems.

  9. Dec 11, 2011 #8


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    Physicist do use symmetry arguments, mostly intuitively. I can not help you to find a rigorous prof now.
    See Noether's theorem http://en.wikipedia.org/wiki/Noether's_theorem if you want to dive in the problem. It is very interesting that every symmetry involves a conservation law.

  10. Dec 11, 2011 #9
    ehild, well , probably my arguments ultimately follow from Noether's theorem. This theorem is not taught widely in physics courses, which is sad. Its the symmetry which gives lot of elegance to physics.......
  11. Dec 11, 2011 #10


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    You are right. Also study Group Theory.:smile:

  12. Dec 11, 2011 #11
    Hello ehild

    But it seems that the symmetry in current distribution doesn't always go into the symmetry of the magnetic field. Consider infinite uniform surface current [itex]\mathbf{K}=K\;\hat{\mathbf{x}}[/itex]
    flowing over the xy plane. here the magnetic field would be

    [tex]\mathbf{B}=(\mu_o/2)K\;\hat{\mathbf{y}} \quad\mbox{for}\;z<0[/tex]

    [tex]\mathbf{B}=-(\mu_o/2)K\;\hat{\mathbf{y}} \quad\mbox{for}\;z>0[/tex]

    So we can see that the current has reflection symmetry in xz plane but the magnetic field doesn't have such reflection symmetry. So what do you think the reason is ?

    But if we look at the vector potential in this case,

    [tex]\mathbf{A}=-\frac{\mu_o K}{2}\lvert z \rvert \hat{\mathbf{x}}[/tex]

    it does have the symmetry of the current. So what do you think ?
  13. Dec 11, 2011 #12


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    For keeping symmetry during some operation, the operation itself has to by symmetrical.
    If two things like charge and electric field or current density and magnetic field are related with some operator u=Hv, and there is a symmetry operation A so that Ax=x, u will be unchanged under the operation A if H itself has the symmetry:

    Au =u => A(Hx)=AHA-1Ax=AHA-1(Ax)=AHA-1x,

    AHA-1 is the transformed of H, and it should be equivalent with the original H.

    The relation between current density i and magnetic field B/μ is given by the Maxwell equation curl(B/μ)=i . The curl operation is not symmetric for reflection.

    In case of electric field and charge distribution ρ, div(εE)=ρ, the E has the symmetry of ρ if ε is also symmetric, as the divergence operation is a symmetric.

    All what I wrote can be wrong. I am very sorry, I do not know enough for such discussion. I used to know more but it was very long time ago. Start a new thread in General Physics, you certainly will find people much better than me to discuss this topics. They do not join to this thread here in Homework section as they think you get enough help from me. But this is not homework any more.

  14. Dec 12, 2011 #13

    should I start a new thread or can somebody just move it to general physics. ?
  15. Dec 12, 2011 #14


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    The whole point of using symmetry principles to deduce solutions is that you don't have to do it mathematically.

    But if you do want to justify the symmetry argument with math, it's basically what ehild wrote in the next-to-last post. You would start by enumerating the transformations "of interest," which in this case are those transformations that leave the charge density field [itex]\rho(\mathbf{r})[/itex] invariant: translation and rotation in the x-y plane, plus reflection in the z axis. But we don't yet care that they leave [itex]\rho[/itex] invariant; for now they are just transformations.

    Each of these transformations has a corresponding operator which implements it. Let's call it [itex]A[/itex], defined such that the transformation takes [itex]\mathbf{r}\to A\mathbf{r}[/itex].

    Now, as a scalar field, the charge density transforms as [itex]\rho(\mathbf{r})\to\rho(A^{-1}\mathbf{r})[/itex], and as a vector field, the electric field transforms as [itex]\mathbf{E}(\mathbf{r})\to R(A)\mathbf{E}(A^{-1}\mathbf{r})[/itex] where [itex]R(A)[/itex] is some sort of rotation matrix. Besides those two, the only other piece involved in Gauss' law is the scaled divergence operator [itex]\epsilon\nabla\cdot[/itex] which I'll denote [itex]\mathbb{D}[/itex]. [itex]\mathbb{D}[/itex] can be considered a dual vector field (it acts on a vector to produce a number), so it transforms as [itex]\mathbb{D}(\mathbf{r})\to \mathbb{D}(A^{-1}\mathbf{r})R^{-1}(A)[/itex].
    Thus, under the transformation, Gauss' law goes from
    [tex]\mathbb{D}(\mathbf{r})\mathbf{E}(\mathbf{r}) = \rho(\mathbf{r})[/tex]
    [tex]\mathbb{D}(A^{-1}\mathbf{r})R^{-1}(A)R(A)\mathbf{E}(A^{-1}\mathbf{r}) = \rho(A^{-1}\mathbf{r})[/tex]
    which is just the expression of Gauss' law at the point [itex]A^{-1}\mathbf{r}[/itex].

    So far I haven't involved symmetry in any way, or chosen any specific transformations; the operator [itex]A[/itex] could be any old transformation operator, like a translation at 20 degrees from the z axis. All this inherently implies is that, if Gauss' law is satisfied at one point, then if you transform the coordinate system using [itex]A[/itex] then it will continue to be satisfied at the same point under the new coordinates.

    Where the symmetry comes in is that when [itex]A[/itex] represents a transformation operator that corresponds to one of the symmetries of the charge density, you have the equivalence [itex]\rho(A^{-1}\mathbf{r}) = \rho(\mathbf{r})[/itex]. This means that two points which can be mapped on to each other by the symmetry operator must have the same charge density. Now, the operative question is, does the dual vector field [itex]\mathbb{D}[/itex] share the same symmetry? Because if it does, you can write
    and since you know that
    this implies that [itex]\mathbf{E}(A^{-1}\mathbf{r}) = \mathbf{E}(\mathbf{r})[/itex]. (unless the action of the dual vector is non-unique)

    Since you mentioned the magnetic field, I will say that if you do the same thing with Ampère's law,
    [tex]\frac{1}{\mu_0}\nabla\times\mathbf{B}(\mathbf{r}) \equiv \mathbb{D}'(\mathbf{r})\mathbf{B}(\mathbf{r}) = \mathbf{J}(\mathbf{r})[/tex]
    you get a sign flip because [itex]\mathbb{D}'[/itex] is a dual pseudovector, which transforms as [itex]\mathbb{D}'(\mathbf{r})\to -\mathbb{D}'(A^{-1}\mathbf{r})R^{-1}(A)[/itex].
  16. Dec 12, 2011 #15
    ehild, thanks for inviting diazona.

    diazona, that is very detailed explanation. thank you. But I have not studied it in my own education. So please provide references in mathematical literature so that I can learn it. Does J.D.Jackson use such kind of reasoning ?

    It should be possible to use the reasoning given by you in undergraduate texts. But David Griffiths doesn't use it in his book on EM. He uses other arguments , which use symmetry but its not very clear what he is trying to say (Intro to Electrodynamics- Griffiths 3 ed. Example 5.8, page 226). Your arguments are crystal clear once we understand the necessary math.
  17. Dec 13, 2011 #16


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    Yes, it is possible to use this reasoning in undergrad-level textbooks, although one would have to keep in mind that the formal study of symmetry transformations tends to come late in the undergrad curriculum, typically after one would have completed an E&M course at the level of Griffiths. So using a detailed mathematical description of symmetry as I've done here would go over most students' heads, and for the vast majority of students there would be no point in going into such detail because the intuitive reasoning used in Griffiths' example 5.8 (for instance) is perfectly sufficient.

    I don't have my copy of Jackson on hand but I don't remember him going into this level of detail about symmetry principles. Then again I don't remember the entire contents of the book so it may well be in there.

    As for references, what I wrote is pieced together mostly from various aspects of differential geometry as it is used in general relativity (for dual vectors) and representation theory as it is used in quantum field theory (for transformation operators). So to understand the relevant math, I would refer you to textbooks of your choice on those subjects.

    By the way, one correction to my previous post: a pseudovector or dual pseudovector only transforms with a sign flip under reflection, not under an arbitrary transformation. (Actually, come to think of it, I believe the curl operator is technically a dual tensor.)
  18. Dec 13, 2011 #17
    Even I don't recall Jackson teaching on those lines in his book. But thanks for the math references. I will look into it.