Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stupid question about invariance of maxwell equations

  1. Feb 18, 2009 #1
    Hi,

    as you all know one can write the Maxwell-equations in covariant form, namely:

    [tex]\partial_a F^{ab} = \frac{4\pi }{c} j^{b} [/tex]

    and

    [tex]\partial_a G^{ab}=0[/tex]

    where [tex]\textbf{G}[/tex] is the dual Tensor to [tex]\textbf{F}[/tex].

    Now the two simple questions.
    I can see that they are invariant, because I have a 4-Vector on both sides, and so the rhs and lhs
    will transform in the same way, right ?
    So the equation will have in another frame exactly the same form.

    But on the other hand this equations would be invariant under all such transformations, not only
    Lorentztransformations ?

    I don't get it..

    Thanks
     
  2. jcsd
  3. Feb 18, 2009 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    All "such" transformations? Do you mean all linear transformations?

    [tex]\partial'_a F'^{ab}=\Lambda_a{}^c\Lambda^a{}_d\Lambda^b{}_e \partial_c F^{de}=\delta^c_d\Lambda^b{}_e \partial_c F^{de}=\Lambda^b{}_e \partial_c F^{ce}=\Lambda^b{}_e j^e=j'^b[/tex]

    The second equality only holds for Lorentz transformations.
     
  4. Feb 18, 2009 #3

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    What do you mean by "all such" transformations?

    What is meant by "such"? What do you refer to? example?

    Lorentz transformations is boosts in space-time, rotations, parity and time-reversal.

    Secondly, the quantities you wrote are not invariant under LT, they transforms as Lorentz 4vectors. The invariants under Lorentz transformation are Lorentz scalars, such as:
    [tex]F^{\mu\nu}F_{\mu \nu}[/tex] and
    [tex]x^{\mu}x_{\mu}[/tex] .

    Those are invariants under Lorentz transformations. A 4vector such as the 4current
    [tex]j^\alpha[/tex] transforms:
    [tex]j'^\alpha = \Lambda ^\alpha_\beta j^\beta \neq j^\alpha [/tex]
    So the 4-current is not a Lorentz invariant.

    But you are correct that the "equations are invariant".
     
  5. Feb 18, 2009 #4

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I've been trying to sort this issue out as well.

    Some facts that play a role in this puzzle.
    - The above equations (or something close to it) are associated with terms like "Generally Covariant Form" or "pre-metric".
    - The metric plays a role in defining the "duality" relationship between F and G, which is associated with the constitutive relations. (Do you use E,B or E,B,D,H?)
     
  6. Feb 18, 2009 #5
    Yes I mean linear transformations, sorry.
    My problem is, what can I say about the [tex]\Lambda^b{}_e[/tex]-matrices ?
    I can put anything I want in the lambda's, because the equations (not the 4-vectors) will
    remain invariant.

    I'm not sure if the second equality just holds for LT. Isn't this just a general property
    of tensor-transformation ?

    Sorry for my sloppiness, yes I mean linear transformations and of course the invariance of
    the equations and not 4-vectors or the field-strength tensor or something like that.


    I use E and B like in Jackson. It's just all a little bit too short in Jackson (for my purpose).
    So would it be right to say the above two equations are invariant under all linear transformations which
    include the Lorentz transformations ?

    Thanks for all your help !
     
    Last edited: Feb 18, 2009
  7. Feb 18, 2009 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No, what I used there is the definition of a Lorentz transformation:

    [tex]\Lambda_a{}^c\Lambda^a{}_d=\delta^c_d[/tex]

    This might look more familiar to you if you multiply by [itex]\eta_{bc}[/itex]

    [tex]\Lambda_a{}^c\eta_{bc}\Lambda^a{}_d=\delta^c_d\eta_{bc}=\eta_{bd}[/tex]

    and then rewrite the left-hand side as

    [tex]\Lambda_a{}^c\eta_{bc}\Lambda^a{}_d=\Lambda_a{}_b\Lambda^a{}_d=\Lambda^c{}_b\eta_{ca}\Lambda^a{}_d[/tex]

    So now we have

    [tex]\Lambda^c{}_b\eta_{ca}\Lambda^a{}_d=\eta_{bd}[/tex]

    which clearly is just the bd component of the matrix equation

    [tex]\Lambda^T\eta\Lambda=\eta[/tex]
     
    Last edited: Feb 18, 2009
  8. Feb 18, 2009 #7
    Ok, thanks Fredrik.

    I still could say one sums over the index 'a' so I don't have to transform that as a
    Tensor with three indices but as a "normal" 4-vector.

    [tex]\partial'_a F'^{ab}=j'^b[/tex]

    But if I would take an arbitray linear transformation for that I wouldn't get the desired result,
    because the primed quantities on both sides wouldn't have terms that cancel (if I would expand them into
    their relations with the unprimed ones), so that one would
    get the unprimed quantites on both sides as expected.
    That was pretty much where I got stuck. And I think that this is
    what is really meant by an equation is lorentz invariant.

    But how is it possible to see, that this transformation will turn out, that the
    extra terms on both sides cancel ? I mean on the right I have a 4-vector, on the left I have the
    divergence of a second rank tensor, which is also (a different) 4-vector.

    Edit:

    I mean I can not see that the divergence of the field strength tensor on the left
    and the 4-vector on the right have a priori this transformation relation.

    Thanks
     
    Last edited: Feb 18, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Stupid question about invariance of maxwell equations
Loading...