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Maxwell Equations Lorentz Invariance - Notation

  1. Jul 31, 2013 #1
    [This is mostly about notation]

    I was working on a problem where I had to prove that [itex] div(B) [/itex] remains invariant under lorentz transformations. That was not too hard, so I came up with

    [itex] div(B) = \partial_{\mu} B^{\mu} [/itex]

    must equal

    [itex] div(B) = \partial'_{\mu} B'^{\mu} [/itex]

    so I did a lorentz transformation on both [itex]B[/itex] and [itex] \partial [/itex].

    [itex] \partial'_{\mu} = \widetilde{L}^{\nu}_{\ \mu} \partial_{\nu} [/itex]

    and

    [itex] B'^{\mu} = L^{\mu}_{\ \nu} B^{\nu} [/itex]

    where [itex]L[/itex] is the lorentz transformation matrix and [itex] \widetilde{L} [/itex] its inverse.

    Now I simply plug in both of them into the first equation (changing [itex]\nu[/itex] to [itex]\alpha[/itex])

    [itex] div(B) = \widetilde{L}^{\nu}_{\ \mu} \partial_{\nu} L^{\mu}_{\ \alpha} B^{\alpha} [/itex].

    Because [itex] \partial_{\nu} [/itex] applies to [itex] B^{\nu} [/itex] I can change the order.

    Then I get [itex] div(B) = \widetilde{L}^{\nu}_{\ \mu} L^{\mu}_{\ \alpha} \partial_{\nu} B^{\alpha} [/itex].

    Now I use [itex] \widetilde{L}^{\nu}_{\ \mu} L^{\mu}_{\ \alpha} = \delta^{\nu}_{\alpha} [/itex] which changes the previous equation to [itex] \delta^{\nu}_{\alpha} \partial_{\nu} B^{\alpha} [/itex] and then I get [itex] div(B) = \partial_{\alpha} B^{\alpha} [/itex].

    This is the same equation as on the top -> QED.

    1) Is this okay (notationally)?

    2) I tried the same for the curl of E

    [itex] curl(E)_i = \varepsilon_{ijk} \partial_{j} E_{k} [/itex] ( in old-school notation )

    But the problem is that I do not really know how to write it down correctly.

    Is it [itex] \varepsilon_{ijk} [/itex] or [itex] \varepsilon^{ijk} [/itex]?

    I am also not sure about if the curl of E (or E itself) is contravariant or covariant and how to transform them correctly to get the same I got for div(B).

    The problem was from Cheng's "Relativity, Gravitation and Cosmology" (problem 2.4).
    He does not use this type of notation in the solutions.

    Thank you in advance.

    First post :)
     
  2. jcsd
  3. Jul 31, 2013 #2

    WannabeNewton

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    ##(\nabla \times E)^{i} = \epsilon^{ijk}\partial_{j}E_{k}## should be fine. It doesn't really matter in the end as far as showing Lorentz invariance goes.
     
  4. Jul 31, 2013 #3
    thank you very much.

    Now I would just do the same as I did for div(B).

    ##curl(E')^{i} = \epsilon^{ijk}\partial'_{j}E'_{k}##

    Does ##\epsilon^{ijk}## change under lorentz transformation? I think it should not, so the above equation should be correct.

    Now I plug in ##\partial'_{j}## and ##E'_{k}## as I did for div(B) and get

    ##\epsilon^{ijk} \widetilde{L}^{\nu}_{\ j} \partial_{\nu} \widetilde{L}^{m}_{\ k} E_{m}##

    But there are two ##\widetilde{L}## and they do not cancel.

    Does this have to do anything with the fact that I left out the other part of the Maxwell equation?
    I think it should be possible doing lorentz invariance term by term.
     
  5. Jul 31, 2013 #4

    WannabeNewton

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    Wait are you trying to show that ##(\nabla \times E)^{i'} = (\nabla \times E)^{i}## under a Lorentz transformation? Because this is not true. The above is a 3-vector for one (not even a 4-vector) and not a scalar. The only reason ##\partial_{\mu}E^{\mu} = \partial_{\mu'}E^{\mu'}## is because this is a scalar constructed out of Lorentz covariant 4-vectors hence it transforms trivially under a Lorentz transformation. I thought you were trying to show that Maxwell's equations were Lorentz invariant (i.e. Lorentz covariant).
     
  6. Jul 31, 2013 #5
    1c75ec5bd337504c9bb2333bb301a7ab.png

    I was trying to show lorentz invariance for this equation, doing it term by term.

    ##(\nabla \times E)^{i'} + ( \frac{\partial B}{\partial t} )^{i'} = 0##

    After getting the transformation for curl(E) and doing the transformation for ##\partial_{t} B## and adding them I should get 0?

    ##\partial'_{t} B'^{\mu} = \widetilde{L}^{\nu}_{\ t} \partial_{\nu} L^{\mu}_{\ \alpha} B^{\alpha}##

    Adding them...

    ##\epsilon^{ijk} \widetilde{L}^{\nu}_{\ j} \partial_{\nu} \widetilde{L}^{m}_{\ k} E_{m} + \widetilde{L}^{\nu}_{\ t} \partial_{\nu} L^{\mu}_{\ \alpha} B^{\alpha} = 0##

    I am not sure about the constant index ##t## and how to continue from there.
     
    Last edited: Jul 31, 2013
  7. Jul 31, 2013 #6

    dextercioby

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    But rot(E)+∂tB=0 comes from ∂μFμν = 0 which is a tensor equation, hence true in every coordinate system.
     
  8. Jul 31, 2013 #7
    thanks. I know :)
    I wanted to do this without tensors first.
     
  9. Jul 31, 2013 #8

    PeterDonis

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    You appear to be mixing up equations written using two different conventions, which is only going to cause confusion.

    If you are going to write Maxwell's Equations in 3-vector form, such as this...

    ...then you shouldn't be trying to transform them using a 4-dimensional transformation matrix (i.e., one which includes the ##t## index as well as the spatial indexes) like you are doing. You would need to apply the Lorentz transformations in 3-vector form, as given, for example, here...

    https://en.wikipedia.org/wiki/Class...rmation_of_the_fields_between_inertial_frames

    ...because when you write the equations in 3-vector form, ##t## is not an vector index, it's a parameter. You can show that Maxwell's Equations are Lorentz invariant this way, but it's somewhat laborious.

    An alternative, and easier, way to show that Maxwell's Equations are Lorentz invariant is to write them in explicitly covariant form, using 4-vectors and tensors:

    $$
    \partial_a F_{bc} + \partial_b F_{ca} + \partial_c F_{ab} = 0
    $$
    $$
    \partial_a F^{ab} = 4 \pi J^b
    $$

    where ##J^b## is the charge-current density 4-vector, whose components are ##(\rho, J^x, J^y, J^z)## (##\rho## is the charge density and ##J^i## is the current 3-vector), and ##F^{ab}## is the electromagnetic field tensor, whose components in a given frame are

    $$
    F^{0i} = E^i
    $$
    $$
    F^{ij} = B^k
    $$

    where the spatial indexes ##ijk## are an even permutation of ##123## (I.e., ##F^{12} = B^3##, ##F^{23} = B^1##, ##F^{31} = B^2##). ##F^{ab}## is an antisymmetric tensor, so ##F^{ba} = - F^{ab}## and the components along the diagonal are all zero. (Also, the index numbers ##0, 1, 2, 3## correspond to the coordinates ##t, x, y, z##.)

    If you write out the above equations by components, and group the terms appropriately, you will see that they are the same as the 3-vector Maxwell's Equations you are familiar with. Once you've verified that, since any equation that can be written using 4-vectors and tensors like that is automatically Lorentz covariant, your proof is essentially done.
     
  10. Jul 31, 2013 #9
    Oh yes, I know.
    I already did it in one problem of Dan Fleisch's "Students Guide to Vectors and Tensors".
    Using 4-vectors and tensors it is quite easy.

    I just started doing serious General Relativity going with Cheng and Inverno.
    After not really liking the way he did it, I tried that way (see post #1).

    Thanks to both of you.
     
  11. Aug 2, 2013 #10
    Couldn't one just argue that ##(\nabla \cdot \mathbf{B}) = 0## is a scalar and thus invariant under lorentz transformations?
    If you think about it (the EM-way), it makes sense that there are no magnetic monopoles in all systems.

    What about ##\nabla^{2} \mathbf{E}##?
    The Laplacian would also produce a scalar.

    Scalars themselves and scalarfields are lorentz invariant, but what about divergences and laplacians and others?

    For ##(\nabla \times \mathbf{E}) + \frac{\partial \mathbf{B}}{\partial t} = 0## you just use the transformed ##\mathbf{E}## and ##\mathbf{B}## and check if it is ##0##.

    I kind of missed the fact that ##\mathbf{E}## and ##\mathbf{B}## are just 3-vectors and ##L^{\mu}_{\nu}## is a ##4\times4## matrix. Next time I should not rush and think more...
     
  12. Aug 2, 2013 #11

    PeterDonis

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    Not the way you've written them, because, again, you are using 3-vectors instead of 4-vectors. The 4-divergence of a 4-vector, ##\partial_u A^u##, and the 4-D analogue of a Laplacian (which is called, I believe, a d'Alembertian), ##\partial_u \partial^u \phi##, are Lorentz scalars. But if you write those out in components, you don't just get the 3-divergence or the 3-Laplacian.
     
  13. Aug 2, 2013 #12
    Oh, of course.

    So basically, ##ds^2 = \eta_{\mu\nu} dx^{\mu} dx^{\nu}## is invariant.

    If you do ##\eta_{\mu\nu} \partial^{\mu} \partial^{\nu} \phi## you get ##\partial_{\mu} \partial^{\mu} \phi##, which is just ##\frac{\partial ^2\phi}{\partial t^2} - \frac{\partial ^2\phi}{\partial x^2} - \frac{\partial ^2\phi}{\partial y^2} - \frac{\partial ^2\phi}{\partial z^2}## (wave equation).

    Which is just the "d'Alembertian" as you mentioned...

    118edea659f876d0d4519c84d6e2d9c5.png

    Unfortunately Cheng does not get very deep into relativistic EM.

    4-vectors are just rank 1 tensors and things like ##p^{\mu} p_{\mu}## or ##x^{\mu} x_{\mu}## are ALWAYS invariant (as long as they are tensors/4-vectors)?

    So if you take a rank 2 tensor and do something like ##T_{\mu\nu} T^{\mu\nu}##, which should be the same as ##\eta_{\alpha\mu} \eta_{\beta\nu} T^{\alpha\beta} T^{\mu\nu}##, you also get an invariant?
     
  14. Aug 2, 2013 #13

    WannabeNewton

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    Yes. As I said, if you form scalar fields out of Lorentz covariant tensor fields, you will get Lorentz invariants. Just don't mix up 3-vectors with 4-vectors because 3-vectors are not Lorentz covariant objects.
     
  15. Aug 2, 2013 #14
    Okay, that made it clear.
    Thanks again.
     
  16. Aug 2, 2013 #15

    WannabeNewton

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    No problem. It might serve you less confusion in the future if you used Roman letters like ##i,j,k## etc. for space indices and Greek indices like ##\mu, \nu, \gamma## etc. for space-time indices. Then ##\partial_{i}A^{i}## would represent the 3-divergence (which is not a Lorentz invariant) whereas ##\partial_{\mu}A^{\mu}## would represent the 4-divergence (which is a Lorentz invariant).
     
  17. Aug 2, 2013 #16
    Actually I know that convention.
    I just have no idea why I did not use it here.
    The biggest problem was just the 3d/4d confusion, I guess..
     
  18. Aug 2, 2013 #17

    WannabeNewton

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    Yeah but I don't blame ya. When I first saw your thread my mind immediately went to the same train of thought as your own. It's just what happens if you're so used to the Lorentz covariant 4-tensor form of Maxwell's equations as opposed to the usual form. Everything is so much more elegant and easier to read off from the former; the differential forms version of Maxwell's equations is more elegant still.
     
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