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Field Strenght Tensor and its Dual (in SR)

  1. Apr 10, 2014 #1

    Zag

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    Hello everyone,

    I have recently read a puzzling statement on my Electromagnetism (Chapter on Special Relativity) material regarding the Field Strength Tensor, [itex]F^{\mu\nu}[/itex], and its dual, [itex]\tilde{F}^{\mu\nu}[/itex]. Since I've been thinking about this for a while now, and still can't understand it, I was hoping to hear your thoughts about it.

    I believe we all agree that a possible definition for the Dual Tensor in terms of the original Field Strength Tensor is the following:

    [itex]\tilde{F}^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\sigma\rho}F_{\sigma\rho}[/itex]

    Having that in mind, the text states the following:

    "It can be shown that this is the only way in which we can construct a Lorentz-invariant four-tensor involving the fields that is independent of the original field strength tensor."

    I can't understand why should this be the case, nor can I come up with a proof for this statement. Any thoughts on this matter would be greatly appreciated!

    Thank you very much,
    Zag
     
  2. jcsd
  3. Apr 10, 2014 #2

    Matterwave

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    "Independent of the original field strength tensor" doesn't sound right to me...obviously the field strength and its dual hold the exact same information, and there is a 1-1 correspondence between the two with a simple way to go from one to the other. Perhaps the book is using the word "independent" in a way that I'm missing right now...?
     
  4. Apr 10, 2014 #3

    dextercioby

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    I'm sure you can prove it: [itex] \tilde{F}^{\mu\nu} = A^{\mu\nu\rho\sigma}F_{\rho\sigma} [/itex]. [itex] A=A\left(\delta,\epsilon,g\right) [/itex].
     
  5. Apr 10, 2014 #4

    Zag

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    Thank you for your replies, everyone.

    Maybe the author meant "linearly independent"? I'm not sure.

    Could you elaborate more on this dextercioby? I am not sure if I am able to follow your notation here.
     
  6. Apr 10, 2014 #5

    Ben Niehoff

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    I've always thought this style of reasoning somewhat bizarre. "What tensors can you make out of X, Y, and Z?". I think it misses the point. We're not interested in tensors because they have such-and-such transformation properties; we're interested in tensors because they represent real, measurable objects and not coordinate artifacts.

    The physical significance of the dual field tensor is quite clear: it exchanges the roles of ##\vec E## and ##\vec B##. So if ##F## describes a bunch of electric charges, then ##\tilde F \equiv \star F## describes a bunch of magnetic monopoles.
     
  7. Apr 10, 2014 #6

    Bill_K

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    I believe he's pointing out that this is not so much a property of the electromagnetic field as it is a property of four dimensions. He's saying, "*F must contain the same components as F, rearranged in some linear fashion, hence it can be obtained from F by multiplication with some constant tensor A," and asking, "What are the only rank four tensors A which have the desired symmetry properties [antisymmetric on the first pair of indices and the last pair] and can be built using only the Kronecker delta δ, the epsilon tensor ε and the metric g"?
     
  8. Apr 10, 2014 #7

    Matterwave

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    Anyone know what the author meant by "independent"? It's irking me that I can't figure out that statement...it can't be linear independence right since any symmetric tensor is linearly independent from an anti-symmetric tensor...
     
  9. Apr 10, 2014 #8

    samalkhaiat

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    They have different transformation properties under the Lorentz group [itex]SL(2, \mathbb{C})[/itex]. In terms of the (self-dual) [itex](1,0)[/itex] and the (antiself-dual) [itex](0,1)[/itex] representations, they can be expressed as
    [tex]F_{ \mu \nu } = ( F^{ ( 1 , 0 ) } + F^{ ( 0 , 1 ) } )_{ \mu \nu } ,[/tex]
    [tex]^{*}F_{ \mu \nu } = ( F^{ ( 1 , 0 ) } - F^{ ( 0 , 1 ) } )_{ \mu \nu } .[/tex]
    They also are linearly independent: [itex]a_{ 1 } F_{ \mu \nu } + a_{ 2 } \epsilon_{ \mu \nu \rho \sigma } F^{ \rho \sigma } \neq 0[/itex] for non-zero [itex](a_{ 1 }, a_{ 2 })[/itex].
     
    Last edited: Apr 10, 2014
  10. Apr 10, 2014 #9

    Matterwave

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    I'm not saying that they aren't linearly independent, I'm just saying that the dual tensor can't possibly be the only rank 2 tensor that's linearly independent... there should be 14 other ones from a simple counting of components.
     
  11. Apr 11, 2014 #10
    The quote alluded to the independence of the rank 4 tensor.
     
  12. Apr 11, 2014 #11

    Matterwave

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    The totally anti-symmetric tensor is the only totally antisymmetric rank 4 tensor in 4 dimensions, but there are still 253 linearly independent other rank 4 tensors in 4 dimensions...? I'm still so confused by that statement lol.
     
  13. Apr 11, 2014 #12

    robphy

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    Possibly helpful...

    By googling your quoted sentence without the quotes, the second link leads to
    http://arxiv.org/pdf/1404.0409 (see page 2)

    (What is the original source of your quote? What is the context?)
     
  14. Apr 11, 2014 #13

    Bill_K

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    The point is that the rank-4 tensor must be built from quantities intrinsic to the 4-geometry. As stated above, these are only the Kronecker delta δ, the metric g, and the epsilon tensor ε. Yes there are many other possible rank-4 tensors, but the additional structure necessary to build them would represent some physical quantity occupying the spacetime and breaking Lorentz invariance.
     
  15. Apr 11, 2014 #14

    samalkhaiat

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    Can you state a QUESTION instead of throwing meaningless numbers at me? In 4-dimensional Minkowski space-time, there are two Lorentz-invariant tensors, the metric [itex]\eta_{ \mu \nu }[/itex] and the totally antisymmetric tensor [itex]\epsilon_{ \mu \nu \rho \sigma }[/itex]. From these you can show that the only rank-2, antisymmetric tensor that can be constructed from [itex]F_{ \mu \nu }[/itex] and linearly independent of it, is the dual tensor [itex]\epsilon_{ \mu \nu \rho \sigma } F^{ \rho \sigma }[/itex].
     
  16. Apr 12, 2014 #15

    Matterwave

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    Thanks. The quote in that paper makes sense to me.
     
  17. Apr 12, 2014 #16

    Zag

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    Thank you for your reply, guys. It makes more sense now thinking in these terms. I really appreciate your help. :)
     
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