(Alpha) Does an EM-Field contain a vorticity

Discussion in 'Physics & Math' started by Green Destiny, Oct 17, 2010.

  1. Green Destiny Banned Banned

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    I've invoked the alpha rules simply because too many threads are being lost to squabblings.

    I have seen written work suggesting that the EM field might have a vorticity, however, I have not found anything conclusive, except for two equations which attempted to describe a vorticity in the field, but it wasn't very satisfying.

    It is said from the equations and language of Maxwell that the electromagnetic field contains a momentum given as:

    \(P=Mv +qA\)

    (the momentum coupling equation)

    where A is the potential M is for mass and P is the canonical momentum. Vorticity as I understand it, should really arise in any field associated to moving objects. I have often pondered the exact mathematical nature in which you can describe this for an EM-field, and the best I personally came up with was a modification,

    \(P=M \sigma +qA=M (\nabla \times v) + qA\)

    Afterall, the vorticity is simply the curl of the velocity vector. If anyone knows whether the EM field is said to have an inherent vorticity, can I see some examples of it, and if that be the case, then can someone properly define the differences between the vorticity and the viscosity, as they seem very similar.
     
    Last edited: Oct 18, 2010
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  3. AlphaNumeric Fully ionized Registered Senior Member

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    Citation needed as that expression is not gauge invariant.

    You can't protect your attempts at pet theories by using Alpha rules unless you're willing to stick to the rules yourself. Provide, in detail, a rigorous and justified method which lead to your 'personal modification'. If you can't then you violate your own rules.

    I suggest starting by defining what \(\sigma\) is, as any new notation should be defined. As any textbook or paper will do, if you hadn't noticed.
     
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  5. 1100f Banned Registered Senior Member

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    This is not the momentum of the electromagnetic field. This is the momentum of a particle with mass M, velocity v, electric charge in an magnetic field given by its potential vector \(A\)
    This is derived by the Lagrangian \( L = {1\over2}({\bf v}-q{\bf A}).({\bf v}-q{\bf A})\)
    (The \(v\) should be in boldface since it represents a vector, and the TeX command "\bf" does not seem to work here).
    Anyway, from this Lagrangian you get the Lorentz force on a particle in the presence of a magnetic field. If you perform the minimal substitution relativisticaly, you will get the Lorentz force on a charged particle in the presence of an Electric and magnetic field.
    The momentum that you wrote is not the momentum of the electromagnetic field, but this is the Cannonical momentum of the particle, after the Legendre transformation which takes you from Lagrangian formalism to Hamiltonian formalism.

    If you want to know the momentum of the Electromagnetic field, you should start by writting the full Lagrangian of an electromagnetic field and derive from there the cannonical momentum of the E.M field. (hint - google "Poyinting vector").


    If you wrote this in order as a training in writing TeX equations, then I guess it's OK, or else, it is nonsense
     
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  7. Green Destiny Banned Banned

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    http://en.wikipedia.org/wiki/Momentum

    There is mention of it here.

    Well let us assume for the moment that the EM field does contain a momentum. If the first equation describes it correctly, then the second equation is purely logic, and the symbol \(\sigma\) is \(\nabla X v\). But the second equation is not important - not really. I just want the just of whether the field contains a vorticity or not.
     
  8. Green Destiny Banned Banned

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    See, I know of the Poyinting vector. My question is truely not about momentum as such. It is ''if there is a momentum in the field, then there may be, or must be(?) a vorticity, no?
     
  9. 1100f Banned Registered Senior Member

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    807
    Let me see if I understand your question.

    You say that by changing the velocity term (in a particle momentum) into the curl of the velocity (which is btw inconsistent from the dimensional pov), you ask if this will give a vorticity of the field.
     
    Last edited: Oct 17, 2010
  10. kurros Registered Senior Member

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    I used to share an office with guy who studied such vorticity (although he studied them quantum mechanically). As a simple example, take a simple two pinhole interferometer (i.e. get some cardboard, poke two (tiny) holes in it, and shine a nice laser field onto it). You will see series of fringes (similar to a two slit experiment) like those in fig 6 of his paper (http://www.physics.monash.edu.au/people/students/physreve-75-066613.pdf).
    If you then poke a third hole in the cardboard so you make a nice triangle then you get an interesting lattice of nodal points in the field (bottom of fig 6). These nodal points are vortices in the field; if you calculate the momentum flow of the field it actually swirls around these points. It is pretty cool. He went off and did all these nice simulations of vortex dynamics in bose-einstein condensates after that, they were super cool also, he had all these pretty movies he made. It makes my work seem so bland...
     
  11. AlphaNumeric Fully ionized Registered Senior Member

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    It gives no citation and given its lack of gauge invariance I'm dubious of it. I can see where someone might think it's derived from but that isn't really a viable expression.

    Well done, I ask you to explain what \(\sigma\) is and then you give its 'definition' in terms of two more things you don't explain or define. It seems like you're just parroting out equations, hoping that I'll see something I recognise and assume that you know its meaning too.
     
  12. Green Destiny Banned Banned

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    Well, whether you think it's dubious or not, you asked for a citation, and I provided one, so stop whining.

    And you asked me to define the equation, it's obviously the vorticity, if you recognize the equation. Again, stop whining over nothing.
     
  13. Green Destiny Banned Banned

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    That's very interesting.
     
  14. Green Destiny Banned Banned

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    Now, just to make clear, the last equation was an attempt to understand how vorticity can be viewed for a field. If the first equation holds, then by all means the second equation would hold
     
  15. Green Destiny Banned Banned

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    Precisely.
     
  16. 1100f Banned Registered Senior Member

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    Again, you want to change v in the expression of the particle momentum, and you ask if this will change A.

    The answer is no. when you change v, it does not change A, it changes v.

    Furthermore, in the expression curl(v) which you want to replace the velocity, v is not a field. it is the velocity of a point particle. How can you take the curl of a particle velocity?

    And finally, if you look at Maxwell's equations, you can see that for static situation, curl(E) = 0, curl(B) is proportional to the generating currents. So that in the static case, B has vorticity, while E does not.
    In the case of non static fields, the Maxwell equation show that both have some vorticity.
    There is no need to invent things that are not related like you did
     
    Last edited: Oct 18, 2010
  17. AlphaNumeric Fully ionized Registered Senior Member

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    A citationless Wiki mention is not entirely a good 'citation'. At worst you should cite Wikipedia's citation, ie the paper or book from which the result is quoted.

    What you wrote down was \(\nabla X v\), which someone familiar with general relativity might via as something like a format-less \(\nabla_{X}v\), which is the covariant derivative of vector v in the direction of vector X. What you should have written down is \(\nabla \times v\). The central character is not an X, it is a times sign, or rather more specifically its the cross product sign. This is basic notation which is covered in electromagnetism, quantum mechanics, relativity, fluid mechanics and basic vector calculus. You have just demonstrated you're unfamiliar with the cross product, despite claiming you're familiar with electromagnetism (the 3+1 dimensional Maxwell equations involve the cross product!!) and this entire thread is on vorticity, which is the curl of the flow vector v, where 'curl' is defined in terms of the cross product.

    That is why I asked about it, because you had botched such a basic bit of notation that it was evident you have no experience with it. Yet another piece of evidence you're throwing around buzzwords and terms you know nothing about. How many times are we going to go through this?
     
  18. Green Destiny Banned Banned

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    Oh I see. No, it's meant to be the curl of the velocity vector. So, it's simply a vorticity.

    I never meant it to come across as a times-sign. I know it's obviously not. Curls are not inherently difficult to understand.
     
  19. Green Destiny Banned Banned

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    And it should be clear I didn't mean it as a multiplication sign, considering the choice of topic in the OP.
     
  20. Green Destiny Banned Banned

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    Exactly. It doesn't change \(A\), but that was not in question. Nevertheless, apparently fields contain momentum, even in static fields. Just a pointer.
     
  21. Green Destiny Banned Banned

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    I've reconfigured the latex to suit your point.
     
  22. Green Destiny Banned Banned

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    Did he ever come to a conclusion what caused the vortices?

    Did he even come to a conclusion what effect this has one particles, or maybe perhaps better worded, what attributes particles have which can be explained by vortices in the electromagnetic field?

    For instance, this would act like a noisy background for say, an electron - it would be like trying to move your legs through swirling cheese.
     
  23. 1100f Banned Registered Senior Member

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    So what is your question?
     
    Last edited: Oct 18, 2010

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