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Magnetic bivector.

  1. Jun 30, 2013 #1
    What would happen if you take the wedge of the magnetic field and the velocity of a charged particle, you get the force or the current correct? So the equation looks like:

    ## \mathbf{J} = \mathbf{v} \wedge\mathbf{B}## This equation is true right?

    I mean since ##\mathbf{J} \perp \mathbf{F}## and when you take the velocity vector and the magnetic field and wedge them you get something which is parallel to the current(but maybe my math is wrong and I am missing a minus sign.)

    well if the velocity vector is is at some angle to the magnetic vector
     
  2. jcsd
  3. Jun 30, 2013 #2

    WannabeNewton

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    The wedge product of two vectors is not another vector; it is a bivector. You are probably thinking of the cross product not the wedge product. ##v\times B## is proportional to the magnetic force (the Lorentz force in the absence of an electric field).

    ##B## has SI units of ##N*s*C^{-1}*m^{-1}## and ##v## has units of ##m*s^{-1}## so the product would have the units of ##N*C^{-1}## i.e. the units of force per unit charge. The current density on the other hand has units of amperes per square meter. The relationship between current density and the magnetic field (in magnetostatics!) is given by Ampere's law ##\nabla \times B = \mu_0j##.
     
  4. Jun 30, 2013 #3
    No, I am thinking of the bivector D: I know that the wedge product give you a bivector I was wondering if there existed a "current" bivector, because it has an orientation in the direction of a current but the others aren't D: so I am not exactly sure how to picture this... or at least geometrically :/
     
  5. Jun 30, 2013 #4

    WannabeNewton

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    Ah ok. Given any axial vector ##a## (also called a pseudovector) we can use the hodge dual operator ##\star## to form a bivector ##A = \star a##. Geometrically axial vectors and bivectors are oriented line and plane segments respectively. The magnetic field is an axial vector but the current density is a pure vector, not an axial vector. As such I have never seen such current bivectors used in vector calculus EM myself. There is a way to write down Maxwell's equations using differential forms and over there the current 3-form, which is gotten by applying the hodge dual operator to a current 1-form, is very important but this is more advanced than what you are speaking of.
     
  6. Jun 30, 2013 #5
    I have seen those also, So does this mean you can not find the wedge between the magnetic field and the velocity of a charge?
     
  7. Jun 30, 2013 #6

    WannabeNewton

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    No that is fine. The hodge dual operator will tell us what that is in fact because ##\star(v \times B) = v\wedge B## as one can easily show. The wedge product is defined for all vectors in the given space and will give us a bivector whereas the hodge dual applies to the axial vectors (oriented line segments), in this case the cross product of ##v## with ##B##, and gives us a bivector (oriented plane segment). The motivation for the hodge dual comes from the concept of an orthogonal complement.
     
  8. Jun 30, 2013 #7
    I didn't know Hodge dual cam from the orthogonal complement, that is rather interesting. Although I thought ##\star(v\wedge B) = v \times B## but this was my original motivation for doing this, an attempt to see if the wedge of these products can produce a significant item.
     
  9. Jun 30, 2013 #8

    WannabeNewton

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    I've never seen it used in vector calculus EM myself although it does show up in more advanced treatments that go beyond vector calculus. There is also the differential forms treatment of EM in which the hodge dual is important when writing down Maxwell's equations in terms of differential forms on space-time, as mentioned earlier, but this goes beyond what we're talking about here.
     
    Last edited: Jun 30, 2013
  10. Jun 30, 2013 #9
    Yeah, indeed I am just trying to look at the magnetic field from a different perspective. Although, I have little training in multi-linear algebras I thought it would have been a good idea for me to start working on something I have no idea where to start. OTL Fail.
     
  11. Jun 30, 2013 #10

    WannabeNewton

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    Well the subject of geometric algebras (which is what we are dealing with here) is quite extensive. It would be worthwhile studying it once you have a more extensive mathematical background as far as algebra goes. With regards to electromagnetism, you can certainly express Maxwell's equations using spacetime algebras (specific kind of geometric algebra) but for all practical purposes you would probably not see this in an introductory vector calculus treatment of electromagnetism. See here for more: http://www.av8n.com/physics/maxwell-ga.htm Using the language of geometric algebras we can cast the electromagnetic field as a bivector and write Maxwell's equations in a ridiculously elegant form.
     
    Last edited: Jun 30, 2013
  12. Jul 1, 2013 #11
    Agreed. Geometric algebra is a rich topic and it would be better understood if I could find text which can help me understand this abstract topic, do you think you could recommend to me a book or website which talks about multi-linear algebra? Also this link was very helpful
     
  13. Jul 1, 2013 #12

    WannabeNewton

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    I think you would get many more responses if you asked in the academic guidance forum so that more people can see your request and you could elucidate on what you already know because I don't know what your background is exactly in linear algebra. People who frequent AG but not classical physics will be able to give you more recommendations (since your request is for a math book). Cheers.
     
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