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Del dot B = 0

  1. Jul 28, 2009 #1
    Another way of stating it is

    the sum of the dot products of vectors B and vectors A = 0

    Is this because the dot product of two vectors is 0 if they are 90 degrees to each other. So this is just adding up a certain amount of these two vector systems, and since magnetism is always normal to the A, you're just summing up a bunch of zeros, equaling 0?
  2. jcsd
  3. Jul 28, 2009 #2
    It means

    [tex]\frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}= 0 [/tex]
  4. Jul 28, 2009 #3
    What does your equation have to do with the angles between vectors? That is the meaning of the dot product? But you've just given me another puzzle and not answered my original "word" question.
    I would like an explanation for "Del Dot E =0" in extremely precise terminology using multiple two vector systems and the angles between the each system of two vectors, such that the sum of all the two vector systems =0.
    I think it means that for any infinitely small moving charge, vector V, interacting with an infinitely small vector B, the angle between V and B is 90 degrees and so the abs value of V time B times the cos of the angle between them is 0. So if you add up a crap load of these, you still get zero. Is this right?
  5. Jul 28, 2009 #4


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    Del is not a vector in the usual sense. It is an operator. So it doesn't make sense to talk about the angles between vectors. It is just a useful mnemonic to think of it as a vector dot product.
  6. Jul 28, 2009 #5
    You can understand[tex]\nabla[/tex] as [tex]\frac{\partial}{\partial x}i[/tex]+[tex]\frac{\partial}{\partial y}j[/tex]+[tex]\frac{\partial}{\partial z}k[/tex], it explains why
    [tex]\nabla[/tex][tex]\bullet[/tex][tex]B[/tex]=[tex] \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}[/tex]。
    But it seems your problem is that you insist finding a intuitive interpretation for "nabla" and take it as a nomal vector, but the thing is it isn't, it's an operator.
  7. Jul 28, 2009 #6
    So are you saying that Del dot B = 0 has absolutely no relation to some vectors and their relative angles? as in the dot product?
  8. Jul 28, 2009 #7
    Nabla is the sum of partial derivatives, I understand that. But in other forms, it relates to the dot product of two vectors. That's what I'm trying to bridge.
  9. Jul 28, 2009 #8
    Em...I never use the word "absolutely",but in this case pretty much I mean that.
  10. Jul 29, 2009 #9


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    Since Del dot B = 0 is a relation regarding one vector (B) there are not "some vectors" with "relative angles" that can be referred to.
  11. Jul 29, 2009 #10
    Dude, this is like the third thread like this. If you want to understand physics then sit down and understand physics. A HUGE part of that is learning the relevant math. You're not going to get 5 pages into a physics text if you don't have the math and starting a new thread every page to try and make us teach every single concept to you is time consuming and pointless for both of us. If you've bitten off more than you can chew with the math then LEARN THE MATH. There are literally HUNDREDS of textbooks and web sources that teach basic vector calc very well. As for EM in general, just grab a copy of griffiths or something and start reading. Read the first two chapters, do all the problem questions and then come back here if you have questions and we'd be happy to answer them for you to the best of our ability.

    However, to answer this particular question you really can't relate it to angles. Del is not a vector, it's an operator and ultimately Del dot B is kind of a wishy washy notation. Del is not a vector and thus it really can't be part of the vector inner product space. This makes the 3D geometrical interpretation of [itex]A \det B = |A| |B| cos\theta[/itex] not apply.
  12. Jul 29, 2009 #11
    I have an understanding of the underlying math. That is not the problem. If you were to ask me to write down the first two equations of Maxwell, I can do it. I can even tell you kind of what it means. So what? If don't really know what it means, I'm must blindly manipulating math.
  13. Jul 29, 2009 #12
    But it is a bad understanding of the divergence, even Nabla is not just "the sum of partial derivatives"... there is no "bridge" with angles between vectors here since the divergence is an operation in a vector field giving an scalar, Nabla or Del is a differential vectorial operator, not a vector. The divergence is associated with the presence of sinks or sources of a field, in this case for magnetic field [tex]\nabla \cdot \vec{B}=0[/tex] indicates the nonexistence of magnetic monopoles and in general that the lines of magnetic field closes and [tex]\nabla \cdot \vec{E}=\frac{\rho }{\varepsilon _{0}}{\qquad }[/tex] indicates that the electric field diverges.
  14. Jul 29, 2009 #13
    I think the "what it means" you said indeed means "what it resembles", and I don't think it's a good way of learning physics.
  15. Jul 29, 2009 #14
    Being able to write down the equations means you can copy a bunch of symbols. Try finding the B-field at an angle theta off the central axis of a conducting ring of radius R (I'm serious). Solve problems! That's how you will understand.
  16. Jul 29, 2009 #15

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    No, no you don't. If you did, you wouldn't be posting what you are posting. (An example of the http://www.apa.org/journals/features/psp7761121.pdf" [Broken]: people can be unskilled and unaware of it)

    Let me suggest again a book by Schey: Div, Grad, Curl and All That. It will help. Really.

    But, I wore the juice!
    Last edited by a moderator: May 4, 2017
  17. Jul 29, 2009 #16
    OK I'll look for the book.
  18. Jul 29, 2009 #17
    Del dot B = 0

    A dot B = 0

    LaTeX Code: \\frac{\\partial B_x}{\\partial x} + \\frac{\\partial B_y}{\\partial y} + \\frac{\\partial B_z}{\\partial z}= 0
  19. Jul 29, 2009 #18
    what do the above three equations have in common?
  20. Jul 29, 2009 #19
    Maverick StarStrider:
    Try finding the B-field at an angle theta off the central axis of a conducting ring of radius R (I'm serious). Solve problems! That's how you will understand.

    If I could demonstrate this repeatedly I could still not understand what a magnetic field is. Should I be in a philosophy forum?
  21. Jul 29, 2009 #20

    Perhaps you should. The purvue of science (and, of course, specifically physics) is not to answer the WHY. Physics can never tell you WHY the universe follows these guidelines. You can show things like "this dynamic is ultimately the result of minimizing action" and condense/amalgamate understanding. However, we can never aswer WHY the universe should behave in such a way that it seeks to minimize action. However, although it is commonly considered the purvue or philosophy and religion to tell the WHY, ultimately, they could no more provide proof to their assertion then anyone/anything else. Physics determines the WHAT, the HOW. Classical EM was completed over a 100 years ago and it is truly amazing. It can correctly predict virtually everything of the macroscopic universe and that which it can't has all been shown to be a result of quantum peaking its head at a macroscopic level. One can answer questions like "what is an electric field" by exactly modeling their dynamics and providing a framework to predict, correctly, the outcome of any observation of the physical/natural world possible. However, such a thing could never tell you why things are the way they are, religion and philosophy can't either. Why do things have inertia? Because they have mass. Why do they have mass? Potentially because of the Higgs Boson. Why does the higgs boson exist? Well then the common tact is to split in two directions. Either you say "that's just the way it is" or you say "it must have been created that way". But of course then if you take the latter someone asks "well who created the creator or why is the creator" and then you still get to the answer "that's just the way it is".
  22. Jul 29, 2009 #21
    It means, by definition

    [tex]\nabla \cdot B \equiv \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}\ .[/tex]

    The same is true of the curl.

    Neither are operations between vectors--but they can be redefined as such, unless I'm misremembered something, where the partial derivatives are directional derivatives, and members of a vector space in their own right.
  23. Jul 29, 2009 #22
    Not quite. Firstly, the definition of Nabla is independent of coordinate system (as is all of vector calculus). Secondly, something like [itex]\nabla \cdot B[/itex] is really nothing more that an abuse of notation just like [itex]\frac{dy}{dx}dx=dy[/itex]. In my example, in reality, what you are doing is [itex]\int \frac{dy}{dx}dx[/itex]. Similarily, with something like [itex]\nabla \cdot B[/itex] it has no mathematical significance beyond a short hand notation for what you wrote (albeit, independent of coordinate system). The dot product is an example of a inner product, which forms an inner product space (add in a defined norm that is a metric with which all Cauchy sequences converge and you've got a Hilbert space). Part of the requirements for an inner product (space) are, what I call, adjoint commutivity (those there are lots of names). i.e. <a,b> = <b,a>*. However, nabla is an OPERATOR, and thus [itex]B \cdot \nabla[/itex] is entirely undefined. It is just an abuse/condensing of notation, it is not a true dot product. By commuting it you obtain an operator where before you had a scalar function. This is nonsense as far as an inner product is concerned.
  24. Jul 29, 2009 #23
    The dot product was asked as an elementry question. I was keeping things simple. We could use different coordinates, more dimensions, curved manifolds, and so forth.

    But if you like, more generally

    [tex]\nabla \cdot V = \partial_{\mu} V^{\mu} ,[/tex]

    but not for curved manifolds.

    I reviewed what I'd learned and forgotten about tensor calculus. I find that I don't have details I'd really needed to examine this. You many not either. If so we can both be equally confused. Anyway, when you brought up

    [itex]\frac{dy}{dx}dx=dy ,[/itex]

    I thought at first that you were objecting to the definition of the dual vector (well, one way to define it),

    [itex]\omega = \omega_{\mu} \hat{\theta}^{(\mu)} [/itex]

    [itex]\omega = dx^{\mu} \frac{ \partial }{ \partial x^{\mu} } [/itex]

    This looks alot like it occupies the same vector space as [itex]\nabla \cdot[/itex] . I can't really tell.
    Last edited: Jul 29, 2009
  25. Jul 29, 2009 #24
    I'm not entirely sure if I'm on the same page as you but yes, now that I think about it more, one can have an inner product space which takes operators (like your dual vector), like bra-ket notation. However, the dot product is not one of them.
  26. Jul 30, 2009 #25
    In Cartesian coordinates

    [tex]B_i = g_{ij}B^j \ .[/tex]

    The metric is flat,

    [tex]g_{ii}=1[/tex] and [tex] g_{ij}=0 , where \ i \neq j \ .[/tex]

    The elements of vectors and their dual vectors are equal,

    [tex]B_i = B^i[/tex]

    So the dot product

    [tex]B \cdot B = B^i B^i = B^i B_i = \left< B,B \right>[/tex]

    In other coordinate systems

    [tex]B_i \neq B^i \ ,[/tex]

    so additional factors are picked up that change the algebraic equation for the dot product.
    Last edited: Jul 30, 2009
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