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Why is del a vector?

  1. Mar 16, 2005 #1
    I have been thinking about this for a while, but why is the del operator a vector?? The book i have states no reason why and i was thinking if you guys could tell me why.

    Thanks....
     
  2. jcsd
  3. Mar 16, 2005 #2

    robphy

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    Del is not an ordinary vector. It's an example of a vector differential operator.
     
    Last edited: Mar 16, 2005
  4. Mar 16, 2005 #3

    HallsofIvy

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    Strictly speaking, it isn't. "grad f", at a given point, is a vector (by definition). Del is a notation for the operator that take a real valued function into that vector. It is, more correctly, a differential operator, as robphy said, that takes real valued functions to vector valued functions.
     
  5. Mar 16, 2005 #4
    hmmm, but from what I know, the definition of del is,

    [tex] \nabla = \frac {\partial}{\partial {x}} \hat {x} +\frac {\partial}{\partial {y}} \hat {y}+\frac {\partial}{\partial {z}} \hat {z}[/tex]

    Why the need for the unit vectors? And how does del differ from the total differential of a function?
     
  6. Mar 16, 2005 #5

    Tom Mattson

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    Well, if the unit vectors weren't there, [itex]\nabla[/itex] wouldn't be a vector differential operator. It would just be a plain vanilla differential operator (sans "vector").

    And it should be clear how [itex]\nabla[/itex] differs from the differential of a function: One is an operator, and one is a differential. On the other hand, you can express a differential of a function as [itex]df=\nabla f \cdot d \mathbf{r}[/itex].
     
    Last edited: Mar 17, 2005
  7. Mar 16, 2005 #6

    jcsd

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    yep, it operates on a scalar field to give a vector field, infact really it gives a covector field.
     
  8. Mar 16, 2005 #7

    dextercioby

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    The del is a one-form.It acts on an (n,m) tensor field -------->(n+1,m) tensor field.It is called the "covariant derivative operator"...Quite useful in GTR...

    Daniel.
     
  9. Mar 16, 2005 #8
    Hmmm, actually one thing which puzzles me is the gradient operator,

    if the gradient operator operates only on scalars and the del is a scalar but a vector differential operator. Why is the del being a vector differential operator allowed to work on a scalar function? And why is the gradient using the del of a scalar function a vector?

    Is my understanding wrong? I can't understand the terminology of tensors yet though, sorry, so i was wondering if this can be explained in a way without tensors.
     
  10. Mar 16, 2005 #9

    Hurkyl

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    Riddle me this: if I define [itex]\vec{f}(s) = (s, s^2, s^3)[/itex], why is [itex]\vec{f}[/itex] a vector? Doesn't the same answer apply to del?
     
  11. Mar 16, 2005 #10
    That vector f has a direction? Or because of the arrowhead? I'm really not too sure.....
     
  12. Mar 16, 2005 #11

    robphy

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    Note that [tex]\vec \nabla[/tex] satisfies the Leibniz rule:
    [tex]\vec\nabla(\psi\phi)=\psi\vec\nabla(\phi)+\phi\vec\nabla(\psi)[/tex]
    whereas a vector [tex]\vec v[/tex] does not
    [tex]\vec v(\psi\phi)=\vec v \psi\phi[/tex]
     
  13. Mar 16, 2005 #12

    cepheid

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    I'm pretty sure the example Hurkyl gave is a vector because he defined it that way: it is a vector-valued function that has three components. Each is a different function of s.
     
  14. Mar 17, 2005 #13
    misogynisticfeminist, vectors are not actually "quantities with magnitude and direction". It is more true that (some of) such quantities can be expressed as vectors.

    An n dimensional vector is an n-tuple (i.e. a set of n ordered elements) which obeys certain transformations.

    Formally, Del is a vector, and (to your satisfaction) you will in your studies see a distinction made between these vectors and the ones you are familiar with now.
     
  15. Mar 17, 2005 #14
    And note that vectors don't even necessarily have to be representable as n-tuples. Calling something a vector just means that it's an element of some vector space.
     
  16. Mar 17, 2005 #15
    hmmm that kinda helped. And thanks for the help also everyone.
     
  17. Mar 17, 2005 #16

    matt grime

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    But what are the coefficients by the unit vectors? are they real numbers?
     
  18. Mar 17, 2005 #17

    HallsofIvy

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    Yes, it appears your understanding is very wrong! No one said "del is a scalar". "del" is just a symbol-that is used for a variety of similar things. "gradient" is a differential operator whose domain is "scalar valued functions" and range is "vector valued functions". Does that make sense?

    We don't need the unit vectors- that just makes it easier to write in terms of that specific coordinate system. The important thing about vectors- and why vector "operators" are useful- is that they don't depend on any one coordinate system (or "basis").

    Again, I think you should be saying "gradient" rather than "del"- "del" is just a symbol- it is the gradient that is the operator. If f(x,y,z) is a scalar valued function of three variables, then its gradient is [tex]\nabla f= \frac{\partial f}{\partial x}\vec i + \frac{\partial f}{\partial y}\vec j+ \frac{\partial f}{\partial z}\vec j[/tex]

    The "total differential" is the dot product of that with the "differential position vector"
    [tex]\frac{\partial f}{\partial x}\vec i + \frac{\partial f}{\partial y}\vec j+ \frac{\partial f}{\partial z}\vec j \bullet dx\vec i+ dy\vec j+ dz\vec k[/tex].
     
  19. Mar 17, 2005 #18
    This statment is contradictory. All vector spaces have basis, and any element of a n-dimensional vector space can be written as an n-tuple.
     
  20. Mar 17, 2005 #19

    jcsd

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    Though the dimension of a vector space though needn't be finite or even countable, but as you say under the axiom of choice all vector spaces have a basis.
     
  21. Mar 17, 2005 #20

    mathwonk

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    Well I know I am old and stupid, but I can't understand any of these answers to the opriginal question, at elast not taken together. i recommend the questioner to just read a good book like spivaks' calculus on amanifolds.

    no offense meant here to the man well intended attempts to answer, some correct, but it is very confusing. the language is hard to get straight, for one thing,k i.e. the main thing. people are calling del something different from d, when apparebntly thet are intended to be essentially the same thing.

    people seem to be confusinf at least in their language if not in their minds, d with df, with dp with dfp. Even if the posters are not confused i would not blame the questioner if he/she were so by this discussion. I am, and I have taught this stuff for decades, and feel entirely comfortable with using it.

    some of the confusion my stem from learning these ideas from inadequate sources such as bachman's book, a few of whose many imprecisions and errors I have pointed out elsewhere.

    but so what, learning is a process, and apparently some of this helped misogynistf. i am just suggesting she/he not feel alone if she is still puzzled.
     
    Last edited: Mar 17, 2005
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