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Is multiplication associative in physics?

  1. Aug 4, 2015 #1
    Work = Force X Distance.

    ? = Distance X Force

    How do you make sense of the second equation?
  2. jcsd
  3. Aug 4, 2015 #2
    For a little displacement ##\vec{dl},## the work the force ##\vec{F}## does is ##\vec{F}\cdot\vec{dl},## which is identical to ##\vec{dl}\cdot\vec{F},## for the commutativity of the vector's scalar product.
    But it doesn't happen everywhere in physics. For example, the non-commutativity of matrices finally implies the uncertainty principle.
    Edited for getting uncleared at the same time.
  4. Aug 4, 2015 #3


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    1) You're asking about the commutativity property, not associativity, which is ## (a \times b)\times c=a \times (b\times c)##.
    2) The multiplication defined for real numbers is commutative, doesn't matter in what field of science you're considering it. But the more general definition of work is through ## W= \vec F \cdot \vec D ##. So we should talk about the inner product defined on vectors. That is commutative too and again it doesn't matter in what field of science you're considering it.
  5. Aug 4, 2015 #4


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    Do you mean commutative? (Associative -> a+(b+c) = (a+b)+c, commutative a*b=b*a). https://en.wikipedia.org/wiki/Commutative_property

    And physics is associative/commutative when the mathematics you are using is. The rules don't change when you're doing physics. If you multiply scalars, it is associative and commutative. If you are multiplying matrices, it is not commutative in general.
  6. Aug 4, 2015 #5
    Yes, I mean commutative.
  7. Aug 5, 2015 #6


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    Then the answer is yes and no.

    A cross product is not commutative. A dot product is.

    A vector product is not commutative. A scalar product is.

    And the multiplication of operators need not be commutative.

    This should be in Math, not Physics.

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