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How to differentiate by a vector ?

  1. Mar 8, 2014 #1

    dyn

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    Is it possible to differentiate a scalar( in this case a Lagrangian) by a vector ?
    If the Lagrangian is r.A ie. the scalar product of vectors r and A what is ∂L/∂r ? My notes say it is A but how ?
    My notes also say that ∂L/∂r is the same as ∇L. Is this correct ? or is it sloppy notation ? or is it because the vector is treated as a generalised coordinate which acts as a scalar ?
    Thanks
     
  2. jcsd
  3. Mar 8, 2014 #2

    Fredrik

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    Sounds like they're just defining ##\frac{\partial L}{\partial\vec r}## to mean ##\left(\frac{\partial L}{\partial r_1} ,\frac{\partial L}{\partial r_2},\frac{\partial L}{\partial r_3}\right)##.
     
  4. Mar 8, 2014 #3

    Vanadium 50

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    Fredrik's right - that's what they mean.
     
  5. Mar 8, 2014 #4

    dyn

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    Thanks. Is it technically possible in precise terms to differentiate by a vector ? As far as I understand grad div and curl isn't differentiating by a vector.
     
  6. Mar 9, 2014 #5

    UltrafastPED

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    For a function f that exists over a vector domain R the vector derivative is defined as:

    df(R)/dA = lim [f(R + eps*A) - f(R)]/eps

    Note that the vector A must be a unit vector here; the vector derivative is a directional derivative in the direction A.

    You can easily prove that this can be calculated via:
    A (dot) grad (f).
     
  7. Mar 9, 2014 #6

    Fredrik

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    Do people use the notation df/dA for that? (I don't have a problem with it. I'm just not used to seeing it). I would use something like ##D_Af##.

    I wouldn't call this "differentiating by a vector". It's just the directional derivative of f in the direction of the unit vector A. It's the ordinary derivative of the function ##t\mapsto f(x+tA)##, which is a function from ##\mathbb R## into ##\mathbb R##.
     
  8. Mar 9, 2014 #7

    dyn

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    When all this occurs with the Euler-Lagrange equations is the vector just treated as a scalar ?
     
  9. Mar 9, 2014 #8

    Fredrik

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    I'm not sure I even understand what it would mean to treat a vector as a scalar.
     
  10. Mar 9, 2014 #9

    UltrafastPED

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    Yes, this notation is used, especially in older texts.

    And yes, the directional derivative has been called "differentiating by a vector" ... the terminology is consistent with the Leibniz notation.

    I like your notation better; it is more modern-looking. I'll use it next time and be more hip!
     
  11. Mar 10, 2014 #10
    I don't think it's sloppy notation. It's just different notation. Although, it's probably better to include tensor indices when written that way.
     
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