Differentiating a Vector Field

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  1. HappyHammer15's Avatar
    1 18-04-2012  12:18
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    Differentiating a Vector Field
    Hi,

    Just need a bit of help with differentiating a vector field.

    F= (ax^2 +bxy +cy^2 -2x)\bold{i} +(x^2 +xy-y^2 +bz)\bold{j}+(2y+2z)\bold{k}

    I need to find \nabla F

    I am unsure what to differentiate with respect to as i need to have a vector as the outcome. Do I differentiate w.r.t x for i, w.r.t y for j, and w.r.t to z for k?

    Any help is appreciated.
  2. nuodai's Avatar
    2 18-04-2012  13:00
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    Re: Differentiating a Vector Field
    If F = \begin{pmatrix}F_1 \\ F_2 \\ F_3 \end{pmatrix} is a vector then \nabla F = \begin{pmatrix} \nabla F_1 \\ \nabla F_2 \\ \nabla F_3 \end{pmatrix} is a matrix, not a vector, where (for instance) \nabla F_1 = (\partial_x F_1, \partial_y F_1, \partial_z F_1).

    If you were differentiating the \mathbf{i}-component w.r.t. x and the \mathbf{j}-component w.r.t. y and so on then what you'd have would be \nabla \cdot F and not \nabla F.
    Last edited by nuodai; 18-04-2012 at 13:01.
  3. HappyHammer15's Avatar
    3 18-04-2012  13:11
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    Re: Differentiating a Vector Field
    (Original post by nuodai)
    If F = \begin{pmatrix}F_1 \\ F_2 \\ F_3 \end{pmatrix} is a vector then \nabla F = \begin{pmatrix} \nabla F_1 \\ \nabla F_2 \\ \nabla F_3 \end{pmatrix} is a matrix, not a vector, where (for instance) \nabla F_1 = (\partial_x F_1, \partial_y F_1, \partial_z F_1).

    If you were differentiating the \mathbf{i}-component w.r.t. x and the \mathbf{j}-component w.r.t. y and so on then what you'd have would be \nabla \cdot F and not \nabla F.
    Okay, thank you. Thats a big help because the question is to find the values a,b,c such that F is soleneidal. i.e. \nabla \cdot F=0
  4. nuodai's Avatar
    4 18-04-2012  13:13
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    Re: Differentiating a Vector Field
    (Original post by HappyHammer15)
    Okay, thank you. Thats a big help because the question is to find the values a,b,c such that F is soleneidal. i.e. \nabla \cdot F=0
    Right. Bear in mind that \nabla \cdot F is the sum of the respective derivatives of the components, so it's not a vector. (I didn't make this clear in my last post.)
  5. HappyHammer15's Avatar
    5 18-04-2012  13:28
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    Re: Differentiating a Vector Field
    (Original post by nuodai)
    Right. Bear in mind that \nabla \cdot F is the sum of the respective derivatives of the components, so it's not a vector. (I didn't make this clear in my last post.)
    Okay thank you, however, in this example i do not get a variable c in the \nabla \cdot F. Thank you for your help thus far.
  6. around's Avatar
    6 18-04-2012  14:01
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    Re: Differentiating a Vector Field
    (Original post by HappyHammer15)
    Okay thank you, however, in this example i do not get a variable c in the \nabla \cdot F. Thank you for your help thus far.
    So?
  7. HappyHammer15's Avatar
    7 18-04-2012  14:05
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    Re: Differentiating a Vector Field
    (Original post by around)
    So?
    I get a=-0.5 and b = 2. Is c an arbitrary constant then?
  8. Jake22's Avatar
    8 18-04-2012  15:07
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    Re: Differentiating a Vector Field
    (Original post by nuodai)
    Right. Bear in mind that \nabla \cdot F is the sum of the respective derivatives of the components, so it's not a vector. (I didn't make this clear in my last post.)
    To add to this: remember that the notation is suggestive and \nabla \cdot F is formally like a dot product. i.e. think of \nabla = (\frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}) and and F=(F_{\textbf{i}},F_{\textbf{j}} ,F_{\textbf{k}}) where in your example F_{\textbf{k}} = 2y+2z and so on.

    Then, you form the 'dot product'

    \displaystyle \nabla \cdot F = \left( \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \cdot (F_{\textbf{i}},F_{\textbf{j}},F _{\textbf{k}}) = \left( \frac{\partial}{\partial x}F_{\textbf{i}}+\frac{\partial} {\partial y}F_{\textbf{j}}+\frac{\partial} {\partial z}F_{\textbf{k}} \right)
    where you interpret the 'product' of, say \frac{\partial}{\partial x} with F_{\textbf{i}} in the obvious way i.e. you take the partial derivative of F_{\textbf{i}} w.r.t. x
    Last edited by Jake22; 18-04-2012 at 15:22.
  9. Jake22's Avatar
    9 18-04-2012  15:09
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    Re: Differentiating a Vector Field
    (Original post by HappyHammer15)
    I get a=-0.5 and b = 2. Is c an arbitrary constant then?
    Exactly, if it doesn't contribute anything then you can take any value.

    e.g. Find all pairs (a,b) such that a = 0.
  10. HappyHammer15's Avatar
    10 18-04-2012  15:18
    • Junior Member
    • Posts: 45
    Re: Differentiating a Vector Field
    (Original post by Jake22)
    To add to this: remember that the notation is suggestive and \nabla \cdot F is formally like a dot product. i.e. think of \nabla = (\frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}) and and F=(F_{\textbf{i}},F_{\textbf{j}} ,F_{\textbf{k}}) where in your example F_{\textbf{k}} = 2y+2z and so on.

    Then, you form the 'dot product'

    \displaystyle \nabla \cdot F = \left( \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \cdot (F_{\textbf{i}},F_{\textbf{j}},F _{\textbf{k}}) = \left( \frac{\partial}{\partial x}F_{\textbf{i}},\frac{\partial} {\partial y}F_{\textbf{j}},\frac{\partial} {\partial z}F_{\textbf{k}} \right)
    where you interpret the 'product' of, say \frac{\partial}{\partial x} with F_{\textbf{i}} in the obvious way i.e. you take the partial derivative of F_{\textbf{i}} w.r.t. x

    If i do it in this way i get a very complicated solution that is quite difficult to reduce. Will the way in which i first did it suffice?, i.e F_{\textbf{i}} differentiating w.r.t x, F_{\textbf{j}} differentiating w.r.t y and so on.
  11. Jake22's Avatar
    11 18-04-2012  15:21
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    Re: Differentiating a Vector Field
    (Original post by HappyHammer15)
    If i do it in this way i get a very complicated solution that is quite difficult to reduce. Will the way in which i first did it suffice?, i.e F_{\textbf{i}} differentiating w.r.t x, F_{\textbf{j}} differentiating w.r.t y and so on.
    I never wrote that you should calculate it in any other way... I said to do exactly that i.e. differentiate the i component wrt to x, the j component wrt y etc. and then sum them up. All I wrote was just a convinient way of remembering what the operation \nabla \cdot F gives you. You shouldn't get anything different at all.

    The point is to think of \nabla as a vector and \nabla \cdot F as a kind of dot product of vectors whose formula in components you are (or at least should be) very familiar with. The reason that it isn't a 'true' dot product is that \nabla is a vector in a different vector space to F i.e. the dual space.

    Note: there was a mistake in my initial post which I have now corrected - I put commas instead of + signs on the right hand side but this shouldn't have affected how you would calculate.
    Last edited by Jake22; 18-04-2012 at 15:28.
  12. HappyHammer15's Avatar
    12 18-04-2012  15:30
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    • Posts: 45
    Re: Differentiating a Vector Field
    (Original post by Jake22)
    I never wrote that you should calculate it in any other way... I said to do exactly that i.e. differentiate the i component wrt to x, the j component wrt y etc. and then sum them up. All I wrote was just a convinient way of remembering what the operation \nabla \cdot F gives you. You shouldn't get anything different at all.

    Note: there was a mistake in my initial post which I have now corrected - I put commas instead of + signs on the right hand side.
    Okay, i see now. I was reading \nabla =\nabla F and i was trying to compute \nabla F \cdot F
    This seems like such a silly mistake to make now. Thanks for all of your help though.
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