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Is the gradient really just a first-order approximation?

  1. May 14, 2009 #1
    In physics texts, its customary to write (and even to define the gradient as) the following:

    [tex]dT = (\nabla T) \cdot dl[/tex]

    Working in Cartesian coordinates, we can expand this as follows:

    [tex]dT = \frac{\partial T}{\partial x} dx + \frac{\partial T}{\partial y} dy + \frac{\partial T}{\partial z}dz[/tex]

    But the above equation is really just a "first-order" approximation for the differential dT, is it not? Working just in the differential dx, a more "complete" way to write this expression would be:

    [tex]dT \approx \frac{\partial T}{\partial x} dx + \frac{1}{2} \frac{\partial ^2T}{\partial x^2} (dx)^2 + \frac{1}{6} \frac{\partial ^3T}{\partial x^3} (dx)^3 + ... [/tex]

    along with the additional, commensurate expansions in terms of the y and z variables as well.

    So while we can define equations in terms of gradient, and work with gradients, etc, is it fair to say that all these gradient expressions are really just giving us answers which are first-order approximations? And that if we desired more precise answers, we would have to delve deeper into the higher-order terms?
     
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  3. May 14, 2009 #2

    Hurkyl

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    Nope; [itex]
    dT = (\nabla T) \cdot d\ell
    [/itex] is an exactly correct equation of differential forms.

    You're thinking about differential approximation, which says [itex]
    \Delta T \approx (\nabla T) \cdot \Delta \ell
    [/itex], or more explicitly,

    [tex]T(\ell + v) - T(\ell) = \nabla_v T(\ell) + \frac{1}{2} \nabla_v \nabla_v T(\ell) + \cdots[/tex]
     
  4. May 14, 2009 #3

    Hurkyl

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    You have to remember that differential forms are not numbers (or number-valued fields); they're differential forms. Their very definition is arranged so that these "first-order" equations are exactly correct. If you want to pass back into the world of "ordinary" mathematical objects, you have to use some sort of integral.

    That said, people do often use expressions like dT when they really mean [itex]\Delta T[/itex].
     
  5. May 14, 2009 #4

    dx

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    They are first order approximations of ∆T, not of dT.

    dT is exactly equal to [itex] \partial_{x}T dx + \partial_{y}T dy + \partial_{z}T dz [/itex]
     
  6. May 14, 2009 #5
    Where can I read up on the difference between dT and [tex]\Delta T[/tex]?

    I don't think I've ever (formally) encountered that distinction before..
     
  7. May 14, 2009 #6
    So in my first equation above, what I should have written, should have been:

    [tex]\Delta T \approx \frac{\partial T}{\partial x} \Delta x
    +\frac{1}{2}\frac{\partial^2 T}{\partial x^2}\Delta x^2
    + \frac{1}{6} \frac{\partial^3 T}{\partial x^3} \Delta x^3 + ...
    [/tex]

    Yes?
     
  8. May 14, 2009 #7

    dx

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    Yes.

    ∆T is an actual difference of T between two points. dT on the other hand is what is called a 1-form. You can read about 1-forms in books about differentiable manifolds and differential geometry.
     
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