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Total derivative

  1. Jul 28, 2008 #1


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    I'm trying to study some basic tensor analysis on my own for practical purposes, but I'm having some problems. More specifically I'm rather puzzled over the concept of total derivative in curvilinear coordinates (well, to be exact, as I've got little experience with differential geometry, it's all a bit hazy).

    I have a function f, which depends on the position vector r, which in turn is a function of time t i.e. f = f(r(t)). Now what I'd want to do is to take the time derivative of f: df/dt. Normally I'd write it as [tex]\frac{df}{dt} = \sum_i \frac{\partial f}{\partial \mathbf{r}_i}\frac{d\mathbf{r}_i(t)}{dt}[/tex]. I however have doubts in my mind whether this is the right way to go about when I'm dealing with curvilinear coordinates. Should I replace the partial derivative with the gradient (covariant derivative)?

    I hope my question is making any sense. Please direct me to a good book on tensors if it is not (I'd prefer mathematical ones, but I'm in a hurry to apply the mathematics, so one for engineers/physicists would be well appreciated).
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  3. Jul 28, 2008 #2
    Hello :)

    first I guess this is Analysis and not Tensor analysis. ;)

    Do you mean the toal derivative or the total differential? (I aks because my bad English could mistake the terms)

    Well, if you have a total derivative you have the normal latin [tex]d[/tex] instead of the partial [tex]\partial[/tex] or if you only have a infenitesimal displacement you use the greek [tex]\delta[/tex] like in the principles of virtual displacement.

    Now the total derivative is the derivative in every dimension! To signalize what you are doing you schould use the latin "d", but it don't would be mathematical wrong if you use the partial d but you have to operate the parial derviative in every dimension. Don't use the greek d for derivative, because this would be wrong. Our mathematical shortcuts here stands for a differentialquotient which are building a limes. The greek delta don't implicit this limes.

    If you have another coordinate system the theorie don't change! But depending on what your coordinates are is the derivative not a priori trivial but it don't changes the upper sense. In general you have to use quotient, product and chainrule for the derivative.

    I hint from me. If I have a cordinate system I try to build a transfer matrix in a more acceptable system. If you find the base vectors of your coordinate systems is the rest quite easy.

    I hope that helps you :)

  4. Jul 28, 2008 #3


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    I'm quite sure I'm dealing with differential geometry here. Rather than a function in general, I should've stated that f is a scalar field.

    Indeed I should've written the equation like (with the upper indices, I believe):
    [tex]\frac{\operatorname{d}f}{\operatorname{d}t} = \sum_i \frac{\partial f}{\partial r^i}\frac{\operatorname{d}r^i}{\operatorname{d}t} [/tex]

    My question was if I should write the equation as above or more like the following:
    [tex]\frac{\operatorname{d}f}{\operatorname{d}t} = \sum_i \nabla_i f \frac{\operatorname{d}r^i}{\operatorname{d}t} [/tex]

    (assuming I got the notation right this time).
  5. Jul 28, 2008 #4


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    It might make more sense to write that last equation as a "dot" product:
    [tex]dr= \nabla f \cdot \vec{r}[/tex]
  6. Jul 28, 2008 #5


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    Sure, but notation isn't the problem I need help with. Rather I wish to know whether to use partial differentation in the "usual" sense or the covariant derivative. I'm sorry if I wasn't making this clear enough.
  7. Jul 28, 2008 #6
    No, you were pretty clear. and the answer is: "partial differentiation as you have described it, is the way to go. Don't worry about the covariant derivative."

    The covariant derivative is needed when one is trying to differentiate a vector field with respect to a particular direction. In this case, though, you simply have a function defined along a curve in n-space and are interested in the value of this function as time passes (i.e., as one moves along the curve with respect to the variable t according to the parametrization t---> r(t)).
  8. Jul 29, 2008 #7
    oh, okay, in this case I missunderstood you. Sorry.

    Well, as you write it I wouldn't mean that it is wrong. I always use the upper form, but beacuase of the indice i it should be the same.

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