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Calculus Please recommend a textbook containing a rigorous proof of ...

  1. Jul 24, 2017 #1
    2mnji1f.png

    This is excerpted from George E. Hay's "Vector and Tensor Analysis". The author gives the statement that the limit is equal to 1 without any explanation, perhaps because he thinks it does not belong to the contents of vector analysis. I can see it intuitively, but I want a rigorous mathematical proof of this limit. So, do you know of any textbook that contains a rigorous mathematical proof of this limit? In particular, Is such a proof in Erwin Kreyszig's "Differential Geometry" or Michael Spivak's "Calculus on Manifolds" if you happen to have once read them?

    Thanks a lot.
     
  2. jcsd
  3. Jul 24, 2017 #2

    vanhees71

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    But that's indeed trivial by definition of ##s##. Let ##\vec{x}(t)## the parametrization of your curve with an arbitrary parameter ##t##. then by definition
    $$\frac{\mathrm{d} s}{\mathrm{d} t}=\dot{s}=\sqrt{\dot{\vec{x}}^2}.$$
    Now you can take ##s## as the parameter, and you get
    $$\left (\frac{\mathrm{d} \vec{x}}{\mathrm{d} s} \right)^2=\left (\frac{\mathrm{d} t}{\mathrm{d} s} \dot{\vec{x}} \right)^2=\left (\frac{\mathrm{d} t}{\mathrm{d} s} \frac{\mathrm{d} s}{\mathrm{d} t} \right )^2=1.$$
     
  4. Jul 24, 2017 #3
    Wrong.
     
  5. Jul 24, 2017 #4

    vanhees71

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    Can you elaborate? It's a definition. How can it be wrong?
     
  6. Jul 24, 2017 #5
    Definition of what? $s$ is the length of the arc, x in bold is the position vector. No definition of them can account for the second = in your reply. You must have mixed them up, so I said you are wrong.
     
  7. Jul 24, 2017 #6

    vanhees71

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    Well, what I wrote is exactly what I just described in formulas. I've no clue how you come to the idea that ##s## has anything to do with the second. You should learn elementary differential geometry first!
     
  8. Jul 24, 2017 #7
    I have no clue what you are talking about either. More often than not, to understand a question is harder than to answer it in a haste. You should also learn it.
     
  9. Jul 24, 2017 #8

    mathwonk

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    there may a little confusion based in varying choices of definition of path length. It seems to me that vanhees' derivation is ok assuming his definition of the arc length. however on its face that definition depends on a choice of the parameter t, hence requires some argument that it in fact is independent of choice. For that reason most authors i have seen give a different definition, that of archimedes, according to which the arc length is defined originally as the limiting value of the perimeter of an oriented polygon formed by a choice of a sequence of points on the curve. It is then proved that the formula vanhees is using does calculate that limit, by appealing to the mean value theorem of differential calculus. this is done e.g. on page 277 of the first volume of courant's calculus. The point is to relate the limiting value of lengths of secants which occur in archimedes' definition (and in zzzhhh's desired limit), to that of the lengths of tangent vectors which occur in vanhees formula.

    now i have also found a book where the formula of vanhees is given as a definition, namely the calculus book of Joseph Kitchen. Kitchen however precedes his definition by an argument that the definition is indeed independent of parameter, provided only that certain axioms hold, among which interestingly is precisely the limiting formula desired by zzzhhh. So Kitchen proves, also using the mean value theorem, that any definition of arc length satisfying his axioms must be computed by the formula of vanhees.

    So it seems to me there is some work to be done either to justify the formula vanhees is applying, or to prove the desired limit directly from archimedes' definition of arc length.

    I.e. the desired limit relates the arc length to the secant length. since velocity vector length is defined in terms of limits of secant lengths, the gap is filled by relating arc length to velocity vector length. Vanhees formula takes that relation for granted. To justify it, one may consult the discussion in courant, pages 277-279. Spivak leaves the topic of arc length to the last few problems in his chapter 15, suggesting one use the MVT. Apostol gives a detailed discussion of arc length deriving all formulas needed here on pages 529-534 of his volume one, in chapter 14. This may be the most detailed and rigorous treatment, as is often the case with Apostol.
     
  10. Jul 24, 2017 #9
  11. Jul 24, 2017 #10

    mathwonk

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    that linked argument is not rigorous.
     
  12. Jul 25, 2017 #11

    vanhees71

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    The pathlength is of course NOT dependent on the parametrization. Length is a scalar quantity:
    $$s=\int_{t_1}^{t_2} \mathrm{d} t \sqrt{\dot{\vec{x}}^2(t)}.$$
    It's a 1st-rank homogeneous functional in ##\dot{x}## and thus parametrization independent. To prove that consider another parametrization ##t=t(\lambda)##. For convenience make it monotonously increasing. Then you get
    $$s=\int_{t_1}^{t_2} \mathrm{d} t \sqrt{\dot{\vec{x}}^2(t)} =
    \int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \frac{\mathrm{d}
    t}{\mathrm{d} \lambda}
    \sqrt{\dot{\vec{x}}^2(t)}=\int_{\lambda_1}^{\lambda_2} \mathrm{d}
    \lambda \frac{\mathrm{d} t}{\mathrm{d} \lambda} \sqrt{\left
    (\frac{\mathrm{d} \vec{x}}{\mathrm{d} \lambda}\right )^2} \left
    (\frac{\mathrm{d} \lambda}{\mathrm{d} t} \right) =
    \int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{\left
    (\frac{\mathrm{d} \vec{x}}{\mathrm{d} \lambda}\right )^2}.$$
     
  13. Aug 9, 2017 #12

    mathwonk

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    thanks for this vanhees71. this is the "justification" I referred to from page 279 of courant, but thank you for providing it here so people do not have to look it up. It is always clearer to give an argument than a reference. cheers!
     
    Last edited: Aug 10, 2017
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