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Div and curl

  1. Feb 4, 2016 #1
    I'm learning vector calculus and am wondering how general it is. The appear to be using a smoothness condition, but what is it? Certainly the functions are required to have two derivatives. That is, the partial derivatives can be taken twice. Are they further required to have an infinite number of derivatives for those identities to hold? Or what?
     
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  3. Feb 4, 2016 #2
    Different theorems have different conditions.

    The definitions are definitions.

    Being a physicist, I realize most physical phenomena satisfy all the conditions of being continuous and differentiable. Some theorems require more like being conservative.

    If you are a mathematician, you may need to pay more attention to the fine print and consider what kinds of functions don't satisfy the conditions of a given case. As a physicist, I've done OK dwelling less on the fine print like whether a function is infinitely differentiable.
     
  4. Feb 4, 2016 #3

    Krylov

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    In general, for a function to have an ##n##th derivative, it is necessary but not sufficient that the ##n##th partial derivatives exist.
    Nobody can tell without knowing what identities you are talking about. The degree of smoothness required depends on the particular identity or theorem. You cannot make a statement like: "In vector calculus, all functions must be at least twice differentiable".

    Of course, you can guess that if you have an identity with first order differential operators such as div and curl that are only applied once to each term, then typically the functions involved need to be once differentiable or continuously differentiable, etc. Sometimes, less is needed, if the identities are interpreted in a weak or distributional sense, for example.

    When you are learning vector calculus with the intent to apply it, it is probably fine to (initially) assume that everything is ##C^{\infty}## (which is weaker than analytic). However, if you are interested in proofs, then you will see that this assumption is gross overkill.

    Also, note that even innocently looking PDE with ##C^{\infty}## coefficients may admit solutions that become non-smooth in time. The development of such a singularity often has a physical significance, e.g. formation of a shock in gas dynamics. Therefore, it is certainly not true that all phenomena of physical significance necessarily have smooth mathematical manifestations.
     
  5. Feb 4, 2016 #4
    Yes, I am interested in proofs. What particularly interests me is the assumption that the div and curl functions return the same results regardless of choice of basis, which is what the formulas are based on. It's not even clear to me that if partials exist for one basis then they exist for all choice of basis. I think I could cook up a function such that partials exist only in the subspaces of the basis vectors. So I'm guessing they are using a smoothness condition to exclude that.

    In Wikipedia there is a definition of curl as a line integral over a circle which seems to me more general, though not all that handy in simple cases. But I'm not sure of anything at this point.

    Would div and grad be defined for a shock in gas dynamics? I would think not.
     
  6. Feb 4, 2016 #5
    Shock fronts are not truly discontinuous.

    "A shock front, however, is not truly singular. Viscosity softens the would-be discontinuity and replaces it with a steep transition over a finite distance."

    From https://www.crcpress.com/Physics-of...-and-Everyday-Phenomena/Lautrup/9781420077001
    p. 257

    If you think a physical phenomena is truly singular, zoom in by a factor of 1 million, and consider the question again. Then zoom in by another factor of a million and consider the question again. If it still seems singular, it might be.
     
  7. Feb 4, 2016 #6
    I got out my textbook and looked up the definitions. The answer is in their definition of differentiability. A vector valued function is differentiable if and only if an NxN matrix of partial derivatives exists that smooths things out so that the derivative can come from any path in the space. This matrix is unique. I presume that they mean it is unique for each basis, and there are transforms from each basis to the other that transform the matrix. Div is the trace of that matrix. Curl seems to be setting the diagonal to zero, then taking the product of a row vector of the basis and the matrix. Not quite, there is some fiddling with signs that I have yet to figure out. How this is equivalent to that Wikipedia line integral is beyond me.

    Grad is the matrix if the function is scalar valued instead of vector valued.
     
  8. Feb 4, 2016 #7

    Krylov

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    Thank you, this reminds me of using viscosity as a means of regularization of solutions of certain classes of PDE, also see my response to you below.
    What happens smoothly as a microscopic level may however manifests itself in a mathematically non-smooth way at the level of the macroscopic equation. Hence, if we ignore non-smooth solutions of certain PDE, I believe we may very well be missing indications of interesting physical phenomena.

    You could then argue that the macroscopic PDE (e.g. Burgers' equation without viscosity) admitting the non-smooth solutions is no good, because it doesn't include viscosity effects. I could very well agree with this, but even then the study of non-smooth solutions has taught us something, namely that the model is defective in this regard and that viscosity (although very small in gases) may not be neglected. In that sense, I believe that questions of smoothness should be an intrinsic part of modeling and not just of the mathematical analysis.
     
  9. Feb 4, 2016 #8

    Krylov

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    I don't know what this says precisely.

    What I would says is that ##f : \Omega \subseteq \mathbb{R}^n \to \mathbb{R}^m## defined on the open set ##\Omega## is differentiable on ##\Omega## if, by definition, for each ##x \in \Omega## there exists a linear operator ##Df(x) \in L(\mathbb{R}^n,\mathbb{R}^m)##, called the derivative of ##f## at ##x##, such that
    $$
    f(x + h) - f(x) - Df(x)h = o(\|h\|) \qquad \text{as}\,h \to 0
    $$
    (Loosely: "##f## can be linearly approximated at every point of ##\Omega##.") Note that ##Df : \Omega \to L(\mathbb{R}^n,\mathbb{R}^m)## is defined in a coordinate-free way.

    The standard result is then that if all partial derivatives of ##f## exist as continuous functions on ##\Omega##, then it follows that ##f## is differentiable (in fact: continuously differentiable or: ##C^1##) on ##\Omega## and the matrix representation of ##Df(x)## w.r.t. the standard basis is exactly the matrix of partials with each of them evaluated at ##x##.
    Yes. In fact: the linear operator is unique. Surely it has different matrix representations, depending on the basis.

    Regarding the definition on Wikipedia as the limit of a line integral, I have not seen this before, but maybe it is done like this to appeal to physical intuition? (More than, say, defining the curl as the determinant of a Cartesian coordinate matrix.) Perhaps someone else knows better?
     
    Last edited: Feb 4, 2016
  10. Feb 4, 2016 #9
    Looking at it again, I did not get this correctly. It isn't the curl that is a line integral, it is the dot product of the curl with a surface. The curl w.r.t. a surface is the line integral.

    I think I'm getting the hang of it. Maybe the curl is a pseudovector that one may think of as representing the "angular momentum" of the field around a point. Wikipedia was taking the dot product of this pseudovector with another pseudovector that is the normal to the locally planar surface. The line integral is equivalent to the dot products of the duals of the pseudovectors. I think.
     
  11. Feb 4, 2016 #10
    Good points.

    I'm happy enough when a mathematical model makes accurate predictions of measurable quantities, I don't need to labor over whether or not the singularities in the model function or a derivative are real. When pushed to experimental tests, many singularities are not real.

    Consider the potential of the hydrogen atom. Is it really -1/r ? Of course not. The proton has a wave function and the charge density is really smeared out over a small, but finite volume. Consequently, somewhere at the bottom of a -1/r potential there is a floor. It's not a real singularity if you zoom in enough. But the effect on the energy levels of hydrogen is very small. The mistake is thinking that getting very accurate energy levels validates the existence of a true singularity.
     
  12. Feb 4, 2016 #11

    Krylov

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    I think you are right. I like the "laboring" quite a bit, firstly of course because I do mathematics, but also because as I argued the defects of a model often appear already through the singular behavior of solutions. However, the ultimate test for any model is of course the physical experiment. Otherwise, it should not pretend to be a model in the first place.

    I believe I agree to this as well.
     
  13. Feb 5, 2016 #12

    PeroK

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    I'm not sure the Wikipedia page is a good place to learn about div and curl. That definition of curl relies on a significant amount of mathematics comapred to the natural definition in terms of partial derivatives.

    Stokes's theorem equates the line integral of a function round a closed curve with the surface integral of its curl over the surface bounded by the curve, but that's not normally the definition of curl. Similarly, you have the divergence theorem, but that's not normally used to define.

    You do, however, seem to be confused between "basis" and coordinates. Neither, in fact, affect the differentiability of a function. Grad, curl and div are normally defined in terms of partial derivatives with respect to the coordinates x, y and z. If you change coordinates, then the format of grad, div and curl will change - e.g. in spherical coordinates. But, if the derivatives of your new coordinates are well-defined with respect to x, y, and z, then by the chain rule, div and curl can be expressed in your new coordinates.

    Basis vectors do not enter into this.
     
  14. Feb 5, 2016 #13

    Krylov

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    Suppose you consider a function ##f## of the type I introduced in post #8, with ##m = n##. Then basis vectors do enter, in the sense that every matrix representation of ##Df(x)## is with respect to some basis, and the standard basis is not always the most appropriate or convenient one.

    Incidentally, as the OP pointed out,
    so it follows that the divergence of a vector field is invariant under a change-of-basis.
     
    Last edited: Feb 5, 2016
  15. Feb 5, 2016 #14

    PeroK

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    I would say div is invariant under a change of basis because it's (normally) defined in terms of the xyz coordinate basis. A change of basis is irrelevant. That just changes the way vectors are labelled.

    And the Wikipedia definition in terms of the limit of a volume integral, in fact, defines Div independent of basis and coordinates. How could a change of basis change the value of a volume integral?
     
  16. Feb 5, 2016 #15

    Krylov

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    Yes, a change of basis is indeed irrelevant for the value of the divergence, and in that sense a property of the operator ##Df(x)##, but this is not obvious from how one would normally calculate it. In post #8 I advocated to first define ##Df## as an operator-valued map. Then it follows that this map admits different matrix-valued representations.

    In this way, I hoped to eliminate the OP's confusion between the uniqueness of the derivative itself and the different basis dependent guises that it can take.

    Incidentally, thank you for pointing out how that definition of curl and div on Wikipedia arises. It was staring me in the face, but I overlooked it.
     
  17. Feb 5, 2016 #16
    My problems are all solved. Thank you, gentlemen. I couldn't have done it without you.

    I was indeed confused about how vector calculus may be "coordinate free." This is free in that one may choose freely any reasonable basis and in the end get the same result. It is not free in that one must still choose such a basis in order to calculate any specific values.

    Finding the definition of the vector calculus derivative tucked away in the "technical" section of my ancient textbook was a key step. It was defined in terms of a matrix instead of an operator. Hmph.

    If you are curious, what I was doing was to extend the notion of curl to N spatial dimensions. The surprise in the result is that to do this you get a curl that is an operator on bivectors. Only in 3D may a bivector represent a locally flat surface. So outside of 3D I'll have to do a little more work to use curl bivectors in the context of an n-dimensional manifold. But that will have to wait for tomorrow.
     
    Last edited: Feb 5, 2016
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