Fundamental confusions of calculus

Discussion in 'Physics & Math' started by arfa brane, Feb 11, 2012.

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  1. arfa brane call me arf Valued Senior Member

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    Penrose discusses what he (or one of his colleagues) calls "the first and second fundamental confusions" of calculus.

    These appear to hang on the definition of "smoothness" for a curve or a surface, and how coordinates are defined for either. He says: "... we may want to consider a quantity \( \Phi\; =\; f(x,y) \), defined on the surface, but expressed with respect to various different coordinate systems. The mathematical expression for the function \( f(x,y) \) may well change from patch to patch, even though the value of \( \Phi \) at any specific point of the surface 'covered' by those patches does not change."

    He also describes partial differentiation of a function like \( f(x,y) \) as having one variable held constant. This must mean that x and y are treated as being independent variables in f. Does this mean that x and y are independent, or is that only the case when a differential operator is applied to one of them?

    Another point he makes is that \(\frac {\partial} {\partial x} \) and \(\frac {\partial} {\partial y} \) can be interpreted as 'arrows' pointing along coordinate lines. This suggests that a system of coordinates is not necessarily a collection of points, yes? Differential operators as arrows also suggests direction vectors. Multiplying a partial differential operator by a real (or complex) number gives you a vector.

    So what's the second fundamental confusion of calculus? This has to do with transforming (the function f(x,y)) to another coordinate system (say, f(X,Y)) and failing to note that y = constant might not agree with Y = constant. That is, y and Y might have different gradients even though they overlap the same region of a surface.

    I think Tach (not sure about Pete) is making these mistakes in the debate about Lorentz transformations and angles that are invariant (transform conformally). But that's another time and place (or coordinate system). Another point Penrose seems to be making is in respect of the importance of establishing a Cauchy-Riemann condition, or "structural integrity" perhaps, between the systems of coordinates.
     
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  3. Pete It's not rocket surgery Moderator

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    Yeah, I don't know if I'm making mistakes either. It's been a long time since I've had to do any multivariate calculus, and that was only introductory course level.
    It seems to me that in \(f(x,y)\), x and y are independent variables.
    When the differential operator is applied to one, the other is treated as a constant - ie it's not a variable at all for the purpose of the operator.

    Not sure how you get that suggestion?
    If f(x,y,z) is a scalar field, then \frac{\partial}{\partial x}[/tex] gives the rate of change of the field in the direction parallel to the x-axis, right?
    It seems to me that the directional vector is a property of the coordinate system that the partial differential uses, rather than something that emerges from the differential operator, if that makes sense.

    I don't understand how?
     
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  5. Tach Banned Banned

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    Really? Where did you learn such a thing? Because it is nonsense.


    I think it is more likely one of your misconceptions like the one above. It is quite clear that you don't understand what Pete and I are talking about in that thread.
     
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  7. arfa brane call me arf Valued Senior Member

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    The part about \(\frac {\partial} {\partial x} \) being interpreted as a direction along a coordinate line is nonsense?

    I think you are grossly overestimating your understanding of calculus. I also think your continual put-downs are just a way to reassure yourself, if no-one else.

    It's why your "debate" with Pete has, after several pages and as many months, achieved so little that it amounts to nothing.
     
  8. Tach Banned Banned

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    The part about multiplying an operator by a scalar and getting a vector , which is exactly what you posted, IS nonsense.


    Doesn't change the fact that you obviously misinterpret the subject of the discussion.
     
  9. arfa brane call me arf Valued Senior Member

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    Not really. If you include the requirement that the scalar is a value of a real (or complex) function of x.

    An astute mathematician might have pointed that out instead of saying it's nonsense.

    If you understood calculus as well as you like to tell everybody constantly, you could have pointed out the mistake. The operator \(\frac {\partial} {\partial x} \) points along the x-axis, if you give it a magnitude which is also a function of x, how isn't that a vector? If you have the same thing along a y axis, how isn't their composition another vector?
     
  10. QuarkHead Remedial Math Student Valued Senior Member

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    Oh dear. Look.....

    Given a differential n-manifold, the tangent vector space at any point consists of objects of the form \(v = \sum \nolimits_{i=1}^n \alpha^i \frac{\partial}{\partial x_i}\) where the \(\alpha^i\) are scalar, the \( x_i\) are coordinate functions and the \(\frac{\partial}{\partial x_i\) are considered as basis vectors.

    Thus any basis vector can be written as \(1\cdot \frac{\partial}{\partial x_i}\), so arfa branes claim has limited validity. I am not familiar with his source, but I suspect it is using the so-called "summation convention" which he failed to pick, and also possible thereby that he may not fully understand index notation

    But in view of the definition of a tangent vector I offered, it seems that Tachs riposte of "nonsense" is preposterously pedantic. yes, the claim is not true in general, but not that far off.

    I wonder what the problem is - is it possible that these 2 posters have some historical baggage? I can assure them, this is of absolutely no interest to a casual reader like me
     
  11. Tach Banned Banned

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    It is very simple, really. Operators aren't vectors. Operators multiplied by scalars (be they real, or as you claim, imaginary) aren't vectors either.
    If you start with \(f=f(x,y,z)\) for example, then \((\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})\) points in the direction of the normal (gradient) to the surface described by f. This doesn't make \((\frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z})\) a vector. Even if you try to multiply it with imaginary (or real) numbers.
     
    Last edited: Feb 11, 2012
  12. przyk squishy Valued Senior Member

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    The OP isn't really about general calculus. It's actually a (rather confused) allusion to some conventions in differential geometry. In differential geometry it turns out that a very elegant way to define the notion of a vector or vector field on a manifold is to identify vectors with differential operators.

    If you've got a coordinate system \(x^{i}\) defined on a manifold, you can decompose any differential operator \(v\) in terms of the partial derivatives \(\frac{\partial}{\partial x^{i}\). With the Einstein summation convention you can generally write something like
    \(v \,=\, v^{i} \frac{\partial}{\partial x^{i}} \,.\)​
    The \(\frac{\partial}{\partial x^{i}}\) naturally fill in the role of basis vectors, and the \(v^{i}\) (the "real (or complex)" numbers arfa brane was talking about) are the vector components.

    If you compare the decomposition of \(v\) in two different coordinate systems:
    \(v \,=\, v^{i} \frac{\partial}{\partial x^{i}} \,=\, w^{j} \frac{\partial}{\partial y^{j}} \,,\)​
    then using the chain rule it's easy to show that this implies the usual tensor transformation rule for the components of \(v\):
    \(w^{j} \,=\, \frac{\partial y^{j}}{\partial x^{i}} v^{i} \,.\)​

    That said, I haven't been following the debate about Lorentz transfrorms, and I have no idea how - if at all - any of this is supposed to be relevant to that discussion.
     
  13. Tach Banned Banned

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    It is totally irrelevant.
     
  14. QuarkHead Remedial Math Student Valued Senior Member

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    Umm, well... the set of linear operators acting on a vector space is itself a vector space (the proof is elementary - try it). So unless you claim that the elements of a vector space are not vectors, then this assertion is not correct

    Sure it doesn't, in fact your notation makes no sense.

    I suggest you review the definition of the gradient - you may find a few "+'s" in there!

    Look. I told you, przyk told you that that the differential operators \(\frac{\partial}{\partial x_i}\) are the basis for some tangent space

    Question: do you doubt that elements in the set of basis vectors for some vector space are themselves vectors?
     
  15. prometheus viva voce! Moderator

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    It certainly is a vector. It even has it's own symbol that has the little vector sign over it: \(\vec{\nabla}\)
     
  16. Tach Banned Banned

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    talk to arfa brane on the subject .

    Thank you, it is exactly what I posted it is.

    That isn't the issue. Multiplying \(\frac{\partial }{\partial x}\) by any scalar doesn't make it into a vector. I suggest you re-read the OP.

    I don't doubt it at all since I know it to be true. By the same token, this is not the subject of my objection to the nonsense in the OP.
     
    Last edited: Feb 11, 2012
  17. arfa brane call me arf Valued Senior Member

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    ??
    Why talk to me about it? Is that a lame attempt at deflection from your mistakes?

    But \(\frac{\partial }{\partial x}\) is a vector, if you multiply it by 1, as QH pointed out, it's still a vector. That's a basic mistake you're making there.

    As to the relevance of any of this to the debate, I think it shows that you're nowhere near the level of ability you keep insisting on.

    No, it's really about chapter 10 of "The Road to Reality". It's about how Penrose explains some aspects of Riemannian geometry. What caught my eye, so to speak, was that he also explains independent variables which is also relevant to the so-called debate.
     
    Last edited: Feb 11, 2012
  18. QuarkHead Remedial Math Student Valued Senior Member

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    This is getting weird!!

    1. Tach claims that an operator is not a vector;

    2. I claimed that the set of all linear operators on a finite-dimensional space is a vector space and invited him to prove it (no result);

    3. Tach accepts that an element in a vector space is a vector;

    It will suffice to disprove (2) for Tach's assertion "an operator is not a vector" to be true
     
  19. AlphaNumeric Fully ionized Moderator

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    In the context given by QH it is because in the tangent space \(\frac{\partial}{\partial x^{i}}\) is another way of writing \(\mathbf{e}_{i}\).

    We're all familiar with writing the usual Cartisian basis elements as \(\mathbf{e}_{x}\) etc or \(\mathbf{e}_{i}\). If you approach the same objects from a formal point of view then the Cartesian coordinates (x,y,z) of \(\mathbb{R}^{3}\) induce a coordinate basis on the associated tangent space \(T\mathbb{R}^{3}\) of \(\partial_{x}\) etc.

    This is actually a very nice notation because when you include the cotangent space basis \(\textrm{d}x^{j}\) and use the fact \(\textrm{d}\lambda(v) = v(\lambda)\) for the action of the dual elements then we get the nice behaviour \(\textrm{d}x^{j}(\partial_{i}) = \partial{i}x^{j} = \delta_{i}^{j}\). Hence writing a basis in terms of partial derivatives is a very nifty, if not initially slightly odd, way of doing things.

    Hence \(\partial_{x}\) can, in the right circumstance, be viewed in the same way as \(\mathbf{e}_{x}\) would in usual geometry. This is why QH said your \((\partial_{x},\partial_{y},\partial_{z})\) thing was meaningless. It is entirely equivalent to someone writing a vector \((\mathbf{e}_{x},\mathbf{e}_{y},\mathbf{e}_{z})\). If a vector is written as \(v = v^{i}\mathbf{e}_{i}\) then you write it in component form as \((v^{1},\ldots)\), which is really a list of the coefficients, not the vector basis elements.


    The nice thing is that now you can do all kinds of wonderful differential geometry and p-form calculus with some slick notation.

    Reading the OP it would seem pretty clear to me that whatever resource AB is reading that is what it is referring to. Obviously he's not sufficiently familiar with the material to get his head around it off the bat but that's no crime. However, you don't appear to be sufficiently familiar with it either, even after several people have laid it out for you. I did stick my head into this thread earlier, after QH's first post and thought "Well now that he's brought it to Tach's attention there's no need for me to say anything". Obviously I was mistaken. Please don't like this descend into another case of multiple people walking you through standard literature again. And try to dial it down a bit.
     
  20. Pete It's not rocket surgery Moderator

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    Thanks, gentlemen.
    I can't pretend to any great understanding, but I like to think I'm learning something.

    The relevance to the debate between Tach and I (I really don't recommend trying to wade through the whole thing) is in using \(\frac{\partial \vec{r}}{\partial \theta}\) to determine the tangent to a rolling wheel.

    Arguments have developed over the total derivative. If Tach is making the mistakes I think he is, then you could probably get the gist from the last post. But if not, then specific things Tach has said which seem wrong to me include:
    post 179
    (It seems to me that if a and b are independent, then da/db is meaningless).
    post 189
    (It seems to me that this is \(\frac{\partial f}{\partial \theta}\), because t is treated as a constant).
    post 193
    (It seems to me that if t is independent of \(\theta\), then \(t(\theta)\) is meaningless.)
    off site document linked in post 195
    (It seems to me that this is the definition of the partial derivative.)
    In the (currently) last post, post 197 Tach spells out what I suspect is his misunderstanding about when the total derivative of a multi-variable function is meaningful:
    I've put off replying, partly because if I'm wrong, then I don't understand why, and if I'm right, I don't think that anything I say will change Tach's mind.
     
    Last edited: Feb 12, 2012
  21. Tach Banned Banned

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    The difference between total derivative and partial derivative is really simple, Pete, and it was spelled out quite clearly in the post you quoted: partial derivative is taken with respect to \(\theta\) being an explicit argument, total derivative is the sum of the partial derivative (if it exists) plus the chain derivatives due to certain arguments being functions of \(\theta\). You do not need to spread the same question over three threads in order to understand this.
    In math terms: if \(f=f(\theta, u(\theta),v,w,...)\)

    then:


    The total derivative wrt \(\theta\)is:


    \(\frac{df}{d \theta}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial u} \frac{du}{d \theta}\)

    while the partial derivative wrt \(\theta\) is:

    \(\frac{\partial f}{\partial \theta}\)

    Example:

    \(f=3 \theta+ sin^2(\theta)+ln(x)\)

    \(\frac{df}{d \theta}=3+2 sin(\theta) cos (\theta)\)

    \(\frac{\partial f}{\partial \theta}=3\)
     
    Last edited: Feb 12, 2012
  22. prometheus viva voce! Moderator

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    I want to be very clear about this: You're saying that \(\frac{\partial }{\partial \theta}\left(3 \theta+ sin^2(\theta)+ln(x) \right)=3\)?
     
  23. prometheus viva voce! Moderator

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    Let's cut to the chase: you are wrong. With your definitions of total derivatives (which I don't remember and am not going to assume are correct) the total and partial derivatives of f are the same in the example you give. If you suppose \(x = x(\theta)\) then \(\frac{df}{d\theta} = \frac{\partial f}{\partial \theta} + \frac{1}{x(\theta)} \frac{dx}{d\theta}\)
     
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