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Closed Curves.

  1. Jul 6, 2009 #1
    Hey, I am wondering if anyone can help me understand a mathematical explanation as to how they work.

    From what I understand, the area under a closed curve is the same, independent of the path taken. So when doing an integral you only need to take the initial and final into account. There have been 2 a few situations so far when I have come across these curves. Particularly when dealing with conservative forces, i.e. Gravity, Electricity. So with these forces, it is saying the work done is the same independent of the part taken. But what actually makes it a closed curve? What part of it all makes it so the path does not matter when getting the area under the curve?
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  3. Jul 6, 2009 #2
    It looks like you are talking about line integrals. But, line integrals do not find the area under the curve that you integrate over. One interpretation (the one it looks like you are talking about) is that if f is a force field, then the line integral of f over a curve is the amount of work done in transporting a particle along the curve.

    Line integrals are independent of parameterization of the curve. If f is a complex-analytic function (a conservative force field), then the line integral does depend only on the end points, but if the curve is closed, the integral is zero. For instance:
    [tex]\int_C f = F(b)-F(a)[/tex]
    where F'=f and a is the starting point of C and b is the ending point. So if C is closed, a=b and the integral is zero.
  4. Jul 6, 2009 #3
    Ah, so how does a line integral work then, if it does not find the area under the curve? Does it use the fact that the integral is always equal to zero in a closed curve?
  5. Jul 6, 2009 #4
    Do you have a particular example that you want clarified? Wikipedia's http://en.wikipedia.org/wiki/Line_integral" [Broken] seems to do a decent job of explaining it.
    Last edited by a moderator: May 4, 2017
  6. Jul 6, 2009 #5
    Yeah, that Article does a pretty good job of explaining it. It made a lot of sense once I saw it. So from what I understand at the moment, it sounds like a line integral, in a vector field, is when you do the dot product between the 2 vectors at each little point then add them all up. Is that the right idea? or have I missed a detail out?
  7. Jul 6, 2009 #6


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    The area under a closed curve? Do you mean the area inside a closed curve? And what "path" are you talking about? If you are trying to calculate the area contained in a closed curve the only "path" that makes any sense is the closed curve itself- and that is certainly NOT "independent of the path taken".

    Do you mean integrating along a path from one point to another? Or along a closed path from a given point back to itself? In either case, the result is "independent of the path" if and only if the integrand itself is an "exact" differential. That's not true in general.
    (And, if you are integrating around a closed path, the integral of an exact differential is 0- definitely NOT the area!)

    You seem to have memorized a number of words without assigning any meaning to them! A "closed curve" is simply any curve that goes back to its beginning point. that has nothing to do with "the path does not matter". You seem to be confusing the path with the integral. In any case, "what makes it so the path does not matter" is given in any Calculus text: the integral [itex]\int f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz[/itex] is "independent of the path" if and only if there exist a function F(x,y,z) so that dF= f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz. In that case, [itex]\int f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz= \int dF= F[/itex] evaluated at the two end points just as in Calculus I.
  8. Jul 6, 2009 #7
    Hahaha it seems that I do not know what I am talking about when I say area under the curve. I have only seen closed curves from a physics point of view, but in terms of maths, I am not really sure what it actually means. I understand if I was to talk about a closed path, it is a path that meets back where it started. But other than that, I am not too sure about how they work.

    Also when I am integrating with this closed curve, I have not been told if I am finding the area, I just have a habit of saying that when I do integrals. I guess since I am generally dealing with a vector field, I would use a line integral? I am sure I am being confusing and unspecific but that is because I am not completely sure about what is going on, when dealing with closed paths and integrating.

    Oh and if a path is different to a curve, please let me know. I probably should be using path.
  9. Jul 6, 2009 #8
    It seems HallsofIvy just repeated exactly what I had said. In terms of math, line integrals are defined. Then they are given physical interpretations. They can have different meanings based upon the interpretation. See my post https://www.physicsforums.com/showpost.php?p=2146029&postcount=6" there too). I'm sure there are others as well.

    For instance. If f is a complex valued function and [itex]\gamma(t)[/itex] is a curve where [itex]t\in[a,b][/itex], then the line integral is defined to be
    [tex]\int_\gamma f(z)\,dz = \int_a^b f(\gamma(t)) \gamma'(t) \,dt . [/tex]
    This is just the definition. As I explained above and in the post I linked to, this can be given different interpretations depending on what you treat f as. The line integral is just another version of an integral, and you don't want to always think of integrals computing areas. I think if you view them as summing something up, then this works better for the physical interpretations. You just need to determine what you are summing.

    I used a complex function for my definitions because it is the simplest in notation. You can think of a complex function as a vector field (it takes a point in [itex]\mathds{R}^2[/itex] and assigns a point in [itex]\mathds{R}^2[/itex]).

    Curve and path are often used interchangeably. There are some books though that make a small technical distinction between them.
    Last edited by a moderator: Apr 24, 2017
  10. Jul 6, 2009 #9
    Cool thanks, I have a good idea of what is going on. I think it is kind of deceptive that I had never been told about the line integral, no wonder it did not make sense to me.

    So with a complex function, why would you not use a matrix? Is it not always linear or something?

    Yeah, I had the idea they were almost the same, but wanted to be careful because I remember reading a text book I had awhile back, and it talked about how they were different, but could not remember how.
  11. Jul 7, 2009 #10


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    Where in the world did you get that idea? Or, perhaps, exactly what do you mean by "complex function"?

  12. Jul 7, 2009 #11
    Now if I was to take a vector and change it around, ie. translate it, change length, or rotate it, and it was linear, I would use a matrix. Of course because I am not familiar with a complex function, I could imagine that the transformation is not always linear. So I am wondering how they work when getting a vector at one point and assigning it to another point in the same vector space.

    I am also told that I can think of a complex function as a vector field. Which means I am probably over complicating things by bring in the idea of matrices. But to me they seem related in the fact that they manipulate vectors.

    I guess another way of dealing with how they move is to use a parametric system, in which you have the components of the vectors as a function of t.
  13. Jul 7, 2009 #12
    I remember having the same questions about what a line integral means and is used for, and I am still learning. What classes are you taking? Are you a physics or math major? Also, complex functions aren't always linear and so can't always be represented by a matrix, although some can (see below).
    http://books.google.com/books?id=og...&dq=complex functions vector fields&pg=PA450" is a good one page explanation with some pictures. The complex numbers or complex plane is just R^2. A complex variable can be represented by z=x+iy, where x and y are real numbers. This can also be represented by z=(x,y), a point in R^2. A complex function takes this point and assigns to it another point f(z) = f(x,y) = u+iv = (u,v), which can be viewed as a vector emanating from the point (x,y). This is a little of what the link explains. So a complex function is just a vector field because it takes points in R^2 and assigns vectors to the points.

    Complex functions (and thus the vector fields associated with them) can be manipulated by fractional-linear transformations, which can be represented by matrices. They can perform translations, dilations, rotations, and inversions (or any combination thereof) which is what you are referring to matrices changing vector fields. These are complex valued functions, but not all complex valued functions are fractional-linear transformations and thus can't always be represented by matrices.

    I don't know what class you are taking, but this is why complex variables is one of the most applicable areas of math. If you view complex functions as vector fields or some other physical interpretation, the theory gives you powerful results.
    Last edited by a moderator: Apr 24, 2017
  14. Jul 7, 2009 #13
    I am a math and physics major. But first year in both. I end up learning most of this stuff next year so I am in no hurry to become an expert in it all, how ever it is nice to understand how things work, and properly. I have definitely dealt with the complex plain before, and used complex numbers as vectors. Hahaha I just did not think that you would use vectors with complex numbers. I know complex numbers have lots of useful properties to do with waves and so on, but did not expect vectors to find them useful.

    I have been told before that you can treat complex numbers like vectors but there are some small things that are not the same. For example I am not sure if you can use a vectors reflection along the y axis like you can with complex numbers and their conjugate. So I understand that you have to be careful with small things.
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