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The Vector Calculus Thread

  1. Jun 28, 2006 #1


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    This isn't directly a request for homework help, since classes won't be starting for another two months, but I suppose it will be helpful to homework because I'll be taking Applied Analysis, Mechanics, and Electromagnetism, all of which include vector calculus.

    Given time, I will try to work through a Vector Calculus book here.

    I'm currently reading "Div, Grad, Curl, and All That" and already having trouble intepreting some of the equations in chapter 1, even though I've taken all the required courses for it. It may have slipped my mind, but here goes the questions.

    (i use vector signs for hats)
    F(x,y) = \vec{i} x + \vec{j} y

    I'm not able to coneptualize how the above equation appears as vectors always flowing away from origin (the corresponding graph shown in the book). I'm guessing that vector function is not simply a vector like I'm assuming it is, but either way, there should be two points right? One for the base of the vector and one for the head. But I can only conceptualize this if the origin is always the base (which doesn't seem the case in the illustration)

    The other one:

    G(x,y) = \frac {\vec{-i} y + \vec{j} x}{\sqrt{x^2+y^2}}

    which is supposed to represent a set of vectors flowing radially, counterclockwise. I can see the presence of a circle in the equation, but I'm still a bit sketchy as to how you would work out the shape/size of the arrow based on numbres substituted in for x and y.
    Last edited: Jun 28, 2006
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  3. Jun 28, 2006 #2


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    In cartesian form, the i j k are unit vectors parallel to the coordinate axes.

    eg. 2i+3j would have a 'head' at (2,3), 'base' at (0,0) - in your words.
  4. Jun 28, 2006 #3


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    That would be one representation of 2i+ 3j. A vector can be "moved" anywhere.

    In particular, the vector field xi+ yj should be visualized as vectors with tail at the point(x,y), pointing directly away from the origin and length the same as the distance from (0,0) to (x,y). All vectors point away from (0,0) and get longer as you get farther from (0,0).
  5. Jun 28, 2006 #4


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    Yes! I actually dug up my Calculus book. It seems my Calc III class ended right before the Vector Calculus chapter. I remember him assigning some problems, but not requiring a turn-in, so as you can imagine, end of school year, never got done.

    The process for solving vector equations, for those of you who are learning is to:

    a) take a number for each x and y, and that point would be the base of your vector arrow.

    b) put the x and y in the vector equation to find the vector (direction and distance) that sprouts from that particula point (x,y) that you arbitrarily chose in part a) above.

    so for an example, given the equation:

    ix + jy

    if we chose (x,y) as (1,3) then the simple vector equation (substituted) would be:

    i1 + j3 which we inteprate as a vector arrow starting at 1,3 and ending at a point 1 over (to the right) and 3 up. This length would be equivalent to a line going from 1,3 to 2,6 which can be found with pythagoras:

    (2-1,6-3) = (1,3) => sqrt(1+9) = sqrt(10)

    please alert me to any mistakes, thanx!
  6. Jun 28, 2006 #5

    George Jones

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    Looks good.

    An extended line that passes through (1 , 3) and (2 , 6) also passes through the origin, so the vector points directly away from the origin.

    What about G?
  7. Jun 29, 2006 #6


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    Yes, I should have seen that F is a sort of linear combination (i think) of the distance between the origin and whatever point you arbitrarily chose.

    Chosing the base-point of G to be (2,4),
    so the vector:

    G(2,4) = -4i/sqrt(20) + 2j/sqrt(20)

    = -4i/4.47 + 2j/4.47

    = -.89i + .45j

    So the vector would be an arrow from (2,4) to (1.11,4.45), which does indeed point in counterclockwise direction.

    Part of my problem visualizing this originally was in part because the author of Div, Grad, Curl made an essentially perfect circle of tangent arrows. At the end of my 2 ton calculus book, however, I found the same equation with a more random (in appearance) set of arrows that made it easier for me to digest.

    I guess sometimes when my assumptions are threatened, I become timid towards a particular technique or concept.

    Thanks for the help so far. I'll probably get back on this after fourth of July if I'm not inspired before the weekend.
  8. Jul 8, 2006 #7


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    A refresher problem from the book (this problem looked simple, but it took a lot more energy and time to work out than I thought. I'm glad I'm doing this before the school year):

    We can visualize both of these particles as points on the x-axis, across the origin from each other and y is a line passing between them.

    we want to figure out the sum of the distances between y axis and each point. This is where I slowed down a bit (hopefully a consequence of psychological shock and fear, and not that a learning limit), trying to remember the exact configuration of the distance formula in three dimensions. Google's first hit returned http://library.thinkquest.org/2647/geometry/measure/measure.htm.

    duh (i.e. technical retention fail)

    so, we can obviously ignore the third dimension, and plug our remaining sets of points into the formula:

    sqrt{(1-0)^2 + (0-y)^2} for +1 charge
    sqrt{(-1-0)^2 + (0-y)^2} for -1 charge



    The distances are the same (which we could have found by just taking one particle and using symmetry). The Electric field can be found with the equation E(r) = (q/r^2)*u-hat.

    Total field E = E1 + E2

    = 1/1+y -1/1+y
    = 0

    The forces cancel each other out.

    I'm a bit confused about the u-hat and how to incorporate it. I know it's the unit vector, but the specific application eludes me. I don't have a more specific question about it, but a brief explaination might help with that.
    Last edited by a moderator: Apr 22, 2017
  9. Jul 8, 2006 #8
    The u-hat that the book mentions is a unit vector as you say. Specifically, it points from the current location to the origin. If you let the current position be defined by [tex]\vec{r}[/tex], then

    [tex]\vec{\widehat{u}} = -\frac{\vec{r}}{|\vec{r}|} [/tex]

    The reason for the minus sign is beacuse [tex]\vec{r}[/tex] points away from the origin, and we want [tex]\vec{\widehat{u}}[/tex] to point towards the origin.

    Some authors go so far as to define the electric field in the following way

    [tex]\vec{E}(\vec{r}) = \frac{q}{|\vec{r}|^2}\vec{\widehat{u}} = - \frac{q}{|\vec{r}|^2}\frac{\vec{r}}{|\vec{r}|} [/tex]
    \vec{E}(\vec{r}) = -\frac{q}{|\vec{r}|^3}\vec{r}[/tex]

    Which can be very confusing the first time you see it thrown down.

    By the way, you'll probably want to take a look at this.

    Change the display to "field vectors" and the setup to "user defined field". Enter the "i" and "j" parts of the equation and press reset. That should let you draw custom vector fields.

    This applet is the number one way to learn vector calculus. Check out the "curl detection" feature once your comfortable with it.
    Last edited: Jul 8, 2006
  10. Jul 10, 2006 #9
    Whenever you see:
    in vector calculus, this strongly suggests polar coordinates. Lots of people forget that a lot of problems can be simplified greatly by a change of variables (to polar, cylindrical, spherical, etc)
  11. Jul 17, 2006 #10


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    a powerufl tool, indeed, and fun.

    I will have to look into the rest of your post once I have the pencil and paper out. I remember the formula for the unit vector, I guess I'm just not sure of its purpose and real-life (or visual) application.
  12. Aug 1, 2006 #11


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    OK, I was way off in my answer.

    The answer in the back of the book is [tex] \frac{-2\mathbf{i}}{(y^2+1)^{3/2}}[/tex]

    so, a refresher of the question:

    a +1 charge is at (1,0,0)
    a -1 charge is at (-1,0,0)

    the question is the electric field at an arbitrary point along the y axis (0,y,0)

    so for charge 1)


    first we tackle r-r_1, which would be the test charge (0,y,0) minus the +1 charge (-1,0,0):


    Assuming the bars are for magnitude, |r-r1| =

    [tex]1 + y^2[/tex]

    so now, in all, we have:

    [tex] \frac {1}{(1+y^2)^2} \mathbf {U} [/tex]

    now for U =

    [tex] - \mathbf {\frac {r}{|r|} } = - \frac {y\mathbf{i + j}}{1+y^2} [/tex]

    so the charge on a point y from the +1 charge is:

    [tex] - \frac {y\mathbf{i + j}}{(1+y^2)^3} [/tex]


    I get the same answer for the -1 charge thread, using the same techniques, and when I add them, the yj's cancel and the i's enforce each other. I get:

    [tex] \frac {-2\mathbf{i}}{(1+y^2)^3} [/tex]

    how does the bottom get rooted?
    Last edited: Aug 1, 2006
  13. Aug 1, 2006 #12


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    Electrostatic force exerted by point charges in 3-space

    always, but

    ought to be

    [tex]|r - r_1|=\sqrt{1 + y^2} = (1 + y^2)^{\frac{1}{2}}[/tex]​

    by the pythagorean theorem (the distance formula.)

    with corrections, this is

    [tex] \frac {1}{\left( (1 + y^2)^{\frac{1}{2}} \right) ^2} \mathbf{U} = \frac {1}{1 + y^2} \mathbf {U}[/tex]

    as with the eariler magnitude, need a root in there, and the charge is one unit out on the x-axis but y-units out on the y-axis, so:

    [tex]\mathbf{U} = - \mathbf {\frac {r}{|r|} } = - \frac {\mathbf{i}+y\mathbf{j}}{\sqrt{1+y^2}} [/tex]

    should be

    [tex] \frac {1}{1 + y^2} \left( - \frac {\mathbf{i}+y\mathbf{j}}{\sqrt{1+y^2}}\right) = - \frac {\mathbf{i}+y\mathbf{j}}{(1+y^2)^{\frac{3}{2}}}[/tex]

    Good observation, do you see it, or did you calculate it?

    it's there...


    P.S. You had except for the root thing, and that is no biggie: keep it up :smile:
  14. Aug 1, 2006 #13


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    the point made so well by halls is the key confusion in using vectors. most books do not explain this well at all. again spivaks calculus on manifolds nails it perfectly.

    the fact is that a vector is given by two points a tail and a head. or a foot and a head. if q is the head and p is the foot, then the vector can be written as q-p. i.e. vector is the difference of two points. and hence a point can be written a s the sum of a vector and a point, the foot plus the vector yielding the head.

    hence a quantity involving only one point is not a vector. so there is intrinsically no such thing as a position vector, except in relation to some fixed but arbitrary origin point.

    thus one should distinguish between quantities that are naturally represented by vectors, like velocity, and those not so, like position.

    this is especially important in pohysics where one wants to understand the true physics independent of coordinates. this is why i rant so much against speaking of tensors as sets of corrdinates that transform in a certain way, raher than as quantities with intrinsic properties that truly mirror those of tensors.

    i.e. just a true vector must be represented by two points ratehr than one, a 2-tensor must be aquantity which involves bilinearity intrinsically, not just something that can be written with two indices, possibly in an artificial way.

    i.e. with coordinates one can write a "velocity vector" as (x,y,z), maybe with an arrow over it, and also one can write a position as (x,y,z) but one should ideally know what this means. when written this way they look the same, but they are not.

    is this making any sense? i have tried for years here to amke this point about tensiors without success. maybe the place to begin is merely to understand the diference between a point and a vector as you are doing. inded you saw it immediately with halls clear explanation.

    maybe halls or someone else can do a better job than i have on the tensor version.
  15. Aug 1, 2006 #14


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    actually the much vaunted representation of homomorphisms as (1,1) tensors is not quite as natural as advertised either.

    i.e. a finite sum of expressions like vtensorw* only represents homomorphisms from W to V whose image vector is in the span of the finite set of v's occuring in the summaton. hence this does not even give the identity map in infinite dimensions.

    for this reaswon althugh the map from such tensors of tyope (1,1) to homomorphisms is naturally defined without coordinates, the inverse map is not, since it does not exist in general.

    how do you like them apples, all those people who think a tensor is a family of coordinates with multiple indicies? and if you think infinite dimensions is for pointy headed mathematicians, remember , feynman integrals are based on infinite dimensional manifolds of arcs.
    Last edited: Aug 1, 2006
  16. Aug 1, 2006 #15


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    Well, somewhat... unfortunately I don't know exactly what a tensor is yet. I'll be starting applied analysis this semester (in about a month) and mechanics, and e&m. I'm guessing I'll see tensors there. I also don't know what a homomorphism is. (I could easily google these and scrape the surface, but to be comfortable with it, I'd like to work with them)

    Describing things as transforms always confused me. I really disliked my linear algebra class, because the teacher programmed gene lines in matlab (as a profession). She had very little physical representation for what we were doing throughout most of the class.
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