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Mathematica Vector Fields Explaination.

  1. Apr 14, 2012 #1
    I was about to do an experiment in Wolfram Mathematica like drawing electric field lines around a charged body and other arrangements. So i saw this nifty little Function for that very purpose called VectorPlot

    My Problem is that i dont know what the function does exactly i went through examples in the Wolfram Mathematica documentation but it did not mean anything to me.

    I Read This Article : http://mathworld.wolfram.com/VectorField.html

    All i saw was that vector field meant something about the x axis and some function on it

    My Questions Are:

    How Exactly are those small vector lines represented ?

    Like In this Example where I was playing with the Function

    http://puu.sh/pxch [Broken]

    what is happening to the x and y and what is making the vector lines change Size?

    What is Giving it direction?

    Is there an equation that is defining the vectors.?

    Could any one explain this to me....Please......BTW im a high school student so can you explain it in a way that is undergrad level...
     

    Attached Files:

    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 14, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey the-ever-kid.

    For this think of the (x,y) representation of the function at a particular point and how that corresponds with firstly the direction and secondly the magnitude.

    As an example look at vectors at each point in your lattice (i.e. the lattice is just the points on the grid where your vectors are drawn from for each tail of the vector) and calculate on a calculator or otherwise the direction and the magnitude and compare it to what you see on screen.
     
  4. Apr 14, 2012 #3
    thank you chiro BTW i actually figured it out a little while ago.....

    its like each point(x,y) become the origin and a vector like f(x)i+f(y)j is formed using it as the tail right?

    i did that for my point charge and got points

    [tex]f(x,y)=\frac{kQx}{(x^2 + y^2)^{3/2}},\frac{kQy}{(x^2 + y^2)^{3/2}}[/tex]

    Thanks,
     
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