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Potential and conservative

  1. Jul 10, 2011 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=37087&stc=1&d=1310357380.jpg

    2. Relevant equations



    3. The attempt at a solution
    The answer should be [itex]-(x^{2}-1)cos^{2}\theta[/itex]
    But I get a constant -1.
    Why?

    attachment.php?attachmentid=37086&stc=1&d=1310357380.jpg
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

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  2. jcsd
  3. Jul 11, 2011 #2

    lanedance

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    can you explain what you are attempting to do? I would approach as follows

    say phi exists then you know
    [tex] F_x = \frac{\partial\phi}{\partial x}[/tex]
    [tex] F_y = \frac{\partial\phi}{\partial y}[/tex]

    so integrating the first gives
    [tex] \phi = \int F_x dx= \int 2x cos^2y dx =?[/tex]
     
  4. Jul 11, 2011 #3
    Sorry that I skip some steps and forget to give explanation.

    attachment.php?attachmentid=37101&stc=1&d=1310366476.jpg

    This is my approach.


    This yields [tex] x^2 cos^2y+C[/tex]

    But how to find the constant?
     

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  5. Jul 11, 2011 #4

    lanedance

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    not quite as you are only integrating w.r.t. to x, so c could be a function of y
    [tex] \phi = \int F_x dx= \int 2x cos^2y dx = x^2cos^2y + c(y) [/tex]

    now also do it from the other direction
    [tex] \phi = \int F_y dy= ?[/tex]
     
  6. Jul 11, 2011 #5
    [itex](x^2+1)\frac{cos2y}{2}+C(x)[/itex]

    it can then simplfied as
    [itex](x^2+1)\frac{2cos^2y-1}{2}+C(x)[/itex]
     
  7. Jul 11, 2011 #6

    lanedance

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    i can't quite follow your simplification, you should equate
    [tex] \phi = \int F_x dx = \int F_y dy[/tex]

    also I would use a different constant function for the 2nd integral (say c(y) for first and d(x) for the second)

    and from there you should be able to deduce withther it is possible to to solve c(x) & d(y)
     
  8. Jul 11, 2011 #7
    Last edited by a moderator: Apr 26, 2017
  9. Jul 11, 2011 #8
    It's my first time to solve the problem like this.
    So I stuck in here attachment.php?attachmentid=37106&stc=1&d=1310371962.jpg


    I would like to ask, what's wrong with my method above?
    I have solved a lot of problems by using that method.(I learn it from mathematical methods in the physical sciences Chapter 6 sec8)
     

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  10. Jul 11, 2011 #9

    lanedance

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    Last edited by a moderator: Apr 26, 2017
  11. Jul 11, 2011 #10

    lanedance

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    stick to whatever you're comfortable with... took me awhile to figure out what you're doing as there's a few jumps (also the big pics are difficult to read on a laptop screen), but i think you set phi=0 at the origin and do a path interval over F to find the form of phi if it exists, in essence i think they're the same thing though yours has the advantage of setting one value to zero in the integral
     
    Last edited: Jul 11, 2011
  12. Jul 11, 2011 #11

    lanedance

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    also note the potential is only unique upto a scalar additive constant ie (phi) or (phi +1) give the same force field
     
  13. Jul 11, 2011 #12

    lanedance

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    this is how i would approach it using my method (i re-named c & d as f & g to be clear they are functions)
    [tex]\phi(x,y) = \int F_x dx = \int 2x cos^2(y) = x^2cos^2(y) + f(y)[/tex]
    [tex]\phi(x,y) = \int F_y dx = \int -(x^2+1) sin(2y) dy = (x^2+1) \frac{1}{2}cos(2y) +g(x) = (x^2+1) (cos^2(y)-\frac{1}{2})+g(x)[/tex]

    in the last line we may as well group all the x only terms in a function, so let
    [tex] h(x) = +\frac{1}{2}(x^2-1)+g(x)[/tex]

    equating and re-arranging a little gives
    [tex] x^2cos^2(y)+ cos^2(y)+h(x) = x^2cos^2(y) + f(y)[/tex]

    then we can read off (up to an additive constant)
    [tex] h(x) = 0 [/tex]
    [tex] f(y) = cos^2(y) [/tex]

    so we get
    [tex] \phi(x,y) = (x^2+1)cos^2(y)+ c[/tex]

    where c is any constant
     
  14. Jul 11, 2011 #13

    lanedance

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    after all that... and re-reading the posts i think it was only post 11 you needed!

    you can only ever determine phi upto a additive scalar constant, as the constant will always be sent to zero in the differentiation to get the field
     
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