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Astrophysics

  • one year ago

Flux

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  1. Astrophysics
    • one year ago
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    Find a flux of the vector field \[\vec F(x,y,z) = \frac{ m \vec r }{ |\vec r|^3 }\], where m = constant. \[\vec r = x \vec i + y \vec j + z \vec k\] out of a sphere od radius 1 centred at origin.

  2. Astrophysics
    • one year ago
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    Don't really remember how to do this, so thought I'd ask for some help, I believe if F is a continuous vector field defined on an oriented surface S with a normal vector n, then the surface integral of F over S is \[\int\limits \int\limits_S \vec F \cdot d \vec S = \int\limits \int\limits \vec F \cdot \vec n dS\]

  3. Astrophysics
    • one year ago
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    @Empty @ganeshie8

  4. Astrophysics
    • one year ago
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    So I guess \[x^2+y^2+z^2=1\] then we can use a parametric representation?

  5. Astrophysics
    • one year ago
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    I think the troubling part is when we cross multiply, not really sure how to visualize it and what way it should be

  6. Astrophysics
    • one year ago
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    |dw:1441476251414:dw|

  7. IrishBoy123
    • one year ago
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    use a Gaussian surface as it's totally symmetrical or stuff it into spherical if you want a slightly harder time

  8. Astrophysics
    • one year ago
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    Hmm interesting, never really worked with gaussian surfaces, I will look into it, thanks @IrishBoy123

  9. hwyl
    • one year ago
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    that's what you do anyway start from cube then use the same for curved

  10. hwyl
    • one year ago
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    |dw:1441476650300:dw|

  11. IrishBoy123
    • one year ago
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    attached is the spherical coords way you have an inverse square law here and you can use spherical and gaussian surfaces for all that stuff, even for gravity!! Lewin's lectures on this are epic

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  12. Astrophysics
    • one year ago
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    Thanks @hwyl Awesome, this is one reason I want to take E&M haha, thank you very much @IrishBoy123

  13. hwyl
    • one year ago
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    |dw:1441476754530:dw|

  14. hwyl
    • one year ago
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    apply gauss law

  15. hwyl
    • one year ago
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    ty for your time

  16. beginnersmind
    • one year ago
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    You can use the divergence theorem:|dw:1441477056233:dw| div F is particularly simple in this example

  17. Astrophysics
    • one year ago
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    Thanks everyone, nice @beginnersmind I will try your approach as well! :)

  18. IrishBoy123
    • one year ago
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    well divergence theorem sounds good in theory, but in spherical is it really any easier than cranking out the surface integral in rectangular, yuck.

  19. beginnersmind
    • one year ago
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    Yeah, scratch that. Probably easiest to use the fact that F is normal to the surface and equal on the whole surface. So I guess the solution is just |F| at the surface times the surface area of the sphere?

  20. IrishBoy123
    • one year ago
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    @beginnersmind that is just so neat! simplifies down all the other Gaussian ideas. i think i may have let my enthusiasm get the better of me - again!

  21. beginnersmind
    • one year ago
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    I mean |F| is equal on the whole surface. It has different directions (always perpendicular to the surface) but the same magnitude.

  22. anonymous
    • one year ago
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    bae is here XD

  23. Jhannybean
    • one year ago
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    *

  24. anonymous
    • one year ago
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    who is "bae"

  25. Jhannybean
    • one year ago
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    Ohh is this a conservative vector field?

  26. beginnersmind
    • one year ago
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    Well, it's like the electric field of a positive charge so it better be :)

  27. Jhannybean
    • one year ago
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    This might help? http://ltcconline.net/greenl/courses/202/vectorIntegration/vectorFields.htm

  28. dan815
    • one year ago
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    divergence theorem cleans it up

  29. dan815
    • one year ago
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    |dw:1441507539703:dw|

  30. dan815
    • one year ago
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    |dw:1441507619509:dw|

  31. dan815
    • one year ago
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    you can do surface integral if u must practice

  32. beginnersmind
    • one year ago
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    Why is div F = 3 though?

  33. dan815
    • one year ago
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    oh oops that r looks like an f to me

  34. beginnersmind
    • one year ago
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    @Astrophysics do you still need this, or have you figured it out?

  35. dan815
    • one year ago
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    hmm im not sure how well F dot N ds really simplifies it doesnt seem liek it would simplify that well but

  36. beginnersmind
    • one year ago
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    dan: F is m in the direction of the normal. So F dot dn is m. You integrate that over that over the whole surface, you get m times the surface area. So 4m*pi

  37. dan815
    • one year ago
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    ya i noticed it simplfied after i tried it earlier xD

  38. dan815
    • one year ago
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    i took F=m*r_, since mag of r is just 1 everywhere on surface of sphere we can cancel out the |r_|^3 unit vector n = r_/|mag r_| F dot n ds m*(x^2+y^2+z^2) ds since its on surface thats just m =integral m ds

  39. Astrophysics
    • one year ago
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    I got it, thanks everyone :)!

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