## Astrophysics one year ago Flux

1. Astrophysics

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

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

@Empty @ganeshie8

4. Astrophysics

So I guess $x^2+y^2+z^2=1$ then we can use a parametric representation?

5. Astrophysics

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

|dw:1441476251414:dw|

7. IrishBoy123

use a Gaussian surface as it's totally symmetrical or stuff it into spherical if you want a slightly harder time

8. Astrophysics

Hmm interesting, never really worked with gaussian surfaces, I will look into it, thanks @IrishBoy123

9. hwyl

that's what you do anyway start from cube then use the same for curved

10. hwyl

|dw:1441476650300:dw|

11. IrishBoy123

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

12. Astrophysics

Thanks @hwyl Awesome, this is one reason I want to take E&M haha, thank you very much @IrishBoy123

13. hwyl

|dw:1441476754530:dw|

14. hwyl

apply gauss law

15. hwyl

16. beginnersmind

You can use the divergence theorem:|dw:1441477056233:dw| div F is particularly simple in this example

17. Astrophysics

Thanks everyone, nice @beginnersmind I will try your approach as well! :)

18. IrishBoy123

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

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

@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

I mean |F| is equal on the whole surface. It has different directions (always perpendicular to the surface) but the same magnitude.

22. anonymous

bae is here XD

23. Jhannybean

*

24. anonymous

who is "bae"

25. Jhannybean

Ohh is this a conservative vector field?

26. beginnersmind

Well, it's like the electric field of a positive charge so it better be :)

27. Jhannybean
28. dan815

divergence theorem cleans it up

29. dan815

|dw:1441507539703:dw|

30. dan815

|dw:1441507619509:dw|

31. dan815

you can do surface integral if u must practice

32. beginnersmind

Why is div F = 3 though?

33. dan815

oh oops that r looks like an f to me

34. beginnersmind

@Astrophysics do you still need this, or have you figured it out?

35. dan815

hmm im not sure how well F dot N ds really simplifies it doesnt seem liek it would simplify that well but

36. beginnersmind

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

ya i noticed it simplfied after i tried it earlier xD

38. dan815

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

I got it, thanks everyone :)!