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The Most Beautiful Equation?

  1. Nov 19, 2008 #1
    Lets elect World's most beautiful equation!

    Two categories:

    1.Mathematics
    2.Physics

    My personal favourites would be:

    1. e[tex]^{i\pi}[/tex]+1=0 (Do i need to give an argument?)

    2. E=mc[tex]^{2}[/tex] (I know its mainstream.. but i doesent get much more simple and general than this!)
     
    Last edited: Nov 19, 2008
  2. jcsd
  3. Nov 19, 2008 #2
    [tex]
    i\hbar\frac{\partial\Psi}{\partial t} = \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi
    [/tex]
     
  4. Nov 20, 2008 #3

    vanesch

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    I'd say: 1 = 0. From this one, you can derive everything :biggrin:
     
  5. Nov 20, 2008 #4
    Maxwell's equations of EM?

    [tex]\nabla \cdot D= \rho [/tex]
    [tex]\nabla \cdot B=0[/tex]
    [tex]\nabla \times E=- \partial B/ \partial t[/tex]
    [tex]\nabla \times H=J+ \partial D/ \partial t[/tex]
     
  6. Nov 20, 2008 #5
    [tex] p = \frac{h}{\lambda} [/tex]
     
  7. Nov 21, 2008 #6

    robphy

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    I think you are missing a minus sign.
     
  8. Nov 21, 2008 #7
    1/. cos²(x) + sin²(x) = 1
    2/. ω² = k/m
     
  9. Nov 21, 2008 #8

    Borek

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    Isn't it only an approximation? :wink:
     
  10. Nov 21, 2008 #9
    FD=½ρv2ACD

    Some engineering fudgeamatics
     
    Last edited: Nov 21, 2008
  11. Nov 21, 2008 #10
    For me, it's:
    i2=-1
     
  12. Nov 21, 2008 #11
  13. Nov 21, 2008 #12

    Borek

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    As far as I remember mc2 is only a first term of the power series. Next terms are smaller by at least c2 factor (or even c4, my memory fails me here), so they can be safely ignored, but E=mc2 is still only an approximation.
     
  14. Nov 21, 2008 #13

    I think you refer this equation:

    E2=m2c4+p2c2
     
  15. Nov 21, 2008 #14

    Borek

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  16. Nov 21, 2008 #15

    robphy

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    You may referring to the Taylor expansion (with respect to the velocity) for the relativistic energy:
    http://en.wikipedia.org/wiki/Kinetic_energy
    expressed as
    [tex]E_{rel}=m_{rel}c^2=m_0c^2 \frac{1}{\sqrt{1-(v/c)^2}}=m_0c^2\bigg(1+\frac{1}{2}(v/c)^2+\frac{3}{8}(v/c)^4+\ldots\bigg) \approx m_0c^2 \bigg( 1+\frac{1}{2}(v/c)^2 \bigg)\mbox{[for small (v/c)]}[/tex]

    The rest energy [tex]E_0=m_0c^2[/tex] is a Lorentz invariant, and [tex]E_{rel}[/tex] and [tex]m_{rel}[/tex] are observer-dependent quantities.

    From a special-relativistic viewpoint, these are exact relations.

    From a Newtonian-physics viewpoint, one often refers to some of these terms as "relativistic corrections".
     
  17. Nov 21, 2008 #16
    E= ir !
     
  18. Nov 22, 2008 #17
    My bad

    [tex] i\hbar\frac{\partial\Psi}{\partial t} = - \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi [/tex]
     
  19. Nov 22, 2008 #18
    Well, it may be hard to say. I am a junior in high school, so my physics knowledge is limited, however, I am fond of

    i^2 = -1

    Imaginary x Imaginary = Real. Seems silly but awesome.

    Hayley
     
  20. Nov 22, 2008 #19
    Maxwell`s velocity distribution formula

    f(v) = 4 [tex]\pi[/tex] [ m / 2 [tex]\pi [/tex] k T ]^(3/2) v^2 e^(-m v^2 / 2 k T )

    The elegance because he derived this formula just by logical reasoning
    almost without calculus or any kind of information. Intellectual wizardy.
     
    Last edited: Nov 22, 2008
  21. Nov 22, 2008 #20
    zeta(s)=1+(2^-s)+(3^-s)+(4^-s)+...

    - the zeta function from which the Riemann hypothesis derives.
     
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