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I Euler's Formula - Moving Particle Argument (Needham)

  1. Jul 29, 2016 #1
    I am reading Tristan Needham's book "Visual Complex Analysis" and am currently focussed on Chapter 1, Section II Euler's Formula ... in particular I am trying to follow Needham's heuristic argument in support of, or justifying, Euler's formula - Needham calls it 'the moving particle argument' ... ...

    In Chapter 1, Section II (2), Needham considers a particle moving along a curve:

    ##Z(t) = e^{it}##

    So he finds that:

    ## \frac{dZ}{dt} = ie^{it} = iZ ##

    Needham then argues that each new velocity, instant by instant, will be at right angles to its new position vector ( ??? ... presumably because the function ##iZ## causes a rotation of 90 degrees - is that right??? ... ) ...

    He then asserts that "it is clear" that the particle will travel around a unit circle ... ... BUT ... ... why exactly is this the case? ... ... indeed, how do we rigorously prove that this is the case? ... ...



    The relevant part of Needham's text is as follows:

    ?temp_hash=9849f1a93823a8981dbef052e780d165.png
    ?temp_hash=9849f1a93823a8981dbef052e780d165.png
    ?temp_hash=9849f1a93823a8981dbef052e780d165.png



    Now ... just a further question based on a claim in the last paragraph above ... ...

    Needham writes:

    " ... ... Thus after time ##t = \theta## the particle will have travelled a distance ##\theta## round the unit circle, and so the angle of ##Z( \theta ) = e^{i \theta }## will be ##\theta##. This is the geometric statement of Euler's formula ... ... "


    My question is ... ... in what way ... that is how/why is this the geometric statement of Euler's formula ... ...

    Help will be appreciated ...

    Peter
     

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    Last edited: Jul 29, 2016
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  3. Jul 29, 2016 #2

    BvU

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    1) If you believe the 'at right angles' you have no tangential acceleration, so constant ##|\vec v|##. Since ##\vec v =i \vec Z##, it must be that ##|\vec Z| ## is also a constant. Circular motion. ##|\vec Z(0)| = 1##, so unit circle.

    2) $${d\over dt}\vec Z^2 = 2\vec Z\cdot \vec v$$ right angles ##\Rightarrow## inner product is zero ##\Rightarrow\ \ \vec Z^2 ## is constant
     
  4. Jul 29, 2016 #3

    micromass

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    If you're looking for rigor, then Needham's book is not what you're looking for. Don't get me wrong, it's an excellent work. But it's far from rigorous and not always easy to supply the rigor yourself.
     
  5. Jul 29, 2016 #4

    Thanks BvU ... most helpful ...

    But ... I am still puzzled regarding how Needham's argument supports/proves Euler's formula ...

    Thanks again!

    Peter
     
  6. Jul 29, 2016 #5

    Thanks for a VERY helpful remark, Micromass ...

    I must, in the light of what you have said, reevaluate my study of Needham's book ...

    Peter
     
  7. Jul 30, 2016 #6

    BvU

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    I agree with ##\mu m## but sometimes it's good not to worry about the foundations too much to become familiar with a large structure. You can always return to the basement in a later stage ...
     
  8. Jul 30, 2016 #7

    BvU

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    So you have ##|\vec Z|## constant, 1 and ##\vec v## ... -- hey, I'm repeating Needham. $$|\vec Z|= 1 \quad \& \quad \angle \vec Z (= \arg {\bf Z}) = \theta \quad \Rightarrow \vec Z = \cos\theta\; \hat\imath + \sin \theta\; \hat\jmath \Rightarrow {\bf Z} = \cos \theta + {\bf i} \sin \theta $$
     
  9. Jul 30, 2016 #8
    Hi BvU ... Thanks for both your new posts ...

    VERY thoughtful point ... getting an idea of the big picture and a basic understanding of how things work in general is so important...

    You have caused me to think yet again ...

    Peter
     
  10. Jul 30, 2016 #9

    micromass

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    I think it is still worth doing Needham. Needham contains so much information and cool background that you won't find in any other book in complex analysis. Sure, the book is not rigorous at all. And I think many of the arguments in Needham can't even be made rigorous. But Needham is also an amazing read and gives you so much intuition. It's really worth it.
     
  11. Jul 30, 2016 #10

    lavinia

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    ##d/dθ(cos(θ) + isin(θ)) = -sin(θ) + icos(θ)## and this equals ##i(cos(θ) + isin(θ))## so it differs from ##e^{iθ}## by a constant. But they are both equal to ##1## for ##θ=0## so they are equal for all ##θ##. But ##cos(θ) + isin(θ)## describes uniform circular motion so geometrically Euler's formula says the ##e^{iθ}## describes uniform circular motion as well.
     
    Last edited: Jul 30, 2016
  12. Jul 30, 2016 #11
    Equality (10) follows directly from the Taylor expansions of the left and right sides. So why are all these wimp explanations needed?
     
  13. Jul 30, 2016 #12

    micromass

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    Because it gives a lot of neat intuition about the Euler formula. If you think intuition is not necessary, then Needham is not the book for you.
     
  14. Jul 31, 2016 #13

    lavinia

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    Perhaps it would be instructive(though very wimp :nb)) to work out the solution to the differential equation ##dy/dθ = iy##

    ##y## is complex so ##y = u + iv## where ##u## and ##v## are real valued functions of ##θ##

    ##dy/y = idθ## and ##dy/y## = ##(du + idv)/u + iv = ((udu + vdv) + i(udv - vdu))/(u^2+v^2)##

    The solution of ##∫dy/y = i∫dθ## is purely imaginary so udu + vdv = 0 or ##u^2 + v^2 = C## for some constant ##C##.
    For ##θ=0## ## u=1## and ##v=0## so ##C =1## and ##u^2 + v^2## = 1 .

    Since the real part of ##dy/y## is zero the equation becomes ##∫udv - vdu = ∫dθ##

    Since ##v = \sqrt {1-u^2} ## , ##udv = -u^2/ \sqrt {1-u^2}du## and ##vdu = (1-u^2)/ \sqrt{1-u^2} du## so the left hand integral is ##∫-/ \sqrt{1-u^2}du## which integrates to the inverse cosine.

    Given that ##u= cos(θ)## the differential equation gives ## v= sin(θ)##
     
    Last edited: Aug 4, 2016
  15. Aug 3, 2016 #14

    lavinia

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    Yes.
    ##iZ## is therefore perpendicular to ##Z##

    One way to think of it is that if the velocity vector has no component pointing away or towards the origin then the distance of the curve to the origin can not change. If a vector is perpendicular to a radial line segment it has no radial component.
     
    Last edited: Aug 4, 2016
  16. Aug 6, 2016 #15
    Thanks so much Lavinia ... appreciate your help ...

    Peter
     
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