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I Inversion of functions that aren't 1-1

  1. Nov 16, 2016 #1
    If ##t## is a function of ##r##, then we may in theory find ##r## as a function of ##t##, as claimed in the last paragraph of the attachment below. My issue is this is only true if ##t## is a 1-1 function of ##r##. Otherwise, suppose ##t=r^2##. Then ##r=\pm\sqrt{t}##, which isn't a function.

    I understand that physically it makes no sense for ##t## not to be a 1-1 function of ##r##, because we would then have two different values of ##r## at a particular time and an object can't be at two places at once. But mathematically, I don't see how (6.15) shows ##t## is a 1-1 function of ##r##.

    Screen Shot 2016-11-17 at 4.00.50 AM.png
     
  2. jcsd
  3. Nov 16, 2016 #2

    FactChecker

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    When you are dealing with multiple-valued functions, you can define different "branches" of the function in which it is 1-1. Then you can select one of the branches, called the "principle branch", and use it for a 1-1 function. In your case, you might start with the branch that makes r positive. Typically an (r,θ) designation has r>0. As you integrate r, it may smoothly become negative. That is a problem where you will have to decide whether you should keep r positive or not. If the integral goes negative, you might want to keep r positive and add 180° to θ. Those discontinuities of θ can cause problems, depending on how you are using it.
     
  4. Nov 16, 2016 #3

    mfb

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    This is a classical physics problem - you can be sure the object won't be at two places at the same time (otherwise your solution is wrong), going from t(r) to r(t) is no problem.
     
  5. Nov 17, 2016 #4
    How about going from ##\theta(r)## to ##r(\theta)##? ##\theta(r)## may not be a 1-1 function since it is possible for an orbit to have 2 values of ##r## with the same ##\theta##.
    Screen Shot 2016-11-17 at 4.31.06 PM.png
     
  6. Nov 17, 2016 #5

    mfb

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    In nonrelativistic mechanics, there is no orbit with two values of r for the same value of θ. If you would have that, you could limit the inversion to a single orbit. The problem doesn't occur however, because there are two values of θ for the same r - you have to limit the analysis to half an orbit in an earlier step already.
     
  7. Nov 17, 2016 #6
    Below is an orbit with two values of ##r## for the same value of ##\theta##. Look at the dotted line just left of the arrow ##r_1##. The line cuts the orbit twice or more times, indicating there are two or more values of ##r##.
    Screen Shot 2016-11-18 at 1.24.25 AM.png
    How do you "limit the inversion to a single orbit"? What's the single orbit in this case?
     
    Last edited: Nov 17, 2016
  8. Nov 17, 2016 #7

    mfb

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    The single orbit ends at the point where the line intersects itself: θ changed by 2 pi along one revolution.
    If you allow θ values larger than 2 pi, you can fix this issue - the two points then differ by 2 pi in θ.

    As this particular motion seems to have a shorter period in r-oscillations, you can study the trajectory for θ from 0 to 2pi/3 only if you want.
     
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