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Algebra /derivatives/ chain rule/

  1. Mar 23, 2015 #1
    1. The problem statement, all variables and given/known data

    ##J=r^{2}\dot{\phi}## [1]
    ##\dot{r^{2}}=E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r}##. [2]
    (the context is geodesic equation GR, but I'm pretty sure this is irrelevant).

    where ##u=r^{-1}##

    Question: From these two equations to derive ##(\frac{du}{d\phi})^{2}=\frac{1}{J^{2}}(E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r})##

    2. Relevant equations

    As above.

    3. The attempt at a solution

    I know that ##(\frac{du}{d\phi})^{2}=(\frac{\dot{u^{2}}}{\dot{\phi^{2}})}##.

    The ##J^{2}## in the denominator is throwing me, I see from [1] that ##\dot{\phi}=Jr^{-2}## so there's some relation between ##\dot{\phi}## and ##J##, but what about a ## r^{-4}## term with the ##J^{2}##, since the long term in brackets remains from [2].

    I'm guessing I'm missing something simple but I've spent a long time looking at it and no luck.

    Cheers.
     
    Last edited: Mar 23, 2015
  2. jcsd
  3. Mar 23, 2015 #2

    HallsofIvy

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    There is NO "u" in either of the two equations, so your result makes no sense.

     
    Last edited by a moderator: Mar 23, 2015
  4. Mar 23, 2015 #3
    cheers, edited.
     
  5. Mar 27, 2015 #4
    Bump. Anyone please, I'm still stuck?
     
  6. Mar 30, 2015 #5

    Fredrik

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    I don't understand this problem. ##r## and ##\phi## are coordinates, right? So r isn't an invertible function, and 1/r isn't a function of ##\phi##. So what do you mean by ##r^{-1}##? I think you will need to include a lot more information.
     
  7. Mar 30, 2015 #6
    Original post, first lst line under the attempt at the solution
    We have both ##\phi## and ##r## differentiated with respect to some paramater, denoted by a dot, so dividing these two has gave ## du / \ d \phi ##
     
  8. Mar 30, 2015 #7

    Fredrik

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    All you're saying there, is that if ##u## is a function of ##\phi##, and ##\phi## is a function of ##t##, then
    $$\frac{du}{dt}=\frac{du}{d\varphi}\frac{d\varphi}{dt}.$$ This is just the chain rule. But you haven't posted a meaningful definition of ##u##. You said that ##u=r^{-1}##, but the map that takes ##t## to ##r## isn't invertible so did you really mean the inverse of that non-invertible function? If you meant ##1/r##, then ##r## has to be a function of ##\phi## (if it's not, then ##u## isn't, and the chain rule can't be applied as written).

    You need to post the full problem statement.
     
    Last edited: Mar 31, 2015
  9. Apr 1, 2015 #8
    I'm looking at Tod and Hughston introductuction to GR and it says introducing a new variable ##u=r^{-1}## from [1] and [2] we can solve for ##u## as a function of ##\phi##.

    This is all the book says
     
  10. Apr 1, 2015 #9

    Mark44

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    I'm pretty sure that u = r-1 means u = 1/r in this context.

    Let u = 1/r, so r = 1/u
    From this we have dr/dt = -1/u2 ( du/dt)

    Dispensing with the dot notation, equation [2] is ##(\frac{dr}{dt})^2=E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r}##
    Hence with the u substitution, we have ##(\frac{-1}{u^2})^2(\frac{du}{dt})^2=E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r}##
    Multiply both sides by u4 to get
    ##(\frac{du}{dt})^2=u^4(E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r})##
    Multiply both sides by ##(\frac{dt}{d\phi})^2## and the desired result can be quickly obtained.

    From the OP, I am inferring that both ##\phi## and r are differentiable functions of t, otherwise ##\dot{\phi}## and ##\dot{r}## wouldn't make sense.
     
  11. Apr 2, 2015 #10

    Fredrik

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    OK, if u=1/r, and there's a constraint that ensures that if you know the value of one of the variables ##r##, ##\phi## and ##t##, you can calculate the values of the other two, then the proof is a straightforward calculation. The word "geodesic" was mentioned, so I guess these variables must be sufficiently constrained by the assumption that the curve we're dealing with is a geodesic. I haven't tried to prove that, but it sounds reasonable when you think about straight lines in ##\mathbb R^2## for example.

    I would do this calculation by proving that ##\frac{du}{dt}## is equal to both ##-u^2\dot r## and ##\frac{du}{d\phi}Ju^2##. This implies that ##\big(\frac{du}{d\phi}\big)^2=\frac{\dot r^2}{J^2}##.
     
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