1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Algebra/differentiation/trig expansions/ small approximation

  1. Mar 23, 2015 #1
    1. The problem statement, all variables and given/known data
    ##\frac{\partial u}{\partial \phi}=-\frac{Me}{J^{2}}sin\phi + \frac{M^{3}}{J^{4}}(-\frac{1}{e}sin\phi+e^{2}sin2\phi+3e\phi cos \phi )##.

    Assume this to be ##0## at ##\phi=\pi+\epsilon##.
    Find ##\epsilon##?

    2. Relevant equations
    The only method I can think of is to expand out ##cos(A+B), sin(A+B)## terms and then use that for ##x## small ##sin x = x## and ##cos x=1## (so neglected terms of order ##x^{2}##).

    3. The attempt at a solution
    Okay so doing the above I get ##0=-\frac{Me}{J^{2}}sin(\pi + \epsilon) + \frac{M^{3}}{J^{4}}(-\frac{1}{e}sin(\pi+\epsilon)+e^{2}sin(2\pi+2\epsilon)+3e(\pi+\epsilon) cos (\pi+\epsilon)##

    ##=-\frac{Me}{J^{2}}(sin(\pi)cos( \epsilon) + cos(\pi)sin(\epsilon))+ \frac{M^{3}}{J^{4}}(-\frac{1}{e}(sin(\pi)cos( \epsilon) + cos(\pi)sin(\epsilon))+e^{2}(sin(2\pi)cos( 2\epsilon) + cos(2\pi)sin(2\epsilon))+3e(\pi+\epsilon) (cos \pi cos\epsilon-sin\epsilon sin\pi))##.

    ##=\frac{Me}{J^{2}} -\epsilon+ \frac{M^{3}}{J^{4}}(\frac{1}{e}(\epsilon)+e^{2}(2\epsilon)+3e(\pi+\epsilon)) ##

    The answer is however ##\epsilon=\frac{3M^{2}}{J^{2}}\pi##.
    (source An introduction to GR, L.P Hughston and K.P Tod).

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

    wabbit

    User Avatar
    Gold Member

    Your solution is unduly complicated by using ##\sin(\pi+\epsilon)=\sin(\pi)\cos(\epsilon)+\cos(\pi)\sin(\epsilon)##. This simplifies to the identity ##\sin(\pi+\epsilon)=-\sin(\epsilon)##. What do you get when using that instead ?
     
  4. Mar 23, 2015 #3
    the same? (after the appoximation ##-\sin(\epsilon)=-\epsilon##)
    or should i not be using this approximation?
     
  5. Mar 23, 2015 #4

    wabbit

    User Avatar
    Gold Member

    Should be the same yes, I am just too lazy to check the long formulas:)

    One other thing, your first line has an unmatched patenthesis, it's not clear how it should be read.
     
  6. Mar 23, 2015 #5

    Mark44

    Staff: Mentor

    In fact, there are two unmatched parentheses.

     
  7. Mar 23, 2015 #6
    Apologies, ta, edited.
     
  8. Mar 23, 2015 #7

    wabbit

    User Avatar
    Gold Member

    You write
    ##0=\frac{Me}{J^{2}} -\epsilon+ \frac{M^{3}}{J^{4}}(\frac{1}{e}(\epsilon)+e^{2}(2\epsilon)+3e(\pi+\epsilon))##
    But this should be
    ##0=\frac{Me}{J^{2}} \epsilon+ \frac{M^{3}}{J^{4}}(\frac{1}{e}(\epsilon)+e^{2}(2\epsilon)+3e(\pi+\epsilon))##
    But after that you don't try to solve for ##\epsilon## so I don't know what your solution is.

    Also, what you give as the answer doesn't seem correct.
     
    Last edited: Mar 23, 2015
  9. Mar 23, 2015 #8
    Thats a multipication by ##-\epsilon## not an addition ! (due to the ##cos(\pi)##.

    Whilst I havent tried to solve for ##\epsilon##, it is clear that the error occured in an earlier stage as the expression is not linear in ##e## and so it's not going to cancel- the actual solution is indepedent of ##e##....
     
  10. Mar 23, 2015 #9

    wabbit

    User Avatar
    Gold Member

    Well I give up, you keep moving signs and parentheses around, and writing products as sums, I can't keep track.
     
  11. Mar 27, 2015 #10
  12. Mar 27, 2015 #11

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Have you tried inserting the given solution into the expression and seeing whether it works out as 0? If it does you have actually proved it, but comparison with your work might indicate where it went wrong.

    Perhaps first check the book to be sure what you have reported as problem really is what they are saying?

    Why would anyone change a variable φ to π + ε when that just changes sin φ into - sin ε etc. and doesn't look as if it can advance any argument? (You didn't need to use sine addition formulae!)

    This is very easy to solve quite directly in a small angle approximation if φ is close to 0 or to (maybe was meant?) to π/2, and otherwise i think impossible to do so as a useful formula. Does not look right offhand.

    If I understood right this is really an ordinary d.e. which is quite easy to solve?
     
    Last edited: Mar 28, 2015
  13. Apr 1, 2015 #12
    Im not looking for any old proof of it, I'm looking to show and explain it.


    There's also a none sin/cos term in the expression?
    ..But the context here is GR comparing Newtonian orbit zero solution which occurs at ##\pi## and it is justified that we expect the zero solution in GR to occur close to this so plus a small term - ##\epsilon##. Hence we plug in to solve for this small term.


    well i dont know how to solve it..which is why i posted it here.
    But I thought small approximations sounds like the right thin to do in this situation rather than solving directly.
    In fact,the book does a similar exercise a few pages on (for timelike geodesics instead of null) and so the equations differ very slightly, the mathematical method is the same, and by using small approximations in a similar way as discussed in the OP I could explain ##\epsilon##.
     
  14. Apr 1, 2015 #13

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    I have no chance of understanding anything from GR, but the math you present makes so little sense I expect only someone knowing GR could guess what the real problem was.
    If it's so difficult you'd better use the possibility you do have of making sure it's right before spending any more effort on it.

    If LHS is zero then the equation gives φ = (A sin φ + B sin 2φ)/cos φ
    and the RHS is unchanged when you change φ into π + φ, so it's a periodic function so if it's satisfied by π + ε so it is by ε as well as by nπ + ε, so there would have been no point in this change of variable.


    I see only two variables u and φ there which I think makes this an ode.
    On the RHS you have only one very slightly difficult term φ cos φ which = d(φ sin φ)/dφ - sin φ, so you have on RHS all functions you know the integrals of.

    I do not have the impression I have solved anything but that you have misdescribed or misunderststood the problemε,.
     
    Last edited: Apr 1, 2015
  15. Apr 2, 2015 #14

    wabbit

    User Avatar
    Gold Member

    As I understand the problem statement (it seems clear to me, not sure what I miss), there is no ODE here, just an equation of the form A×epsilon+B=0, and the solution is epsilon =-B/A. That is, after the correct approximations sin(pi+epsilon)=-epsilon etc. have been made, and if the equation is derived correctly without replacing at some stage a multiplication C×(-epsilon) by a difference (C-epsilon), which was OP's error.
     
    Last edited: Apr 2, 2015
  16. Apr 2, 2015 #15

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    It is easy enough to do a small angle calculation of the given equation with LHS = 0. It gives φ = J/M X something, quite unlike the answer given, and π never comes into it, unless in the ± trivial way I mentioned before.
    Maybe some brackets have been missed or misplaced?

    The d.e. Is incidental to answering the question asked, but I said this looks like an ode and the OP did not deny it, but then it is so easily solved you wonder why they would do approximations, another reason to suspect misquotation.

    This is enough exegesis to try and find out what the real problem was and question should have been (I am often doing it) and the OP should reproduce the relevant passages from the book to get any help.
     
  17. Apr 2, 2015 #16

    wabbit

    User Avatar
    Gold Member

    Nothing to add to that:) that was pretty much my sentiment in post#9.
     
  18. Apr 4, 2015 #17
    In addition to post 12 on the context of the question, here is the book, pretty much word for word:

    ##\frac{\partial u}{\partial \phi}=-\frac{Me}{J^{2}}sin\phi+\frac{M^{3}}{J^{4}}(\frac{-1}{e}sin \phi + e^{2}sin 2\phi+3e\phi cos \phi)## [1]

    This is zero at ##\phi=0##, and then the next zero is not at ##\phi=\pi##, as it is Newtonian theory but at say ##\phi=\pi+\epsilon##. Substituting into [1] we find ##\epsilon=\frac{3M^{2}}{J^{2}}\pi## **

    My working:

    Using:

    ##cos(\pi+\epsilon)=-cos(\epsilon)##
    ##sin(\pi+\epsilon)=-sin(\epsilon)##
    ##sin(\epsilon) = \epsilon##
    ##cos(\epsilon)=1 ##

    The last two approximations for some small quantity.

    ##0= \frac{Me}{J^{2}}sin\epsilon-\frac{M^{3}}{J^{4}}(\frac{-1}{e}sin \epsilon + e^{2}sin 2\epsilon-3e( \pi+ \epsilon) cos \epsilon))
    =\frac{Me}{J^{2}}\epsilon-\frac{M^{3}}{J^{4}}(\frac{-1}{e}\epsilon + e^{2} 2\epsilon-3e( \pi+ \epsilon) ))##

    Which is not linear in ##e##, I can't see any more approximations to use, and so don't get answer ** as the book.
     
  19. Apr 4, 2015 #18

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    You have an e written small and written large. I have been assuming they are the same thing, are they?

    A way it would begin to make sense would be if the second and third terms can be assumed negligible and then you do get the stated result, among others.
    Difficult for (-1/e + 2e2) to be negligible (<< 3πe) it seems to me.
     
    Last edited: Apr 5, 2015
  20. Apr 4, 2015 #19

    wabbit

    User Avatar
    Gold Member

    @binbagsss, you say the equation you arrive at is not linear in ## e ## - yes, but that doesn't matter : your problem statement says "Find ## \epsilon ## ",so you are looking to solve for ## \epsilon ## ,not ## e ## . so what is the equation for ## \epsilon ##? Can you regroup the terms that have the same power of ## \epsilon ##? What do you get ?

    You really need to think a bit more carefully, read your problem statement, derive your solution with care, and you will find that this is far easier than you expect. And if you do that and find an answer that is different from the book, check all your steps, make sure you didn't make a mistake, redo the calculation if you find a mistake, and then tell us what you found. We can tell you then if there is a mistake in the book or if you missed something. Show us your result, don't just say "I get something different."
     
    Last edited: Apr 4, 2015
  21. Apr 5, 2015 #20

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Mods, maybe this should be transferred to the S&G Relativity or Advanced Physics forums where there is surely someone familiar with what these authors are telling or similar things and what the point is?

    Student who has made some effort should not be left stuck on one formula (and I wouldn't mind seeing the explanation, and is it true this is just a solvable ode?).
     
    Last edited: Apr 5, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Algebra/differentiation/trig expansions/ small approximation
  1. Differentiating trig (Replies: 6)

  2. Trig Differentiation (Replies: 5)

Loading...