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Classical mechanics: Central potential, trajectory

  1. Jun 19, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Determine the possible trajectories of a particle into the following central potential: [itex]U(r)=U_0[/itex] for [itex]r< r_0[/itex] and [itex]U(r)=0[/itex] for [itex]r>r_0[/itex].


    2. Relevant equations
    Not sure. What I used: Lagrangian+Euler/Lagrange equations.


    3. The attempt at a solution
    I used polar coordinates but I'm not sure it's well justified. Since we're dealing with a central potential problem, isn't the motion restricted to a 2 dimensional plane? If so, then I think it's safe to use polar coordinates.
    Ok so my Lagrangian is [itex]L=T-V=\frac{m(\dot \theta r^2 + \dot r ^2 )}{2}-U(r)[/itex]. I notice that [itex]\theta[/itex] is cyclic, thus the angular momentum is conserved.
    Using Lagrange's equations for [itex]\theta[/itex], I reached the differential equation [itex]\dot r r = cte[/itex]. For [itex]r[/itex], I reach [itex]\ddot r - \dot \theta r =0[/itex].
    I know I made an error, I reach the same motion equation regardless what U(r) thus r is.
    Hmm... Since [itex]\theta[/itex] is cyclic, does this imply that [itex]\dot \theta =0[/itex]?
    If so, then [itex]\ddot r =0[/itex]. But this is senseless since it would imply that [itex]\dot r[/itex] is also a constant, and thus that r is also a constant from my first use of Lagrange equations...
    It might be a little too late to do physics for me now, but I'm willing to put the effort.
    Any help will be appreciated. Thanks a lot!

    Edit: I notice an error for the 1st Euler/Lagrange equation conclusion. I do reach that r is constant instead of [itex]r \dot r[/itex], which still doesn't make sense to me.
     
    Last edited: Jun 19, 2011
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  3. Jun 20, 2011 #2
    There is a silly mistake at T (kinetic energy). Check that :smile:
    (by the way, [itex]d\theta/dt=0[/itex] is meaningless. That would mean the particle's motion, including the initial state, is restricted to radial direction)
     
  4. Jun 20, 2011 #3

    fluidistic

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    Thanks for pointing this out. Indeed.
    T is in fact [itex]\frac{m(\dot \theta ^2 r^2 + \dot r ^2 )}{2}[/itex].
    The equations of motion I reach are now [itex]2r \dot r \dot \theta + \ddot \theta r^2=0[/itex] and [itex]\ddot r - \dot \theta ^2 r =0[/itex].
    I also know that [itex]r^2 \theta = constant[/itex].
    I don't know if I can divide any of these equations by r. As far as I know, r could be worth 0 and thus I can't divide by r.
    I don't know how to proceed in order to reach the trajectories; or in other words, how to solve the motion equations.
    Also I notice that there's no difference at all for r>r_0 or r<r_0 in my equations. It seems wrong to me.
     
  5. Jun 20, 2011 #4

    George Jones

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    The [itex]\partial L / \partial r[/itex] term involves [itex]\partial U / \partial r = d U / dr[/itex]. What is [itex]d U / dr?[/itex] It isn't zero everywhere.
     
  6. Jun 20, 2011 #5

    fluidistic

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    Ah yes you're right. It's worth 0 everywhere except in [itex]r=r_0[/itex] where it's worth [itex]\pm \infty[/itex] depending upon if [itex]U_0[/itex] is greater or lesser than 0.
    So U(r) isn't analytic. How do I deal with this problem?!
     
  7. Jun 20, 2011 #6

    George Jones

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    Are you familiar with the Dirac delta "function"?
     
  8. Jun 20, 2011 #7

    fluidistic

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    A bit. I've seen some relations and the formal definition of it (if I remember well it was the limit of a sequence) but I'm not so comfortable with it.
    Out my memory, I remember that [itex]\int g(x) \delta (x-a) dx=g(a)[/itex] where the integral bounds cover of course an interval containing x=a.
    I didn't check in wikipedia so my memory might fail. Also a friend of mine once told me it was a double abuse of notation to write the integral I just wrote, but I didn't really understood why.
    So in my example, [itex]a=r_0[/itex]. And [itex]g(x)[/itex] would be [itex]U(r)[/itex]. So basically... err no that can't be... Hmm.
     
  9. Jun 20, 2011 #8

    George Jones

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    Have you seen the connection between the Heavisde step function and the Dirac delta function?

    With experience, it is actually possible to see the answer to the original problem without doing any calculations (somewhat similar to billiard ball collision problems). We should go through the calculations, though.
     
  10. Jun 20, 2011 #9

    fluidistic

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    About the Heaviside function: Heard of it before but never checked it in details. I just did in wikipedia and its connection to the Dirac Delta function seems to be that [itex]H(x)=\int _{-\infty} ^x \delta (t)dt[/itex]. So in my case it'd be worth 0 for [itex]x<r_0[/itex] and 1 for [itex]x>r_0[/itex]. But I don't see how this can help. I've checked in Golstein's book (1st edition) on Classical dynamics and to my surprise there's a short comment on Dirac's Delta and nothing on Heaviside function.
    So the problem seems simple but it requires the use of Dirac's delta due to the singularity of the potential function. I'll help for further help here as I'm in unknown territory (I'll still try to figure out things if I can). And thank you so far.
     
  11. Jun 21, 2011 #10

    George Jones

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    Can you invert this?
    Consequently, [itex]U \left( r \right) = U_0 H \left( r_0 - r \right)[/itex]. What does this give for [itex]dU / dr[/itex]?

    Look up "impulsive force" in Goldstein or your text or ...
     
  12. Jun 21, 2011 #11

    fluidistic

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    Do you mean that I find the inverse of the Heaviside function?


    The first edition of Golstein as only 1 exercise about impulsive forces. It seems rather tough to me and doesn't seem to involve any special function.
    Ok back to your expression. Now I understand that [itex]U(r)=U_0 H(r_0-r)[/itex]. :smile:
    I also understand I must find [itex]\frac{d U}{dr}[/itex]. If I'm not wrong, I think this is simply worth [itex]U_0 \delta (r_0 -r)[/itex].
    So as I had, [itex]\frac{d U}{dr}=0[/itex] for [itex]r\neq r_0[/itex] which makes sense.
    Now in [itex]r=r_0[/itex] it's worth [itex]\infty[/itex] as it should I guess.
    How would I continue? I'm not seeing where we're going.
     
  13. Jun 22, 2011 #12
    That expression [itex]U(r)=U_0H(r_0-r)[/itex] along with one ODE and the expression [itex]r^2\dot{\theta}=C=const[/itex] will give you an ODE of r. As you want to stay cautious with dividing by r, you may simply consider ODE at [itex]r \neq 0[/itex]. You will get something like this:
    [itex]f(\ddot{r},\dot{r},r)=\delta(r-r_0)[/itex]

    Here's the trick: observe that the change only happens at [itex]r=r_0[/itex] (this is what George meant by impulsive force: an immediate change without change in position). So deal with the ODE within the interval: [itex](r_0^-;r_0^+)[/itex]. Notice that: [itex]\int^{0^+}_{0^-}\delta(x)dx=1[/itex]
     
  14. Jun 22, 2011 #13

    fluidistic

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    Thanks guys for all the help so far.
    To hikaru, could you be a little more specific about the ODE's I should get? I've redone all the algebra. I do reach [itex]\dot \theta r^2 = constant[/itex]. I do reach [itex]\ddot \theta r ^2 + 2 \dot \theta \dot r r =0[/itex]. Is this the first ODE you talk about?
    I don't see how I can "mix them" along with [itex]U(r)=U_0H(r_0-r)[/itex] in order to get a single ODE depending on r.
    And ok for your trick, I'll use it.

    Edit: If [itex]\frac{dU}{dr}=U_0 \delta (r-r_0 )[/itex] then I reach the ODE [itex]\dot \theta ^2 r + U_0 \delta (r-r_0)-\ddot r =0[/itex]. I have no idea if this is right.
     
  15. Jun 22, 2011 #14

    George Jones

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    Good! Are there some missing m's?

    Now use your other equation to eliminate [itex]\dot \theta[/itex].
     
  16. Jun 22, 2011 #15

    fluidistic

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    Thanks for helping me. :)
    Yes I miss one m (my draft hasn't this error so I suppose it's a typo error). Edit: OH actually 2 I see.
    This should be [itex]m \dot \theta ^2 r + U_0 \delta (r-r_0)-m \ddot r =0[/itex](*).
    Now I'm not sure what equation to use. I assume you mean [itex]\dot \theta r^2=K_1[/itex] but I don't see how I can use it in order to give rid of [itex]\dot \theta[/itex] in (*).

    Edit2: My bad nevermind, I think I know how!
     
  17. Jun 22, 2011 #16

    fluidistic

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    I get [itex]\frac{mK_1 ^2}{r^3}+U_0 \delta (r-r_0)-m \ddot r =0[/itex]. This is true for [itex]r\neq 0[/itex].
    I need to solve this equation for r... Ok I'll think about how to do but I'm not that confident.
    Edit: Ok it doesn't look that bad. I still have a doubt about the limits of some integrals (funnily I've helped someone on this forum yesterday night about the limits of an integral...); let me show you.
    (*) is equivalent to [itex]\ddot r = \frac{K_1 ^2}{r^3} + \frac{U_0 \delta (r-r_0)}{m}[/itex].
    I integrate with respect to time in order to get [itex]\dot r[/itex] (with the idea of doing it again to get r).
    So I have [itex]\dot r=K_1 ^2 \int \frac{dt}{r^3} + \frac{U_0}{m} \int \delta (r-r_0)dt[/itex]. My guess is that the limits of the integrals are [itex]-\infty[/itex] and [itex]+\infty[/itex].
    I'd like a confirmation if I'm on the right track and the limits are correct.
     
    Last edited: Jun 22, 2011
  18. Jun 22, 2011 #17

    George Jones

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    Maybe I shouldn't give this hint, but .... reduction of order.
     
  19. Jun 22, 2011 #18

    fluidistic

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    Oops I didn't see this message. I edited my above post and was looking for a different approach. Could this work too?

    Edit: If reduction of order is the only "easy" way to go then I'm already lost on what is the characteristic polynomial.
     
    Last edited: Jun 22, 2011
  20. Jun 23, 2011 #19

    George Jones

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    This equation is correct, but since [itex]r \left( t \right)[/itex] is to be determined, we don't know how to do the integrals.
    See Type 2 (example almost halfway down the page) from

    http://www.cliffsnotes.com/study_guide/Reduction-of-Order.topicArticleId-19736,articleId-19724.html [Broken].
     
    Last edited by a moderator: May 5, 2017
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