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Transformation of variables in definite integral limits

  1. Aug 16, 2012 #1
    Hello,

    I am going through Whittaker's treatise on Classical Mechanics. In chapter 3 he derives the equation of motion for a simple pendulum, and I have a question about his method.

    Starting from the general form for the equation of energy (s is the path):

    [itex]\frac{m}{2}\dot{s}^2 = \int\limits_{s_0}^{s} f(s)\, ds + C[/itex]

    and using

    [itex]s = a\cdot θ[/itex]

    and

    [itex]f(s)=-m\cdot g\cdot sin(θ)[/itex]


    The equation of energy is transformed to:

    [itex]\frac{m}{2}{(a\cdot \dot{θ})}^2 = \int\limits_{θ_0}^{θ} -m\cdot g\cdot sin(θ)\,\cdot a\cdot dθ + C[/itex]

    But if I remember my Calculus correctly, the transformation should be:


    [itex]\frac{m}{2}{(a\cdot \dot{θ})}^2 = \int\limits_{a\cdotθ_0}^{a\cdotθ} -m \cdot g\cdot sin(θ)\,\cdot a\cdot dθ + C[/itex]

    Notice the limits of integration. Where am I going wrong here? This is on page 72 here: http://archive.org/stream/treatisanalytdyn00whitrich#page/n87/mode/2up/search/pendulum

    Thanks for your help.
     
    Last edited: Aug 16, 2012
  2. jcsd
  3. Aug 16, 2012 #2
    No, it's written correctly. You don't keep the limits equal; you transform them according to the equation of the transformation. In this case, [itex]s = a\theta[/itex], so you have to find the new limits by [itex]\theta = s/a[/itex].

    Think about if you were going from lengths to angles. If you kept the limits the same, you'd be integrating with respect to an angle with limits that are lengths. That's nonsensical.
     
  4. Aug 16, 2012 #3
    Hello Muphrid,

    It is precisely because I believe you have to "transform the limits by the equation of transformation" that I am stuck!

    [itex]s = a\cdot θ[/itex] - The equation of transformation.

    Perhaps you mean something like this, first taking the upper limit, given by

    [itex]s[/itex]

    Now if we use our equation of transformation we have

    [itex]s=a\cdot θ[/itex]

    So plugging in for [itex]s=s[/itex] and solving for θ:

    [itex]θ=\frac{s}{a}[/itex]

    but now we use our equation of transformation again to get:

    [itex]θ=\frac{a\cdot θ}{a} = θ[/itex]

    Now the lower limit:

    [itex]s_0[/itex]

    Plugging in our equation of transformation and solving for θ

    [itex]θ=\frac{s_0}{a}[/itex]

    which is just a constant we can call [itex]θ_0[/itex].

    Is it that simple? How the world did I miss using the equation of transformation twice, LOL?

    Thanks.
     
    Last edited: Aug 16, 2012
  5. Aug 16, 2012 #4
    It's not that you're using the transformation twice so much as applying it in slightly different ways. With [itex]ds[/itex] you apply it "forward", saying [itex]s = a \theta[/itex] so [itex]ds = a d\theta[/itex]. With the limits, you apply it "backward," saying [itex]s_0 = a \theta_0[/itex] so [itex]\theta_0 = s_0/a[/itex].

    The basic idea here is that though you may change the coordinates (see, the limits do change), the differential's size shouldn't change (you still have [itex]ds[/itex], just expressed in another way).
     
  6. Aug 16, 2012 #5
    I don't see how you get around using the transformation twice, at least on the upper limit. Just solving for θ (the first use of the transformation) would give:

    [itex]θ=\frac{s}{a}[/itex]. This is the "backward" use of the transformation, as you call it.

    But we don't want this in our new limit. So we remember that [itex]s=a\cdot θ[/itex] and plug this in (the second use of the transformation) to get:

    [itex]θ=θ[/itex]. Now that is what we want as our new upper limit.

    Basically, when we change limits we are asking, "What is θ when [itex]s=s[/itex]?" Well, duh, if [itex]s=s[/itex] then [itex]θ=θ[/itex] because:

    [itex]s=s[/itex] → [itex]a\cdot θ=a\cdot θ[/itex] → [itex]θ=θ[/itex]

    (Still using the transformation twice, once on each side!)

    At least that is what makes sense to me. Thanks again.
     
    Last edited: Aug 16, 2012
  7. Aug 16, 2012 #6
    Okay, there are a couple things going on here because you slightly abused some notation. Let's go back to your original integral.

    [tex]\frac{1}{2} m \dot s^2 = \int_{s_0}^s f(s) \; ds[/tex]

    This is an abuse of notation because you have [itex]s[/itex] as one of the integration limits and as the dummy variable of integration. Let's write it a bit more clearly.

    [tex]\frac{1}{2} m \dot s^2 = \int_{s_0}^s f(w) \; dw[/tex]

    Now, this means there are two changes of variables going on. First, taking [itex]s = a \theta[/itex]:

    [tex]\frac{1}{2} m a^2 \dot \theta^2 = \int_{s_0}^{a \theta} f(w) \; dw[/tex]

    In the limits, you do a substitution to get from [itex]s[/itex] to [itex]a \theta[/itex]. This has nothing to do with the change of variable of the integral, since we haven't even done that yet.

    Now do the change of variables in the integral, taking [itex]w = a \phi[/itex]:

    [tex]\frac{1}{2} m a^2 \dot \theta^2 = \int_{s_0/a}^{\theta} f(w(\phi)) \frac{dw}{d\phi} \; d\phi[/tex]

    Here, you do a substitution on the differential and divide both limits of the integral by [itex]a[/itex].
     
  8. Aug 16, 2012 #7
    LOL, I am using (The Great) Whittaker's notation! You can see for yourself in the book I linked to above, so don't complain to me! ;-)

    That aside, what you said makes sense, I appreciate it.
     
  9. Aug 16, 2012 #8
    Yeah, I mean, as abuses of notation go, what you (or Whittaker) did is not a real crime (I've probably done it myself now and then); i just felt that it wasn't clear that there are two separate transformations taking place without pointing it out.

    What's important is that you've got a grasp of it, and that's good.
     
  10. Aug 16, 2012 #9
    Thanks.

    I have another question about this problem. In thinking about the pendulum, why don't we count the force exerted on the particle by the beam? Whittaker says the only force acting on the particle is the force of gravity, but this is strictly false. The beam is constraining the particle to move in a circular arc, and thus exerting a force on the particle. I have seen this in other places; we don't count the constraining force, why?

    Is it because the constraining force isn't causing the particle to move? Yet it is causing it to accelerate, right? If it were not, then gravity would take the particle straight down.

    Thanks.
     
  11. Aug 16, 2012 #10
    You've already imposed the constraint by parameterizing the motion by only an angle. To find the force used to keep the constraint from being violated would require Lagrange multipliers.
     
  12. Aug 16, 2012 #11
    So if I may rephrase, you are saying that the effects of the beam's constraining force on the particle's path of motion have already been taken into account by the fact that we have constrained that motion to the path which would result from such a force. Now when we look at the effects of the force of gravity on the particle's motion, we are looking at how gravity affects that motion along that path.

    Is that correct? That makes a bunch of sense to me!

    So forces can both constrain the path of particle motion and cause the particle to move along that path. As long as we take all forces into account, either in describing the shape of the path or the particle's motion along the path, we are fine and all is accounted for, correct?
     
  13. Aug 16, 2012 #12
    Exactly right. You can do the problem without explicitly imposing the constraint by, for example, creating a modified Lagrangian:

    [tex]L = \frac{1}{2} m (\dot x^2 + \dot y^2) - mgy + \lambda(\sqrt{x^2 + y^2} - a)[/tex]

    Taking the Euler-Lagrange equations for variables [itex]x, y, \lambda[/itex] generates the equations of motion with additional force terms describing the force exerted to maintain the constraint.

    When you're not interested in finding the constraint force, though, explicitly reducing the system's complexity to fewer coordinates (1 instead of 3, in this case) is almost always easier, but the power to find those constraint forces is there if you choose to pursue it.
     
  14. Aug 16, 2012 #13
    Thanks so much, Muphrid. A lot of this is familiar to me, as I have a degree in Physics but haven't used it in several years. Now I am planning on going back for my PhD and I figure I should put myself through a refresher. I am finding that some things that I just accepted on faith when getting my degree now don't make sense as I think about them more carefully (initially like the constraining force thing).

    Anyway, I appreciate the info. I'll probably be back with more questions as I go through this book and my old Marion Thornton 4th edition.

    Cheers!
     
    Last edited: Aug 16, 2012
  15. Aug 16, 2012 #14
    No problem at all. If the PhD program you enter is anything like mine, you might get pounded with coursework right off the bat, so brushing up is definitely a smart idea. I probably pulled as many all-nighters my first year as I did in all of undergrad. Best of luck!
     
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