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First Integral Problem

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex]r^2 \dot{θ}[/itex] is a first integral in any central field.


    2. Relevant equations
    for a first integral it suffices that [itex]L_v f = 0[/itex]
    central field:[itex]\ddot{\vec{r}}=f(r)\vec{r}[/itex] where f(r) is an arbitrary function of r.

    3. The attempt at a solution
    Physically it seems quite obvious. θ is implicit in the Newtonian equation, thus the Lagrangian is invariant under the transformation group of rotation, hence the conservation law.

    But for a more mathematical solution, directional direvative [itex]L_v[/itex] of [itex]r^2 \dot{θ}[/itex] only gives [itex]2r\dot{r}\dot{θ}, r^2\ddot{θ}[/itex] which alone does not guarantee its value to be zero. And how should I build connections between the given first integral and the Newtonian equation? Thanks guys.
     
    Last edited: Jan 21, 2013
  2. jcsd
  3. Jan 21, 2013 #2
    Vector-multiply the Newtonian equation with the r vector.
     
  4. Jan 22, 2013 #3
    Thanks for the reply but I am sorry that I cannot figure out your hint and in the end I get around using another method: since the acceleration is only radial, the angular component of the acceleration must be zero. Making use of this expression and substitute in the time derivative of the given first integral which then equals to zero. Although it sounds too "physical" for a mathematical book..
     
  5. Jan 22, 2013 #4
    You have ## \displaystyle \ddot{\vec{r}} = f(r)\vec{r} ##. Multiplying, we get ## \displaystyle \vec{r} \times \ddot{\vec{r}} = \vec{r} \times f(r)\vec{r} = f(r) \vec{r} \times \vec{r}##. What can be said about the right hand side?
     
  6. Jan 22, 2013 #5
    I get it would be zero, but this seems to be unrelated to the first integral or its derivative. I think that is used to prove the conservation of [itex][\vec{r}, \vec{\dot{r}}][/itex] which is not totally the same with this one...sorry if I am being slow..
     
  7. Jan 22, 2013 #6
    So we get ## \displaystyle \vec{r} \times \ddot{\vec{r}} = 0 ##. Can you see that on the left hand side we have a time-derivative of ## \displaystyle \vec{r} \times \dot{\vec{r}} ##?

    What does that imply about the plane of motion? Can you really talk about a "plane of motion'?
     
  8. Jan 22, 2013 #7
    eh yes I know that, but like I said above the conservation of [itex][\vec{r}, \vec{\dot{r}}][/itex] as a vector quantity is not totally equivalent to the conservation of [itex]r^2 \dot{w}[/itex] as a scalar.
     
  9. Jan 22, 2013 #8
    So what about the plane of motion?
     
  10. Jan 22, 2013 #9
    It means the moving body always remain in that plane made up by speed and coordinate since the vector is constant?
     
  11. Jan 22, 2013 #10
    Exactly. Since we have ## \displaystyle \vec{r} \times \dot{\vec{r}} = \vec{l} ##, the latter being a constant of motion (angular momentum), ## \vec{r} ## must be at all times be perpendicular to a fixed direction, which means it lies in a plane. By the same token, the velocity also lies in the same plane.

    Now, because we have a plane of motion, the problem is reduced to 2D, and you can introduce some (arbitrary) vector in the plane, and then you will have angle ## \theta ##, which is the the angle between that vector and ## \vec{r} ##, and which is always measured in that plane.

    ## \dot{\theta} ## is angular velocity.

    Can you represent ## \dot{\vec{r}} ## as a sum of radial (parallel to ## \vec{r} ##) and tangential velocities?
     
  12. Jan 22, 2013 #11
    Do you mean to obtain both the tangential and radial velocity and differentiate it with regards to time to obtain the tangential acceleration? It is my original method: because the tangential acceleration is clearly 0 by the Newtonian equation, and substitute that into the derivative of [itex]r^2\dot{θ}[/itex] to realise that the latter is zero which proves its first integral property.
     
  13. Jan 22, 2013 #12
    No, I do not mean this. ## \displaystyle \dot{\vec{r}} = \vec{\rho} + \vec{\tau} ##, the sum of radial and tangential velocities, respectively. Now substitute this into the ## \displaystyle \vec{r} \times \dot{\vec{r}} ##.
     
  14. Jan 22, 2013 #13
    [itex][\vec{r}, \vec{\dot{r}}]=[\vec{r}, \dot{r} \vec{e_r} + r\dot{θ}\vec{e_θ}] [/itex] where [itex]\vec{e_r}, \vec{e_θ}[/itex] being the tangential and radial direction vector. differentiate it with t gives [itex]r^2\dot{θ}[e_r, e_θ][/itex] thus proves its conservation.
     
  15. Jan 22, 2013 #14
    But why do you need to differentiate this? You already have ## \displaystyle \vec{r} \times \dot{\vec{r}} = r^2\dot{\theta}\vec{e}_n = \vec{l} ##.
     
  16. Jan 22, 2013 #15
    Ehh sorry don't know what is in my head when I typed that.. And thanks for your time on this.
     
  17. Jan 22, 2013 #16
    You are welcome. I hope you will memorize that the integral (a.k.a Kepler's second law) and planar motion are a direct consequence conservation of angular momentum in a central field.
     
  18. Jan 22, 2013 #17
    Yes and angular momentum differs from it as a geometric quantity by its mechanical parameter m. Well I was initially working on a "purely" mathematical proof because it shows up in an ODE book on first integrals.. it gets quite confusing when known physical concepts come into relation to pure mathematics
     
  19. Jan 22, 2013 #18
    There is no "purely" mathematical proof because the question itself is based on certain physical assumptions. Namely, that the motion can be described by just two variables, a distance and an angle. And this follows only when you prove that angular momentum is conserved. Well, one could, of course, just ignore that that quantity has a "physical" name, but that is as "mathematical" as it gets.
     
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