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Derivation of the Lagrangian for Rotating Polar Coordinates

  1. Aug 28, 2014 #1
    I'm reading Leonard Susskind's The Theoretical Minimum Vol. 1.

    1. The problem:
    I'm on the section in which he asks the readers to derive the Lagrangian for a particle on a rotating carousel in polar coordinates.
    2. Relevant ideas:
    The same Lagrangian in Cartesian coordinates is given as $$\mathcal{L}=\frac{m}{2}(\dot{x}^2 + \dot{y}^2) + \frac{mω^2}{2}(x^2+y^2) + mω(\dot{x}y-\dot{y}x)$$
    3. The attempt at a solution:
    According to the book, the required Lagrangian is $$\mathcal{L}=\frac{m}{2}(\dot{r}^2 + ω^2 r^2)$$But merely substituting $$x= r\cos {ωt}$$ and $$y=r\sin {ωt}$$ renders $$\mathcal{L}=\frac{m}{2}(\dot{r}^2)$$Maybe, I've done the algebra wrong. But that possibility is a bit unlikely because I have checked my work many times.

    The official solution on this page http://www.madscitech.org/tm/slns/l6e4.pdf mentions a derivation that involves the use of separate kinetic and potential energy terms (the potential term is arbitrary V(r)). One can find similar derivations online. So, why is my approach wrong?
     
    Last edited: Aug 28, 2014
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  3. Aug 28, 2014 #2

    ShayanJ

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    The point is, [itex] x=r\cos{\omega t} [/itex] and [itex] x=r\sin{\omega t} [/itex] are wrong. Because it means the particle is stuck to the axes and so its angular motion is the same as the axes but that's not the case. Instead of [itex] \omega t [/itex], you should write [itex] \omega t + \varphi [/itex] where [itex] \varphi [/itex] is the azimuthal angle.
     
  4. Aug 28, 2014 #3

    nrqed

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    There is something wrong here. In Cartesian coordinates the lagrangian of a free particle moving in two dimensions is simply
    $$\mathcal{L}=\frac{m}{2}(\dot{x}^2 + \dot{y}^2) $$
    That's all! Then we get the correct answer in polar coordinates.
    Your initial expression in Cartesian coordinates is therefore simply incorrect. It describes something else. To me it looks like the lagrangian attached to a rotating sphere (in which case there is also a depending on [itex] \phi [/itex] as Shyan pointed out).
     
  5. Aug 28, 2014 #4
    The correct answer in polar coordinates that you refer to here is simply [itex]\mathcal{L}=\frac {m} {2}(\dot{r}^2)[/itex]. Right?

    What I really want to ask is this: if a particle's Lagrangian in Cartesian coordinates is: [itex]\mathcal{L}=\frac{m}{2}(x^2+y^2) + \frac{mω^2}{2}(x^2+y^2) + mω(\dot{x}y−\dot{y}x)[/itex]then how does one go about finding the equivalent Lagrangian in polar coordinates? Because, I've tried and failed by simple substituting x=r cosωt and y=r sinωt. It's because I ignored [itex]\varphi[/itex], right?
     
  6. Aug 28, 2014 #5

    nrqed

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    No. Just plug in your x and y expressions and you will get
    [itex]\mathcal{L}=\frac {m} {2}(\dot{r}^2 +\omega^2 r ^2)[/itex]
    Just give it a try.

    As I mentioned, this is a different physical system, in which case we would indeed have to take into account the azimuthal dependence. So there should actually be a z dependence as well. And then we would have to go to spherical coordinates and include the [itex] \varphi [/itex] dependence as well, yes. Where did you get that Lagrangian? What do they say about what it represents?

    Regards,

    Patrick
     
  7. Aug 28, 2014 #6

    vanhees71

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    The original Lagrangian is indeed the one for free particles measured in a frame rotating around the 3-axis of an inertial frame.

    To see this just let [itex](x',y')[ /itex] cartesian coordinates in the inertial frame. Then
    [tex]\vec{x}'=\begin{pmatrix}
    \cos(\omega t)& -\sin(\omega t)\\
    \sin(\omega t) & \cos(\omega t)
    \end{pmatrix} \vec{x}.
    [/tex]
    Now calculate
    [tex]L=\frac{m}{2} \dot{\vec{x}}'{}^2[/tex]
    and you'll get the original Lagrangian.

    Edit: I don't know, why the latex doesn't work as usual. Is it, because I'm posting via my android tablet (within Chrome browser)?
     
    Last edited: Aug 28, 2014
  8. Aug 28, 2014 #7

    nrqed

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    Ah! It is in terms of non inertial coordinates! Thanks for clearing this up
    But the Lagrangian they give as the answer is in terms of the inertial coordinates, and this is what threw me for a loop.

    So the answer to the OP is: first go from the non inertial coordinates to the inertial coordinates, where the Lagrangian is simply ## m (\dot{x}^2 + \dot{y}^2)/2 ## and then the rest is direct.
     
  9. Aug 29, 2014 #8

    vanhees71

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    Ok, here's my posting, which for some strange reason didn't work. I guess, it's because I posted it using my Android tablet instead of a real (linux) PC. It's strange, because this shouldn't happen, because I use the Chrome browser both on the Android as on the PC, and there was never a problem posting in these forums ;-).

    The original Lagrangian is indeed the one for free particles measured in a frame rotating around the 3-axis of an inertial frame.

    To see this just let [itex](x',y')[/itex] cartesian coordinates in the inertial frame. Then
    [tex]\vec{x}'=\begin{pmatrix}
    \cos(\omega t)& -\sin(\omega t)\\
    \sin(\omega t) & \cos(\omega t)
    \end{pmatrix} \vec{x}.
    [/tex]
    Now calculate
    [tex]L=\frac{m}{2} \dot{\vec{x}}'{}^2[/tex]
    and you'll get the original Lagrangian.

    Just let me add that you can directly get also the form in polar coordinates by setting in the non-inertial frame,
    [tex]x=r \cos \varphi, \quad y=r \sin \varphi[/itex]
    plug this in the above given formula with the time-dependent rotation describing the transformation from the accelerated to the inertial fram. Then you get very simple expressions for the rotation, which is immediately intuitive, and it's also very easy to get the Lagrangian from this.
     
  10. Aug 29, 2014 #9
    @nqred: I did plug in [itex]x=\cos{ωt}[/itex] and [itex]y=\sin{ωt}[/itex] and got the right answer. Thanks.

    @vanhees71: Your formula didn't parse because you used the wrong tags. You should use itex & /itex with [] enclosing them. Thank you for your help.

    I still have one question. Is the potential energy [itex]V(r)[/itex] absolutely necessary? The solution that I linked to makes use of it.
     
  11. Aug 29, 2014 #10

    vanhees71

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    But, in my notation, you should get
    [tex]x'=r \cos(\varphi+\omega t), \quad y'=r \sin(\varphi+\omega t).[/tex]
    This you plug into
    [tex]L=T=\frac{m}{2} (\dot{x}'+\dot{y}')[/tex]
    to get the Lagrangian in terms of the rotating reference frame coordinates.
     
  12. Aug 29, 2014 #11

    Orodruin

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    I agree with vanhees, the Lagrangian in the original post is already given in a coordinate system that is rotating. This means that there is not really any need to go back to the inertial coordinate frame using ##\cos\omega t## and ##\sin\omega t##. You can just as well just introduce the polar coordinates on the already rotating frame:
    $$
    x = r\cos\phi, \quad y = r\sin\phi.
    $$
    This should give the result immediately.
     
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