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Motion in a Central Field

  1. Oct 26, 2004 #1


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    In describing the motion (orbit) of a particle of mass m moving in a plane in a potential V(r), my prof wrote the Lagrangian as:

    [tex] \mathcal{L} = \frac{1}{2}m|\dot{\vec{r}}|^2 - V(r) [/tex]

    Without further ado (or explanation of any kind), he wrote this as:

    [tex] \mathcal{L} = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\phi}^2) - V(r) [/tex]

    Where [itex] (r,\phi) [/itex] are the coordinates in the orbital plane.

    I understand that he is expressing the magnitude of r'(t) in terms of its components, and that [itex] \dot{r} [/itex] is the velocity in the radial direction, whereas the second term should represent tangential velocity. But that part bothers me. After all...how did he arrive at this second term? It seems to me that he assumed r was constant, otherwise the second term is invalid. Yet if r were constant, then [itex] \dot{r} [/itex] would be zero!

    Shouldn't it be done as follows?

    [tex] |\dot{\vec{r}}|^2 = \left(\frac{ds}{dt}\right)^2 = \left(\frac{d(r \phi)}{dt}\right)^2 = (\dot{r}\phi + r\dot{\phi})^2 [/tex]

    What am I doing wrong, and how did he get his formula?
  2. jcsd
  3. Oct 26, 2004 #2


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    s is not equal to phi*r unless it is a circular motion (constant r). In polar coordinates,

    [tex]ds=\sqrt{(dr)^2+(rd\phi)^2} [/tex]. See attached picture. The yellow thing is a rectangular triangle when the displacement is very small.

    Last edited: Jun 29, 2010
  4. Oct 26, 2004 #3


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    He's computing .5*m*v^2 to get the kinetic energy for the first term.

    In cartesian coordinates, it's .5*m*(vx^2 + vy^2).

    To duplicate his result, you can either

    1) note that the vectors [tex]\frac{d \hat r}{dt}[/tex] and [tex] \frac {d \phi}{dt}[/tex] are orthogonal, so you just add the squares of the two velocites, or

    2) you can make the coordinate transformation

    x = r(t)*cos(phi(t))
    y = r(t)*sin(phi(t))

    use the chain rule to compute dx/dt and dy/dt

    square the results and add them together

    but the first way is much easier. (The second is not too bad if you have a symbolic algebra program).
  5. Oct 31, 2004 #4


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    Hmm...I think I've got it. However, it is somewhat confusing when two people give slightly different answers. However, am I right in saying that pervect's method number 1 is the same as ehild's method? The only difference is that pervect assumed that I would be able to get the two expressions for radial and tangential velocity components just like that, whereas ehild showed me how to do it on a diagram. Another thing that bothers me: to get the desired answer from what ehild showed on the diagram, you have to do this:

    [tex]ds=\sqrt{(dr)^2+(rd\phi)^2} [/tex]

    [tex] \frac{ds}{dt} = \frac{\sqrt{(dr)^2+(rd\phi)^2}}{dt} [/tex]

    [tex] = \sqrt{\frac{(dr)^2+(rd\phi)^2}{(dt)^2}} [/tex]

    [tex] = \sqrt{\frac{(dr)^2}{(dt)^2}+ \frac{(rd\phi)^2}{(dt)^2}} [/tex]

    [tex] = \sqrt{\left(\frac{dr}{dt}\right)^2 + r^2\left(\frac{d\phi}{dt}\right)^2} [/tex]

    [tex] = \sqrt{\dot{r}^2 + r^2\dot{\phi}^2} [/tex]

    So, we get the right answer, but I'm wondering about the mathematical validity of the manipulations I performed. I understand that there must be some validity to algebraic manipulations of differentials (i.e. infinitesimals), because it is done all the time in integral calculus. But it seems to me that physicists are a lot more free and easy with differntials than mathematicians would say is permissible.

    My question is, are we allowed to interpret them the way we do, as actual infinitesimal quantities, that can be squared, grouped together etc. Can we say that the derivative is a ratio of infinitesimal changes in (i.e. "elements" of or "little bits" of) two variables? This is the method of presenting calculus in Silvanus P. Thompson's Calculus Made Easy, and it is totally intuitive to me. He even gives reference to a mathematician (I can't remember who) who wrote a paper asserting (with some proof) that infinitesimals are legitimate quantities in mathematics. However, I have seen people on this site freaking out about that. My calculus prof would also have none of it. "You could interpret the derivative (etc.) that way," he said, "but you would merely be restating the same thing a different way. A limit is a well defined concept, and the definition of a derivative is given in terms of one."

    In fact, isn't the definition of a differential a limit as well? I haven't looked it up, but I think it goes something like, for a differentiable function [itex] y = f(x) [/itex], on an interval [itex] \Delta x [/itex], we can make the following definition:

    [tex] dy = \lim_{\substack{\Delta x \rightarrow 0}}f'(x) \Delta x [/tex]

    I guess I could have derived the equation above more rigorously with that in mind. Still, clarification on this very confusing issue would be appreciated.
  6. Oct 31, 2004 #5


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    This sort of thing is a real quagmire, because a great many number of concepts all get similar notation because they all "act like infintiessimals".

    Infinitessimals can, in fact, be used rigorously, but you have to learn non-standard analysis. For example, the derivative is defined by:

    \frac{d}{dx} f(x) := \mathrm{Std} \frac{f^*(x^* + \Delta x) - f^*(x^*)}{\Delta x}

    Where [itex]\Delta x[/itex] is any nonzero infinitessimal, Std means to round to the nearest standard number, and the superscript * means to use the nonstandard version of that symbol. (It's sort of like distinguishing between 1 the integer and 1 the rational number)

    However, you're probably better off translating your intuition to the standard way of doing things than trying to do nonstandard stuff intuitively.

    Differentials, like f(x) dx, are simply functions on paths. That is, given a differential, f(x) dx, you can apply it to a path, [itex]\gamma[/itex], to get a number: [itex]\int_{\gamma} f(x) \, dx[/itex].

    However, I don't know if the "arithmetic" used in computing ds is an actual arithmetic, or just a formal symbol that means "do the right thing".
  7. Nov 1, 2004 #6


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    Yes, it is. In a polar coordinate system the base vectors (angular and azimuthal) can be written in terms of Cartesian coordinates as

    [tex] \vec e_{r} = \cos\phi \vec i + \sin \phi \vec j \mbox { and }\vec e_\phi= -\sin \phi \vec i + \cos \phi \vec j[/tex].

    The position of a point in vector form is

    [tex] \vec r = r \vec e_r[/tex]

    The velocity is the time derivative of the position vector.

    [tex] \vec v = \dot \vec r = \dot r \vec e _r +r \dot \vec e_r[/tex]

    We have to find the time derivative of the radial unit vector.

    First, as

    [tex]\vec e \cdot \vec e = 1\rightarrow \vec e \cdot \dot \vec e =0[/tex]

    the time derivative of an unit vector is normal to the original vector, which means it is parallel to [tex] \vec e _\phi[/tex].

    The magnitude of an unit vector [tex]\dot \vec e [/tex] can be found from a geometrical approach. In the limit when

    [tex]\Delta\phi \rightarrow 0[/tex]

    [tex] \frac{\Delta |\vec e|}{\Delta t}=\frac{\Delta \phi}{\Delta t} \rightarrow \dot \phi[/tex]

    If you do not like this method, you can derive the radial unit vector written in Cartesian components

    [tex]\dot \vec e_{r} = -\sin\phi *\dot\phi *\vec i + \cos \phi *\dot\phi*\vec j =\dot \phi * \vec e_\phi[/tex]

    We get the velocity now in polar coordinates :

    [tex] \vec v = \dot r \vec e _r +r\dot \phi \vec e_\phi[/tex].

    Take the square :

    [tex] (\vec v) ^2 = (\dot r )^2 + (r \dot \phi)^2 [/tex]

    As for your problem with ds, how did you learn the line integral? How do you calculate the lenght of a line y=f(x) from A to B?

    [tex] \int _A^B ds [/tex]

    You replace ds with

    [tex]\sqrt{dx^2+dy^2}=\sqrt{ 1+(f'(x))^2}dx[/tex],

    don't you? This is the same method with infinitesimal quantities I used to get ds from dr and d(phi).

  8. Nov 2, 2004 #7


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    How did we get this step? Sorry, probably something really easy I'm missing.

  9. Nov 2, 2004 #8


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    It is differentiation of a product.
    If you do not like vectors,

    [tex]( \vec e )^2= \vec e \cdot \vec e = = e_x^2+e_y^2=1 [/tex]
    [tex]\frac{d(\vec e)^2}{dt}= 2 (e_x*\dot e_x+e_y*\dot e_y) = 0[/tex]
    Notice that
    [tex](e_x*\dot e_x+e_y*\dot e_y)=\vec e \cdot \dot \vec e[/tex]

    dot product of he unit vector and its derivative, and this is zero, which means that they are orthogonal.

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