1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relative motion of two particles in newton's field theory of gravitation.

  1. Sep 3, 2009 #1
    I don't understand the physical meaning of one of my dimensions/variables.

    Let there be a gravitational potential [tex]\phi (x_a), a=1,2,3.[/tex]

    Equation of motions of a freely falling particle is:
    [tex] \frac{d^2 x_a}{d t} = - \frac{\partial \phi}{\partial x_a}.[/tex]

    If there are 2 particles falling, family of paths
    [tex]x_a = x_a(u,v)[/tex] (u can be considered the newtonian absolute time). u is the curve parameter and "v labels the paths".

    There are tangent vector fields:
    Single particle motion: [tex]U_a = \frac{\partial x_a}{\partial u}[/tex]
    and relative motion of the two particles [tex]V_a=\frac{\partial x_a}{\partial v}[/tex].
    After this we came up with a couple of equations and the class ended.

    I understand u, which could be understood as time. But what is v? What does [tex]\frac{\partial x_a}{\partial v}[/tex] mean? I need some real physical explanation instead of "it's a parameter" or something similar. Just like with u, instead of "curve" I now have time.

    And what is this uv-space? How is v related to [tex]x_a[/tex]s.
    Last edited: Sep 3, 2009
  2. jcsd
  3. Sep 3, 2009 #2
    Consider one particle path first. It requires only a single parameter, u, to describe it geometrically; that is, u counts in some chosen way as you move along the path of the particle. You can then go on to describe the path geometrically, like tagging on tangent vectors at various points of the path, which are labeled by the parameter u. But to make calculations, we like to introduce coordinate systems, so that [tex] x_{a}(u) [/tex] (with a=1,2,3) are "coordinate functions" of the parameter u; they describe the embedding of the particle path in the coordinate frame. The tangent vector at parameter value u' is then [tex]\left(dx_{a}/du\right)|_{u=u'}\equiv U_{a}[/tex].

    Now more specifically onto your question. When we introduce a second particle path, we can parametrize it with u as well. So now we have two paths that we can follow along as we count using the single parameter u. But we can also ask about the relative motion of the two paths at any value of u. To do this we introduce a parameter v that labels the paths according to: Path 1 has v=0, Path 2 has v=1. Therefore, the two paths can be described by coordinate functions of the two parameters u and v: [tex]x_{a}(u,v)[/tex]. Just to be explicitly clear, [tex]x_{a}(u,0)[/tex] is the coordinate description of path 1 while [tex]x_{a}(u,1)[/tex] is the coordinate description of path 2. We can then consider the quantity [tex]\left(\Delta x_{a}/\Delta v\right)|_{u=u'}[/tex], which is just the coordinate separation vector from path 1 to path 2 for a fixed value of u=u'. Since v only takes values 0,1 (and 2,3,4,...,n if there were n particle paths) we can sum all that info up by writing this vector as [tex]\partial x_{a}/\partial v\equiv V_{a}[/tex]; it's still just the separation vector pointing from path 1 to path 2 for each fixed value of u.

    The "u-v space" as you call it is a parameter space for the two paths, nothing really special about it.
    Last edited: Sep 3, 2009
  4. Sep 4, 2009 #3
    Actually, I didn't write all things up:
    We also had a separation vecto called [tex]\rho _a[/tex]. [tex]V_a[/tex] was introduced as relative motion, not as a separation vector. The two paths that we had were [tex]v[/tex] and [tex]v+\Delta v[/tex]. Afterwards we made [tex]\Delta v[/tex] infinitesimal, so that [tex]\rho _a = V_a \Delta v[/tex] would be infinitesimal too.

    I understood [tex]V_a[/tex] as speed (how fast the two paths approach each other in our coordinate space [tex]x_a[/tex]) rather than as a separation vector. And if this is the case, then why is [tex]\rho _a[/tex] defined the way it is?
  5. Sep 4, 2009 #4
    I'm going to stand by what I wrote, but let me add some clarification, and I'll be using a bit of my own notion. Instead of v=0,1 that labeled two paths we can take the neighboring paths at v=0 and [tex]v=0+\Delta v[/tex] where [tex]\Delta v[/tex] can be taken to be arbitrarily small. In general, we could consider a whole family of paths labeled by v, separated by the infinitesimal delta-v from their neighbor. Then [tex]\Delta x_{a}=V_{a}\Delta v[/tex] is the *coordinate displacement* vector between neighbors at a given value of the parameter u. Now, the velocity of the neighboring path relative to the original will be [tex]\nabla_{U}V[/tex], where U and V are the vectors formed from the components we've previously been talking about (think about this geometrically with (1) the tangent vector to the path and a (2) discrete displacement vector for the change in (capital )V...if those vectors (1) and (2) are orthogonal, the paths are not moving relative to each other)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Relative motion of two particles in newton's field theory of gravitation.