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How do we decide if a trajectory is 1D, 2D or 3D?

  1. Aug 24, 2012 #1
    Hello Forum,

    In kinematics we study motion and the trajectories of moving bodies.

    The trajectory is a line (straight or curved) that joins all the positions occupied by the object in the various instants of time. A trajectory has an equation that contains only spatial coordinates (not time t).

    For example, a particle moving in a circle in the 3D space: the trajectory equation can be x^2+y^2=16 in Cartesian, r=4 in polar, etc....

    Is this trajectory and this motion 2D, 1D or 3D? How do we decide?

    There seem to be only one independent variable in x^2+y^2=16 ......

    Is a curve always a 1-dimensional object, manifold that lives in a higher dimension space?

    thanks
    fisico30
     
  2. jcsd
  3. Aug 24, 2012 #2
    All curves are one dimensional, surfaces are two dimensional and solids three dimensional.

    the test is : How many parameters are required for a parametrisation?

    Your curve as f(x,y,z) can be parameterised as x(t), y(t), z(t)

    For a surface you require two parameters say, (t,s)

    Does this help?
     
  4. Aug 24, 2012 #3
    A trajectory is a 1D object intrinsically. A trajectory in a 3D space, however, may be studied in a 2D or 1D subspace if the trajectory is "flat enough". Specifically, each trajectory is characterized by its "curvature" and "torsion". If torsion is zero, then the trajectory is in a plane; if curvature is zero, then it is a straight line.
     
  5. Aug 28, 2012 #4
    Thanks everyone.

    From Dr. Math:
    ".... Although the sphere is a subset of three-dimensional space, it is a two-dimensional object. A circle can exist in either two-dimensional or three-dimensional space (or even higher-dimensional spaces), but it is a one-dimensional object.

    A line segment, which is one-dimensional, can be deformed into a circle. We can think of the line as elastic. A circular disk, which is two-dimensional, can be deformed elastically into a sphere.
    If you know about describing curves and surfaces parametrically, then, with certain restrictions, the dimension of the object is equal to the number of parameters required in its description....."


    So a trajectory is a one dimensional object. But if the points of the trajectory all belong to a straight line then the motion is commonly called one-dimensional. If the points belong to the same plane it is the 2-dimensional and if they belong to a 3D space the trajectory is 3-dimensional. So the statement "the dimension of the object is equal to the number of parameters required in its description" is what it is often meant in kinematics to define the dimension of the trajectory.

    Parametrization? The trajectory is either an equation involving only space variables or it is given by by the point of the vector [x(t), y(t), z(t)]. The 3 vector coordinates are functions of time t. Does that mean parametrization?
    The 3 space location variables could be functions of the travelled space s along the trajectory itself, as x(s), y(s), z(s), correct?
    How would we move from the time parametrization [x(t), y(t), z(t)] to the space parametrization [x(s), y(s), z(s)]?

    thanks
    fisico30
     
  6. Aug 28, 2012 #5
    Yes. In this case t is a parameter, for each value of which you can produce a point of the trajectory.

    You would need to find the "line element" ds; it is expressed via the derivatives of x(t), y(t), z(t). Which is going to be a differential equation with s and t. Which you can solve for s(t) and t(s).
     
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