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Vector analysis in cylindrical coordinate system

  1. May 14, 2010 #1
    if a point in cylindrical cordinates is A=(r1,[tex]\theta[/tex]1,z1) and another point is
    B=(r2,[tex]\theta[/tex]2,z2)
    (if ar,a[tex]\theta[/tex],az are unit vectors)
    then the position vector of A is = r1<ar>+[tex]\theta[/tex]1<a[tex]\theta[/tex]>+z1<az>
    and the position vector of B is =
    r2<ar>+[tex]\theta[/tex]2<a[tex]\theta[/tex]>+z2<az>
    so vector BA=
    (r1-r2)<ar>+([tex]\theta[/tex]1-[tex]\theta[/tex]2)<a[tex]\theta[/tex]>+(z1-z2)<az>
    am i right upto here??
    what is magnitude of vector BA??i think to calculate magnitude we have move back to Cartesian coordinates,am i right??
    In cylindrical coordinates at every point we have new unit vectors(i mean in direction) how this is going to be useful for us,than having a fixed unit vectors in Cartesian coordinate system??
    why cylindrical coordinates are preferable in case of cylindrical symmetry because even though we use them in order to calculate vectors,distances we have to switch back to Cartesian coordinates??!!
    i know i totally mis understood the concept but i was not able where i was wrong about cylindrical coordinate system...plz some body help me through this..:confused:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 14, 2010 #2

    diazona

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    Actually no, it doesn't work that way. The coordinates [itex]r,\theta,z[/itex] of a vector in cylindrical coordinates tell you what distance to move away from the z-axis, how far around the z-axis to rotate, and how far to move up the z-axis to get from the origin to the identified point. Specifically, the angular coordinate [itex]\theta[/itex] doesn't identify a distance, and it isn't something you can multiply by a unit vector to get some component of the vector. So basically, in cylindrical coordinates, there isn't any set of basis vectors such that you can multiply [itex]r[/itex] times the first one, [itex]\theta[/itex] times the second, and [itex]z[/itex] times the third and wind up with the position vector for [itex]r,\theta,z[/itex].

    You could think of it this way: a position vector always points from the origin to some other point. So it really involves (at least) two different positions: the origin, and the other point. Now, if you want to express the position vector in terms of a set of basis vectors, you need basis vectors that are the same at the origin, and at the other point, and everywhere in between. Except for certain points, that isn't the case in a cylindrical coordinate system. For example, if you could write a position vector for some point as [tex]3\hat{r} + 2\hat{\theta} + 5\hat{z}[/tex], which [tex]\hat{\theta}[/tex] should you use? The one at the origin or the one at the point you're trying to identify? Not to mention, [tex]\hat{\theta}[/tex] isn't even defined at the origin... so for several reasons, it doesn't work.
    No, you can't just subtract the coordinates like that with cylindrical coordinates. It only works for Cartesian coordinates (or any system in which the unit vectors are independent of position).

    Part of what makes these coordinate systems useful (or, I guess, keeps them from being completely useless) is that subtraction of vector components like that does work for infinitesimal distances, in all coordinate systems. But only for infinitesimal distances. That's because, when you're trying to express the relative position vector between two points and those two points are infinitesimally close, the unit vectors at the two points (and everywhere in between) are the same. (More precisely, any variation in the unit vectors over an infinitesimal distance is small enough to ignore) So it winds up working like Cartesian coordinates, over these small distances.
     
    Last edited: May 14, 2010
  4. May 14, 2010 #3
    thank u diazona:smile:,, so there is no way to get vector between two points in this system( i mean in terms of basic vectors)??
    i had a problem here...
    Ques:Express the unit vector which points from z=h on z axis towards(r,[tex]\theta[/tex],0) in cylindrical coordinates(figure related to question is attached)
    Ans:vector between these points is
    R(let)=r<ar>- h<az>
    and the unit vector is
    aR(let)=R/mod(R)
    =r<ar>-h<a> / squareroot(r2+h2)

    my question is how did he expressed it in terms of ar,both points can have different ar in fact ar at z=h is not defined..how did he arrived to this result??

    and where did the component of theta gone in the vector R ??

    same kind of analysis is used in many problem to obtain unit vector in case for obtaining electric fields...
     

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  5. May 14, 2010 #4

    diazona

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    This is one of those special cases where the unit vectors are the same at both endpoints and everywhere in between. Obviously the unit vector [itex]a_z[/itex] (in your notation) is the same everywhere, and I hope you can tell that the unit vectors [itex]a_r[/itex] and [itex]a_\theta[/itex] are the same on any radial half-plane (that is, any half-plane that is bounded by the z-axis and extends out to infinity in one direction). It happens that the two positions in this question lie on the same half-plane, so you can use the same [itex]a_r[/itex] for both of them, and that makes it possible to write the different in position as a vector.

    You're right that [itex]a_r[/itex] is not well defined on the z axis, but what that really means is that it is multivalued. In this case, it can point in any direction perpendicular to the z axis, and you can choose one of those directions to use for a particular problem if it makes sense to do so. The solution you've given does exactly that, it chooses the unit vector [itex]a_r[/itex] on the z axis to be the one that points toward the point [itex](r,\theta,0)[/itex] (although it really should explicitly say that that's what it's doing). This is equivalent to choosing the value for [itex]\theta[/itex] along the z axis to be the same as the value of [itex]\theta[/itex] at the given point [itex](r,\theta,0)[/itex].

    The fact that there is no [itex]\theta[/itex] component in the answer simply means that the vector points radially outwards from the z axis. This will be true of any position-difference vector in which one of the points is on the z axis.
     
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