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Representing displacement vectors in cylindrical coordinates

  1. Dec 18, 2013 #1
    Hello,
    In Cartesian coordinates, if we have a point P(x1,y1,z1) and another point Q(x,y,z) we can easily find the displacement vector by just subtracting components (unit vectors are not changing directions) and dotting with the unit products. In fact we can relate any point with a position vector by drawing a vector from the origin to the point.
    Now when it comes to cylindrical coordinates for some reason the two books I have don't seem to take this approach to find the position vectors of any two points P(r1,θ1,z1) and Q(r,θ,z). What they seem to do is go through Cartesian coordinates, i.e. write each point with its equivalent Cartesian coordinates representation and using a transformation matrix we get the displacement vector in cylindrical coordinates.

    So given that a vector in space doesn't know (or care) about the coordinate system used a point P which is given by (x,y,z) in Cartesian coordinates can be transformed to cylindrical coordinates (r,θ,z) using x=r*cos(θ), y=r*sin(θ), z=z. So a position vector in Cartesian is <x,y,z>.

    The questions:
    1. I want to know how to represent any vector in cylindrical coordinates (eg. A=2[itex]\hat{r}[/itex]+3[itex]\hat{θ}[/itex]+5[itex]\hat{z}[/itex].) I'm asking this because a point P(2,3,5) in the cylindrical system does not correspond to a vector A=2[itex]\hat{r}[/itex]+3[itex]\hat{θ}[/itex]+5[itex]\hat{z}[/itex].
    2. I want to ignore Cartesian coordinates and use just the pure geomtry of the cylindrical system to write the displacement vector.

    Any explanation or material is welcomed.
     
  2. jcsd
  3. Dec 18, 2013 #2
    In cylindrical coordinates, a position vector from the origin is given by [itex]r\hat{r}+z\hat{z}[/itex]. So, a position vector to point 1 is [itex]r_1\hat{r_1}+z_1\hat{z}[/itex] and a position vector to point 2 is [itex]r_2\hat{r_2}+z_2\hat{z}[/itex]. So a displacement vector from point 1 to point 2 is given by [itex]r_2\hat{r_2}-r_1\hat{r_1}+(z_2-z_1)\hat{z}[/itex]. Now [itex]\hat{r_1}[/itex] and [itex]\hat{r_2}[/itex] are not the same vectors because they are pointing in different directions in space. However, either of them can be expressed in terms cosine and sine of (θ21), and in terms of [itex]\hat{r}[/itex] and [itex]\hat{θ}[/itex] at the other point. So you can get the displacement vector exclusively in terms of the unit vectors at one of the points, or exclusively in terms of the unit vectors at the other point.
     
  4. Dec 18, 2013 #3
    Two questions:
    1. Why did you just consider [itex]\hat{r}[/itex] and [itex]\hat{z}[/itex]. What about [itex]\hat{θ}[/itex] ?
    2. I have already said that I know that unit vectors [itex]\hat{r}[/itex] and [itex]\hat{θ}[/itex] change. What I was asking was what point does this position vector (A=2[itex]\hat{r}[/itex]+3[itex]\hat{θ}[/itex]+5[itex]\hat{z}[/itex]) correspond in cylindrical coordinates.
     
  5. Dec 18, 2013 #4
    Here's whats troubling me also. P1(r1,θ1,z1) and P2(r2,θ2,z2). Finding the displacement vector P1P2. Its converted to Cartesian and back again by the transformation matrix above. I already understand how the matrix was formed. However in this case I don't know the values of [itex]\varphi[/itex] in the coefficient matrix. I know [itex]\varphi[/itex]1 and [itex]\varphi[/itex]2 but what is [itex]\varphi[/itex] ?

    http://imageshack.com/a/img850/200/1nr8.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Dec 18, 2013 #5
    Because every position vector drawn from the origin is perpendicular to the θ direction. Just draw a diagram and see what I mean.

    This can't be a position vector drawn from the origin, because it has a θ component. If this is a displacement vector, then you need to identify the value of θ that the unit vectors in the radial- and cylindrical directions are referenced to.
     
  7. Dec 18, 2013 #6
    Thanks I see it now. In fact I should have thought so because θ does not have distance units !! I'm really surprised that my book talks nothing of position vectors.
    What about the φ in the transformation matrix ?
     
  8. Dec 18, 2013 #7
    I don't understand the matrix equations that you wrote, and I don't understand what the φ stands for. If I were determining a displacement vector between two points in cylindrical coordinates, I would do it directly, without going through cartesian coordinates. I would use the method I alluded to in post #2.

    Chet
     
  9. Dec 18, 2013 #8
    The formula in the picture is for conversion between Cartesian and cylindrical. The column vector on the extreme right is displacement vector of two points given by their cylindrical coordinates but expressed in the Cartesian form.
    Its like dx=x2-x1= r2cosφ2 - r1cosφ1 . . . and so on
    So the displacement vector in catersian is : P1P2 = dx[itex]\hat{x}[/itex] + dy[itex]\hat{y}[/itex] + dz[itex]\hat{z}[/itex]

    Usually A use the transformation matrix like this : A vector is given in its cartesian form and so you need to find its cartesian representation at point (x,y,z). The φ can be calculated just the usual way using arctan(y/x). Then just simple substitution will give you the cylindrical components.
     
  10. Dec 19, 2013 #9
    Thanks. I can see what's going on with your matrix equation now. Since the unit vectors in the radial direction and in the circumferential directions are functions of θ, φ is being specified to select the specific value of θ that these unit vectors are being referenced to. The components of A on the left are the components that correspond to the chosen value of θ=φ.
     
  11. Dec 19, 2013 #10
    I'm sorry but I'm still confused about the concept of position vector in cylindrical coordinates. You say that only [itex]\hat{r}[/itex] and [itex]\hat{z}[/itex] matter when defining position vectors in cylindrical coordinates. It seems fine since this [itex]\hat{r}[/itex] vector can span all the rφ plane and [itex]\hat{z}[/itex] can give the elevation.

    I was looking more into the subject when I saw a book on GoogleBooks in which the definition of a position vector is not defined without [itex]\hat{φ}[/itex] component. For some reason the position vector depends on φ. How can that be when we already know the coordinates of the point in space. I still can't really understand vectors in other coordinate systems.

    Please take a look at : http://books.google.com.sl/books?id...=onepage&q=nathan ida position vector&f=false

    The page is 49 and its example 1.21
     
  12. Dec 19, 2013 #11
    In the example you cited, they are not using the unit vectors corresponding to each of the two points to resolve the position vectors into component form at these points. They are using the unit vectors at some arbitrary angle θ=φ with respect to the x coordinate axis to resolve the position vectors into component form. See what happens if you substitute φ=30 degrees into their equation for the first position vector, and φ= 0 degrees into the equation for the second position vector. Then the θ components will be zero. But if you resolve the position vectors into component form using the unit vectors at some other θ angle different from that of the position vectors, then of course there will be a θ component.

    I hope that this makes sense.
     
  13. Dec 21, 2013 #12
    It's been a while since I worked coordinate system changes, but if my math is correct, I think what medwatt was looking for is something like the following. Note, if I'm completely off base, please disregard this comment and sorry for any added confusion.

    Given [tex]\hat{v_{1}}=r_{1}\hat{r}+θ_{1}\hat{θ}+z_{1}\hat{z}[/tex] and [tex]\hat{v_{2}}=r_{2}\hat{r}+θ_{2}\hat{θ}+z_{2}\hat{z}[/tex] find the displacement vector in cylindrical coordinates using the cylindrical values. In which case:[tex]\hat{d}=\hat{v_{2}}-\hat{v_{1}}=\sqrt{r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos(θ_{2}-θ_{1})}\hat{r}+\cos^{-1}(\frac{r_{1}-r_{2}\cos(θ_{2}-θ_{1})}{\sqrt{r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos(θ_{2}-θ_{1})}})\hat{θ}+(z_{2}-z_{1})\hat{z}[/tex]
     
  14. Dec 21, 2013 #13
    The reference medwatt provided is very clear on what they were doing. This doesn't seem to be it. Medwatt's reference expresses all vectors in terms of the cylindrical coordinate unit vectors at one specific angle θ=phi, so that they can add and subtract the vectors on a common basis.

    Chet
     
  15. Dec 22, 2013 #14
    Actually, that's exactly what it means if you are thinking in cylindrical coordinates. You go 2 units in the r direction (outward in units of distance), 3 units in the angular direction (around in units of degrees or radians), and 5 units in the z direction (up-down in distance).

    That is incomplete and therefore incorrect.

    That would be somewhat true if you were dealing with multiple coordinate system spaces, but still incomplete.

    You wouldn't be defining those r-hat vectors in terms of the difference of the theta-hat components across the spaces.

    Here you are overlapping the concept of unit vectors and basis vectors. You mean "unit vectors aligned with the basis vectors of each space"

    This still doesn't address the original questions, and only offers a partial implicit solution, not the full explicit form

    You still use arctan(y/x) to find the φ, and in this case as you stated, the right column vector holds Cartesian displacement values. That is, [tex]x=r_2\cos(θ_2)-r_1\cos(θ_1)[/tex] and [tex]y=r_2\sin(θ_2)-r_1\sin(θ_1)[/tex]. Therefore, I think [tex]φ=\tan^{-1}(\frac{r_2\sin(θ_2)-r_1\sin(θ_1)}{r_2\cos(θ_2)-r_1\cos(θ_1)})[/tex]

    Again, that is the difference between the unit vectors of one traditional cylindrical space and multiple spaces with unit vectors aligned to basis vectors. The angle of φ is usually defined as perpendicular to r in the x-y plane.

    No it doesn't, you suggested that.

    Lastly, I revisited the direct displacement conversions, and instead of using Law of Cosines, just did the calculations into Cartesian then back into Cylindrical while leaving things in the cylindrical variables. I think the result is more intuitive and accurate. I'm more sure of the theta component this time.[tex]\hat{d}=\hat{v_2}−\hat{v_1}=\sqrt{r_1^{2}+r_2^{2}−2r_1r_2\cos(θ_2-θ_1)}\hat{r}+\tan^{-1}(\frac{r_2\sin(θ_2)-r_1\sin(θ_1)}{r_2\cos(θ_2)-r_1\cos(θ_1)})\hat{θ}+(z_2-z_1)\hat{z}[/tex]

    Hope this helps.
     
    Last edited: Dec 22, 2013
  16. Dec 22, 2013 #15
    I stand by everything I said. I've had lots of actual real world practical experience working with cylindrical coordinates (and vectors in cylindrical coordinates) on problems for which there was real money riding on the outcome of the work: e.g., automobile tire composite structural behavior under large loads, groundwater flow in underground geological formations to and from wells). I'm not familiar with every last detail of your mathematical rhetoric, but I am totally confident in everything I said in my posts.

    Chet
     
  17. Dec 24, 2013 #16

    vanhees71

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    Well, I don't understand what you are disagreeing about. Also the linked page of the book is not very clear to me. Fortunately the question is well formulated. What seems to be asked is to find the difference between the two position vectors [itex]\vec{r}_j=\overrightarrow{OP_j}[/itex]. For that purpose we have to express everything in Cartesian coordinates, because you need to express everything in components with respect to the same basis vectors, and Cartesian coordinates are fixed everywhere in space, while in cylindrical coordinates [itex]\hat{r}[/itex] and [itex]\hat{\varphi}[/itex] depend on the point in space you are sitting.

    A position vector is always given as
    [tex]\vec{r}= r \hat{r}+z \hat{z}=r \cos \varphi \hat{x} + r \sin \varphi \hat{y} + z \hat{z}.[/tex]
    With this formula you get (using [itex]\cos(\pi/6)=\sqrt{3}/2, \quad \sin(\pi/6)=1/2[/itex])
    [tex]\vec{r}_1=\frac{\sqrt{3}}{2} \hat{x} + \frac{1}{2} \hat{y} + \hat{z},[/tex]
    [tex]\vec{r}_2=2 \hat{x}+2 \hat{z}.[/tex]
    the difference is
    [tex]\vec{r}_1-\vec{r}_2=\frac{\sqrt{3}-4}{2} \hat{x} + \frac{1}{2} \hat{y}-\hat{z}.[/tex]
    The book's solution is not referring to the usual definitions of cylindrical coordinates. I don't understand the solution they give, particularly you must distinguish between the basis vectors for [itex]\vec{r}_1[/itex] and [itex]\vec{r}_2[/itex], which are of course different since you have
    [tex]\hat{r}=\cos \varphi \hat{x} + \sin \varphi \hat{y}, \quad \hat{\varphi}=-\sin \varphi \hat{x} + \cos \varphi \hat{y},[/tex]
    which implies that
    [tex]\hat{r}_1=\frac{\sqrt{3}}{2} \hat{x} + \frac{1}{2} \hat{y}, \quad \hat{\varphi}_1=-\frac{1}{2} \hat{x} + \frac{\sqrt{3}}{2} \hat{y}[/tex]
    and
    [tex]\hat{r}_2=\hat{x}, \quad \hat{\varphi}_2=\hat{y}.[/tex]
    So the whole solution in the book doesn't make sense. Particularly they don't say which [itex]\hat{r}[/itex] and [itex]\hat{\varphi}[/itex] they use. Is it at [itex]P_1[/itex] or at [itex]P_2[/itex] or even somewhere else?

    Maybe they define cylindrical coordinates in an unusual way before. This I don't know, and I also don't know, why one should do that. Is it to confuse students? At least I'd be very careful to use that book!
     
  18. Dec 24, 2013 #17
    I agree. I don't like the way the OP's book did it either. But, I've been able to make some sense out of what they did, and maybe this will help. Instead of using the usual x and y cartesian axes, they use a set of cartesian axes and unit vectors oriented at an angle phi to the usual cartesian axes. Everything is referenced to these axes and their unit vectors. The two position vectors of interest are resolved into components with respect to these unit vectors, and they can then be subtracted from one another unambiguously. The difference corresponds to these particular axes and unit vectors. These unit vectors also correspond to the cylindrical coordinate unit vectors at the angle θ = phi. So the difference between the two position vectors of interest is expressed in terms of the r and θ unit vectors at the angle θ = phi (for whatever that's worth). I hope that this makes some kind of sense. I would never do it that way.

    Chet
     
  19. Dec 25, 2013 #18

    vanhees71

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    But that doesn't make any sense either. If you just use a rotated Cartesian system, you have just another Cartesian system and just use it. Then it's much simpler to use the original Cartesian system and then use the corresponding rotation matrix (perhaps followed by a space reflection if they also choose to change the orientation of the coordinate system, which I would avoid except when really necessary ;-)).

    I don't know the book in detail, but I've the impression that it is rather misleading. I fear, it doesn't make the proper carefull distinction between vectors (tensors) and vector (tensor) fields. To introduce (normalized) curvilinear orthogonal coordinates, in my opinion, makes only sense when applied to fields.

    Take a vector field [itex]\vec{A}(\vec{r})[/itex]. Then you define its components wrt. the curvilinear coordinates at the given position vector [itex]\vec{r}[/itex], and then the vector components and the basis vectors are uniquely defined. The vector field itself is, of course, an inveriant object. You can express with any coordinates you like:
    [tex]\vec{A}(\vec{r})=A_x(x,y,z) \hat{x} + A_y(x,y,z) \hat{y} + A_z(x,y,z) \hat{z}=A_r(r, \varphi,z) \hat{r} + A_{\varphi}(r,\varphi,z) \hat{\varphi} + A_{z}(r,\varphi,z) \hat{z}.[/tex]
    The only caveat is that curvilinear coordinates generally do not cover the entire [itex]\mathbb{R}^3[/itex]. E.g., the here considered cylinder coordinates are singular along the [itex]z[/itex] axis, because there the Jacobian determinant between Cartesian and cylinder coordinates, which is [itex]r=\sqrt{x^2+y^2}[/itex], vanishes.

    This is the more important in physics, where the fundamental laws are formulated as local (!) field theories.
     
  20. Dec 25, 2013 #19
    What can I say? I totally agree. As an engineer, I thought that the issues raised by the OP were just a "mathematician thing." Apparently not.

    Chet
     
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