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Symmetries and shift of coordinates in 3D

  1. Oct 6, 2009 #1
    Hello,
    given a vector x=(a,b) in 2D, and considering another vector obtained by shifting cyclically the coordinates of x, we get x'=(b,a). It is straightforward to prove that x and x' are simply the reflection of each other on the line k(1,1).

    Now let's suppose we are in 3D space.
    Given a vector x=(a,b,c) we can form other two vectors:
    x' =(c,a,b)
    x''=(b,c,a)

    What is the symmetric relationship between such vectors?
     
  2. jcsd
  3. Oct 6, 2009 #2

    tiny-tim

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    Hello mnb96! :wink:

    Hint: the operation (x -> x') cubed takes (a,b,c) to … ? :smile:
     
  4. Oct 7, 2009 #3
    oh :) thanks for the hint!
    If we apply the operation to any x three times we always get: (xxx)=x.
    I can also notice that if we apply the operation to any vector n=(a,a,a) we get [itex]n^k=n[/itex] for any integer k.

    Should I deduce that this operation defines 120 degrees symmetries across the axis (1,1,1) ?

    And even if so, the reasoning I used was not too rigorous, so I was wondering if there is a more 'algebraic' approach to find the symmetry axes/planes, given three vectors x' x'' x'''.
    Thanks again.
     
  5. Oct 7, 2009 #4

    tiny-tim

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    Hi mnb96! :smile:

    I know it looks like a shortcut, but it's actually quite rigorous …

    you know that R3 always (a,b,c) -> (a,b,c), so the rotation must be 2πn/3 for some value of n not divisible by 3, which basically means n = ±1. :wink:

    For any three vectors x' x'' x''' (assuming they are at 120º), symmetry means you can find the axis simply by adding them … in the original case, you get (a+b+c,a+b+c,a+b+c). :smile:
     
  6. Oct 7, 2009 #5
    Hi!
    thanks for the answer!
    It seems clear now that the reflection axis is again (1,1,1) also in 3D!
    However, I am not an expert with the topic, and I feel you are going faster than I can :)
    I still need two points to be clarified. The first is, how can you state the following without proving it?

    [itex](\mathbf{x}^3 = \mathbf{x})[/itex] [itex]\Rightarrow[/itex] "the operation is a rotation"

    In my opinion, in order to conclude that we are dealing with a rotation we should at least show that:
    [itex]<\mathbf{x}^p, \mathbf{x}^{p+1}> = <\mathbf{x}^q, \mathbf{x}^{q+1}>[/itex] for any p,q Am I right? Basically, the angle produced by each application of the operator must be always the same.

    The second point is:
    "symmetry means you can find the axis simply by adding them"
    Is this a basic definition? I find it obvious when dealing with only two reflected vectors. For example, if x and x' have been reflected across a versor n, then obviously x+x' must be the reflection axis. But what if we have three vectors x , x', x''? If you sum them all we can't get a reflection axis, but we can get a rotation axis. I am bit confused on this point.
     
  7. Oct 7, 2009 #6

    tiny-tim

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    Hi mnb96! :smile:
    No, if R3 = 1 (and assuming it's linear), then the angle thing works anyway.
    Yes, only a rotation axis …

    the 2D reflection in an axis was really a 3D rotation about that axis … so finding the axis that way worked because it works as a rotation.

    Why does it work? If there's an axis, then it's just obvious! :wink:
     
  8. Oct 7, 2009 #7
    Hi again :)
    You convinced me about the point (2).
    I have still some doubts about the point (1).

    you said: "if [itex]R^3[/itex]= 1 (and assuming it's linear), then the angle thing works anyway"

    But, what would you answer to a dumb student like me who asks you "I don't believe it! you have to prove that!" :)
    Seriously, I intuitively and visually understand it, but I was just curious to see how to prove that given a vector space, and a linear operator [itex]R[/itex] such that [itex]R^k[/itex] is the identity, then R is a rotation.

    Thanks!
    Oh...BTW, what is the branch of mathematics in which these things are studied in detail?
     
  9. Oct 7, 2009 #8

    tiny-tim

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    Hi mnb96! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    I'm assuming that R preserves shapes (unlike a shear transformation, which is still linear, but "squashes" one dimension, like tilting a rectangle into a parallelogram), and that it doesn't turn space inside out (a 3D reflection).

    Then, if Rk = 1, then R also preserves sizes.

    If it preserves both shape and size, then Euler proved in 1776 that,if you fix one point, then a whole line must be fixed … see http://en.wikipedia.org/wiki/Euler's_rotation_theorem#Euler.27s_theorem_.281776.29" :wink:
    Either three-dimensional geometry or linear algebra. :smile:
     
    Last edited by a moderator: Apr 24, 2017
  10. Oct 7, 2009 #9
    Hello,
    everything became pretty obvious when I thought of the transformation R in matrix form.
    It is a bit painful to write it down in latex but it is essentially the identity matrix with all the columns shifted by one to the right (the last column becomes the first). Then I observed that [tex]R^T R = I[/tex] which clearly proves that it is indeed a rotation.

    The funny thing is that in order to find the rotation axis in n-dimensions I found myself reproducing the matrix proof on wikipedia that you just mentioned 5 minutes after :)
    The rotation axis is always (1,1,1,...,1) !


    I agree that here we have been just using linear algebra and euclidean geometry. However I know there are more specialized branches of algebra which specifically study and classify symmetries, although I dont know exactly which ones, but I was refering to those.


    Now, one very last thing:
    Do you know if it is possible to deduce that this operation R is indeed a rotation and then find the rotation axis without resorting to linear algebra, but just using 'Geometric Algebra'?

    Thanks a lot.
     
  11. Oct 7, 2009 #10

    tiny-tim

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    Hello mnb96! :smile:

    In 4 or more dimensions, it's more complicated …

    foe example, in 4 dimensions, you can rotate y and z while keeping x and w the same … or you can even have one rotation for y and z and a different rotation for x and w.

    The general study of symmetries is Group Theory …

    but there are more specialised branches such as Exterior Calculus (Exterior Forms).
    You can prove it's a rotation using that theorem of Euler, and then find the axis just using components. :smile:
     
  12. Oct 7, 2009 #11

    D H

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    In general, a permutation of elements in an n vector can be described by a permutation matrix. A permutation matrix is a binary matrix (elements are either zero or one). The determinant of a permutation matrix is either +1 or -1. If the determinant is +1 the permutation is a proper rotation.

    There is an easy way to determine the determinant of a permutation matrix. Swap pairs of elements in the vector {x,y,z,...} until you arrive at the permuted vector. Tthere are more than one way to do this. However, all paths from the original vector to the permuted vector will have one thing in common: The number of swaps will always be even (or always be odd). The determinant is +1 and the permutation is a proper rotation if the number of swaps is even, -1 and an improper rotation if the number of swaps is odd.

    Example 1: {x,y,z} -> {z,x,y}.
    Swap 1: Swap x,y. {x,y,z} -> {y,x,z}
    Swap 2: Swap y,z. {y,x,z} -> {z,x,y}
    Two swaps, so this is a proper rotation.

    Example 2: {x,y,z} -> {z,y,x}.
    Swap 1: Swap x,z. {x,y,z} -> {z,y,x}
    One swap, so this is an improper rotation.

    Example 3: {x,y,z,w} -> {w,x,y,z}.
    Swap 1: Swap x,y. {x,y,z,w} -> {y,x,z,w}
    Swap 2: Swap y,z. {y,x,z,w} -> {z,x,y,w}
    Swap 3: Swap z,w. {z,x,y,w} -> {w,x,y,z}
    Three swaps, so this is an improper rotation.
     
  13. Oct 7, 2009 #12
    Hello,
    thanks you both. Your answers are very helpful.

    I was thinking now how to approach the problem in n-dimensions.
    I figured out that if R is the NxN orthogonal transformation matrix I mentioned in the previous post (the NxN identity matrix shifted), it is possible to prove that:

    1) R has always det(R)=1 when N is odd, and det(R)=-1 when R is even
    2) R is an orthogonal matrix
    3) when N is odd, R has only one real eigenvector (with corresponding eigenvalue=1)

    Is there a geometric interpretation given these observations?
    There is a subspace of dimension 1 (an axis) which is clearly invariant under the transformations R. Is then allowed to talk about rotation? If not, then what are we doing?
     
  14. Oct 7, 2009 #13
    ...actually it is straightfoward to prove also a 4th property, that is:

    4) for N odd, given any transformed vector x'=Rx, the distance between x' and the eigenvector-axis is always the same.

    Should we really conclude that when we are in a space of dimensions N (where N is odd), the transformation R is indeed a rotation across (1,1,1,...,1) ?
     
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