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Scaling and Translation Invariance

  1. Sep 21, 2009 #1
    Let's say I have a vector x in [tex]\mathcal{R}^3[/tex]. Let's also suppose that any vector x undergoes the transformation x' = kx (where k is a positive real).
    Obviously, normalizing the vector will give us a quantity which is invariant to uniform scaling. In fact, [tex]\frac{\mathbf{x'}}{|\mathbf{x'}|} = \frac{k\mathbf{x}}{|k\mathbf{x}|} = \frac{\mathbf{x}}{|\mathbf{x}|}[/tex].

    Now, is it possible to find another quantity which is invariant to both translation and uniform-scaling?
    The vector x would now undergo the transformation x' = kx+a, where k is a constant scalar, and a is a constant vector in [tex]\mathcal{R}^3[/tex].

    Thanks!
     
  2. jcsd
  3. Sep 21, 2009 #2

    Ben Niehoff

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    The unit tangent vector to a curve is translation- and scale-invariant. I.e., the vector

    [tex]\vec t = \frac{d \vec r / d \lambda}{\left| d \vec r / d \lambda \right|}[/tex]

    evaluated at some particular [itex]\lambda[/itex].
     
  4. Sep 22, 2009 #3
    ok,
    but now you gave me a parametric curve in space, while in my case the situation is different.
    In fact, the original (and unknown) vector was simply [tex]\mathbf{x}=x_1\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3}[/tex].
    Then, such vector underwent the transformation x' = kx+a, and the vector x' is the only think which is known.

    The problem is: What are the possible quantities which are invariant to euclidean transformations?

    - for uniform scaling, it is x' normalized
    - for rotation, it is the module of x'
    - for translation ???
     
  5. Sep 22, 2009 #4

    Ben Niehoff

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    There is no position vector which is invariant under translation. This should be pretty clear, considering that translation changes every vector by a nonzero vector a. You are trying to solve the equation

    x = x + a

    which has no solution.

    Relative vectors may be translation-invariant, such as the vector A defined by

    A = x - y

    for some position vectors x and y. Then

    A' = x' - y' = (x + a) - (y + a) = x - y

    and so A is invariant. Consequently, velocity vectors are also translation-invariant, because they are just infinitesimal relative vectors.

    Note: the space of relative vectors A is called an "affine space". This is like a vector space, but without the notion of an origin.
     
  6. Sep 23, 2009 #5
    Hi,
    thanks for the clear explanation.
    You are perfectly right in stating that there is no vector [tex]\mathbf{x}\in \mathcal{R}^3[/tex] which is invariant under translation. We both agree on this.

    However, I must point out that you again put several unwanted constrains to the original problem, and then came to the obvious conclusion that it is impossible to solve.

    In fact, in my first post I was talking about a quantity which is invariant under translation. I carefully avoided the word vector because I am not necessarily interested in finding invariant elements in [tex]\mathcal{R}^3[/tex], the invariant quantities may be as well elements of [tex]\mathcal{R}^2[/tex] or even [tex]\mathcal{R}[/tex].
    I think you should rather formulate the problem as:

    f(x) = f(x+a)
    and try to find a suitable function f.

    For example, for uniform-scaling we took an [tex]f:\mathbf{R}^3 \rightarrow \mathbf{R}^3[/tex], which was precisely [tex]f(\mathbf{x})=\frac{\mathbf{x}}{|\mathbf{x}|}[/tex], and that made the job.

    For rotation invariance we took instead an [tex]f:\mathbf{R}^3 \rightarrow \mathbf{R}[/tex] which was [tex]f(\mathbf{x})=|\mathbf{x}|[/tex].

    In your reasoning, you are making the double assumption that f is of the kind [tex]f:\mathbf{R}^3 \rightarrow \mathbf{R}^3[/tex] and it is moreover the identity function! That's why the problem was clearly impossible to solve.
    Under such assumption, even with the uniform-scaling you are not gonna get anywhere because the equation x=kx clearly doesn't have solutions.
     
  7. Sep 23, 2009 #6

    Ben Niehoff

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    It's not my fault you were unclear. Solve it yourself.
     
  8. Sep 23, 2009 #7
    I am sorry you took it so roughly.
    I appreciated anyways your help and interest with your answers, and obviously I did not mean to make any kind of remarks about you.
    In my last post I just wanted to make things very clear (perhaps too much?).

    Besides, after reading my first post, I honestly don't think it was so unclear; surely it lacked rigor, but I also gave a specific example which clearly could not work with your reasoning.

    And by the way, (if I did not do any mistake) I proved that this problem is not possible to solve unless you put a constrain on the direction of translation...but I hardly believe anyone is interested in this.
     
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