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Homework Help: Why does translation work with the extra dimension (Homogeneous coordinates.)

  1. Feb 8, 2017 #1
    1. The problem statement, all variables and given/known data


    I have read a lot on homogenous coordinates
    and I feel like I now have a solid foundation.

    However none of the videos or books I have read give an explicit reason as to why translation with the extra dimension works(i.e. it does not result in scaling).

    Here 's what I have come up with. The extra dimension allows for the entire plane to be moved as a whole i.e. instead of moving the vectors within the plane in which they exist the extra coordinate moves the entire plane in the dimension in which it exists. This then results in a translation.

    Is this wrong? If so, why does translation work with the extra dimension?

    Why does translation result in scaling without the extra dimension?

    2. Relevant equations

    3. The attempt at a solution
    Here 's what I have come up with. The extra dimension allows for the entire plane to be moved as a whole i.e. instead of moving the vectors within the plane in which they exist the extra coordinate moves the entire plane in the dimension in which it exists. This then results in a translation.
    Last edited by a moderator: Feb 9, 2017
  2. jcsd
  3. Feb 9, 2017 #2


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    You are asking one of those "why" questions which really doesn't have a clear context. "why does 2+2=4?" I can give you various contexts for this matrix representation of translations which may satisfy your need for "why-ness" but the direct answer to any "why" question, since we're talking pure math here is "because the axioms so imply!".

    First point, There actually is scaling but not of what you're paying attention to... Note that the matrix rep of translations does change the length of the representation vector (x,y) --> (x+ay,y). It does not map e.g. a unit vector to a unit vector which, for example a rotation does. We simply pay attention to the effect on a subspace (x --> x+ay) and choose our original vector (so y=1) to, by construction, represent translation this way. It is a matter of appreciating that --as you say-- shears translate whole planes onto themselves.

    Second point, The translations form a continuous group (Lie group) and thus have a linear representations (in fact an infinite number of them). Any such group will have linear representations, its all a part of Lie theory. (linear representation = representation in the linear group = group of invertible matrices under multiplication).

    Third point, [Which is probably more confusing than enlightening but let me give it a go...] You can embed the translation group in a natural way withing a larger (pseudo)orthogonal group. You can view the translations group as an extension of a rotation group, specifically with the group of rotations in d-dimensions, SO(d) = special orthogonal group in d dimension, you can extend this group to include the translations as well, ISO(d) = inhomogeneous special orthogonal group = the Euclidean group in d-dimensions. Both ISO(d) and SO(d+1) can be embedded in an indefinite case SO(d+1,1) which itself has a (d+2)x(d+2) matrix representation. They are subgroups leaving a given subspace invariant and thus will each have a d+1 dimensional representation if we choose the basis carefully and drop the invariant one. The inhomogenous case appears as a boundary to the homogeneous group's embedding (and there's an indefinite homogeneous case on the other side of this boundary.) To see the inhomogeneous boundary case you vary which 1-dim subspace you leave invariant and note the null vectors are on the boundary between "space-like" and "time-like" cases. The inhomogenous case is the subgroup leaving a null vector invariant.

    In short there is a specific natural way to embed the inhomogeneous groups (which contain translations) within a higher dimensional homogeneous group (provided we allow indefinite metrics) which in turn has a fundamental matrix representation. This is to say translations are singular deformations of rotations. Geometrically we see this every day. We walk on Earth and imagine it is a translation but we are actually rotating about the Earth's center. The singularity is in the scale of our motion relative to the Earth's radius. The rotations expressed with trigonometric functions involve very small angles so we take the small angle limit and get translation. cos(x)=1, sin(x)=x = tan(x))

    It manifests also in our imagining that boosting in space-time is simply a translation in the velocity space. It is in fact a pseudo-rotation expressible using hyperbolic trigonometry but the singularity is in our small distance scale relative to our time units (meters are a very small fraction of light-seconds) thus we can use the small pseudo-angle limit of the hypertrig. cosh(x)=1, sinh(x)=x=tanh(x).

    You could create a specific boundary case, where you have a laser beam emitted from the north pole and you pick combinations of rotations about the center of the Earth and boosts (i think along the polar axis) such that the laser beam appears unchanged both in direction and frequency. That set of transformations would leave the event of t=0 at the center of the Earth invariant but the formal structure of that group would be the same as ISO(2) translations and rotations in some abstract plane. You would thus be expressing translations using 3x3 matrices (the 3-rotations and 3-boosts leaving t=0, r=0 invariant within the Lorentz group).
    [I think I could further clarify this example by having the radius of the Earth move but... I have to get to work now.]
  4. Feb 9, 2017 #3
    Context, I talking about 3D geometry, linear transformations using matrix representations and homogenous coordinates.
  5. Feb 9, 2017 #4
    Could someone please provide a more detailed and less technical explanation.
    To rephrase my questions:
    1. Why does trying to represent a translation without homogeneous coordinates result in scaling.(e.g. in computer graphics using a 2 by 2 matrix to represent translation in a 2D space scales the object, but using a 3 by 3 matrix works )?

    2.How does the extra dimension perform the translation of an object?
  6. Feb 9, 2017 #5


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    If you try to use only 2x2 matrices you of necessity will leave the 2-dim zero vector alone. ([itex] M\cdot \vec{0} =\vec{0}[/itex]) .
    The existence of that extra dimension allows one to perform a shear in that plane: (x,y,z) --> (x+az,y+bz,z) which for z=1 gives you translation by distance a in the x direction and b in the y direction.

    [tex] \left(\begin{array}{ccc} 1 & 0 & a\\ 0 & 1 & b\\0 & 0 & 1\end{array}\right)\left( \begin{array}{c} x\\y\\1\end{array}\right)=\left( \begin{array}{c} x+a\\y+b\\1\end{array}\right) [/tex]

    Again your "How does..." is answered by simply doing the math and seeing what happens. Are you asking to see the math worked out?

    Geometrically a shear is the linear transformation that maps say a rectangle (with corner at the origin) to a parallelogram of the same height (and area). Picture a deck of cards stacked vertically but then you slide each card a distance proportional to its height so that you get a parallelogram cross section. The fact that each coordinate's change is a linear combination of other coordinates is why it can be expressed as a linear transformation, i.e. matrix multiplication. Translations as we generally express them are changes of our coordinates that do not depend on the coordinates we are using. To express them linearly we must have an auxiliary coordinate which itself remains unchanged but off of which we can "push" via a matrix to translate the others
  7. Feb 11, 2017 #6
    Thanks for the help.
    I still don't quite get it but thanks anyway I appreciate the help.
  8. Feb 11, 2017 #7


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    Consider the simple plane linear transformation (x,y)→(x,x+y)
    Look at what it does to the line (1,y). It translates to (1, y+1).
    Similarly, if we embed a plane in 3 space as (x, y, 1) then there are linear transformations, such as quoted by jambaugh above, that act as translations on that plane.
  9. Feb 13, 2017 #8
    Thanks for the help everyone.
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