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A question on the commutativity of finite rotations

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  1. Apr 6, 2015 #1
    I was reading a section in my book discussing the commutativity of infinitesimal and finite rotations. In the book the authors try to set up a scenario to explain why finite rotations are not commutative. The following is an excerpt from the book regarding this language:

    "The impossibility of describing a finite rotation by a vector results from the fact that such rotations do not commute, and therefore, in general, different results will be obtained depending on the order in which the roations are made. To illustrate this statement, consider the successive application of two finite rotations described by rotation matrices ##\lambda_1## and ##\lambda_2##. Let us associate the vectors ##\vec{A}## and ##\vec{B}## in a one-to-one manner with these rotations. The vector sum is ##\vec{C}=\vec{A}+\vec{B}##, which is equivalent to the matrix ##\lambda_3=\lambda_2\lambda_1##. But because vector addition is commutative, we also have ##\vec{C}=\vec{B}+\vec{A}##, with ##\lambda_4=\lambda_1\lambda_2##, But we know that matrix operations are not commutative, so that in general ##\lambda_3\neq\lambda_4##."

    Now, to me, this language seems kind of murky so I had to think on this for awhile. Assume you have point P desginated by a vector ##\vec{r}## in a reference frame S. Applying the first rotation ##\lambda_1## can be represented by rotating P in S. Once the rotation is complete, the point is now at P'. If you consider vector ##\vec{A}## the vector that connects point P to P', it would be a representation of the rotation. Is this a physically plausible way to interpret the above explanation if you apply the same reasoning to the case of multiple rotations.
     
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  3. Apr 7, 2015 #2

    Simon Bridge

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    No idea... I get lost in that sort of talk too.
    To see what they are on about, get an object like a book and rotate it two different ways and see if it ends up the same.
    i.e. rotate 90deg about the vertical and then abouth a horizontal and see if it works out the same in the other order.
    Work out how to describe that with vectors.

    This approach is more useful than trying yo figure out what some author was trying to say.
     
  4. Apr 7, 2015 #3

    vanhees71

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    The math behind this is that for a Lie group you can define "infinitesimal transformations", mapping it to a Lie algebra. Formally the Lie algebra is the tangent space at the unit element of the group. From the Lie algebra you come back to the group via the exponential mapping.

    In quantum theory you encounter this mathematics from the very beginning. E.g., the translations in space form an Abelian group, and the infinitesimal generators are the momentum operators. Then there are the rotations, which form a non-abelian group. Here the infinitesimal generators are the angular-momentum operators, which span the Lie algebra su(2) of the rotation group. The commutators define the Lie bracket. The Lie product for the su(2) is thus given by the commutation relations of the angular-momentum operators
    $$[J_j,J_k]=\mathrm{i} \sum_{l=1}^3 \epsilon_{jkl} J_l.$$
    A finite rotation is then given by the operator-exponential function
    $$D_{\vec{n}}(\varphi)=\exp(-\mathrm{i} \varphi \vec{n} \cdot \vec{J}).$$
    Two rotations around two non-collinear axes ##\vec{n}_1## and ##\vec{n}_2## do not commute, although the infinitesimal rotations commute, because vector addition is always commutative.
     
  5. Apr 7, 2015 #4
    Thank you for those responses.

    @Simon Bridge: I did do the book turn thing to get an idea of what they are talking about and it is clear. Right now I am actually just trying to wrap my head around the author's language. Are you saying it is a better idea to just leave the language and take the main idea away that finite rotations don't commute (If I had the money to get a better book, I would. Marion and Thorton is all that I have at the moment).

    @vanhees71: My mathematical knowledge doesn't go past a bit of Linear Algebra. I am working on gaining more, but for now your explanation is a bit beyond my grasp. I do get the basic idea from a few Googlings and from what I can gather from the searches, you're saying that the application of the commutators to the finite rotations (the non-Alebian group) is order dependent and not so for infinitesimal rotations (the Alebian group). Is that in the ballpark?
     
  6. Apr 7, 2015 #5

    vanhees71

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    The point is that, roughly speaking, "infinitesimal" transformations commute to linear order of the parameters, while in general not at higher orders. For rotations, you can indeed make clear that for finite rotations two rotations around different axes do not commute. Simon gave a nice example.

    If you have no money for a good book, I suggest you google for free manuscripts online. There are tons of good ones around about classical mechanics, e.g.,

    http://ocw.mit.edu/courses/find-by-topic/#cat=science&subcat=physics&spec=classicalmechanics
     
  7. Apr 7, 2015 #6
    Do you have any suggestions for one that is of a decent rigor or are the ones on MIT's OC of a decent rigor?
     
  8. Apr 7, 2015 #7

    vanhees71

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    What level do you need?
     
  9. Apr 7, 2015 #8
    I am looking for something that presents the same material you would find in a text use for a two semester undergrad Mechanics class. I have a copy of Marion and Thorton's text which I used when I took the class (as for why I'm trying to "relearn", I'd rather not get into it). It's a good text with good rigor, but the author's explanations sometimes leave me to work out steps to try and see how they got from one conclusion to the next.
     
  10. Apr 7, 2015 #9

    vanhees71

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    I don't know this specific book, but from the table of contents it looks very complete. Is there a chapter about Lie groups and Lie algebras? That would answer your questions. The most natural place is in the chapter about the Hamilton canonical formalism, where you have a direct realization of the Lie algebra of the space-time symmetries in terms of Poisson brackets. It's closely related to Noether's theorems, relating each conserved quantity with a one-parameter subgroup of the symmetry group of the dynamical system. For the space-time symmetries these are energy (time translations), momentum (spatial translations), angular momentum (rotations), and center of mass (energy) position (Galilei/Lorentz boosts). Unfortunately this point of view is rarely found in any texts. I've worked this out in my own lecture notes, but these are in German:

    http://theory.gsi.de/~vanhees/faq/mech/mech.html
    http://theory.gsi.de/~vanhees/faq-pdf/mech.pdf
     
  11. Apr 7, 2015 #10
    The book actually doesn't get into the Hamiltonian formalism until chapter 6. This was actually the division in my courses between Mech 1 and Mech 2 (Mech 2 started with an introduction into Calculus of Variations then went into the Hamiltonian formalism). Nothing on Lie algebras.

    Thank you for the notes too. :)
     
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