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Translational Invariance

  1. Apr 18, 2016 #1
    1. The problem statement, all variables and given/known data

    Consider a system of objects labeled by the index ##I##, each object located at position ##x_{I}##. (For simplicity, we can consider one spatial dimension, or just ignore an index labeling the different directions.) Because of translational invariance

    ##x'_{I}=x_{I}+\delta x##

    where ##\delta x## is a constant independent of ##I##, we are led to define new variables

    ##x_{IJ} \equiv x_{I}-x_{J}##

    invariant under the above symmetry. But these are not independent, satisfying

    ##x_{IJ}=-x_{JI}, \qquad x_{IJ} + x_{JK} + x_{KI} = 0##

    for all ##I,J,K##. Start with ##x_{IJ}## as fundamental instead, and show that the solution of these constraints is always in terms of some derived variables ##x_{I}## as in our original definition. (Hint: What happens if we define ##x_{1}=0##?) The appearance of a new invariance upon solving constraints in terms of new variables is common in physics: e.g., the gauge invariance of the potential upon solving the source-free half of Maxwell’s equations.

    2. Relevant equations

    3. The attempt at a solution

    If ##x_{1}=0##, then ##x'_{1}=\delta x##.

    Not sure where to go from here.
     
  2. jcsd
  3. Apr 18, 2016 #2

    Samy_A

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    If you set ##x_1=0##, how could you define ##x_I## such that the constraints ##x_{I1}=-x_{1I}## are verified?
     
  4. Apr 18, 2016 #3
    Aren't the constraints ##x_{I1} = - x_{1I}## always satisfied regardless of whether ##x_{1} = 0## or ##x_{1} \neq 0##, simply because ##x_{IJ} = x_{I} - x_{J}##?
     
  5. Apr 18, 2016 #4

    Samy_A

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    What the exercise asks you to do is the following:
    Assume that you have the numbers ##x_{IJ}##, and that they satisfy the constraints ##x_{IJ}=-x_{JI}, \qquad x_{IJ} + x_{JK} + x_{KI} = 0##.
    Now show that there exists numbers ##x_I##, such that ##x_{IJ}=x_I-x_J##.

    In other words, the ##x_I## are not given, you have to define them to suit the given ##x_{IJ}##.

    The hint is to set one value, ##x_1=0##.
    Now for an arbitrary index ##I##, how could you define ##x_I## so that they suit the given ##x_{IJ}## and satisfy the constraint ##x_{I1}=-x_{1I}##.
    (I use the index 1 here because you start by setting ##x_1=0##. Of course you will have to show that the ##x_I## you construct satisfy the constraints ##x_{IJ}=-x_{JI}, \ x_{IJ} + x_{JK} + x_{KI} = 0##).
     
  6. Apr 18, 2016 #5
    I need to find the solutions ##x_{IJ}## (in terms of some variables ##x_{I}## and ##x_{J}##) which satisfy the constraints ##x_{IJ}=-x_{JI}## and ##x_{IJ}+x_{JK}+x_{KI}=0##.

    The funny part is that we already know that the solution is ##x_{IJ} = x_{I} - x_{J}##, but we have to assume that we don't know the solution ##x_{IJ}## and set about finding it.



    For ##x_{1}=0## and ##x_{I1} = - x_{1I}##, one possible choice of ##x_{I1}## and ##x_{1I}## are ##x_{I}## and ##-x_{I}## respectively.

    ##x_{IJ}## is antisymmetric, and ##x_{I} - x_{J}## is also antisymmetric, therefore ##x_{IJ} = x_{I} - x_{J}## is a possible solution of the constraints.

    To prove that the solution ##x_{IJ} = x_{I} - x_{J}## satisfies the constraints, we need to plug in the solution into the constraints.



    Am I on the right track?

    My idea is to conjecture a possible form of ##x_{IJ}## and then to check if the conjectured solution satisfies the constraints.
     
  7. Apr 18, 2016 #6

    Samy_A

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    You misunderstand the exercise.
    First they show you how the numbers ##(x_I)_I## allow to define the numbers ##(x_{IJ})_{IJ}## by setting ##x_{IJ}=x_I-x_J##.
    It then appears that these numbers ##(x_{IJ})_{IJ}## satisfy the constraints ##x_{IJ}=-x_{JI}, \ x_{IJ} + x_{JK} + x_{KI} = 0##.

    Now they want you to reverse this process.
    Assume you have numbers ##(x_{IJ})_{IJ}## that satisfy the constraints ##x_{IJ}=-x_{JI}, \ x_{IJ} + x_{JK} + x_{KI} = 0##. No ##x_I## are given or assumed.
    What you are asked to do is: show that you can define numbers ##(x_I)_I## such that ##x_{IJ}=x_I-x_J##.
     
  8. Apr 18, 2016 #7
    I still don't see how my understanding of the exercise differs from yours.

    Perhaps if you provide a few initial lines of the solution, that will clarify things.
     
  9. Apr 18, 2016 #8

    Samy_A

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    Set ##x_1=0##. Now, you need ##x_I## to satisfy ##x_{1I}=x_1 - x_I##. That should give you a clue about how to define ##x_I##.
    Once that is done, you will have to show that ##x_{IJ}=x_I-x_J## holds for all indices ##I,J##.
     
  10. Apr 18, 2016 #9
    Let's define an origin such that ##x_{1} = 0## and let's define a coordinate ##x_{I}## for particle I such that ##x_{I} = x_{I1}##.

    Now, the first constraint ##x_{IJ} = -x_{JI} \implies x_{I1} = -x_{1I} \implies x_{1I} = - x_{I}##.

    Next, the second constraint ##x_{1I} + x_{IJ} + x_{J1} = 0 \implies - x_{I} + x_{IJ} + x_{J} = 0 \implies x_{IJ} = x_{I} - x_{J}##.

    Am I correct?
     
  11. Apr 18, 2016 #10

    Samy_A

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    As the ##x_{IJ}## are given, and you have to define the ##x_I##, it would make more sense to write the last equation as ##x_I=-x_{1J}##. But given that, that is a good definition for ##x_I##.
    Yes.
     
  12. Apr 18, 2016 #11
    I modified my post a little (before I was alerted about your new post). Would you check if my modified post is a better answer?
     
  13. Apr 18, 2016 #12

    Samy_A

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    Yes, totally correct.
     
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