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I Complex Exponential solutions in time invariant systems

  1. May 29, 2016 #1
    Hi there! First Post :D

    In a recent CM module we've been looking at coupled oscillators and the role of time translational invariance in the description of such physical systems. I will present the statement that I am having trouble understanding and then continue to elaborate.

    In stating that a system has a time translational invariance, it follows that

    x(t+c) = f(c)x(t)

    where x(t) is some function of time, f(c) is some function of proportionality dependent on some constant c, and therefore x(t+c) is the function x at some later time.

    After this, it is stated that differentiating with respect to c and setting c=0 gives

    d/dt[x(t)] = (omega) x(t) where (omega) = d/dt[ f(c=0) ]

    It's then given that x(t) = exp^(omega)(t)

    I can clearly see that the exponential form is a solution to differential equation above. My question is how is the differential equation derived with no known form of x(t)? In particular, how does the time derivative of f(c) come into the equation if f(c) is only dependent on c? (I understand that the chain and product rules must be used but wouldn't the time derivative of f(c) return a zero value?)

    As well as this, some friends have alluded to Noether's theorem as the governance of this particular rule; is this warranted? I don't see any particular conservation laws here.

    Any help or insight is much appreciated!
  2. jcsd
  3. Jun 1, 2016 #2


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    Homework Helper

    Hi Dagorodir,
    The first thing you say is done is the differentiating with respect to c.
    If you do this, you should get:
    ##\frac{\partial}{\partial c}x(t+c) = f'(c)x(t)##
    Setting c to zero, you get
    ##\frac{\partial}{\partial c}x(t+0) = f'(0)x(t)##
    Now, notice that if you set dt = dc, then
    ## x(t+dt) = x(t+dc)##
    So, at c=0,
    ##\frac{\partial}{\partial c}x(t+c) = \frac{\partial}{\partial t}x(t+c)##
    Then you can jump to the conclusion:
    ##\frac{\partial}{\partial t}x(t) = f'(0)x(t)##
    Where ##f'(0) = \omega##.
    I think that the only problem in the explanation you posted was in using d/dt (f(c)) to define omega. f is a function of one variable, and you want to find its derivative evaluated at 0.
  4. Jun 28, 2016 #3
    Hi RUber,

    Thanks for your reply!

    Your explanation makes sense to me; I think that I was failing to make the link between the derivatives and therefore using the fact that c is a constant to get over the last hurdle.

    Thanks again
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