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Losing the Units

  1. Oct 2, 2012 #1
    Hi All and sorry if this is too easy a question but here goes....

    Sines, Cosines and the rest of the trig functions are the ratio of two lengths and thus are dimensionless quantities.

    That is if I plug in a value for t in sin(ωt) there are no units.

    For example the solution of

    x'' + ω^2 x = 0 with x(0) = x0 and x'(0) = v0 is given by

    x(t) = x0 cos(ωt) + v0/ω sin(ωt)

    The units come from the initial conditions not the sine or cosine.

    So here is the question...

    Same is true ( I believe ) when using the natural exponential function exp(t).

    How does one simply explain this.

    I tried to reason it out using eulers formula exp(iω) = cos(ω) + i sin(ω) figuring that again we get ratios of lengths,
    however in the case where the real part is non-zero we get another exponential (which is not the ratio of lengths)

    exp(a +ib) = exp(a)(cos(b) + i sin(b))

    Is there a simple explaination as to why we "lose the units" when using the exponential function?
  2. jcsd
  3. Oct 2, 2012 #2
    It is not strictly correct to supply any argument with dimensions to an exponential or trig function. Example: [itex]\omega t[/itex] has no dimensions (frequency is 1/time, and time/time is dimensionless).

    Why is it incorrect? Consider a series expansion of the exponential function.

    [tex]e^x = 1 + x + \frac{1}{2} x^2 + \ldots{}[/tex]

    [itex]x[/itex] must be a dimensionless parameter. If it had, say, dimensions of length, how would we add length to 1 and to length squared?

    Nevertheless, in physics this strict need is sometimes ignored. As an example, you should take as implicit that [itex]e^t[/itex] is in fact [itex]e^{t/\tau}[/itex] where [itex]\tau[/itex] is 1 in whatever units of time you're working with.
  4. Oct 2, 2012 #3
    Yes indeed, and it may be useful to give an example (of not ignoring this):
  5. Oct 2, 2012 #4
    Thank you both for your fast insightful and illustrative replies.
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