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How to do dimensional analysis of exponential functions?

  1. Jun 20, 2012 #1
    This may sound like a homework-type, but it isn't from any piece of homework, I just wanna know how to do dimensional analysis by giving an example.

    Let's have this simple equation:

    [tex]k = a^{b}[/tex]

    Where [tex]k[/tex] is a certain property of an object, [tex]a[/tex] is the mass of the object, [tex]b[/tex] is the length of the object.

    So what should the units of [tex]k[/tex] be?.

    In kilograms only, or something else? I'm not sure how to deal with exponential dimensions.
    Last edited: Jun 20, 2012
  2. jcsd
  3. Jun 20, 2012 #2


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    What is b, why do we care about z and do you have a more real-world example?
  4. Jun 20, 2012 #3
    Any expression at the seat of power is always dimensionless. e.g. x multiplied by x is x^2. Here 2 is a dimensionless constant. In your question if b = 2 then k = a^2. Which implies that units of k are unit of mass square (e.g. kg square). If were three units of k would be mass cube.

    One thing I would like to emphasize here that, in this b has to dimensionless here.
  5. Jun 20, 2012 #4
    Sorry I messed up.

    There is no z. b is in meters.

    And I don't think I have a real-world example
  6. Jun 20, 2012 #5
    i kinda messed up.

    b is not dimensionless.
  7. Jun 20, 2012 #6
    The formula should read:

    k=C a b/b0

    where b0= 1 meter or any characteristic length of the problem
    Now you need another constant (C) that matches the units of k and a.

    The same happens with exponentials, logarithms, sine, cosine, etc. Functions that take as argument a pure number
  8. Jun 20, 2012 #7
    Here C may or may not be dimensionless. Only point to be noted is exponent has to be dimensionless.
  9. Jun 20, 2012 #8
    so i can't guess the units of k.
    I have to know the units of k beforehand, then work out the proportionality constant?
  10. Jun 20, 2012 #9
    You have to know the units of k beforehand
  11. Jun 20, 2012 #10
    i guess the same technique also applies when graphing exponentials and trigo relationships.
  12. Jun 20, 2012 #11

    Stephen Tashi

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    How does dimensional analysis deal with power series?

    If I write [itex] e^x = 1 + x + \frac{x^2}{2} + .... [/itex] then if [itex] e^x [/itex] has a physical unit then (reasoning naively) [itex] 1,x, x^2, x^3 [/itex] each must have the same physical units as [itex]e^x [/itex]. I don't know of any physical unit for which this is possible, so this suggests that [itex] x [/itex] is dimensionless.

    The above reasoning may make physicists happy, but it isn't rigorous mathematical reasoning. (For example, it confuses "true for any finite number" with "true for an infinite number".) Has dimensional analysis been formulated in a rigourous way?
  13. Jun 21, 2012 #12
    well i think most physicists treat infinite quantities as ideal or theoretical. So real world quantities can only approach infinite quantities.
  14. Jun 21, 2012 #13


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    The rigorous way to handle dimensionful quantities is to assign a scaling weight to the units. For the sake of simplicity, suppose that we were only interested in units of energy, which would be the case for physics in so-called natural units. Then under a scaling of the fundamental energy unit, ##E \rightarrow \lambda E##, the set of observables of the theory, ##(A,B,C,\ldots)##, will transform according to their scaling dimensions ##(a,b,c,\ldots)## as

    $$(A,B,C,\ldots) \rightarrow (\lambda^a A,\lambda^b B,\lambda^c C,\ldots).$$

    The scaling dimension of an arbitrary function of the observables can be computed using the "dilatation" operator

    $$ \mathcal{D} = A \frac{\partial}{\partial A} + B \frac{\partial}{\partial B} + \cdots.$$

    Functions of definite energy dimension are eigenfunctions of this operator. They are also homogeneous in ##\lambda## with definite degree. Conversely, we can use the representation spaces of this operator to define irreducible representations to classify functions on the space of observables.

    Associated with the scaling is the notion of projective variables, ##(\alpha,\beta,\gamma,\ldots)##, which correspond to an irreducible set of dimensionless quantities formed from the ##(A,B,C,\ldots)##. A function of dimension ##d## can always be put in the form

    $$ F = A^{d/a} f_A(\alpha,\ldots),$$

    where ##f_A## is dimensionless. More generally, the coefficient will be some monomial in the ##(A,B,\ldots)##. This is a basic point of dimensional analysis. A function with a specific set of units can be written as a basic combination of dimensionful quantities times an arbitrary function of all of the dimensionless quantities that can be formed.

    Now the structure above of functions of definite dimension is analogous to having tensors of definite degree. We can add functions of the same dimension to get another function in the same representation, or we can multiply functions of degree ##d_1## and ##d_2## to get a function in the representation of degree ##d_1+d_2##. We are allowed to consider functions which are monomials in the homogenous coordinates ##(A,B,\ldots)## multiplied by arbitrary functions of the projective variables ##(\alpha,\beta,\ldots)##. In the case of the exponential function, we'd require that the argument be dimensionless, since it doesn't make sense to add objects with different dimensions.
  15. Jun 21, 2012 #14


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    You could assign a meaning to e.g. elength, but I doubt it would have any physical correspondence, and you'd be very limited in what you could do with it.
    In normal dimensions, you can multiply and divide any two terms meaningfully, and you can add and subtract terms with the same dimension. Correspondingly, with exponentiated dimensions, you could multiply terms of the same (exponentiated) dimension and exponentiate by other (non-exponentiated) dimensions. E.g.: (elength)(1/time) = espeed.
    That said, I can't see any use for it.
  16. Jun 21, 2012 #15
    I think it will be fundamentally wrong if we come out with an expression having some exponent with dimensions.
    x multiplied by x equals x ^2. Here u can see 2 is not a physical quantity but a number and therefore dimensionless. In the same manner any expression at the seat of expression has to be dimensionless.
  17. Jun 22, 2012 #16


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    I certainly wouldn't encourage it, and I see no use for it, but if it were done within a framework of rules that did not lead to inconsistencies then you couldn't actually say it was invalid. You've provided an example of not doing it. That does not constitute a demonstration that it cannot be done.
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