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Log(meter) is about 80?

  1. Sep 3, 2008 #1
    If we reject the notion of fundamental dimensionful constants and interpret the constants h-bar, c and G as rescaling constants, an artifact of declaring certain quantities to be incompatible and using inconsistent units (e.g. for space and time we use the meter and second and we assume that space and time are fundamentally incompatible quantities), we can reason as follows.

    The Planck length is

    [tex]l_p = \sqrt{\frac{\hbar G}{c^3}}=1.616 252 10^{-35}\text{ meter}[/tex]

    But in Planck units hbar = c = G = 1, we have:

    [tex]l_p = 1[/tex]

    If we take this literally and assume that [tex]l_p[/tex] is a pure number, we have:

    [tex]1.616 252 10^{-35}\text{ meter} = 1[/tex]


    [tex]\text{meter }= 6.187746\times 10^{34} [/tex]

    Now, this assignment of a pure number to the meter depends, of course, on the way we assign numbers to hbar, c and G when we define Planck units. Assuming that the "correct" assignment will only differ from putting hbar = c = G = 1 by factors of order unity, we can put:

    [tex]\text{meter }= 6.187746\times 10^{34}\lambda [/tex]

    where [tex]\lambda\approx 1[/tex]

    If we substitute this in an expression involving only lengths that is dimensionally correct, the constant lambda will drop out exactly. But we can write down expressions that don't make any sense from a fundamentalistic dimensional point of view, in which the lambda dependence is nevertheless negligible. Consider e.g. Log(meter). We have:

    [tex]\log(\text{meter})= 80.11 + \log(\lambda) [/tex]

    We can, of course, express this in terms of meters by dividing this by 6.187746 10^(34) times lambda, but then the result will strongly depend on lambda.
  2. jcsd
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