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Dimensionally impossible equation

  1. Oct 23, 2015 #1
    I was reading some stuff about dimensionally impossible equations. It was said that the equation

    v = e^bt (b is a constant so that bt is dimensionless)

    was dimensionally impossible. I understand that BUT I don't understand what they said next about the dimension of the right-side. They said it was dimensionless, but I thought it had dimension T (like the left-side has dimension LT^-1)... Can someone explain me why I'm wrong? Thanks :)
     
  2. jcsd
  3. Oct 23, 2015 #2

    jbriggs444

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    If that equation were written as v(t) = v0ebt then it would make more sense.

    For that to make sense, ebt has to be dimensionless.
     
  4. Oct 23, 2015 #3
    You said that ##bt## was dimensionless, so of course ##e^{bt}## is dimensionless also. But this is inconsistent with ##v(t)## having dimensions of length/time, so it is a "dimensionally impossible equation".

    Generally speaking, transcendental functions like ##e^x##, ##\sin(x)##, ##\log(x)##, etc. can never operate on units and are always dimensionless. A small exception is when the inverse function appears in the same equation, like ##f(t) = e^{4 \log t}##. But it's unclear why you would ever write it this way rather than ##f(t)=t^4##.
     
    Last edited: Oct 23, 2015
  5. Oct 24, 2015 #4
    Any such equation has a constant in front, in this case, the constant is 1, in appropriate units. for example, v = e^bt would mean: v = (1 m/s) e^bt, a particular case of the equation written by jbriggs 444 above
     
  6. Oct 24, 2015 #5

    HallsofIvy

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    If b has dimensions of "1 over time", say, sec^{-1}, then bt is "dimensionless". Then ##v(t)= x_0e^{vt}## has whatever units ##x_0## does.

     
  7. Oct 24, 2015 #6
    No, unless the text specifically states that a particular system of natural units are being used, such as atomic units. You would never see an equation like this with implied SI units.
     
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