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Natural unit confusion

  1. Oct 12, 2012 #1
    If we set c=G=1, then c=1 leads to [itex]L=1/T [/itex], the [itex]G=1[/itex] means that [itex] M=L^3/T^2[/itex] and combining the two means that [itex] M=L=1/T [/itex] so far so good. But say if I also want [itex] \hbar=1 [/itex] this seems to imply [itex] M=T/L^2 [/itex] but combining this with [itex]c=1 [/itex] which gave the [itex]L=1/T[/itex] now suggests that [itex] M=1/L=T [/itex]. Which seems inconsistent, what am I missing here?
    Last edited: Oct 12, 2012
  2. jcsd
  3. Oct 12, 2012 #2


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    I've never seen anyone use inverse seconds for length as natural units. I believe the correct formulation would be to say that t = L/c, so that you could use seconds for t and light-seconds for L, or you could use centimeters for L, and (centimeters/c) for time.

    MTW, for instance, uses cm to measure length, time, and mass.

    THis would give planck's constant units of cm^2
  4. Oct 12, 2012 #3
    Sorry that was a typo, should be L=T (from c=1). Then G=1, means M=L^3/T^2, and combinging with this L=T, gives M=L=T. But now if we want hbar=1, this implies M=T/L^2, which combing with what we learnt from c=1,G=1 (such as L=T) seems to suggest M=1/L (as oppose to M=L). All my equals are proportionals really.
  5. Oct 14, 2012 #4
    must be a silly mistake anyone know where?
  6. Oct 14, 2012 #5
    So M=T/L^2, but M=L=T, so M=1/M, or M^2=1, meaning the mass itself has become a dimensionless quantity, as have all others. What this means is that setting a fourth constant equal to one finally sets the scale. in [itex]c=G=1[/itex] units, you could use any length, time, or mass as the fundamental unit of the that all others are just multiples of. You could choose [itex]M_{\text{sol}} = 1[/itex], for instance. Choosing [itex]\hbar=1[/itex] has the same effect, just yielding different numbers.
  7. Oct 14, 2012 #6

    D H

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    I flip between three perhaps contradictory views, but I'm ever more leaning to the third. These views:

    (1) Those natural units still have dimensions. What's really meant when physicists pick a set of units such that c=1 is that c has a numerical value of 1. It still has dimensions of L/T. The same applies to all those other natural units. Those things we pick for natural units still have dimensions. They just have a numeric value of 1 if we choose right.

    (2) Or perhaps c=1 in natural units has dimensions of velocity rather than length/time.Our choice of what constitutes a primitive unit versus a derived unit is a bit arbitrary. This isn't all that different from view #1. Velocity, length, time, energy are still dimensionful. This view just uses a different set of primitives.

    (3) Those natural units truly are dimensionless. Our metric system isn't quite so normalized as we think. The metric system has moved just one quantity (force) from the realm of primitive units to derived units. That's a good start, but it is only just a start. Just as F=kma is better written as F=ma, E=mc2 is better written as E=m. No conversion factor is needed. Mass isn't just convertible to energy. Mass a kind of energy. The k in F=kma is a consequence of not knowing that force is mass times acceleration. Those apparently dimensionful physical constants such as c in E=mc2 are a consequence of not knowing that length is time, mass is energy, etc. Our vaunted metric system is one unit shy of being just as goofy as English units.
  8. Oct 14, 2012 #7


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    If you choose units in which c, G and [itex]\hbar[/itex] are all equal to 1, then everything becomes dimensionless (a pure number). The unit of mass is then [itex]\sqrt{\dfrac{\hbar c}{G}}[/itex]. The unit of length is [itex]\sqrt{\dfrac{\hbar G}{c^3}}[/itex]. The unit of time is [itex]\sqrt{\dfrac{\hbar G}{c^5}}[/itex]
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