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Setting hbar = c = 1 how can you just do that?

  1. May 13, 2009 #1
    I can sort of understand working in a system of units where, say, either [itex]c[/itex] or [itex]\hbar[/itex] takes a numerical value of one. But I read a paper today where BOTH [itex]c[/itex] AND [itex]\hbar[/itex] were set to be 1. I confess I haven't thought about it very carefully...but could someone please help me understand how you can get by with that?
     
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  3. May 13, 2009 #2

    diazona

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    Basically, the way to come up with a system like that is to pick a unit of time, multiply by the speed of light to get a unit of length, and then just pick a unit of mass such that [tex]\hbar[/tex] is equal to the length unit squared times the unit of mass divided by the unit of time. These units of length, mass, and time replace the meter, kilogram, and second from SI.

    In fact, I'll do you one better: imagine a system of units where [tex]c[/tex], [tex]\hbar[/tex], and [tex]G[/tex] (Newton's gravitational constant) are all equal to one! There actually is such a system of units, called Planck units or natural units. The fundamental quantities are the Planck length:
    [tex]l_P = \sqrt\frac{\hbar G}{c^3} = 1.616\times 10^{-35}\ \mathrm{m}[/tex]
    the Planck mass:
    [tex]m_P = \sqrt\frac{\hbar c}{G} = 2.176\times 10^{-8}\ \mathrm{kg}[/tex]
    and the Planck time:
    [tex]t_P = \sqrt\frac{\hbar G}{c^5} = 5.391\times 10^{-44}\ \mathrm{s}[/tex]
    Then the fundamental constants are
    [tex]\hbar = \frac{l_P^2 m_P}{t_P}[/tex]

    [tex]c = \frac{l_P}{t_P}[/tex]

    [tex]G = \frac{l_P^3}{m_P t_P^2}[/tex]

    If you're familiar with linear algebra, you can imagine units as vectors. Call the meter (1, 0, 0), the kilogram (0, 1, 0), and the second (0, 0, 1), and say that adding these vectors corresponds to multiplying units. Then you'll recognize that the units of the three constants [tex]\hbar[/tex], [tex]c[/tex], and [tex]G[/tex] form a basis of this "unit space," and that means you can use them to create a unit system in which the values of all three are equal to one. If you had three different constants which also spanned the "unit space," you could make a unit system out of those other constants as well.
     
    Last edited: May 13, 2009
  4. May 14, 2009 #3
    That's one of the best explanations I've ever read. Thanks very much for your help!

    Am I to infer from this discussion that the chances of finding a system of units in which *4* fundamental constants have a numerical value of 1 would be well-nigh impossible?
     
  5. May 14, 2009 #4

    Born2bwire

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    Depends on how mean I feel in my next publication.
     
  6. May 14, 2009 #5
    This means also that e0u0 = 1. Perhaps the only thing that cannot be reset is the fine structure constant. In grad school, I relied on the multiplicative constants to help me keep track of units in my calculations. C'est la vie.
     
  7. May 14, 2009 #6
    I seem to recall a story that Feynman routinely set h-bar to one in his calculations. Then if his final answer was off by twenty-seven orders of magnitude, he knew he was right. :smile:
     
  8. May 14, 2009 #7

    Nabeshin

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    Fine structure constant is a unitless number, so it has the same value regardless of what units you choose. Anything else that's unitless is going to be the same in all unit systems.
     
  9. May 14, 2009 #8

    diazona

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    Cool, thanks ;-)

    As far as finding a system in which 4 fundamental constants have a value of 1... once you set hbar, c, and G to 1, you determine the value of all constants which have units involving only mass, length, and time. (Basically, you fix the 3D vector space of mass, length, and time units) But you could add in another fundamental quantity, for example electrical charge, and develop a unit system based on the Planck units (from my last post) and the elementary charge (a.k.a. magnitude of the electron charge).

    You could even take it a step further, and develop a unit system in which the Boltzmann constant is equal to 1 (so you get a 5D vector space of mass, length, time, charge, and temperature)... I'm not sure if there are any others, though. At some point you run out of independent units to work with.
     
  10. May 14, 2009 #9
    Not setting all that crap to 1 is one of the most difficult things about teaching!

    Setting hbar = c = k_boltzmann = elementary charge = 1 is more true to the physics then carrying that crap around. There is only one unit: energy = mass = 1/distance = 1/time. We describe a physical quantity as having dimension N e.g. "these quantities have dimension 2, and these have dimension -2" instead of saying historical names of anthro-units.
     
  11. May 15, 2009 #10

    diazona

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    On the other hand, setting them not equal to one (or at least writing them down explicitly) is tremendously helpful in figuring out where you've made a mistake in the math. When I'm getting a result that doesn't make sense, probably almost half the time I'm able to identify it quickly by looking for the line where the units are not balanced. When you use natural units you can't do that.
     
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