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Force constants and semi-natural units

  1. Apr 15, 2003 #1


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    I have been playing a bit with the units of coupling constants. Instead of setting c=1, h=1 (natural units) it is interesting to set c=1 but still to keep h explicit in the formulae.

    You may know that in natural units there is a basic difference between electromagnetism and gravity, namely that gravity is a dimensional constant with dimensions of area, while electromagnetism is not.

    But if we do not put h=1, then other difference appears. In one of them, h is in the numerator, in the other h is in the denominator.

    So I wonder, can we use this to generate four kinds of coupling constant? The two new ones should be

    a) As the gravity, but without the area term: we get a constant force, depending on masses but not in distance.

    b) As the electromagnetism, but with the area term: we get a weaker force, decreasing as the fourth power of distance.

    (a) is not exactly the strong force, because there is not a show of asymtotical freedom, and (b) is not exactly the Fermi weak force. But it is an interesting rule. On the other hand, it should imply that gravity unifies with SU(3)... I have never heard such a thing.
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  3. Apr 25, 2003 #2


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    I'm a naive respondent rather than the Dario type, so perhaps not as useful, however.

    I think of G as equal to Ac^3/hbar, where A is the planck area---the square of planck length. Is it OK to use hbar instead of h?

    So naturally if you set c = hbar = 1 you get that G "is" an area, in fact the Planck area, just like you say.

    Now if you keep hbar alive and well then G becomes A/hbar.
    So, like you say, it is in the denominator. Your post is a kind of
    tutorial for me. Now what do you mean by the coupling constant
    in EM? I assume you mean alpha.

    Or else you mean the Coulomb constant???? (analogous to G, anyway!)

    And indeed alpha is just the VALUE which the Coulomb constant takes on in Planck units with elem.charge e adjoined.

    k_coulomb = alpha hbar c/e^2

    So then when you set c = 1 but keep hbar alive you get
    something with hbar in the numerator, as you said we would. OK

    BTW in Planck units the unit force F = c^4/G and hbar c is equal to the unit force x area quantity FA. But FA/e^2 is the natural unit appropriate to the Coulomb constant. That is why I say that alpha is the value of that constant expressed in P. units.

    Well two masses determine a constant "force x area" product. As the square of the distance is reduced the force increases so that
    the product is always constant. If the area is normalized or set equal to unity somehow, then one does get a constant force which is characteristic of the product of the two masses. Yes, just as you say.

    Now as for EM I do not immediately see what you mean. However please write out the formula you mean, with the fourth power of distance.

    thanks for the ideas, looking forward to more.
    do you by any chance write hbar with an ordinary h?
    or do you mean 2πhbar?
  4. Apr 29, 2003 #3


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    Hmm probably yes, I have been using hbar instead of h. Or taking units where Pi=1 :DDD

    To further obscure the issue, I will repost my usenet comment:
    Code (Text):

    Naively, very NAIVELY, very slowly:

    Electrostatic force is  F= K  q1 q2/ x^2; so lets see its units  
                     M L T-2 = [K] Q2 X-2
                    M L [c]2 = [K] Q2
                [h][c] [e]-2 = [K]
    the natural system is [h]=[c]=[e]=1 so we say that [K] is

    Now, gravity force is     F= G  m1 m2 / x^2
    thus   M [c] T-1= [G] M2 L-2
    and then [G]= M-1 L2 T-1 [c]= [h]-1 L3 T-1 [c]2= [h]-1 L2 [c]3
    put [c]=[e]=1 only, then

    Let me complete the square with another two constants,
           [K]=[h]       [Kw]=L^2 [h]
           [Gs]=1/[h]     [G]=L^2/[h]
    Kw has the units of weak force, as far as I remember. Not sure,
    please check. But it is an adimensional constant divided by a mass^2
    term, so it seems very much as G_F, the fermi constant.
    note that Kw is consistent with a law F= Kw q1 q2/x^4
    Code (Text):

    As for Gs, and here was the motivation of the post, I have never seen
    such a beast. It is adimensional too, but in a different sense that K
    because h is in the denominator. If we try to build a force law for
    this constant, we are forced to postulate
           F = Gs m1 m2
    independent of distance. A force independent of distance is typical
    of the strong force in the standard model, for distances big enough.

    Now, we know that [K] and [Kw] unify at the electroweak scale. So the
    question appears, if we should say the same for [Gs] and [G] at some
    (Planck?) scale.

    Are we just doing some math trick? Maybe. For instance if instead of
    h we use [h]=[x][p] then the constants are
    and we are not justifyed to play the previous game anymore.
  5. May 10, 2003 #4
    A suggestion

    You don't even need h or h-bar in your equations. Every quantity that has m (mass) in it can be factored through by h or h-bar to make a new set of quantities. (I call it getting the h out of Physics.) We only use h and h-bar because we still insist on using outmoded classical dimensional units from a time when we didn't know about Quantum Mechanics. Now we know better and don't have to use them!
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