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Looking at Planck units, there seems to be a curious rule between the dependance in $\hbar$ of a Planck unit and the unit dimensions of the corresponding physical quantity.

Let the dimensions of the physical quantity be ($Q$ being the electric charge unit and $\Theta$ being the temperature unit):

$$ L^l M^m T^t Q^q \Theta^\theta.$$

Then, if : $$l + m + t + q + \theta = 0$$

the Planck unit does not depend on $\hbar$.

This seems to work for all base Planck units, and, consequently, for all derived Planck units.

Is it just chance, or is there a more fundamental reason?

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I wonder if you would consider dimensional analysis a "fundamental reason"...? –  Michael Brown Apr 2 '13 at 11:58
I think that dimensional analysis would not give the answer - because, with this tool, we note some facts, but we can't explain them, that is : why base Planck units dimensions others than ℏ, have the same structure (total exponent = 0) , but maybe I am wrong –  Trimok Apr 2 '13 at 12:21

3 Answers 3

Notice that you measure in the five units corresponding to $L,T,M,Q,Θ$.

Choose one unit, say time $T$. Then firstly

$[c]=L T^{−1},$

translates length to time, then

$[G]=M^{−1} L^3 T^{−2}=TM^{-1},$

translates mass to time, then

$[1/ε_0]=Q^{−2}L^3 M T^{−2}=(TQ^{-1})^2 ,$

translates charge to time, then

$[k_B]=Θ^{−1}L^2 M T^{−2}=TΘ^{−1}$

translates temperature to time, and then

$[\hbar]=L^2 M T^{−1}=T^2$

is the only one with nonvanishing exponent. The point is that you don't use "$L^2 M$" from the power constant as a unit, so this is the quantity which gets left over.

After the elimination process, units are multiples of


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It is correct, but it does not explain why base units others than $\hbar$, have the same structure (total exponent = 0) –  Trimok Apr 2 '13 at 11:37
@Trimok: I think some theorists claim that their theories will eventually be able to determine all of naturals constants in a unique way. In our consideration, this would essentially mean to produce fundamental frequencies, relating energy and time, giving the action $S$ a direct numerical value. Similarly to how the statistical mechanics formulation of thermodynamics is able to explain the entropy by counting, discarding the need for a unit. Motion is explained by geometry, thermodynamics is explained by combinatorics, but nobody understands the path integral $\int \exp{(iS/\hbar)}$ :D –  NikolajK Apr 2 '13 at 12:05

This is not by chance. If you look a bit further down that Wikipedia article, you will see that all base units contain $\sqrt{\hbar}$. So we could rewrite every Planck unit as $L = \sqrt{\hbar}L'$ where $L'$ does not depend on $\hbar$ and so on. Therefore:

$L^lM^mT^tQ^q\Theta^\theta \\=(\sqrt{\hbar}L')^l(\sqrt{\hbar}M')^m(\sqrt{\hbar}T')^t(\sqrt{\hbar}Q')^q(\sqrt{\hbar}\Theta')^\theta \\=\sqrt{\hbar}^{l+m+t+w+\theta}L'^lM'^mT'^tQ'^q\Theta'^\theta$

And hence, if (and only if) $l+m+t+w+\theta=0$ then there is no dependence on $\hbar$.

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Yes, of corse, but this does not explain why the base units (different of $\hbar$) have the same structure. Your observation is correct, but it is not an explanation. –  Trimok Apr 2 '13 at 11:22
@Trimok oh, then I guess I misunderstood your question. So what you're actually asking is, why does every base unit contain $\sqrt{\hbar}$ but different powers of the other normalized constants? –  Martin Büttner Apr 2 '13 at 11:24
Yes, they have the same structure, but why ? –  Trimok Apr 2 '13 at 11:26
@Trimok hm, sorry, am at a loss there. Might as well delete my answer then. Your question has my upvote! ;) –  Martin Büttner Apr 2 '13 at 11:30

Since you are not liking dimensional analysis I'll try a slightly different (but ultimately equivalent) approach by trying to relate the Planck units to physical quantities.

The Planck mass is the mass of the smallest possible black hole. It is also an energy because $c=1$. The Planck length is the size of such a black hole and the Planck time is the light-traversal time over a Planck length. Since the size of a black hole is proportional to its mass, so all three quantities are proportional to each other. Further the proportionality constants can't involve $\hbar$ since the relations between them are essentially classical (the size of black holes, the speed of light). So these three scale the same way in terms of $\hbar$.

Now temperature is an energy and the only energy around is the Planck mass so again you have the connection. You can think of this as the temperature where Planckian black holes are thermally produced in abundance, so it is a sort of maximum temperature a reasonable system could have.

Incidently the Hawking temperature of a black hole is proportional to $\hbar$ since it is a quantum mechanical effect, but it is also inversely proportional to the mass, and since the Plank mass $\sim\sqrt\hbar$, the net scaling for the temperature of a Planckian black hole is $\sim\sqrt\hbar$ the same as the Planck mass. If the Planck temperature is defined this way you may find this less than satisfactory since I didn't really establish the scaling of the Planck mass earlier, only that it scales the same way as the other quantities.

That leaves the Planck charge. This is the maximum charge allowed for a charged Planckian black hole (an extremal black hole). The charge of an extremal black hole is proportional to its mass (I have no intuition for why), so this scales the same way again. You can see this result from the metric of a charged black hole, but I know of no intuitive reason. Perhaps someone else could enlighten us both on this matter.

Huge caveat: none of this takes into account quantum gravity things that surely dominate at this scale. This is a huge pile of handwaving, but apart from dimensional analysis this is what I can give you.

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I agree with your presentation, but I think there is partially a circular argument. I wonder if the answer could be considering the action S which has an $\hbar$ dimension and which is the product, for instance, of time and energy. If we admit a equal repartition of powers of $\hbar$ between time and energy, or length and impulsion, this could be the beginning of an answer. –  Trimok Apr 2 '13 at 16:09

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