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And by nice looking law I mean with no constants. I mean, what would we need to set, so all laws would without those nasty constants in front of them? (I mean all of them, also $\pi$!) What would it mean?.

What I am trying to say with this question is: Is there a universal language, do particles and gravity live in the same space? If gravity and quantum world should go together, there should be one basis, where you don't need to use any constants.

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    $\begingroup$ You can find out yourself: Set all constants to $1$ (except $\pi$ of course!) and see what happens. The laws are fixed, setting the constants just changes our choice of units. [Spoiler: You don't get rid of $\pi$.] $\endgroup$ – ACuriousMind Jun 30 '14 at 12:20
  • $\begingroup$ @ACuriousMind Wouldn't you be able to absorb $pi$ into some constant. $\endgroup$ – jinawee Jun 30 '14 at 13:23
  • $\begingroup$ @jinawee: I would be very impressed if you could do so in a way that makes the $\pi$ in the area or circumference formulae $A = \pi r^2$ or $C = 2\pi r $ disappear (and these are often the underlying reason $\pi$ shows up in physical laws). $\endgroup$ – ACuriousMind Jun 30 '14 at 13:28
  • $\begingroup$ @ACuriousMind Prove it! You don't seem to understand my question - my question is directly aiming to prove that assumptions of all constants equaling one is contradictory to $\pi=1$! $\endgroup$ – user74200 Jun 30 '14 at 13:46
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    $\begingroup$ $\pi$ is what it is, it's a number (properly defined for example as double the first positive zero of the cosine). $\pi = 1$ is simply false. The value of $\pi$ is not tunable. $\endgroup$ – ACuriousMind Jun 30 '14 at 13:50
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Ok, here's why rescaling units won't get rid of $\pi$:

The circumference of a circle is $C = 2\pi r$. Note that $C$ and $r$ are measured in the same unit, both are units of length. Now rescale lengths as $l' = c l$ with an arbitrary constant $c$ (might be $\pi$, but doesn't matter). Plugging in yields

$$C' = cC = c 2 \pi r = 2 \pi c r = 2 \pi r'$$

So you haven't changed the circumference law at all. Thus, you don't get rid of $\pi$.

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  • $\begingroup$ No, you have $C = r'$, which amounts to saying that your new unit of length is the length of the radius of a circle with circumference 1, but if you express the circumference also in your new unit of length, you have $C' = \frac{C}{2\pi} = \frac{2\pi r}{2\pi} = r = 2\pi r'$. If you rescale, you have to rescale all lengths, else you measure circumference and radius in two different units, which is patently absurd. $\endgroup$ – ACuriousMind Jun 30 '14 at 14:11
  • $\begingroup$ Right, that should be $r$ not $r'$. Then it holds. However, that does not prove that by rescaling you won't have for example Einstein's equations without $\pi$. And something else with $\pi$(some other law I mean). $\endgroup$ – user74200 Jun 30 '14 at 14:15
  • $\begingroup$ Please, understand, I am aiming at Physics laws, not mathematical formulae. $C=2\pi r$ is not a physics law. It exists independently from physics. $\endgroup$ – user74200 Jun 30 '14 at 14:23
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    $\begingroup$ The factors of $\pi$ in physics arise mainly out of (solid) angle integrals (which are, in turn, essentially there because we intergrate along the circumference of a circle or the area of a sphere). Their value is also independent of rescalings, by the same argument I illustrated above. The part about gravity and quantum theories you just edited into your question is totally incomprehensible to me, and suggests there is some more fundamental confusion as to what units are going on. $\endgroup$ – ACuriousMind Jun 30 '14 at 14:58
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Planck units is almost (except coefficients like $\pi$) exactly the unit system you want, and it is frequently used in quantum field theory.

According to its definition:

$c = G = \hbar = k_\text{B} = 1 \ $

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  • $\begingroup$ Yeah, I know about the Planck's system and geometrized units and Gauss unit system, my question is directly aiming to prove that assumptions of all constants equaling one is contradictory to $\pi=1$! $\endgroup$ – user74200 Jun 30 '14 at 13:47
  • $\begingroup$ Hint: you contradict yourself if you define $c=1$ and $1/4\pi\varepsilon$=1 $\endgroup$ – user74200 Jun 30 '14 at 14:07
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When you're doing Fermi estimation, it's convenient to use a "system of units" where $$ \sqrt{10} = \pi = 3 = 1 = -1 = c $$ though this is probably more tongue-in-cheek than you were hoping for. For some reason I think of these as "pauper's units," though I'm not sure whether that phrase is my own invention or not.

You might be interested in this other question.

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