# What would/could be the fundamental physical origin of the Gaussian?

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#### zordim

• Jr. Member
• 46
##### What would/could be the fundamental physical origin of the Gaussian?
« on: 27/11/2012 17:37:38 »
Well, the question is in the title.
So, what would/could it be?

#### RD

• Neilep Level Member
• 8171
##### Re: What would/could be the fundamental physical origin of the Gaussian?
« Reply #1 on: 27/11/2012 18:15:25 »
If you mean Gaussian distribution, regular arrays (like the packing of atoms) can produce a normal distribution pattern ... http://en.wikipedia.org/wiki/Bean_machine

#### zordim

• Jr. Member
• 46
##### Re: What would/could be the fundamental physical origin of the Gaussian?
« Reply #2 on: 03/12/2012 16:35:53 »
What do you think about the following derivation of energy distribution of a localized energy whirl:
In any radial direction away from center, the energy decreases. At each point $$r$$, energy has a certain value $$E(r)$$. Along the radius, the energy does change: $$\frac{dE}{dr}$$. This is force, by definition.
At some given radius $$r$$, that force has a certain value $$\frac{dE(r)}{dr}$$. Both force and energy do not change along the circle of that given radius $$r$$. And that is the case in each given value $$r$$ of a radius, along which energy is distributed. Hence:

$$\frac{dE(r)}{dr} = - a(r) \cdot m(r) = - \frac{v^2}{r} \cdot \frac{E(r)}{c^2} \Rightarrow \frac{dE(r)}{E(r)} = - \frac{v^2}{c^2} \cdot \frac{dr}{r}$$
$$v = \omega \cdot r, dv = \omega \cdot dr \Rightarrow \frac{dr}{r} = \frac{dv}{v}$$
$$\Rightarrow \frac{dE(r)}{E(r)} = - \frac{v \cdot dv}{c^2} \Rightarrow \frac{E(v)}{E_0} = e^{ - \frac{1}{2} \cdot \frac{v^2}{c^2}}$$