# Let me create a new constant, and I'll unify all the physical constants.

Discussion in 'Alternative Theories' started by Suomy Nona, Feb 14, 2015.

1. ### Suomy NonaRegistered Member

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I figured out how to unify the family of electromagnetic constants and the family of gravity mass constants into one logical structure. One small problem, it only works if you will let me create a new SI constant. (Beta = 3.364623E-25).

Think of it this way, we got lots of constants, what's one more among friends?

Here is my equation:

(Universal Gravity Constant)=(speed of light)^3*(Elementary Charge)^2*3.364623E-25^2*(ampere)^-2*(vacuum permittivity)^-1*(vacuum permeability)^-1*(reduced planck constant)^-1

Here is a Wolfram Alpha link showing my equation. http://wolfr.am/3d01tUUe (note - sometimes it gives a parsing error on the first try, and you have to press the calculate button)

Using my equation as a starting point, try to imagine any other physical constant that you couldn't graph into my equation by using substitution. Think of it as a central node of a large network.

I'll give you a central node for all the constants, if you will give let me create one new constant.

3. ### DaeconKiwi fruitValued Senior Member

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What function of the Universe could your new constant be describing? Also, if you're just making up the number why do you need to then square it as well?

5. ### rpennerFully WiredValued Senior Member

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What is fundamental about a current of exactly one ampere?

7. ### Suomy NonaRegistered Member

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It produces an attractive force of 2E−7 newton per metre of length between two straight, parallel conductors of infinite length and negligible circular cross section placed one metre apart in a vacuum.

My definition of beta is (planck time)*(ampere)*(Elementary Charge)^-1=3.364623E-25.

Here's a link to wolfram alpha showing beta http://wolfr.am/3g0~slvT

I agree that my beta definition does not seem fundamental but it meets my requirements.

1) It produces the correct answers
2) It will never change as long as we use the S.I. unit system.
3) It unifies the physical constants into one logical family.

It would be nice if beta was some fundamental quantity but I'm more that willing to accept it based on the first three requirements. I think of beta as a type of conversion factor from Planck units to SI units.

As long as the value never changes, can't we use it in our equations?

8. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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So it is not fundamental it is just a convenient.

What do you mean it produces the correct answers? What questions can I use it in? Will this constant allow me to find my missing socks?

9. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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Let me be a bit more direct: show us a correct answer it can produce by using it to solve a problem.

10. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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Agreed. A simple demonstration should clear this up.

11. ### Suomy NonaRegistered Member

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My equation is based on the total number of charge carriers that flow during the duration of a Planck second to produce an attractive force of 2E−7 newton per metre of length between two straight, parallel conductors of infinite length and negligible circular cross section placed one metre apart in a vacuum.

I will come up with a an example where this could be useful.

12. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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A Planck second is the same as a normal second. Do you mean Planck Time? $10^{-44} sec$
Since the magnetic force per unit length on two current carrying conductors is based on the current in the wires it does not matter if you are talking about a duration of 1/1000000 of a second or an hour, the force would be the same.

You don't have a bunch of examples that you have already tested it on? Really? Well, looking forward to your example.

13. ### rpennerFully WiredValued Senior Member

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There is (almost)* NOTHING fundamental about SI units, and your promotion of $\beta$ ignores that there is a physical quantity that makes more sense than your introduction of both a current of 1 ampere and a dimensional constant $\beta$.

$X^2 = \frac{c^3 e^2}{G \epsilon_0 \mu_0 \hbar} = \frac{c^5 e^2}{G \hbar} = \left( \frac{e}{t_P} \right)^2$
$X = \frac{e}{t_P} = \sqrt{\alpha} q_P/t_P \approx 0.0854 q_P/t_P \approx 2.972 \times 10^{24} \, \textrm{amperes}$

http://en.wikipedia.org/wiki/Planck_units

With this current defined, $G = \frac{c^3 e^2}{X^2 \epsilon_0 \mu_0 \hbar}$ which is simpler than the equation in the OP. But it has no informational physics content newer than 1899.

In Planck Units, $G = 1 \ell_P^{3} m_P^{-1} t_P^{-2}, \; c = 1 \ell_P t_P^{-1}, \; e = \sqrt{\alpha} q_P, \; \epsilon_0 = \frac{1}{4 \pi} \ell_P^{-3} m_P^{-1} t_P^{2} q_P^{2}, \; \mu_0 = 4 \pi \ell_P m_P q_P^{-2}, \; \hbar = 1 \ell_P^2 m_P t_P^{-1}, \; X = \sqrt{\alpha} t_P^{-1} q_P$
So the claim is:
$\ell_P^{3} m_P^{-1} t_P^{-2} = \frac{\ell_P^3 t_P^{-3} \; \alpha q_P^2}{\alpha t_P^{-2} q_P^2 \; \frac{1}{4 \pi} \ell_P^{-3} m_P^{-1} t_P^{2} q_P^{2} \; 4 \pi \ell_P m_P q_P^{-2} \; \ell_P^2 m_P t_P^{-1}}= \frac{ \alpha \ell_P^3 t_P^{-3} q_P^2}{\alpha m_P t_P^{-1} q_P^2} = \ell_P^3 m_P^{-1} t_P^{-2}$

So until X has some significance to physics that isn't just juggling physical constants around, it lacks both utility and practicality. The same applies to $\beta = 1 \, \textrm{ampere} / X$.

Finally, since only a few digits of G are known, you can't know $\beta$ to as many decimal places as you propose.

* -- The speed of light is an exact integer in SI units because it was close to that number in the previous system of units. The ampere's current definition is a peculiar one that makes certain electromagnetic quantities rational numbers.

Last edited: Feb 17, 2015
14. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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Indeed, typically the problems/equations are identified before the constants because until you know what problem you are trying to solve and how, you can't have any idea what you need from a constant!

Clearly, this was just pulled out of the air and mashed together and is meaningless.

15. ### Suomy NonaRegistered Member

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rpenner, thank you for taking the time to write out the equations. When I was trying to come up with an example I realized my work was a recursive definition at best.

Clearly analytical mathematics is not my best career choice.

(I keep reinventing other peoples mistakes)

Thank you

16. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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Well I have to say good for you! You investigated some mathematics; learned new things (I assume) and when it was shown to you that one of your deductions from this was incorrect, you looked at the evidence and accepted it. This is what the educational process is all about.

17. ### Suomy NonaRegistered Member

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I have found an use for my equation.