# Interesting find

1. Jun 7, 2005

### Jonny_trigonometry

this is kinda wierd, and I'll have to admit I'm going out on a limb, but I seemingly found a relation between four constants. here it is...

h^4((h^2*e^2)/me^2 - 8(PI)/μ) = -4me^2e^2

me=mass of electron
e=charge of electron
μ=Mu_naught
h=(Planks constant)/2PI= h bar

this relation could be used to gain more accurate values of all four... I think.

using the most accurate values I could find for me, e, and μ, I solved for hbar...

1.0545715964207857*10^-34

it's probably nothing... but what do you think?

2. Jun 7, 2005

### rachmaninoff

Huh? It's not even order-of-magnitude correct, assuming you do in fact mean

$$\hbar^4\left(\frac{\hbar^2 e^2}{m_e^2}-\frac{8 \pi}{\mu_0}\right)= -4m_e^2 e^2??$$

Left side is $$-2.47\times 10^{-129}$$, right is $$-8.52 \times 10^{-98}$$ in whatever the resultant SI units are.

Besides, the units don't agree, it would be a meaningless equation even if it were numerically right.

3. Jun 7, 2005

### Jonny_trigonometry

ya, hehe. whoopse. thats odd, but when I solved for h bar, thats what I got

4. Jun 7, 2005

### Jonny_trigonometry

ahh, ok I guess mathematica has a built in value of h, and when I punched it in to solve for h, it output the stored value as one of the answers. Sorry people

5. Jun 7, 2005

### Jonny_trigonometry

I computed this:

$$\frac{1}{\frac{h^2}{m_e}+\frac{4\pi m_e}{\mu_0e^2 }\sqrt{(1-\frac{\mu_o^2e^4}{4\pi^2h^2})}}$$

and saw that it is incredibly close to the compton radius, so I set it equal to the compton radius:

$$\frac{e^2}{4\pi\epsilon_0m_ec^2}$$

but I've now looked more closely and the two differ at the 10th decimal place...

Last edited: Jun 7, 2005
6. Jun 7, 2005

### ZapperZ

Staff Emeritus
Y'know, you CAN go back and EDIT what you have posted without having to make new postings.

Zz.

7. Jun 7, 2005

### Jonny_trigonometry

either I messed up in simplifying (probably the case), or the small deviation at the 10th decimal place was way more noticable after some cancelations were made

8. Jun 7, 2005

### Jonny_trigonometry

oh, hehe. thanks Zapperz, I see that... (I'm slow)

say, is that your name because you're a frank zappa fan?

9. Jun 7, 2005

### ZapperZ

Staff Emeritus
Good god, no!

Zz.

10. Jun 7, 2005

### Gokul43201

Staff Emeritus
Ooh, that's a priceless exchange !!

Take a look at this :

$$\frac{1}{\frac{4\pi m_e}{\mu_0e^2}}=\frac {\mu_0e^2}{4\pi m_e} = \frac {e^2}{4\pi m_e \epsilon _0 c^2}$$

In other words, you are merely displaying surprise that a binomial approximation works !

11. Jun 7, 2005

### Jonny_trigonometry

you mean that the binomial approximation of $$\frac{e^2}{4\pi\epsilon_0m_ec^2}$$ is $$\frac{1}{\frac{h^2}{m_e}+\frac{4\pi m_e}{\mu_0e^2 }\sqrt{(1-\frac{\mu_o^2e^4}{4\pi^2h^2})}}$$ ?

then how does h show up in the approximation?

I know that $$\frac{h^2}{m_e}$$ is pratically zero, and the $$\sqrt{1-\frac{\mu_o^2e^4}{4\pi^2h^2}}$$ is practically 1, but I don't think they should be dropped.

12. Jun 7, 2005

### Gokul43201

Staff Emeritus
No, I mean that $$\frac{e^2}{4\pi\epsilon_0m_ec^2}$$ is a binomial approximation for $$\frac{1}{\frac{h^2}{m_e}+\frac{4\pi m_e}{\mu_0e^2 }\sqrt{(1-\frac{\mu_o^2e^4}{4\pi^2h^2})}}$$

Actually, the h goes away when you make the approximation. But the 'h' showed up in the first place...because you put it there.

That's what an approximation does, and hence the deviation at the tenth decimal place.

Looks at some orders of magnitude :
$$\frac{h^2}{m_e} \approx 10^{-36}$$ and $$\frac{e^2}{4\pi\epsilon_0m_ec^2} \approx 10^{-26}$$. Do you see the ten orders of magnitude difference there that I do ?

13. Jun 8, 2005

### Jonny_trigonometry

hehe, ya. you're right. thanks for clearing that up for me.

to give more background where I came up with the first eq. I am trying to "re-invent the wheel" because I'm not happy with the bohr atom. I've been working on this over for a few months now, and I think I still want to write a paper on it, so I'm not going to tell you exactly what I'm doing, but alls I can say is that one of the ways I chose to derive it's radius ended up with the above equation, so I just had to post it. obviously, it's bunk.

14. Jun 14, 2005

### James Jackson

Errm, right. Physicists aren't happy with the Bohr atom either. Remember it is 'old' QM... It could model the observed spectra etc of the Hydrogen atom at the time, but is now superceded by 'new' QM.