# Question about fine-structure and free vacuum constants

1. Jul 24, 2008

### nuby

Hello,

When the fine structure constant (a) is calculated from the electric and magnetic constants.

What does each part represent.

(1/c) * (1/electric_constant) * (e^2/(2*h)) = a

i.e. What does (1/electric_constant) represent? The opposite of vacuum permittivity?

And with the magnetic constant: c * magnet_constant * (e^2/(2*h)) = a

What is the significance (physically) of (e^2/(2*h)) and why is it a common denominator between both equations?

Thanks.

2. Jul 26, 2008

### nuby

anyone?

3. Jul 26, 2008

### will.c

4. Jul 27, 2008

### malawi_glenn

In Q.E.D we use interpretation number 4

"a constant representing the strength of the interaction between electrons and photons"

You can't ask what each part represents, since there are many ways to decompose alpha. Also alpha arises in many different forumulas and depending how and from where you derive those formulas, the 'different parts' of alpha comes from different steps in those equations.

The realtion bewteen the magnetic constant and alpha comes from maxwells equations, where the speed of light in vacuum is DEFINED as:
$$c = \frac{1}{\sqrt{\epsilon _0 \mu _0}}$$

So you can just work out the algebra from there to get your:
"c * magnet_constant * (e^2/(2*h)) = a"

5. Jul 28, 2008

### nuby

So there's not a physical representation for e^2/(2*h) which I guess would have units:

(C^2 / J*s) Is this similar to capacitance (farads)? (C^2/J)

6. Jul 28, 2008

### malawi_glenn

You can interprent many things into e^2/(2*h), you have the combination of e^2/(2*h) in thousands of formulas, don't spend your time on this.

And is h similar to E? Units J*s vs. J.. no....

7. Jul 29, 2008

### nuby

Not much time to be spent on this, all I'm looking for is one line answer.

h similar to E? I guess they're related since E = h*f ?

But what would you say these units amount to in my previous post? Just nothing?
C^2/J*s or maybe: (C^2/J) * (1/s)

8. Jul 29, 2008

### will.c

Those are the units of conductance. The conductance quantum is $$\frac{2e^2}{h}$$.

This is nothing more than coincidence, however. You can assign a physical interpretation to the fact the conductance quantum appears in $$\alpha$$ but that doesn't make it a useful or meaningful interpretation.

9. Jul 29, 2008

### nuby

Thanks will.c I was starting to feel like I was asking for a limb, or something.

10. Jul 30, 2008

### malawi_glenn

h is the 'proportionallity constant' between Energy and frequency, just as we have discussed in the thread about heaviest atomic weight you and I.

And C^2/J*s = (C^2/J) * (1/s) = (C/J)*(C/s) = etc. You can combine this in any way you want, it is just units dude.

11. Jul 30, 2008

### nuby

I just thought it was interesting that u * c * (e^2/2*h) can boil down to. u*c (impedance of vacuum) * e^2/2h (conductance, going by units)

12. Jul 30, 2008

### nuby

Am I the only one here that wonders about the significance of these things? lol

13. Jul 30, 2008

### will.c

The problem we seem to be having is that the significance, at least of this thing, does not come from first principles. It is assigned. What does "vacuum impedance" times "conductance" (which it would, of course, have to be - alpha is dimensionless!) have to do with the spacing of atomic energy spectra? The relative contributions of Feynman diagrams in QED? The relative strength of the fundamental forces?

Everyone wonders where, exactly, it comes from and what, exactly, it means, but rearranging units doesn't further that program.

14. Jul 31, 2008

### nuby

Sorry, I'm playing with units again... But could you maybe say (Z*e^2)/(2*h) is a ratio of the force between a electron and photons at 1hz?

15. Aug 1, 2008

### malawi_glenn

what are you meaning by 'is a ratio of the force between a electron and photons at 1hz ' ??

What you have written is just rubbish logically and physically speaking. Maybe you want to reformulate your question.

If you don't have any serious motivations behind your statements, please stop taking up our time. I recall from the 'largest atomic weight'-thread, that you are just guessing and writting a lot of nonsense..

16. Aug 1, 2008

### nuby

Glenn, I'll try to avoid posting random questions and thoughts, sorry about that.

Here is an interesting equation I ended up with while playing with units, alpha, and my limited math skills.

(2*pi*r) * (Pm*c) * (a/h) = Pm/Em = 1836.15

Where,

r = Bohr Radius = 5.2917720859e-11 m
Pm = Proton Mass = 1.6726e-27 kg
Em = Electron Mass = 9.10938188e-31
a = Alpha = 0.0072973525705
h = Planck's = 6.626068e-34

And

a = h / ((Em*c) * (2*pi*r)

I haven't look too closely at this yet, but let me know what it is, or what you think.

17. Aug 1, 2008

### nuby

This may be related to a bohr equation.

Last edited: Aug 1, 2008
18. Aug 2, 2008

### malawi_glenn

Dude who cares about what equation you end up with if you are not presenting the starting points and assumptions and all derivations etc. ?? You results is trival due to the definition of bohr radius and alpha...

And now lets take the Forum rules seriously now. These threads are NOT to present own results etc, but we have a special thread called 'independent research' - which of course have its own rules about what can be posted there etc, so I will refer you to that thread for these things.

Bohr radius is just a constant, a unit. Does not represent anything special. Only in the classical picture of atom orbitals, but not in QM - which is the paradigm of todays physics.

19. Aug 2, 2008

### Count Iblis

If you want to play with units, you should realize that dimensional quantities are just dimensional by convention, i.e. by our choice of our unit system. You can just as well work in natural units and then reinsert c, hbar and G in the equations but now regarded as dimensionless rescaling quantities. Results of nonrelativistic quantum mechanics are then obtained by considering the limit c to infinity.

So, to derive the equation for the ground state energy of hydrogen this way, we can start with the equation:

E = m

Reinsert c:

E = m c^2

This diverges in the nonrelativistic limit c ---> infinity. To cancel the factor c^2 we multiply it by alpha^2, because alpha contains a factor 1/c:

E = alpha^2 m c^2

It turns out that the correct formula is 1/2 alpha^2 m c^2

You can object to this reasoning by noting that alpha being dimensionless should be independent of c. I.e. if you change your units so that the value of c changes, then the other parameters (in particlar the electron charge) will also change so that alpha will remain invariant.

But to get the proper classical limit when electrodynamics is involved you must pretend that you can choose the charge independent of c. So, formally alpha tends to zero, even though in reality it is fixed. This is reasonable since in the nonrelativisctic limit the energy of the electron relative to m c^2 should be zero, but we know that it is 1/2 alpha^2, which is a fixed dimensionless number. Now 1/2 alpha^2 is quite small, meaning that the electron is pretty much nonrelativistic.