Why 1/137 is important?

[Martin]

You can't derive the dimensionless constants this way but you should be able to see consistancy with units (or in other words you should be able to derive the unit dimensions of alpha.

 

Planck units are [math]c = \hbar = 1[/math]' date=' and since you know c and hbar in SI units you can write 1 metre in terms of c etc...

 

Looking at your earlier formulae, your problem could be that for gravity you are using masses in the numerator in one case, which are dimensional, whereas for electromag you are using charges which are not?[/quote']

 

Hi Severian, are you talking to me? I am not sure what the problem you refer to is. Please specify.

 

Planck units are given in terms of kilograms, meters, seconds at the US government NIST site

I will give a link.

Actually Planck himself listed the units with approximately correct values in 1899, in terms of old metric units: centimeter, centigrade, gram etc. but the most convenient source is, I think, the NIST "fundamental constants" site.

 

You will see, if you do the algebra, that in fact the value of the Coulomb constant in terms of Planck units is alpha, almost by definition!

[Martin]

Severian, here you are!

http://physics.nist.gov/cuu/Constants/

 

Go here and click "Universal" and you will get

http://physics.nist.gov/cgi-bin/cuu/Category?view=html&Universal.x=74&Universal.y=9

 

There you will see a menu with

Planck Length

Planck Mass

Planck Temperature

Planck Time

 

If you click, you will get definitions and current best values.

 

In addition to this a value in Coulombs for the fundamental charge, or electron charge, is needed. The NIST site has this as well.

[Severian]

Sorry, I was trying to answer your question. Maybe I misunderstood? I was trying to say that one can at least see that alpha is dimensionless using the formulae which you wrote down earlier.

 

To answer your equstion more specifically:

 

Q: 'Is it obvious to you that in Planck units the value of the Coulomb constant is alpha?'

A: No it is not, because alpha is a dimensionless number and depends solely on the details of the theory. One could imagine having a 'super-electromagnetism' which was ten times as strong as standard electromag and would have an 'alpha' which was approx 1/14.

 

Is this what you were meaning?

[Severian]

I have a paper copy of the Phys.Rev.D Particle Data Book sitting in front of me.

 

You can also get the same stuff at http://pdg.lbl.gov/

[Martin]

Sorry' date=' I was trying to answer your question. Maybe I misunderstood? I was trying to say that one can at least see that alpha is dimensionless using the formulae which you wrote down earlier.

 

To answer your equstion more specifically:

 

Q: 'Is it obvious to you that in Planck units the value of the Coulomb constant is alpha?'

A: No it is not, because alpha is a dimensionless number and depends solely on the details of the theory. One could imagine having a 'super-electromagnetism' which was ten times as strong as standard electromag and would have an 'alpha' which was approx 1/14.

 

Is this what you were meaning?[/quote']

 

I see the problem.

What I say, or meant to say, is

alpha is the number which is the VALUE of the Coulomb constant when that constant is expressed in Planck terms.

 

The units of the coulomb const are F L² C^-2

 

that is, force x length-squared per charge-squared

 

If you write the Coulomb constant in terms of Planck units, using the electron charge as unit, then its numerical value is alpha.

 

this is actually trivial to show, just some algebra

[Martin]

Severian, Let's not get obstructed by different meanings of the word "value"

 

the way I'm talking: I say for instance that the dimensionless number 299792458

is the NUMERICAL VALUE of the speed of light when it is expressed in SI metric terms.

 

the speed of light is a physical quantitity, not a number.

 

however, when the speed of light is expressed in terms of some units, then it has a numerical value associated with it.

 

Planck units are in fact physical quantities (the NIST site or the Particle Data Book or whatever standard reference gives their metric equivalents)

and the numerical value of c, in those units, is !.

that is, light in vacuo goes one Planck length unit per Planck time unit.

 

Now the Coulomb constant is

[math]\frac{1}{4\pi\epsilon_0}[/math]

and that is a physical quantity, with dimension, not a number.

 

But if you express that quantity in terms of Planck units, as they are customarily defined, you will indeed discover that the numerical value is exactly alpha. this is trivial.

 

If you have any doubts about it, I would be happy to derive it for you

(but think you can probably see it for yourself )

[Martin]

Actually why not derive it anyway, since it is easy to do...

 

By the very definition of alpha, which you and I both gave in our posts, we have that

 

[math]\frac{1}{4\pi\epsilon_0} = \alpha \frac{\hbar c}{e^2}[/math]

 

However hbar c is equal to Planck unit energy times Planck unit length.

So if we have adjoined the fundamental charge to Planck units what we see here

 

[math] \frac{\hbar c}{e^2}[/math]

 

is nothing other than the units of the Coulomb constant! So we are done.

[Severian]

OK - sorry, I was misunderstanding. I see what you mean.

 

(I had thought that you were trying to predict the strength of electromagnetism from first principles, which if you could do it would be Nobel Prize time....)

[Martin]

it is a pleasure to talk to you Severian!

(dont worry I'm not going to try to win the Nobel )

BTW here is another simple question about Planck units:

 

for sure you are thoroughly familiar with the Stefan-Boltzmann

constant of the fourth power radiation law, let's call it sigma.

 

what do you think is the numerical value of the S-B sigma

expressed in Planck units?

[Severian]

The Stefan-Boltzmann constant is [math]\sigma=\frac{\pi^2k^4}{60 \hbar^3 c^2}[/math]

 

Natural units has [math]\hbar=c=k=1[/math] so then [math]\sigma=\frac{\pi^2}{60}[/math].

[Martin]

The Stefan-Boltzmann constant is [math]\sigma=\frac{\pi^2k^4}{60 \hbar^3 c^2}[/math]

 

Natural units has [math]\hbar=c=k=1[/math] so then [math]\sigma=\frac{\pi^2}{60}[/math].

 

 

Of course, and it was clearly easy for you. Very happy you are around.

 

I am a physics-watcher (orig. in math) but I will try to answer your question in the other thread about Quantum Gravity as best i can.

I am very interested in QG---especially the explicitly background-independent approaches (like LQG and simplicial QG) because

General Relativity is itself background-independent. The quantization

of GR seems to be speeding up, which makes it fun to watch.