Unitless constants and unitized forces

[enotstrebor]

Coupling constants are unitized or unitless. Forces are associated to unitized coupling constants like charge (e), gravity (G), etc. All unitless coupling constants that I know of, like the fine structure constant (\alpha), or the Lorentzian \beta, etc are ratios where the units cancel out.

As the weak coupling constant is unitless;
Does this mean it is a ratio of forces like the fine structure constant? If not why not.

 

[bcrowell]

Coupling constants are unitized or unitless. Forces are associated to unitized coupling constants like charge (e), gravity (G), etc. All unitless coupling constants that I know of, like the fine structure constant (\alpha), or the Lorentzian \beta, etc are ratios where the units cancel out.

What do you mean when you refer to the "Lorentzian \beta?" To me this suggests the relativistic \beta=v/c, which isn't a coupling constant.

I don't think it makes sense to classify coupling constants as unitful or unitless. The SI has three basic units, for length, time, and mass. You can pick any three universal constants you like and set them equal to 1, and then you get a system of units where everything is unitless. For instance, if you pick c=1, G=1, and h=1, you get a system of units where everything is unitless, and the Planck distance equals 1. Of c, G, and h, one (G) is clearly a coupling constant, another (c) could be considered to be one, and one (h) is not. Suppose I choose to set c=1, h=1, and the mass of the electron=1. Then I get a system of units where G does not equal 1, but it's unitless, because everything is unitless. I could also just set c=1 and h=1, and then G would have units.

As the weak coupling constant is unitless;
Does this mean it is a ratio of forces like the fine structure constant? If not why not.

Above the electroweak unification energy, the weak interaction is unified with the EM interaction, so aren't the coupling constants the same?

 

[xepma]

What do you mean when you refer to the "Lorentzian \beta?" To me this suggests the relativistic \beta=v/c, which isn't a coupling constant.

I don't think it makes sense to classify coupling constants as unitful or unitless. The SI has three basic units, for length, time, and mass. You can pick any three universal constants you like and set them equal to 1, and then you get a system of units where everything is unitless. For instance, if you pick c=1, G=1, and h=1, you get a system of units where everything is unitless, and the Planck distance equals 1. Of c, G, and h, one (G) is clearly a coupling constant, another (c) could be considered to be one, and one (h) is not. Suppose I choose to set c=1, h=1, and the mass of the electron=1. Then I get a system of units where G does not equal 1, but it's unitless, because everything is unitless. I could also just set c=1 and h=1, and then G would have units.



Above the electroweak unification energy, the weak interaction is unified with the EM interaction, so aren't the coupling constants the same?

But the beauty of the coupling constants is that their value does not depend on your choice of units. So in units where h=c=G=1 the value of alpha, the EM coupling constant, is still 1/137 (approx.), and it stays that way whatever your choice of units is.

 

[ansgar]

No coupling constant does not per se be unitless, look at

\lambda \, \phi^3

a fully allowed term in a lagrangian.

is \lambda

unitless here? (in the choice where c = hbar = 1 and unitless)

 

[enotstrebor]

What do you mean when you refer to the "Lorentzian \beta?" To me this suggests the relativistic \beta=v/c, which isn't a coupling constant.
?

My appologies I meant to say unitless constants in the first part.

 

[enotstrebor]

You can pick any three universal constants you like and set them equal to 1, and then you get a system of units where everything is unitless. For instance, if you pick c=1, G=1, and h=1, you get a system of units where everything is unitless

Using c=\hbar=1 does not make the value of a force unitless. For instance if you were really setting c=\hbar=1 then the units of the fine structure constant alpha=e^2/(c*h)=e^2/(1*1) would not be unitless but have the units of charge. The use of thetype conventions are just a mathematical convenience. They do not make things unitless.

The question is, if the weak angle is a unitless coupling constant does that imply that it is the ratio of two forces just like the fines structure constant is ther ratio of two forces (charge and spin).

If not why not?

If so, what are the components of weak angle ratio (what are the two forces?).

 

[Vanadium 50]

The question is, if the weak angle is a unitless coupling constant does that imply that it is the ratio of two forces just like the fines structure constant is ther ratio of two forces (charge and spin).

As was pointed out in https://www.physicsforums.com/showthread.php?t=365156", charge and spin are not forces.