Why are Some Constants Dimensionless?

[physics kiddy]

Why are pure numbers like 1 , 2 , 3 ... dimensionless and Avogadro's Number, Plank's Constant, Gravitational Constant dimensional ?

 

[HallsofIvy]

Because the latter have dimensions! The Gravitational constant, G, appears in F= -GmM/r^2. In the metric system (mks), r has units of meters, both m and M have units of kg so that "mM/r^2" have units of $kg^2/m^2$. F, a force, has units of "$kg m^2/sec^2$". In order to make the units on both sides of the equation the same, G must have units of $1/(kg sec^2)$.

Avogadro's number is the number of molecules per mole. The number of molecules does not depend on any units so Avogadro's number has units of $mol^{-1}$.

Plank's constant is the "h" in $h\nu$ where E is energy, and so has units of $kg m^2/sec^2$ while $\nu$, a frequency, is "number of cycles per second". "Number of cycles", like "number of molecules" is just a number without units. Since we need to have left "$kg m^2/sec^2$" we need . That means that h must have units of $kg m^2$ in the numerator and one "sec" in the denominator: $kg m^2/sec$.

 

[physics kiddy]

Why is 1/2 in 1/2 mv^2 dimensionless ?

 

[aati2sh]

Because mv^2 has the unit same as that of Energy.

 

[physics kiddy]

Because mv^2 has the unit same as that of Energy.

Please elaborate it mathematically.

 

[Whovian]

Let this constant (which turns out to be 1/2) be C. Let k and k' be different dimensionless numbers. It's quite easy to see that we can set up

$$C\cdot k\cdot \left(1\ \mathrm{J}\right)=k'\cdot\left(1\ \mathrm{J}\right)$$

And so

$$C=\dfrac{k'}{k}$$

And so C's dimensionless.

I wanted to point out that c, the speed of light, is dimensionless, equal to 1, in Special (and General, I'd imagine) Relativity. Basically, $299792458\ \mathrm{m}=1\ \mathrm{s}$.

I've always found it useful to think of units as constants that are, in some cases, incompatible with one another, so the simplest form is just their product.