# What are the physycal meaning of universal constants such as the magnetic, electric and gravitational constants?

I know that:

$G=6.67300 × 10^{-11}\dfrac{Nm^2}{kg^2}$(Gravitational constant)

$K_e=9×10^9\dfrac{Nm^2}{C^2}$(Electric constant or Coloumb's constant)

$k_m=1×10^{-7}\dfrac{Ns^2}{C^2}$(Magnetic constant)

But what are the meaning of this values? Would the universe be much different with a little change on this values? What if we hadn't known them?

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2
Closely related to Deriving or justifying fundamental constants and to a lesser degree to units and nature. The short-short version is that the numeric values of dimensional constants have no meaning at all (because they depend of the system of units you choose). However this does not apply to dimensionless constants like the Fine Structure Constant $\alpha(Q^2 = 0) \approx 1/137$. – dmckee♦ Mar 21 '12 at 19:32
– Qmechanic♦ Mar 21 '12 at 20:00
– 0x5f3759df Mar 21 '12 at 21:47
This isn't a duplicate, because it asks about the electric constants, one of which is "too round" to be accidental, and the other is the exact speed of light squared, ignoring the power of 10. – Ron Maimon Mar 22 '12 at 3:03

## 3 Answers

The three constants you give illustrate the arbitrariness of units. The magnetic constant

$$K_m = 10^{-7} {N\over A^2}$$

serves to define the Ampere. The definiton of the Ampere implied by this (defined) constant is that the force between two long wires carrying one Ampere of current at a separation of 1 meter is 10^{-7} Newtons per unit length. If you change this constant, you redefine the Ampere.

the other constant you mention is

$$K_e = 8.996 \times 10^9 {Nm^2\over C^2}$$

This is interesting, because if you just look at the number, forgetting the power of 10, it's the square of the speed of light. The reason for this is that electric and magnetic effects are related by relativity, and the ratio of $K_e$ and $K_m$ is the speed of light squared. Because we choose units where $K_m$ is a power of 10, the $K_e$ is then the same as the square of the speed of light.

Since the speed of light, like $K_m$, is defined, thereby setting the standard definition of the meter, the electric constant $K_e$ is also defined--- it is $10^{-7}$ times the exact speed of light squared. If you vary it, you can only do so in conjunction with varying the definition of the speed of light, and therefore the meter.

The gravitational constant, in an ideal world, would define the unit of mass. But gravity is too weak to measure accurately enough, so we use a block of metal in a vault in Paris for now to define what a "Kilogram" is. This will probably change at some point. But "G" is also a constant that is philosophically incapable of varying, since it defines the system of units.

If you want to ask a sensible question, you should ask them of quantities that don't depend on the units. In practice, you ignore $G,\hbar,c,k_b,K_m$ (your stuff, swapping out $K_e$ for c, plus Planck's constant and Boltzmann's constant). These you set to 1 in a reasonable system of units, and then all other quantities become dimensionless and meaningful, so you can ask about why they have the values that they do.

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The fine-structure constant is a good example of a more sensible question, I guess.. – Manishearth♦ Mar 22 '12 at 3:04
@Manisheath: that, and the mass of the hydrogen atom in Planck units, or the force between two H-atoms at a distance of 10^{25} Planck distances, etc. – Ron Maimon Mar 22 '12 at 3:19
– Manishearth♦ Mar 22 '12 at 3:25

Explaining the meaning of universal constants has been discussed by other posts.

Many of these universal constants are set by attempting to keep as many quantities and units equal to 1 as possible. Having the number 1 in calculations simplifies the math.

If we did not know these universal constants, we would eventually measure them with experiments or prove them from other fundamental equations.

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These three fundamental constants do matter and the universe would be much different when they would be far off the curretn values. Let us look at G as an example.

There is a fundamental property called mass. Mass is a property of objets like particles. Mass manifests itself in two ways: 1) as a force that is required to accelerate a massive object, and 2) a force that acts on a massive object when it is exposed to a gravitational field. $G$ defines and scales the second manifestation: gravity. If you expose an object with mass $m_1$ to a gravitational field produced by a second object $m_2$, the force acting on $m_1$ is:

$$F = G \frac {m_1 \cdot m_2} {d^2}$$

Where $d$ is the distance between the two objects (their center of mass, to be more accurate).

This means that if $G$ suddenly became 10 times larger than it is today, your weight would increase for a factor of ten, making your life quite miserable. And this would even be the least problem.

Note 1: Many people and unfortunately many theoretical Physicists do not make the difference between the value of a constant (which in case of G would be $6.67300\cdot10^{−11} \frac {Nm^2}{kg^2}$) and the numerical value of a constant (which in case of G would be $6.67300\cdot10^{−11}$ when using SI units). This lack of rigour results in various misunderstandings.

Note 2: It is NOT true that the value of a constant depends on the units.

Note 3: It is true that the numerical value of a constant depends on the units. If you increase the unit U for a factor of $a$, the numerical value $n$ must compesate this by decrasing for a factor of $a$. However, this is extremely trivial and was not part of your question.

Note 4: Many people belive that it is possible to "scale" a constant to 1. In case of $G$, this would mean $G = 1$. However, this is a misconception. It is only possible to change the numerical value of a constant to 1, which is written as $\{G\} = 1$. This is trivially done by adapting the unit. In case of $G$, the unit U has to be multiplied by a factor of $6.67300\cdot10^{−11}$ for the numerical factor $n = \{G\}$ to become 1.

Note 5: Here is the correct nomenclature (for any system of units): $$G = n \times U = \{G\} \times [G]$$ For the SI system we get:

• numerical factor $= n = \{G\} = 6.67300 \cdot 10^{−11}$
• unit = U $= [G] = \frac {Nm^2}{kg^2}$
• hence $G= 6.67300 \cdot 10^{−11} \times \frac {Nm^2}{kg^2}$

For a system of natural units:

• numerical factor $= n = \{G\} = 1$
• unit = U $= [G] = 6.67300 \cdot 10^{−11}\frac {Nm^2}{kg^2} = G$
• hence $G = 1 \times G$

This means: in natural units $G$ becomes a unit (not in all systems of natural units, but in many). However, it does not magically "disapear".

Note 6: Natural units is one subject where Wikipedia got it completely wrong.

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-1: I am sorry for downvoting a new user, but when the numerical factor is made equal to 1, G disappears. It is a philosophical thing, but it is clarified by considering the situation where you measure height in feet and horizontal lengths in meters, and you try to do a rotation. The number of units is a construct of macroscopic physics, it has nothing to do with nature. – Ron Maimon Mar 26 '12 at 7:28
"..., but when the numerical factor is made equal to 1, G disappears." - As I have shown in Note 5, this is not true. The numerical factor is 1, but the unit is still there. – Vera Metro Mar 26 '12 at 10:59
You have not shown it, because when the numerical factor is made 1, G disappears. You have not shown anything different. You should sit down and think about it before saying silly things. – Ron Maimon Mar 26 '12 at 16:32
I have shown that if the numerical factor $= n = \{G\} = 1$, then the unit becomes U $= [G] = 6.67300 \cdot 10^{−11}\frac {Nm2}{kg2}$ which is $G$. Thus, G does not disapear, it becomes a unit. I am well aware that this is in contrast with your believe-system and that it is extremely difficult to overcome one's believe-system. But I also think you are smart enough to achieve this. Sit back and take your time to really think about it. In the end it is very trivial. – Vera Metro Mar 26 '12 at 23:02
Think of it this way: another constant quantity, the kg prototype in Paris, also has the numerical factor 1 (per definition). Its mass is $m = 1$kg. If your statement were true, the quantity would also "disappear" and become 1. However, I hope we both agree that the mass of the prototype kg is $m = 1$kg. – Vera Metro Mar 26 '12 at 23:10
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