# CGS and SI Units

1. Aug 4, 2012

### schaefera

So I understand that converting between CGS and SI units is no easy task when you look at the general form of each equation in electromagnetism, as you can't simply relate the two systems given they have different ways of looking at what is a derived unit and what is fundamental. But once you've used either system's set of equations to solve a problem, is there an easy conversion at that point (once all you've got is some number with certain dimensions sitting in front of you)?

Additionally, what are we to make of two systems have completely different sorts of units for quantities like charge or current? It seems very different from the simply types of conversions between mechanical units like length (eg mile to kilometer).

The equations surely tell us the same thing, but if each piece has different units in corresponding equations, isn't that a bit odd?

2. Aug 4, 2012

### cepheid

Staff Emeritus
I think the answer to your first paragraph is "yes." If you have some physical quantity that is expressed as a number along with associated units in cgs, there should be just a straightforward conversion from that to a number with units in mks.

Your second paragraph puzzles me. Specifically I'm unsure of what you mean when you say "different sorts of units." Different unit systems always have "different" units for the same type of physical quantity. In SI units, force is measured in newtons, and in cgs, it's measured in dynes (I think). So...what exactly is the issue?

3. Aug 4, 2012

### AlephZero

There isn't a single set of "CGS units" for electromagnetism. There were at least 4 different sets of units used at different times and for different purposes.

The equations all have the same basic forms, except the constants $4\pi$, $\epsilon_0$, $\mu_0$ and $c$ occur in different places, and $\epsilon_0$ and/or $\mu_0$ are 1 by definition in the some sets of CGS units.

4. Aug 5, 2012

### schaefera

Well, perhaps to give a better example, in Gaussian units resistivity is measured in seconds and resistance in sec/cm. This strikes me as odd in the same way-- the SI units do not imply that resistivity has units of time, and resistivity isn't a measure of time, so it's odd (although I get that the unit system is still consistent) to measure the quantity as such.

5. Aug 5, 2012

### Studiot

Well it's true that many familiar quantiies had different dimensions in the CGS and MKS systems and even different within the different CGS systems.

So you will not find easy conversions.

For example

in the emu system resistance has the dimensions μ0LT-1
but in the esu system it has dimensions ε0-1L-1T

As regards Gauss

In the MKS system the total electric flux across a closed surface is numerically equal to the charge enclosed.

In the esu system the flux is 4π times the charge enclosed.

6. Aug 5, 2012

### schaefera

Unless I'm mistaken doesn't flux in SI units have units of volt*meters? If so, how can flux equal charge?

7. Aug 5, 2012

### cepheid

Staff Emeritus
I'm not intimately familiar with this stuff, because I learned electromagnetism in SI units. But let's say for the sake of argument that you pick a unit system in which the Coulomb constant is unity (or, at least, it is some dimensionless number). In other words, you set things up so that whenever you have a charge in this unit system, the electric field at a unit distance away from that charge is numerically equal to the magnitude of the charge. So, in this unit system, it so happens that electric field has dimensions of charge*length-2. The flux of an electric field across a surface would then have dimensions of charge in this unit system.

If you think that's whack, consider the system of "natural units" used by particle physicists. In this unit system, they arrange things so that many of the fundamental physical constants are equal to unity (for convenience). In particular, one system is set up so that c = $\hbar$ = k = 1.

Setting c = 1 means that length and time have the same dimensions. But setting c = 1 also means that energy and mass have the same dimensions, and setting $\hbar$ = 1 means that energy has dimensions of time-1. So we have energy = mass = time-1 = length-1. So, in this unit system, everything can be measured in the same physical unit: the electron-volt (eV) which is a unit of energy.

8. Aug 5, 2012

### schaefera

So Studiot wasn't talking about SI when he mentions MKS, then?

9. Aug 5, 2012

### cepheid

Staff Emeritus
Uh, sure he was. SI = mks. However I think his statement might also have been wrong. In the mks system, the flux of the electric field through a closed surface is numerically equal to the charge enclosed, divided by ε0. The constant ε0 is not unity in SI units.

10. Aug 5, 2012

### Studiot

I was using the MKS system in its widest sense.

SI is based on the Rationalised MKS system, mainly by some changes to the names of the units.

Let us say that a charge of n coulombs is enclosed by a ball of area a sq metres.

The electric flux density is n/a coulombs per sq metre.

So the total electric flux is n/a *a = n coulombs.

I am sorry that my previous post didn't make it clear I meant the total flux (as required by Gauss, to which I did refer).

Were you referring to the electric field intensity ?

11. Aug 5, 2012

### vanhees71

In the SI Maxwell's equations (in the vacuum) read

$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}, \quad \vec{\nabla} \cdot \vec{B}=0,\\ \vec{\nabla} \times \vec{B} -\frac{1}{\epsilon_0} \partial_t \vec{E}=\mu_0 \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
According to the last equation, by integration over a volume and using Gauß's Theorem,
$$\int_{V} \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot \vec{E} = \int_{\partial V} \mathrm{d}^2 \vec{A} \cdot \vec{E}=\frac{1}{\epsilon_0} \int_{V} \mathrm{d}^3 \vec{x} \rho=\frac{Q_V}{\epsilon_0}.$$
The flux of the electric field through a closed surface equals the charge up to the permeability of the vacuum, $\epsilon_0$ which is an arbitrary conventional factor to match the units of electricity with those of mechanics. This is the very reason, why the SI makes electromagnetics look so difficult ;-)).

12. Aug 5, 2012

### Studiot

What exactly do you mean by this?

the electric flux density is defined as the vector

D = ε E.

Thus the total electric flux is given by the surface integral as

ψ = ∫∫DndS

this may be related by Gauss' theorem to the charge as

ψ = Q, the total charge enclosed by the surface.

13. Aug 5, 2012

### cepheid

Staff Emeritus
I tend to think of the "electric flux" as the flux of the electric field through a surface. So, basically, exactly what vanhees71 computed. If this is just a difference in terminology, then fine.

14. Aug 5, 2012

### Studiot

Yes I asked you in post#10 if you meant the electric field intensity.

D and E are the conventional symbols for these vectors.

D is a confusing name because there is no flow (flux) of anything, but there it is.

However D, unlike E depends only on the free charges. E also depends upon the polarisation charges (in a dielectric of course).

15. Aug 5, 2012

### schaefera

Fair enough, but my question really concerns units. Let's take the natural units example because it shows my point better.

How can we say, for instance, time and length have the same units? We don't measure these dimensions the same way, with the same instruments, or for the same purpose. How can this represent a useful physical idea without taking away a huge amount of information?

16. Aug 5, 2012

### cepheid

Staff Emeritus
Here's the way I think about it. In a unit system where the conversion factor between two types of quantities is 1 in those units, you might as well ignore one of the two units and take the conversion factor to be dimensionless. Then you can just measure both quantities in the same units.

Example: suppose we find a unit system where the unit of distance is judiciously chosen so that it turns out that c = 1 (distance unit)/s. If we're measuring time in seconds, what we're really saying is that when a displacement x = 1 unit, it's actually equal to c*(1 second). It's equal to the distance that light travels in 1 second. So really, we're measuring distance in light-seconds, and the conversion factor is c = 1 light-second/second. But rather than writing 1 light-second/second, it's more convenient to just say that c = 1 second/second = 1 (dimensionless). So we say that we "measure distance in seconds", and it is understood that what this means is that 1 "second of distance" is the distance covered by light in a travel time of one second. So now, anywhere where you see c in an equation, you can just omit it, because it is equal to 1. This is really convenient in relativity, where what you experience entirely as an interval in space (or entirely as an interval in time) could be experienced partially as a space interval and partially as a time interval by another observer who is in motion relative to you. The underlying physical reality is spacetime (both observers agree about the length of the interval in spacetime), so it makes sense to treat distance and time on the same footing and measure them in the same units.

So, for the equation E = mc2, we'd just write E = m, since c = 1. When we say that the rest mass of a particle is 1 GeV, it's understood that what we mean is that the mass of the particle is such that its rest energy would be 1 GeV. Even if c weren't equal to 1, you could always express the rest mass as 1 GeV/c2. So, due to mass-energy equivalence, it's pretty handy (and makes sense) to have a unit system where c = 1, and we don't have to worry about the distinction.

Last edited: Aug 5, 2012
17. Aug 5, 2012

### cepheid

Staff Emeritus
Practical example: in this unit system, all velocities are dimensionless, since both distance and time are measured in seconds. So, let's say we have object that is moving with speed v = 0.45. What does this mean? It means that the object covered 0.45 "seconds of distance" in 1 second of time. 0.45 seconds of distance is the distance that light would have travelled in 0.45 seconds. So, this object took a whole second to cover the distance that light would have covered in only 0.45 seconds. Clearly this object is moving at only 45% the speed of light.

Last edited: Aug 5, 2012