### Gauss's Cube

It has been a busy week, with a talk to the Macarthur Astronomical Society on Monday, and to the Astronomical Society of New South Wales last night. And if you were in Blighty on Wednesday at 4am, you would have heard me join dr karl on Radio 5 Live's graveyard shift to talk about why high speed protons don;t become black holes, but that is the topic for another post.

As I mentioned, I've recently been teaching electromagnetism, and I like to take a little bit of a side-ways glance at derivations and equations. Why? Because sometimes those given in text books can appear a little too idealized or simplified.

One of the things that you have to talk about in electromagnetism is Gauss's law. Mathematically, Gauss's law can seem quite intimidating to a first year student, even those in our advanced class, but let's take a look at what it means in a simplified sense, and then something a little more complicated.

Right, the maths (and to our American cousins, it is maths, not math. Mathematics is the plural of the word mathematic).

If you are feeling mathematically challenged at this point, let's try and understand what we are looking at here. The right had side is the amount of charge in a region (the q) and the thingy on the bottom is a constant of nature, the permittivity of free space. What's the flippy thing on the left. Let's look at this in picture terms.

Thanks to the genius of Michael Faraday, we talk about idea of the electric field, which we think of each charge in the universe being the source (if positive, sink if negative) of electric field lines (the blue lines in the above picture). The number of lines is dependent upon the amount of charge, so doubling the charge makes twice as many blue lines.

So, what is the right side of the equation telling us? Imagine we consider a surface around the charge (the red, badly drawn thing in the picture above), then the what the right hand size effectively does is count the number of blue lines going through the surface; in this case 4.

It should not take a huge amount of mental effort to understand that if you change the shape and size of the red surface, total number of blue lines crossing the surface is 4. There is a little thought needed for this as you can imagine deforming the surface so much that a blue line going out comes back in again, but then goes back out. So really the right hand side is the number of outward going blue lines minus the number of inward heading blue lines.

When students are introduced to this, you usually have the following picture.

Firstly, we make the surface a sphere. We can work out the size of the electric field on the surface of the sphere using Coulomb's law (that's the first bit up there, and yes, I know that I left the unit vector of on the right hand side). The integral simply becomes the size of the electric field over the surface, multiplied by the area of the surface (I've not worried about the vector dot product which we will return to in a minute). The simplicity of choosing a sphere makes the problem, well, simple.

But let's consider something a little more complex. Let's put the charge at the centre of a cube, with a side length of 2a.

It is easiest if we break the problem and consider the number of lines through each side of the box (as we have put the charge in the centre of the box, it should be clear than the number of lines through the total box is 6 times that through one side).

For those that don't like maths, look away now. First, the geometry of the problem

and then we can work out what the integral has to be (note that we cannot ignore the dot product now)

Ugh... Ugly integral time (and not strictly correct, as this is an integral over one face of the cube, and so is not over the entire surface). But we are saved! No more slogging over integrals, now we can use wonderful software like Mathematica to do them for us. So, what do we find?

But this is the integral through one face of the cube, and as we noted above, we need to multiply this by 6 to get the total over the cube. And this is simply the right hand side of where we came in.

Gauss's law works for a cube (which, of course, we knew it would :). Right, back to the grindstone.

As I mentioned, I've recently been teaching electromagnetism, and I like to take a little bit of a side-ways glance at derivations and equations. Why? Because sometimes those given in text books can appear a little too idealized or simplified.

One of the things that you have to talk about in electromagnetism is Gauss's law. Mathematically, Gauss's law can seem quite intimidating to a first year student, even those in our advanced class, but let's take a look at what it means in a simplified sense, and then something a little more complicated.

Right, the maths (and to our American cousins, it is maths, not math. Mathematics is the plural of the word mathematic).

If you are feeling mathematically challenged at this point, let's try and understand what we are looking at here. The right had side is the amount of charge in a region (the q) and the thingy on the bottom is a constant of nature, the permittivity of free space. What's the flippy thing on the left. Let's look at this in picture terms.

Thanks to the genius of Michael Faraday, we talk about idea of the electric field, which we think of each charge in the universe being the source (if positive, sink if negative) of electric field lines (the blue lines in the above picture). The number of lines is dependent upon the amount of charge, so doubling the charge makes twice as many blue lines.

So, what is the right side of the equation telling us? Imagine we consider a surface around the charge (the red, badly drawn thing in the picture above), then the what the right hand size effectively does is count the number of blue lines going through the surface; in this case 4.

It should not take a huge amount of mental effort to understand that if you change the shape and size of the red surface, total number of blue lines crossing the surface is 4. There is a little thought needed for this as you can imagine deforming the surface so much that a blue line going out comes back in again, but then goes back out. So really the right hand side is the number of outward going blue lines minus the number of inward heading blue lines.

When students are introduced to this, you usually have the following picture.

Firstly, we make the surface a sphere. We can work out the size of the electric field on the surface of the sphere using Coulomb's law (that's the first bit up there, and yes, I know that I left the unit vector of on the right hand side). The integral simply becomes the size of the electric field over the surface, multiplied by the area of the surface (I've not worried about the vector dot product which we will return to in a minute). The simplicity of choosing a sphere makes the problem, well, simple.

But let's consider something a little more complex. Let's put the charge at the centre of a cube, with a side length of 2a.

It is easiest if we break the problem and consider the number of lines through each side of the box (as we have put the charge in the centre of the box, it should be clear than the number of lines through the total box is 6 times that through one side).

For those that don't like maths, look away now. First, the geometry of the problem

and then we can work out what the integral has to be (note that we cannot ignore the dot product now)

Ugh... Ugly integral time (and not strictly correct, as this is an integral over one face of the cube, and so is not over the entire surface). But we are saved! No more slogging over integrals, now we can use wonderful software like Mathematica to do them for us. So, what do we find?

But this is the integral through one face of the cube, and as we noted above, we need to multiply this by 6 to get the total over the cube. And this is simply the right hand side of where we came in.

Gauss's law works for a cube (which, of course, we knew it would :). Right, back to the grindstone.

to our British cousins, it's MsP (Members of Parliament), not MPs.

ReplyDeleteMathematics is not the plural of mathematic. "Mathematic" is an adjective, not a noun; ergo, there is no way to pluralize it. The word comes to us from the Latin "mathematica" (which itself comes from Greek mathematike, the feminine inflection of the singular neuter suffix -ikos) via 14th century Middle English. While it appears plural in form, and can, in certain limited situations, be used with a plural verb, the noun "mathematics" is decidedly singular. It is of exactly the same form and etymology as other collective singular nouns referring to groups of scientific fields, modes of manipulation, or areas of study, such as "physics", "athletics", "politics", &c. You don't hear anyone ever saying "I was in physics class, and this one physic REALLY confused me!" or "Politics is pretty interesting, but America's politic is rather dull."

ReplyDeleteThere is still a bit of debate as to whether "mathematics" should be regarded as a strictly singular collective noun, never taking a plural verb, or as a noun that can be both singular collective AND plural, taking either a singular or plural verb depending on context; however, there has never been) any debate over the fact that "mathematic" is not, and never has been, a noun.

As a side note, both the American "math" and the British "maths" are both equally correct. British English and American English are separate, distinct forms of the English language, each with its own slightly different stylistic conventions. Neither is more correct than the other, and, in fact, neither is a strictly prescriptive language. This means that there is no supreme authority on precisely which linguistic features, grammatical forms, syntactic elements, &c. are strictly "right" or "wrong", unlike many other languages. Moreover, there is significantly more difference among many spoken variations of English within England (let alone the U.K.) than there is between written (as well as standard spoken) American English and British English.

To conclude, it would be wise not to tell other countries that they're speaking their own language wrong because they're not following the standards of your language, especially when the standard you're espousing isn't correct in ANY version of the language. :)

Sincerely,

A linguist-turned-physicist.