Maxwell

[ydoaPs]

[math]{\nabla}{\cdot}{E}=\frac{\rho}{\epsilon_0}[/math]

 

[math]{\nabla}{\cdot}{B}=0[/math]

 

[math]{\nabla}{\times}{E}=-\frac{\partial{B}}{\partial{t}}[/math]

 

[math]{\nabla}{\times}{B}={\mu_0}{J}+{\mu_0}{\epsilon_0}{\frac{\partial{E}}{\partial{t}}}[/math]

 

Where [math]\rho[/math] is electric charge density, B is magnetic field, E is electric field, [math]\epsilon_0[/math] is permittivity of free space, [math]\mu_0[/math] is magnetic permeability of free space, J is current density, and t is time.

 

I have a general idea of what divergence and curl are, so I have a basic understanding of what most of the equations mean. I was wondering, though, how are B, E, [math]\rho[/math], and J defined? I have a decent understanding of what B and E are, but do they have functions permanently associated with them like B(x,y,z,t) and E(x,y,z,t) or do the equations used depend on the situation? In general, I'd like to know more about the equations.

[Klaynos]

They're vector fields, which I think normally are x,y,z, and t dependent, but of course you could simplify it to just x...

[ydoaPs]

They're vector fields, which I think normally are x,y,z, and t dependent, but of course you could simplify it to just x...

 

I knew they were vector fields, but I don't know how these fields are defined. For example, the bottom two are partial differential equations. How would I go about solving one of these partial differential equations?

[Fred56]

Maxwell's equations were derived from static field equations. The current I and charge q are defined as sources that allow the "computation" of the time-dependent fields. Any help?

[Klaynos]

Well you'd have some E or B field, so something like (note I've droped some constants so there'd probably be an epsilon and maybe a 4Pi in there) E = q/r^2 And then you can operate on that with Del in spherical polar coordinates and then solve the partial DE how you'd normally solve any partial DE...

[Fred56]

What surface are the charge and current density against?

[Bignose]

I knew they were vector fields, but I don't know how these fields are defined. For example, the bottom two are partial differential equations. How would I go about solving one of these partial differential equations?

 

All 4 are actually partial differential equations. You should probably bold the vector terms in there, just to keep it clear. For example, the diverence of a vector is:

 

[math]\nabla\cdot\mathbf{E} = \frac{\partial{E_x}}{\partial{x}} + \frac{\partial{E_y}}{\partial{y}} + \frac{\partial{E_z}}{\partial{z}}[/math] where [math]E_x[/math] is the x component of the E vector.

 

In general, there is no analytical solution to partial differential equations, that a large reason the computational techniques have gotten so much attention. However, a lot of the times you can make simplifying assumptions. Such as, dependence only on one space coordinate. This is what Klaynos was talking about, if a problem is spherically symmetric, for example, the answers will only depend on the radial direction, not either of the two angles. Then, hopefully, with the simplifications, the pde becomes a form that is easier to solve, such as a separable pde or maybe a solvable ode. I don't know of many examples in electricity/magnetism, but I could give you lots of examples from fluid mechanics, if you were interested. Also, you can consult a text on pdes for solution methods, there are many texts out there.

 

On a broader note, a great text to learn about the vector operators is Schey's Div, Grad, Curl, and All That: An Informal Text on Vector Calculus. It is like it says fairly informal, but at the same time, good at explaining what the operators do, what they mean, how to use them, etc. Very highly recommended.

 

What surface are the charge and current density against?

 

Technically, you can do them over any surface if you set it up correctly. usually, the surface is chosen to make the math easier, such as a spherical surface around a point source (so, like above, only the radial dependence will show up). But, you can pick any surface -- in the derivation of the differential equations, you are taking the limit as the surface area goes to zero anyway.

[ydoaPs]

So, there's no set equations for J, rho, E, and B?

[Bignose]

Well, you are going to solve Maxwell's equations for E & B. That's the point of Maxwell's equations, that the fields E & B have to obey them. The differential equations must be supplemented with the appropriate boundary conditions to completely define a situation. J is the current density that will either be prescribed, or it may obey it's own equation and have to be solved simultaneously with Maxwell's (that is, they could be coupled). Same thing with [math]\rho[/math], it may be prescribed or have to obey it's own equation as well.

 

Maxwell's equations are vector equations. Vector equations are great because they describe the physical situation no matter what coordinate system we choose. That is, nature does not know whether we use Cartesian, cylindrical, spherical, bipolar, or any of the other many coordinate systems to describe the situation. The equations still hold. However, the choice of coordinate system is usually made by people to make the situation easier to describe and analyze. Then, depending on what coordinate system, the appropriate divergence, curl, and gradient operators would have to be used. (see http://mathworld.wolfram.com/CylindricalCoordinates.html equation 32 for the gradient operator in cylindrical coords and compare that with equation 33 of http://mathworld.wolfram.com/SphericalCoordinates.html . In fact, the general gradient operator is equation (2) of http://mathworld.wolfram.com/Gradient.html note the use of the scale factors h_1, h_2, and h_3. Those can be used to find what the gradient of Elliptic Cylindrical Coordinates http://mathworld.wolfram.com/EllipticCylindricalCoordinates.html by using equations 6 through 12 which are the scale factors.)

 

Is this the "set equations" you speak of? I am unsure what you mean by that term. Because Maxwell's equations are the equations for B and E.

[ydoaPs]

I mean, don't you have to have an equation for one to find the other? How would we take, say,

[math]{\nabla}{\times}{E}=-\frac{\partial{B}}{\partial{t}}[/math] and find a value for B at a point?

[Bignose]

Maxwell's equations are coupled. You solve them simultaneously. This is a big reason why, in general, there won't be an analytic solution. Only in very special situations will analytic solutions be available.

[ydoaPs]

So, if I restrict the equations to describing how the fields change with respect to time, I can say the following?

 

[math]{\nabla}{\cdot}{E}=\frac{\partial{E}}{\partial{t}}[/math]

Then using that, I can say:

[math]\frac{\partial{E}}{\partial{t}}=\frac{\rho}{\epsilon_0}[/math]

Then using [math]{\nabla}{\times}{B}={\mu_0}{J}+{\mu_0}{\epsilon_0}{\frac{\partial{E}}{\partial{t}}}[/math] I can say [math]\nabla\times{B}={\mu_0}{J}+{\mu_0}{\rho}[/math]. Now, I have [math]\nabla\times{B}={\mu_0}(J+\rho)[/math]. How would I go from this equation for the curl of B to the equation for B?

 

This isn't really related to the thread, but I wanted to play around a little.

 

Maxwell's equations are coupled. You solve them simultaneously. This is a big reason why, in general, there won't be an analytic solution. Only in very special situations will analytic solutions be available.

 

 

What do you mean by solve them simultaneously?

[Bignose]

Well, at is basis, simultaneously means, at the same time. Obviously, this is harder to do with with pdes than say an algebraic system. Let me an example:

 

5x + 8y = 9

-2x + 2y = -1

 

You can solve these sequentially, by re-writing the second equation: x = y +1/2, plug this into the first equation, solve that first one for y, then solve for x.

 

Or you can solve them simultaneously, by re-writing it in matrix form A*x= b where A = [ 5 8 ; -2 2], x = [x ; y] and b = [ 9 ; 1]. (I don't know how to write matrices in LaTeX, so the ; means new line. That is, A is a 2x2 matrix, and x and b are column vectors). Then compute the inverse of A,front multiply both sides of the equation by A^-1, compute A^-1*b and that is equal to x.

 

Sometimes you can do this with pdes, often you can't. Like I said, analytic solutions are pretty rare.

 

Computational methods can do both. They can turn the pde system and linearly approximate it as a matrix and then invert the matrix, or they can do a sequential system. The sequential system is easy to understand. Make guesses for both E & B, call them E(0) and B(0). (I'm going to stop bolding them, since it is a pain to make each one bolded.) Then, use B(0) to solve for E, call that solution E(1). Then, use that updated solution, E(1) to solve for B, and call that B(1). Then, use B(1) to solve for E(2) and so on until the solution converges.

 

But, the idea is still the same. Because they are coupled together, you have to solve the entire system of equations together. That's what I mean by simultaneously.

 

So, if I restrict the equations to describing how the fields change with respect to time, I can say the following?

 

[math]{\nabla}{\cdot}{E}=\frac{\partial{E}}{\partial{t}}[/math]

Then using that, I can say:

[math]\frac{\partial{E}}{\partial{t}}=\frac{\rho}{\epsilon_0}[/math]

 

How did you get this? Because this doesn't seem right to me at all.

[ydoaPs]

So, basically, what I was trying to do above?

 

btw, I have no idea how to undo the curl operator to get the vector field itself. Is there a way?

 

How did you get this? Because this doesn't seem right to me at all.

 

Earlier, you said [math]\nabla\cdot\mathbf{E} = \frac{\partial{E_x}}{\partial{x}} + \frac{\partial{E_y}}{\partial{y}} + \frac{\partial{E_z}}{\partial{z}}[/math]. If we are only looking at time, then can't we get pretend the other dimensions don't exist? I assumed that was a legitimate operation and then substituted into the first equation.

[Bignose]

Well, what you were trying to do above seems very non-valid to me. Why do you think you can set the divergence equal to the time derivative.

 

Is there an un-curl operator? I don't think so, at least not that I am aware of.

 

Write out the terms of the equation in Cartesian coordinates and you'll see what you get.

 

Are you familiar with vector equations, yourdadonapogos? Do you know what I mean when I say write out the terms? Let me give you can example:

 

I'll write out the [math]\nabla\times\mathbf{E} = -\frac{\partial{\mathbf{B}}}{\partial{t}}[/math] equation

 

[math]\nabla\times\mathbf{E} = ( \frac{\partial{E_z}}{\partial{y}} - \frac{\partial{E_y}}{\partial{z}} )\mathbf{x} + ( \frac{\partial{E_x}}{\partial{z}} - \frac{\partial{E_z}}{\partial{x}} )\mathbf{y} + ( \frac{\partial{E_y}}{\partial{x}} - \frac{\partial{E_x}}{\partial{y}} )\mathbf{z} [/math]

 

[math] -\frac{\partial{\mathbf{B}}}{\partial{t}} = -\frac{\partial{B_x}}{\partial{t}}\mathbf{x} - \frac{\partial{B_y}}{\partial{t}}\mathbf{y} - \frac{\partial{B_z}}{\partial{t}}\mathbf{z} [/math]

 

Now, in vector equations, the terms in front of each base vector must be equal, so combining these two above equations, we actually get 3 separate pdes:

 

[math]( \frac{\partial{E_z}}{\partial{y}} - \frac{\partial{E_y}}{\partial{z}} ) = -\frac{\partial{B_x}}{\partial{t}} [/math]

 

[math]( \frac{\partial{E_x}}{\partial{z}} - \frac{\partial{E_z}}{\partial{x}} ) = -\frac{\partial{B_y}}{\partial{t}} [/math]

 

[math]( \frac{\partial{E_y}}{\partial{x}} - \frac{\partial{E_x}}{\partial{y}} ) = -\frac{\partial{B_z}}{\partial{t}} [/math]

 

These differential equations are coupled and need to be solve simultaneously. You can write out similar equations for the rest of Maxwell's equations by fully writing out the curl and divergence operators. The equations I wrote above are for the Cartesian coordinate system, they would be written out differently for a different coordinate system, but the idea remains the same. You will get differential equations that show the relationship between E_x, E_y, E_z, B_x, B_y, and B_z.

 

*****************

 

if E only varied in time, not space, then [math]\nabla\cdot\mathbf{E} = 0 [/math] not equal to the time derivative.

 

***************

 

yourdad, I don't want to seem snotty, because it is not the tone I want this to be taken in. But, have you studied a vector equation/calculus book? Because this is pretty basic stuff, and rather me typing it out piecemeal on a forum, you'll probably get a lot more out of a more organized presentation.

[mourici]

I send u and internet address u will find inside a pdf paper about maxwell with good explantion

http://mourici.googlepages.com/home

 

and the best book for vector theory is vector analysis by Mrraay R.spiegel -schaum publishing

[ydoaPs]

But, have you studied a vector equation/calculus book? Because this is pretty basic stuff, and rather me typing it out piecemeal on a forum, you'll probably get a lot more out of a more organized presentation.

I'm in the middle of a multivariable calculus book, but I haven't gotten to vector fields yet. Thats why I said I had a basic idea of what the curl and divergence operators mean.

[Severian]

An analogy with fluids is a good place to start I think.

 

A scalar field is a just a number for every point in space and time. For example temperature is a scalar field, since there is a temperature (a single number) depending in where and when you are.

 

A vector field is a number and a direction, a bit like a fluid flowing. Imagine a river; at every point in the river, the water is flowing with a certain speed in a certain direction. It is a vector field, just like the electric and magnetic fields are.

 

The divergence of a field, e.g. [math]\vec{\nabla} \cdot \vec{E}[/math] is a measure of how much of the field (or fluid flow) is coming out of a (infinitesimally small) volume minus the amount to going in. So in a fluid analogy, a negative divergence means there is more going in that coming out, as we have a sink; a positive divergence is a source.

 

So [math]\vec{\nabla} \cdot \vec{E}= \rho/\epsilon_0[/math] is telling us that charge density is a source for an electric field (it is spewing an electric field out) and [math]\vec{\nabla} \cdot \vec{B}= 0[/math] is telling us that there are no magnetic sources (no magnetic monopoles).

 

The curl is basically telling you how much the field curls around on itself (hence the name) a bit like a whirlpool in the water. Maxwell's equations are telling us that if the magnetic field changes very suddenly then it will make the electric field swirl around. The curl of the magnetic field is similarly dependent on how quickly the electric field changes, but also on the electric current passing (which is why an electric current forms a magnetic field swirling around it).

[Bignose]

I'm in the middle of a multivariable calculus book, but I haven't gotten to vector fields yet. Thats why I said I had a basic idea of what the curl and divergence operators mean.

 

OK, you'll get there, then. Be patient, and please do not hesitate to ask if you have any further questions (you can start a diff thread, keep asking here, or feel free to PM me if you'd like -- I'm happy to help when I have the time).

 

I like Severian's example of using fluids (I am much more comfortable and have studied fluids a lot more than electromagnetism).

 

In particular, an incompressible Newtonian fluid (water is a good example) obey the mass conservation law of [math]\nabla\cdot\mathbf{v}=0[/math] where [math]\mathbf{v}[/math] is the velocity field. In words, what this means is that around a control volume, what fluid comes in must be equal to the fluid that comes out.

 

Let's look at a 2-D example:

 

----------N------------

|...........................|

|...........................|

W........................E

|...........................|

|...........................|

-----------S-----------

 

Consider this crudely drawn box to be a control volume. W-E is in the x direction, and N-S is in the y direction. Let u be the velocity in the x direction, and let v be the velocity in the y direction. Δx and Δy represent the length of the face, Δx = distance from W to E, and Δy = distance from N to S.

 

Mass in = Mass out

 

uw*(Δy) + vs(Δx) = ue(Δy) + vn(Δx)

 

now collect terms

 

0 = (ue - uw)*(Δy) + (vn - vs)*(Δx)

 

Divide by (Δy)*(Δx)

 

0 = (ue - uw)/(Δx) + (vn - vs)/(Δy)

 

Now, take the limit as both Δx and Δy do to zero. Those differences over deltas become differentials:

 

[math]\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} = 0 = \nabla\cdot\mathbf{v} [/math]

 

Physically, What this means is that if we took a fluid flow and say somehow the outflow ue became doubled, there would have to be corresponding changes to the other outflows and inflows. That is, the outflow vn would have to decrease, or the inflows uw and vs would have to increase, or most likely, some combination of all three.

 

Same thing with [math] \nabla\cdot\mathbf{B} = 0 [/math]. If the magnetic flux increases through one face of a control volume, then the other fluxes through the face have to change accordingly.

 

The vector notation in this case is just the mathematical way of saying what comes in must come out. Whether that "what" is fluid flow, or magnetic flux.

 

It also is why the different components of the vector become coupled together. From the differential equation (which, again, is just what results when we shrink the control volume to a very small, infinitesimal volume), [math]\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} = 0 [/math], you see how a change in u (x-direction velocity) results in a corresponding change in v (the y-direction velocity). Same thing with the magnetic field. A change in B_x (x-direction magnetic flux) results in a corresponding change in B_y (y-direction magnetic flux).

 

And, if the problem is three-D, the equation looks like [math]\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} + \frac{\partial{w}}{\partial{z}} = 0 [/math], which just says that a change in u (x-direction velocity) results in a corresponding change in v (y-direction velocity) and in w (z-direction velocity). Completely analogous for the 3-D magnetic field equation.

 

So, the result is coupled differential equations. You are solving for each component of each vector simultaneously because when you change one component, the rest of the components will be changed too. That's why you can't have just a "set equation" for them (or, really, Maxwell's equations are that "set"). When you provide the necessary boundary conditions, the set of equations are enough to determine the field. Again, it won't be easy because of all the coupling, but, mathematically a solution does exist (proving certain conditions are met -- that's a whole different branch of mathematics, proof that solutions do exist).

[Mr Skeptic]

OK, you'll get there, then. Be patient, and please do not hesitate to ask if you have any further questions (you can start a diff thread, keep asking here, or feel free to PM me if you'd like -- I'm happy to help when I have the time).

 

Best ask it here, so folks like me who are quietly listening in can learn too. I appreciate your perseverance and patience with this, Bignose.

 

I like Severian's example of using fluids (I am much more comfortable and have studied fluids a lot more than electromagnetism).

 

Me too. I understand that the field lines were originally modeled as fluids as way back then too. In any case, it is a useful analogy and allows visualization of the field and allows for [some] "common sense" like you showed.

[Bignose]

There is one more point I'd like to make, as to why vector equations are written in vector form, and not always written out like I did in the last post. I've talked a little about this, but I'm going to supplement it with a few examples to further drive home to point.

 

It has to do with the fact that the laws of nature do not know what coordinate system human beings choose to use to describe what is going on. In that way, the vector equations are the more general equations.

 

For example, the conservation of mass equation for that incompressible Newtonian fluid in vector form is [math]\nabla\cdot\mathbf{v}=0[/math].

 

In Cartesian coordinates, what I've already written out above, this is:

 

[math]\frac{\partial{v_x}}{\partial{x}} + \frac{\partial{v_y}}{\partial{y}} + \frac{\partial{v_z}}{\partial{z}} = 0 [/math]

 

but in cylindrical coordinates it is

 

[math]\frac{1}{r}\frac{\partial{(r v_r)}}{\partial{r}} + \frac{1}{r}\frac{\partial{v_{\theta}}}{\partial{\theta}} + \frac{\partial{v_z}}{\partial{z}} = 0 [/math]

 

and in spherical coordinates the equation is:

 

[math]\frac{1}{r^2}\frac{\partial{( r^2 v_r )}}{\partial{r}} + \frac{1}{r \sin{\phi}}\frac{\partial{ ( \sin{\phi} v_{\phi} )}}{\partial{\phi}} + \frac{1}{r \sin{\phi}}\frac{\partial{v_{\theta}}}{\partial{\theta}} = 0 [/math]

 

Same equation, [math]\nabla\cdot\mathbf{v}=0[/math], just in three different coordinate systems. So, why use the different coordinate systems? Consider a spherically symmetric situation, where the field would only be a function of r. Then all the [math]\frac{\partial}{\partial{\theta}}[/math] and the [math]\frac{\partial}{\partial{\phi}}[/math] terms go to zero. That third equation suddenly gets a lot simpler. Now, we can describe the situation using Cartesian coordinates. Again, the phenomena is indifferent to what coordinate system human beings pick. But, as humans, we try to pick the coordinates so that the problem is as simple as possible.

 

And, making things simpler allows a greater chance of those analytic solutions that I talked about above to exist.

 

It also makes it easier for professors to have students solve problems for homework and exams

[NeonBlack]

YD, you're funny. I apologize if I'm wrong, but I'm almost certain it was you who, no more than a couple of weeks ago threw these in someone's face and said (I'm paraphrasing; I don't remember your words or the thread) "Shut up and don't open your mouth again until you understand these."

 

Anyway, the equations as they are in general will require a solid understanding of vector calculus. You can get a good idea of what they mean if you simplify them. As bignose suggested, you should work with these in 1 dimension. Look at spherical or cylindrical coordinates with no angular dependence. For example, a charged sphere or a long wire with a current running through it. You will be able to handle these with minimal knowledge of vector calculus.

 

I will give a brief conceptual rundown:

You asked what are E and B- E is the electric field, which is completely analogous to gravitational field when mass is replaced by charge. The only difference between a g field and an E field is that there are two charges, but only one mass.

B field is a bit more complicated. We say that electricity and magnetism are the "same thing" and that's true. I like to think of magnetic fields as a relativistic effect. If you know a little SR, then you know that when you go fast, your spatial coordinates shrink and your time coordinate gets bigger. Look at the lorentz transformations for E&M. E and B are connected in the same way as space and time.

 

(I will try to remember the order in which you wrote the equations)

The first one is Gauss's law for electricity. It works the same way as it does for gravity. It tells us that a charged sphere behaves the same way as a point charge in the center would, also if you are inside a charged spherical shell there will be no electric field.

The second is Gauss's law for magnetism. It's not really too useful when analyzing magnetic fields, but it is important in electromagnetic theory of light. It also tells us that there are no magnetic monopoles.

3. Faraday's law. A changing B field makes a voltage. Principle for how generators and motors work.

4. Ampere's law. Basically, current makes a magnetic field. You can forget about the J part.

 

That's very nice, but who cares?

 

Take out the J part and notice how symmetric these equations look. About 150 years ago or so, there was a young physicist in England named Jimmy. (I think he was also working as a Clerk or something like that. I don't remember.) Anyway, he took these equations and played around with them putting them together and stuff- as you, YD attempted - and he came up with a very familiar partial differential equation. He said "sweet, the wave equation! Maybe this electricity and magnetism is some kind of wave." He looked at the velocity term, [math]\frac{1}{\sqrt{\epsilon_0\mu_0}} [/math]and said "hmm, I wonder how fast an electric wave goes." He punched the number into his calculator and said, "whoa, that's pretty fas- OH MY GOD THE SPEED OF LIGHT!!!!!!!!!!!!!" And thus was born the electromagnetic theory of light, which is why we have radios, televisions, cell phones and wi-fi.

[ydoaPs]

On a broader note, a great text to learn about the vector operators is Schey's Div, Grad, Curl, and All That: An Informal Text on Vector Calculus.

 

It(fourth edition) arrived in the mail today along with Introduction to Electrodynamics Third Edition by David J Griffiths.

 

Take out the J part and notice how symmetric these equations look. About 150 years ago or so, there was a young physicist in England named Jimmy. (I think he was also working as a Clerk or something like that. I don't remember.) Anyway, he took these equations and played around with them putting them together and stuff- as you, YD attempted - and he came up with a very familiar partial differential equation. He said "sweet, the wave equation! Maybe this electricity and magnetism is some kind of wave." He looked at the velocity term, [math]\frac{1}{\sqrt{\epsilon_0\mu_0}} [/math]and said "hmm, I wonder how fast an electric wave goes." He punched the number into his calculator and said, "whoa, that's pretty fas- OH MY GOD THE SPEED OF LIGHT!!!!!!!!!!!!!" And thus was born the electromagnetic theory of light, which is why we have radios, televisions, cell phones and wi-fi.

Do you happen to have the derivation?

[NeonBlack]

Look up electromagnetic radiation on wiki.

It looks like there's something on there, I haven't tried to follow it though.

In my physics 3 class, we only went through a hand-waving motivation for the final result, but the professor said a complete derivation would take no more than half an hour (that is if you're comfortable with some vector calculus).