Maxwell's equations

I guess I forgot about the horizontal bar in equations 1, 3, and 4. This refers to

en.wikipedia.org/wiki/Fraction_(mathematics)
en.wikipedia.org/wiki/Division_(mathematics)

this is what Im talking about
keep going please

>maxwell

lol

you realise theres 150 years of work to catch up on after this baby shit?

You're missing terms

[math]\Box A^\mu = \mu_0 J^\mu[/math]
[eqn]\Box A^\mu = \mu_0 J^\mu[/eqn]

>NAMBLA

How would one use these equations to answer a question?

depends on the question. generally what you would do in a lot of simple cases is integrate these equations over a simple surface or volume. for example if you integrate either of the curl equations you can relate the circulation of one field to the flux of another.

It's just notation you little shit. It has to be very useful for mnemonics and abuse of notation for remembering how to calculate them, but in the end of the day it's just fucking notation.[eqn] \mathrm{div}\ \mathbf{E} = \frac{\rho}{\varepsilon_0} \\ \mathrm{curl}\ \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} [/eqn]Happy now you little shit?

spotted the meme theorist

It happens to be* (and that's why we keep it around)

Is a 'divergent' field one with non-zero divergence?

Does this relate in any way to the usual meaning of 'divergent', which is 'tends to infinity'?

Thanks mans.

A magnetic monopole was experimentally observed

The inverted triangle (called 'del') is a vector of the partial derivatives of the function you're operating upon. In 3D, ∇ = (d/dx, d/dy, d/dz). This doesn't make sense mathematically, but when calculating gradient, divergence and curl it's a very helpful mnemonic since it feels like you're simply operating on two vectors:

∇F(x,y,z) = gradient = (dF/dx, dF/dy, dF/dz)
∇ . F = divergence = dF/dx + dF/dy/ + dF/dz (dot product)
∇ X F = curl =

>Does this relate in any way to the usual meaning of 'divergent', which is 'tends to infinity'?
no

no

[math] \text{d}\!\star\!F ~~= \mu_0J [/math]
[math] \text{d}\!\star\!\text{d}A = \mu_0J [/math]
[math] \star\square A ~~= \mu_0J [/math]

>works on any spacetime manifold

Ok then. I'll do a little at a time over the next few days. Be patient.

LESSON 1: ELECTROSTATICS

Starting with the experimentally determined Coulomb's Law:
[math]\vec{F}=\frac{1}{4 \pi \epsilon_0} \frac{qQ}{\mathcal{R}^2} \hat{\mathcal{R}}
[/math]

It was been experimentally determined that the force on a test charge by many source charges is the vector sum of the forces from each individual source charge:
[math]\vec{F}=\sum_i \frac{1}{4 \pi \epsilon_0} \frac{qQ_i}{\mathcal{R}^2} \hat{\mathcal{R}}
[/math]

The test charge does not depend on the sum, and can be factored out:
[math]\vec{F}=q \frac{1}{4 \pi \epsilon_0} \sum_i \frac{Q_i}{\mathcal{R}^2} \hat{\mathcal{R}}
[/math]

The sum can then be taken to be it's own entity independent of the test charge (barring ontological objection), which we can the Electric Field:
[math]\vec{F}=q\vec{E}, \: \: \: \: \vec{E}\equiv \frac{1}{4 \pi \epsilon_0} \sum_i \frac{Q_i}{\mathcal{R}^2} \hat{\mathcal{R}}
[/math]

It's not difficult to generalize the sum point charges into an integral of continuous charge:
[math]\vec{E} = \frac{1}{4 \pi \epsilon_0} \sum_i\frac{Q_i}{\mathcal{R}^2} \hat{\mathcal{R}} \Rightarrow \vec{E} = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{\mathcal{R}^2} \hat{\mathcal{R}}[/math]

To evaluate this integral, we need to replace the differential charge [math]dq[/math] with a charge density. The substitution depends on the problem at hand and whether it's 1D (line charge), 2D (surface charge), or 3D (volume charge).
[math]dq \Rightarrow \lambda dl \Rightarrow \sigma da \Rightarrow \rho dv[/math]

The most general integral under consideration is over a volume, so we can write:
[math]\vec{E} = \frac{1}{4 \pi \epsilon_0} \int_V \frac{\rho\, dv}{\mathcal{R}^2} \hat{\mathcal{R}}[/math]

This last equation usually inherits the name "Coulomb's Law" because it is used more than the force law.

This is the physics of electrostatics. Next lesson, next time, the mathematical consequences.

Whoops, every factor that I kept in the sums should have a subscript and not just [math] Q [/math] because they depend on the source charge. For example:

[math]\vec{F}=q\vec{E}, \: \: \: \: \vec{E}\equiv \frac{1}{4 \pi \epsilon_0} \sum_i \frac{Q_i}{\mathcal{R}_i^2} \hat{\mathcal{R}_i}[/math]

Also, once again, throughout my derivations I will follow Griffiths and use this notation to represent the seperation between test point [math]\vec{r}[/math] and source point [math]\vec{r'}[/math]:
[math]\vec{\mathcal{R}}\equiv\vec{r}-\vec{r'}[/math]

DUDE MAXWELL EQUATIONS BREAK THE LAWS OF PHYSICS
NON-EINSTEINIAN RELATIVITY LMAO

LESSON 2: ELECTROSTATICS DIVERGENCE

Columb's Law is the physical input for electrostatics.

Let's analyze the the law with our mathematical vector calculus toolbox to see if we can say anything mathematically interesting.

Starting with the electric field of a point charge at the origin, such that [math]\vec{\mathcal{R}}=\vec{r}[/math]:
[math]\vec{E}
= \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}[/math]

We take the divergence:
[math]\nabla \cdot \vec{E}
= \nabla \cdot \left( \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r} \right)
= \frac{q}{4\pi\epsilon_0} \nabla \cdot \left( \frac{1}{r^2} \hat{r} \right)[/math]

To evaluate the divergence of [math]\vec{E}[/math], we can immediately apply the divergence in spherical coordinates to the last factor(look it up in a table):
[math]\nabla \cdot \left( \frac{1}{r^2} \hat{r} \right)
= \frac{1}{r^2} \frac{\partial }{\partial r} \left( r^2 \left( \frac{1}{r^2} \right) \right)
+\frac{1}{r\,sin(\theta)} \frac{\partial}{\partial \theta} \left( sin(\theta) \, (0) \right)
+\frac{1}{r\,sin(\theta)} \frac{\partial}{\partial \phi} (0)
=0[/math]

This indicates that:
[math]\nabla \cdot \vec{E} = 0[/math]

However if we represent the divergence in terms of its equivalent integral equation by integrating over volume we find:
[math]\int_V\nabla \cdot \vec{E} \, dv
= \frac{q}{4\pi\epsilon_0} \int_V \nabla \cdot \left( \frac{1}{r^2} \hat{r} \right) dv
[/math]

To evaluate the last factor, we apply Gauss' Law to transform the volume integral into a surface integral:
[math]\int_V \nabla \cdot \left( \frac{1}{r^2} \hat{r} \right) dv
= \int_A \left( \frac{1}{r^2}\hat{r} \right ) \cdot \vec{da}[/math]

...

...

Choosing to integrate over a spherical shell to make the math easy, we find:
[math]\int_A \left( \frac{1}{r^2}\hat{r} \right ) \cdot \vec{da}
= \int_0^{2\pi} \int_0^\pi \frac{1}{r^2} \hat{r} \cdot \left( r^2 sin(\theta)\,d\theta\,d\phi \, \hat{r} \right)
= \int_0^\pi sin(\theta) \, d\theta \int_0^{2\pi}d\phi
= 4\pi[/math]

Inserting this result into the integral equation:
[math]\int_V \nabla \cdot \vec{E} \, dv
= \frac{q}{\epsilon_0}[/math]

To generalize this to many charges, we simply invoke the argument that the fields obey superposition, and so too must the result of the integral. Thus if we integrate over a volume with containing total charge [math]Q[/math], then we find:
[math]\int_V \nabla \cdot \vec{E} \, dv
= \frac{Q}{\epsilon_0}[/math]

Now total charge can be written in terms of charge density:
[math]Q
= \int_V \rho \,dv[/math]

Inserting this into the integral equation, we find:
[math]\int_V \nabla \cdot \vec{E} \, dv
= \frac{1}{\epsilon_0} \int_V \rho \, dv[/math]

Observing that both sides of the equation must be equal for any volume, it must be true that the integrands are equal:
[math]\nabla \cdot \vec{E}
= \frac{\rho}{\epsilon_0}[/math]

This is a weird result because it seems to contradict the differential equation, while both the differential and integral forms of these equations are supposed to be equivalent. The key here is that we have accidentally divided by zero at a singularity at the origin where the charge is in the case of the differential equation, and have encountered an object known as the Dirac Delta, a "distribution" that lives under integrals and "picks" out values of functions when the delta is evaluated at zero:
[math]\int_V f(\vec{x}) \delta(\vec{r}) \, dx = f(0)[/math]

So actually the divergence of a point particle in general not at the origin is:
[math]\nabla \cdot \vec{E} = \frac{q}{\epsilon_0}\delta(\vec{r})[/math]

Small tyoo on last equation. To locate a charge somewhere not an the origin, you use:
[math]\nabla \cdot \vec{E} = \frac{q}{\epsilon_0}\delta(\vec{\mathcal{R}})[/math]

Next lesson: curl of electrostatics (a much shorted lesson).

Does all of this make sense?

>Maxwell’s equations thread
>Naturally turns into the most civilized and mature thread on Veeky Forums
Why is Maxwell so based that he can make even the monkeyposting board awesome?

Get a decent textbook and learn them yourself nigger

Retarded time/potentials was the best chapter in Griffiths.

Though the "why is the sky blue" part from the radiation chapter is really up there too.

>babby level physics
>mature
Choose 1, brainlet

Yes, but that's only in Cartesian coordinates.

Finished Calc III just now taking pH II and the fact that I can understand these equations is
cool to me

Not a fundamental one. It just looks like it inside a material.

>3: B field which changes with time creates an rotational E field

what designates the polarity of the field (north or south) and what designates the direction and magnitude of the torque (rotation) ?

It was my understanding that much of the difficulty with Maxwell's equations is the result of the orginal equations being in quaternion and everything you are showing is cartesian.

the rotations and torques that are integral to the relationships between electric and magnetic fields are clumsy and cumbersome in cartesian

Right hand rule sets the convention on pseudovectors (the things that depend on choice of coordinate system, specifically, in this case, the results of curl and cross product). Everything else discussed in this thread is a vector, and therefore coordinate system independent.

This is not a stumbling block at all.

LESSON 3: ELECTROSTATICS CURL

We now want to take the curl of an electrostatic field. Once again, we will start with the electric field of a charged particle at the origin (we can generalize later by invoking superposition)
\vec{E}
= \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}

Working with the symmetry of the problem, the curl in spherical coordinates is:
[math]\nabla \times \vec{A}
= \frac{1}{r sin\theta} \left( \frac{\partial}{\partial \theta} (sin\theta A_\phi) - \frac{\partial A_\theta}{\partial \phi} \right) \hat{r}
+\frac{1}{r} \left( \frac{1}{sin\theta} \frac{\partial A_r}{\partial \phi} - \frac{\partial}{\partial r} (r A_\phi) \right) \hat{\theta}
+\frac{1}{r} \left( \frac{\partial}{\partial r} (r A_\theta) - \frac{\partial A_r}{\partial \theta} \right)\hat{\phi}[/math]

Applying this to the E field of a point charge:
[math]\nabla \times \vec{E}
=\frac{1}{r} \left( \frac{1}{sin\theta} \frac{\partial}{\partial \phi} \left( \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \right) \right) \hat{\theta}
+\frac{1}{r} \left(-\frac{\partial}{\partial \theta} \left( \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \right) \right)\hat{\phi}[/math]

Observing the derivatives in the preceding equation are acting on constants w.r.t the differentiation variable, it is easy to see that the curl must be zero:
[math]\nabla \times \vec{E}
=\vec{0}[/math]

We have better check the integral form of this equation to make sure nothing like a Dirac delta pops up like in the case of divergence. Stoke's theorem is the relevant vector calculus identity:
[math]\int_A \nabla \times \vec{E} \,\cdot d\vec{a} = \oint_L \vec{E} \cdot d\vec{l}[/math]

Investigating the connection between the curl of E and line integral of E, we perform a line integral in spherical coordinates:
[math]\oint_L \vec{E} \cdot d\vec{l}
= \oint_L \left( \frac{1}{4 \pi \epsilon_0}\frac{q}{r^2}\hat{r} \right) \cdot
\left(dr\hat{r} + rd\theta\hat{\theta} + rsin\theta d\phi \hat{\phi} \right)[/math]
...

I failed calc 3 times and the fact that I can't understand these equations make me feel like poo

>This is not a stumbling block at all.
because electric and magnetic fields have no rotation or torque component?

...
Most of the terms drop out because the basis vectors have been chosen to form an orthonormal set:
[math]\oint_L \vec{E} \cdot d\vec{l}
= \oint_L \left( \frac{1}{4 \pi \epsilon_0}\frac{q}{r^2} \right) dr[/math]

Evaluating this integral, and noting that we begin and end at the same point, we find:
[math]\oint_L \left( \frac{1}{4 \pi \epsilon_0}\frac{q}{r^2} \right) dr
= \frac{q}{4\pi\epsilon_0}\oint_L \frac{1}{r^2} dr
= \frac{q}{4\pi\epsilon_0} \left. \frac{1}{r} \right|_{\vec{r_0}}^{\vec{r_0}}
= 0[/math]

Thus, no surprises, the curl really is zero, which is somewhat intuitive when you imagine the shape of the field; nothing is curving! This applies to one charge, any number of charges, and charge distributions.

[math]\nabla \times \vec{E}
=\vec{0}[/math]

=========================================

Electrostatics summary:

We have just shown that starting with the electrostatic Coulomb's Law and applying vector calculus, we find that the curl and divergence of the electrostatic field obeys:

[math]\nabla \cdot \vec{E}
=\frac{\rho}{\epsilon_0}[/math]
[math]\nabla \times \vec{E}
=\vec{0}[/math]

An extremely important result of vector calculus is the Helmholtz Theorem (en.wikipedia.org/wiki/Helmholtz_decomposition). The theorem states that any vector field can be fully described by it's divergence and curl provided that the field is smooth and falls off faster than [math]\frac{1}{r^2}[/math], both of which are satisfied by the physics of electrostatics. Thus, with the divergence and curl of [math]\vec{E}[/math] in hand, we can uniquely specify [math]\vec{E}[/math]. The great importance of this is that we have essentially boiled down the global (integral) physics of electrostatics down to the local (differential) physics.

The Helmholtz theorem has a number of corollaries pertaining to divergenceless and curlless fields. They are very useful in EM.

I skipped voltage, but Maxwell's equations don't use potentials.

Onto Magnetostatics.

>because electric and magnetic fields have no rotation or torque component?
Maxwell's equations are riddled with pseudovectors. The curl, a type of cross product, introduces them. Thus the RHR is incredibly important, and without it people would have competing definitions of Maxwell's equations; both equally valid, both equally incompatible!

I mean it's not a stumbling block in the sense that I've never met a person whose had difficulty keeping things straight after a semester or so of physics/engineering/math where they have to use it. Of course people may have trouble visualizing cross products because of their strictly 3D nature, but that's why you (hopefully) have a right hand to mnemonically remember the content of the convention. I see people I work with flap their hands around in the air sometimes when thinking about them, me included.

I've use right-hand motor & left hand generator rules before.
Is a torque a pseudovector?
are cross products part of the divergence?

kill yourself OP. you cared enough to google an image of the equations but wikipedia has an explanation RIGHT FUCKING next to them

I know this doesn't matter but you should use displaystyle because it looks better!

[math]
\displaystyle{\oint_L \left( \frac{1}{4 \pi \epsilon_0}\frac{q}{r^2} \right) dr = \frac{q}{4\pi\epsilon_0}\oint_L \frac{1}{r^2} dr = \frac{q}{4\pi\epsilon_0} \left. \frac{1}{r} \right|_{\vec{r_0}}^{\vec{r_0}} = 0}
[/math]

Griffiths senpai, calm down

SEE, YOU GOTCHU TRIANGLES AND YOUR LETTERS WHERE YOUR NUMBERS ARE SUPPOSED'TA BE.
ITS OK, MY LIL DARLING SALLY-RAE GETS THEM ALL CONFUSED IN ELEMONTRY SCHOOL TOO.
JUST HANG IN THERE, AND DON'T BE AFRAID ABOUT REPEATIN UR GRADE.
NO SHAME IN NOT GRADUATIN GRADE 4.
- REDNECK SCI GUY

This is not true

Upside-down triangle (or del) is a gradient is a dot is beside it, or a curl is the cross is beside it.
E is the electric field.
B is the magnetic field.
Rho is the charge density.
Epsilon sub naught is the electric constant
t is time
Mu sub naught is the magnetic constant
J is the current density

Or del, but yes.

You don't get it, google NAMBLA.

Or just use the [eqn] tags

Oh sheet mang, completely over my head.
I read nabla, didn't even see the m.