What are the operators here and how are these formulas derived?

[Ren_Hoek]

In (23), are grad and div some kind of scalar operators comparing to $\nabla$ and $\nabla\times$?
because tbh I dont know how $\text{curl}(\mu^{-1}\text{curl}\textbf{A})$ turns into $\text{div}\mu^{-1}\text{grad}A_z$ even given $A=(0,0,A_z)^T$.

tbh I don't really get it. First, Is \sigma, which I assume is electrical conductivity, related to static magnetic field?

 

[Borek]

No idea about this particular situation, but typically div stands for divergence and grad for gradient, these have well defined meanings.

 

[renormalize]

In (23), are grad and div some kind of scalar operators comparing to $\nabla$ and $\nabla\times$?
because tbh I dont know how $\text{curl}(\mu^{-1}\text{curl}\textbf{A})$ turns into $\text{div}\mu^{-1}\text{grad}A_z$ even given $A=(0,0,A_z)^T$.
tbh I don't really get it. First, Is \sigma, which I assume is electrical conductivity, related to static magnetic field?

Hint: On Physics Forums, "single dollar signs" don't denote LaTex. Instead, inline formulas are sandwiched between "double pound signs" and display formulas between "double dollar signs". (See the LaTeX Guide at the bottom left of the posting screen.)

Below I've reformatted your text to make the LaTeX readable:

In (23), are grad and div some kind of scalar operators comparing to $\nabla$ and $\nabla\times$?​

because tbh I dont know how $\text{curl}(\mu^{-1}\text{curl}\textbf{A})$turns into $\text{div}\mu^{-1}\text{grad}A_z$ even given $A=(0,0,A_z)^T$.​

tbh I don't really get it. First, Is $\sigma$, which I assume is electrical conductivity, related to static magnetic field?​

​

And to answer at least your first question: yes, $\text{grad}\equiv\nabla$, $\text{div}\equiv\nabla\cdot$, and $\text{curl}\equiv\nabla\times$.

 

[Ren_Hoek]

Hint: On Physics Forums, "single dollar signs" don't denote LaTex. Instead, inline formulas are sandwiched between "double pound signs" and display formulas between "double dollar signs". (See the LaTeX Guide at the bottom left of the posting screen.)

Below I've reformatted your text to make the LaTeX readable:

In (23), are grad and div some kind of scalar operators comparing to $\nabla$ and $\nabla\times$?​

because tbh I dont know how $\text{curl}(\mu^{-1}\text{curl}\textbf{A})$turns into $\text{div}\mu^{-1}\text{grad}A_z$ even given $A=(0,0,A_z)^T$.​

tbh I don't really get it. First, Is $\sigma$, which I assume is electrical conductivity, related to static magnetic field?​

​

And to answer at least your first question: yes, $\text{grad}\equiv\nabla$, $\text{div}\equiv\nabla\cdot$, and $\text{curl}\equiv\nabla\times$.

thank you.
No matter how I calculate (calculating directly or using $\nabla\times (\nabla\times A)=\nabla(\nabla\cdot A)-\nabla^2 A$), it is

$$\text{curl}(\text{curl} A) = (\dfrac{\partial^2 A_z}{\partial x\partial z}, \dfrac{\partial^2 A_z}{\partial y\partial z}, -\dfrac{\partial^2 A_z}{\partial x^2}-\dfrac{\partial^2 A_z}{\partial y^2})^T$$

for me so I don't know what this section of the book is talking about.

 

[Ren_Hoek]

I think grad before $A_z$ seems redundant here, as $\text{curl}(\text{curl} A)_z=-(\dfrac{\partial^2 A_z}{\partial x^2}+\dfrac{\partial^2 A_z}{\partial y^2})$ here, so as long as $\dfrac{\partial^2 A_z}{\partial z^2}=0$ it is much more understandable, but I couldn’t find evidence of it and maybe it's related to strong physics meaning which I'm not good at.

 

[renormalize]

I think grad before $A_z$ seems redundant here, as $\text{curl}(\text{curl} A)_z=-(\dfrac{\partial^2 A_z}{\partial x^2}+\dfrac{\partial^2 A_z}{\partial y^2})$ here, so as long as $\dfrac{\partial^2 A_z}{\partial z^2}=0$ it is much more understandable, but I couldn’t find evidence of it and maybe it's related to strong physics meaning which I'm not good at.

Don't forget that by eq.(28) in your posted text, the permeability can also be a function of the transverse-position (as well as time): $\mu=\mu(x,y,t)$. Just use the basic definition of the curl in cartesian coordinates and "turn the crank" twice to get the result in eq.(22) of your text:$$\mathbf{A}=\left(0,0,A_{z}(x,y,t)\right)^{T}\Rightarrow\nabla\times\mathbf{A}=\left(\partial_{y}A_{z},-\partial_{x}A_{z},0\right)^{T}\Rightarrow\mu^{-1}(x,y,t)\nabla\times\mathbf{A}=\left(\mu^{-1}\partial_{y}A_{z},-\mu^{-1}\partial_{x}A_{z},0\right)^{T}$$$$\Rightarrow\nabla\times\left(\mu^{-1}\nabla\times\mathbf{A}\right)=\left(0,0,-\partial_{x}\left(\mu^{-1}\partial_{x}A_{z}\right)-\partial_{y}\left(\mu^{-1}\partial_{y}A_{z}\right)\right)^{T}=-\nabla\cdot\left[\mu^{-1}(x,y,t)\nabla\left(0,0,A_{z}(x,y,t)\right)^{T}\right]$$No "strong physics" involved!

 

[Ren_Hoek]

Don't forget that by eq.(28) in your posted text, the permeability can also be a function of the transverse-position (as well as time): $\mu=\mu(x,y,t)$. Just use the basic definition of the curl in cartesian coordinates and "turn the crank" twice to get the result in eq.(22) of your text:$$\mathbf{A}=\left(0,0,A_{z}(x,y,t)\right)^{T}\Rightarrow\nabla\times\mathbf{A}=\left(\partial_{y}A_{z},-\partial_{x}A_{z},0\right)^{T}\Rightarrow\mu^{-1}(x,y,t)\nabla\times\mathbf{A}=\left(\mu^{-1}\partial_{y}A_{z},-\mu^{-1}\partial_{x}A_{z},0\right)^{T}$$$$\Rightarrow\nabla\times\left(\mu^{-1}\nabla\times\mathbf{A}\right)=\left(0,0,-\partial_{x}\left(\mu^{-1}\partial_{x}A_{z}\right)-\partial_{y}\left(\mu^{-1}\partial_{y}A_{z}\right)\right)^{T}=-\nabla\cdot\left[\mu^{-1}(x,y,t)\nabla\left(0,0,A_{z}(x,y,t)\right)^{T}\right]$$No "strong physics" involved!

Thank you, I see.
But how does $(A_x, A_y) = (\mu^{-1} \dfrac{\partial^2 A_z}{\partial x\partial z}, \mu^{-1} \dfrac{\partial^2 A_z}{\partial y\partial z})$ turn into (0, 0) or are they simply neglected since we're considering the z axis? or was my calculation wrong?

 

[Ren_Hoek]

I think $\dfrac{\partial A_z}{\partial z} = 0$ as suggested in (28), it is also suggested in the expression $A_z(x,y;t)$ as it does not contain $z$, and will make the second order derivative to 0 in the above case.
But I still don't fully understand the physical meaning of it.

 

[renormalize]

But I still don't fully understand the physical meaning of it.

I can't comment on "physical meaning" since the partial text you posted doesn't describe the configuration that is being analyzed. Maybe it's something like a long, current-carrying solenoid coil or a waveguide? A configuration, relatively long in the $z$- (longitudinal) direction and relatively short in the $x$-,$y$- (transverse) directions, can sometimes be well-approximated physically, at least away from the ends, by postulating that the currents and fields have a specific functional form $\zeta(z)$. Examples are a DC solenoid ($\zeta(z)=\text{constant}$) or an electromagnetic waveguide propagating an oscillating field ($\zeta(z)=e^{ikz}$).

 

[Ren_Hoek]

I can't comment on "physical meaning" since the partial text you posted doesn't describe the configuration that is being analyzed. Maybe it's something like a long, current-carrying solenoid coil or a waveguide? A configuration, relatively long in the $z$- (longitudinal) direction and relatively short in the $x$-,$y$- (transverse) directions, can sometimes be well-approximated physically, at least away from the ends, by postulating that the currents and fields have a specific functional form $\zeta(z)$. Examples are a DC solenoid ($\zeta(z)=\text{constant}$) or an electromagnetic waveguide propagating an oscillating field ($\zeta(z)=e^{ikz}$).

Thank you.

I think it simulates this coil model.

 

[renormalize]

I think it simulates this coil model.

OK, that clarifies your configuration. From Google, there are a fair number of references about modeling C-electromagnets. For example, CERN Dirac C-magnet:

Since the transverse size ($x,y$ directions) is large compared to the $z$-extent, your 2D FEM model is only an approximate representation of the actual magnet. As the reference says:

Here's the full 3D model of the magnetic field:

Of course, in the analysis and model you've cited, you apparently have the additional complication of time-dependent currents and fields. So it's a fairly complex problem.

 

[Ren_Hoek]

OK, that clarifies your configuration. From Google, there are a fair number of references about modeling C-electromagnets. For example, CERN Dirac C-magnet:
View attachment 347767
Since the transverse size ($x,y$ directions) is large compared to the $z$-extent, your 2D FEM model is only an approximate representation of the actual magnet. As the reference says:
View attachment 347768
Here's the full 3D model of the magnetic field:
View attachment 347769
Of course, in the analysis and model you've cited, you apparently have the additional complication of time-dependent currents and fields. So it's a fairly complex problem.

Thank you, I will take a further look into this model.