Questions and answers about electromagnetic general-purpose AE method theory
Question 1. Dimensional soundness of the relational expression between \(A\) and \(E\) claimed by Kubo
I had written the dimension of the electric field \(E\) incorrectly. We modified it and confirmed that the equation \(A = jE/ω\) is dimensionally sound. Attached is the revised document.
Answer 1.
It is clear if you look at the dimensions of the electromagnetic \(AE\) relation \(E=\dot{A}\). Documentation and double-checking of formulations are essential.
Question 2. existence space
Both \(A\) and \(E\) are vectors in real space, but the spaces (real space, imaginary space) that exist are different on the left and right sides of this equation connected by an equal sign. How should we think about this?
Answer 2.
Frequency response is described in complex number space. The electric field charge conservation equation is
$$Electric field E method:$$
$$ divE=ρ/ε$$
$$Electromagnetic A method:$$
$$ divE=ρ/ε \rightarrow E=-\dot{A} \rightarrow$$
$$ divA=jρ/εω (1)$$
The electric field conservation constraint finite element matrix of the divergence theorem on the left side is \([αgg^T]\) for both the electric field \(E\) method and the electromagnetic \(A\) method, and the charge conservation constraint load on the right side is the electric field \(E\) modulus is a real number and the electromagnetic \(A\) modulus is an imaginary number.
Question 3. Vector uniqueness and indeterminacy
The electric field E is uniquely determined, but the vector potential \(A\) has indeterminacy. These are connected by an equal sign on the left and right sides, but how should we think about this?
Answer 3.
It is necessary to understand the meaning of Poincaré's Lemma P1P2 and the Faraday-Maxwell equation.
Lemma P1 is the magnetic flux conservation equation:
$$ If \quad divB=0, then \quad B=rotA \qquad (2)$$
Lemma P2 is a steady (DC) electric field equation:
$$ If \quad rotE_s=0, E_s=grad\phi \qquad (3)$$
Faraday-Maxwell equation:
$$in \ unsteady \ state$$
$$ rotE=-\dot{B} \rightarrow E=-\dot{A}$$
$$in \ steady \ state$$
$$ rotE=0 \qquad (4)$$
From equation (4), the unsteady (AC) electromagnetic general-purpose AE relational expression is
$$rotE=-rot\dot{A} \rightarrow E=-\dot{A} $$
The key point is the asymmetry between the electric field/charge conservation equation and the magnetic flux conservation equation.
Electric field/charge conservation equation:
$$divE=ρ/ε$$
Magnetic flux conservation equation:
$$divB=0 \qquad (5)$$
As mentioned above, the electromagnetic vector potential \(A\)\(\phi\) method formulates both unsteady AC and steady DC, so
$$AC \ electric \ field:$$
$$ rotE=-\frac{d}{dt}rotA \rightarrow E=-\dot{A}$$
$$DC \ electric \ field:$$
$$ rotE_s=rot(-grad\phi)=0 \qquad (6)$$
From equation (6) above, it is assumed that the electric field is indeterminate and not uniquely determined, but this is due to a serious combination of lack of insight into steady and unsteady phenomena, lack of understanding, lack of knowledge, and lack of experience. A unique solution can be achieved by analyzing unsteady AC and steady DC separately and combining the results as necessary.
Additional question 3 about $Question 3.
1) Do you accept that the potential \(A\) is not uniquely determined?
2) If yes: How do you understand that a uniquely determined quantity is equal to a non-uniquely determined quantity?
Answer 3'
In the \(A\)\(\phi\) method,
The scalar potential \(\phi\) is a concept that makes sense when combined with the vector potential \(A\). Since scalar potentials and vector potentials are not unique, they may be determined uniquely by imposing conditions.
The electric field is defined by the following formula based on .
$$E=-grad\phi-\dot{A} \qquad (7)$$
The electromagnetic field problem is as shown in Equations (2) to (4), where the first term on the right side of Equation (7) represents a steady (DC) electric field, and the second term on the right side of Equation (7) represents an unsteady (AC) electric field. Therefore, steady problems and unsteady problems can be uniquely analyzed separately.
In practical electromagnetism, which is not a mathematical proposition, the physical meaning and analytical theory of vector \(A\) and potential \(\phi\), not vector potential \(A\)\(\phi\), are important. Since we cannot solve for stationary and unsteady conditions at the same time, we seem to be obsessed with the \(A\)\(\phi\) joint illusion of mathematical assumptions.
The general-purpose electromagnetic AE method relational expression for frequency \(ω\) is expressed by the following equation from equations (2) and (4).
$$E=-\dot{A} \rightarrow$$
$$ E=-jωA \rightarrow A=\frac{j}{ω}E \qquad (8)$$
In addition, the electromagnetic field frequency equation of motion for the electromagnetic \(A\) method and the electric field \(E\) method is expressed by the following equation.
$$ \frac{1}{μ}rotrotA+σ\dot{A}+ε\ddot{A}=J_0 \rightarrow$$
$$ C(\frac{1}{μ}rotrot+jωσ-ω^2ε)A=J_0 \rightarrow$$
$$ A=C^{-1}J_0 \qquad (9)$$
$$ \frac{1}{μ}rotrotE+σ\dot{E}+ε\ddot{E}=-\dot{J}_0 \rightarrow$$
$$ C(\frac{1}{μ}rotrot+jωσ-ω ^2ε)E=-jωJ_0 \rightarrow$$
$$ E=-jωC^{-1}J_0 \qquad (10)$$
Equation (9) converts \(B=rotA\) to the left side \(H\) of the Ampere-Maxwell equation, and \(E=-\dot{ A}\),\(\dot{E}=-\ddot{A}\) to formulate it. Equation (10) is formulated by time-differentiating both sides of the Ampere-Maxwell equation and substituting the Faraday-Maxwell equation \(\dot{B}=-rotE\) into the left-hand side \(\dot{H}\). do. The frequency responses of the electromagnetic \(A\) method and the electric field \(E\) method can be calculated using equations (9) and (10), the divergence theorem of the electric field/charge conservation equation (13), and the undetermined multiplier law constraint matrix The same response result is obtained from the loading effects of \([αgg^t]\) and charge load \((αQg/ε)\). When the electric field is conserved, the magnetic flux is conserved strictly according to equation (12). For reference, we will reproduce the four Maxwell electromagnetic field equations.
$$Ampere-Maxwell \ equation$$
$$rotH=J+\dot{D} \qquad (11)$$
$$Faraday-Maxwell \ equation$$
$$rotE=-\dot{B} \qquad (12)$$
$$E \ field/charge \ conservation\ equation:$$
$$The source of electric field is charge$$
$$divD=ρ \qquad(13)$$
$$Magnetic flux conservation equation:$$
$$magnetic field has no source$$
$$divB=0 \qquad (14)$$
The current density \(J\) on the right side of equation (11) is expressed as the sum of the wire current \(J_0\) and the displacement current \(J=J_0+σE\).
We will organize Maxwell's electromagnetic field equations for steady DC and unsteady AC.
$$TABLE$$
From the Faraday-Maxwell equation for DC and Poincare Lemma P2, the DC electric field is defined as follows.
$$rotE=0 \rightarrow E_s=-grad\phi \qquad (15)$$
From the magnetic flux conservation equation, Poincaré Lemma P1, and the Faraday-Maxwell equation for alternating current, the alternating current electric field is defined as follows.
$$divB=0 \rightarrow B=rotA \rightarrow$$
$$ rotE=-\dot{B} \rightarrow rotE=-rot\dot{A} \rightarrow$$
$$ E=-\dot{A}$$
$$ \quad where\ rot(-grad\phi)=0 \qquad (16)$$
The electromagnetic vector potential \(A\), which is not uniquely determined, is reduced to a pair of scalars by \(rot\) screening in the fourth equation of equation (16).
-potential \(-grad\)\(\phi\) is eliminated from the vector identity and becomes the fifth vector equation \(E = -\dot{A}\). Since the electric field \(A\) method and the electric field \(E\) method are theoretically equivalent, they can be theoretically integrated as a general-purpose AE method. The general electric field \(E_G\) of the electromagnetic \(A\)\(\phi\) method is determined by equation (7).
be justified.
General electric field = DC electric field + AC electric field
$$E_G=-grad\phi-\dot{A} \quad(7’)$$
The DC balance equation and AC balance equation are expressed by the following equations using the Ampere-Maxwell equation.
$$DC \ equilibrium \ equation:$$
$$ rotH=J_0+σE_s \qquad (17)$$
$$AC \ equilibrium \ equation:$$
$$ rotH=J_0+σE+\dot{D} \qquad (18)$$
From the DC electromagnetic field equilibrium equation (17) and the AC general-purpose \(AE\) method electromagnetic field equilibrium equation (18), the electromagnetic \(A\)\(\phi\) method cross-current analysis of Equation (7') can be performed independently of DC and AC. Analyze and overlay the results.
