Tartalomjegyzék

Electromagnetic field theory

Teljes Matlab script kiegészítő függvényekkel.

Lorentz invariants of the electromagnetic field tensor

File: Lorentz_invariants_of_EMFT.m
Directory: 4_gyujtemegy/13_fizika/QFT
Author: Peter Polcz (ppolcz@gmail.com)

Created on 2020. October 09. (2020b)

Auxiliary variables

syms E_x E_y E_z B_x B_y B_z v c gamma beta real

% beta  = v/c;
gamma = 1/sqrt(1-beta^2);

E = @(F) F(2:4,1) * c;
B = @(F) [
    F(4,3)
    F(2,4)
    F(3,2)
    ];

Electromagnetic field tensor

$F^{\mu\nu} = \left(\begin{array}{cccc} 0 & -\frac{E_{x}}{c} & -\frac{E_{y}}{c} & -\frac{E_{z}}{c} \\ \frac{E_{x}}{c} & 0 & -B_{z} & B_{y} \\ \frac{E_{y}}{c} & B_{z} & 0 & -B_{x} \\ \frac{E_{z}}{c} & -B_{y} & B_{x} & 0 \end{array}\right)$

F = [
    0     -E_x/c -E_y/c -E_z/c
    E_x/c  0     -B_z    B_y
    E_y/c  B_z    0     -B_x
    E_z/c -B_y    B_x    0
    ];

Minkowski metric tensor

Due to the special structure of $g^{\mu \nu}$, the matrix representation of both the contravariant metric tensor $g^{\mu \nu}$ and its inverse, the covariant metric tensor $g_{\mu \nu}$ are the same ($g^{\mu \varrho} \, g_{\varrho \nu} = \delta^{\mu}_{\varrho}$). However, $g^{\mu \nu}$ are $g_{\mu \nu}$ are often distinguished (see, e.g., [1]) as the positions of their indices are different.

g = [
    1  0  0  0
    0 -1  0  0
    0  0 -1  0
    0  0  0 -1
    ];

Covariant electromagnetic field tensor

$F_{\mu \nu} = g_{\mu \varrho} \, g_{\nu \sigma} \, F^{\varrho \sigma}$

coF = g*F*g;

Dual electromagnetic field tensor

$\tilde F^{\mu \nu} = e^{\mu \nu \varrho \sigma} F_{\varrho \sigma}$.

We use block matrix/vector representation to compute this.

$\underbrace{\left(\begin{array}{cccc|cccc|cccc|cccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right)}_{\text{Block matrix representation of $e^{\mu \nu \varrho \sigma}$. The block vector representations of $F_{\varrho \sigma}$ and $\tilde F^{\mu \nu}$ are considered column-wise.}} \cdot \left(\begin{array}{c} 0 \\ -\frac{E_{x}}{c} \\ -\frac{E_{y}}{c} \\ -\frac{E_{z}}{c} \\ \hline \frac{E_{x}}{c} \\ 0 \\ B_{z} \\ -B_{y} \\\hline \frac{E_{y}}{c} \\ -B_{z} \\ 0 \\ B_{x} \\\hline \frac{E_{z}}{c} \\ B_{y} \\ -B_{x} \\ 0 \end{array}\right) = \left(\begin{array}{c} 0 \\ B_{x} \\ B_{y} \\ B_{z} \\ \hline -B_{x} \\ 0 \\ -\frac{E_{z}}{c} \\ \frac{E_{y}}{c} \\ \hline -B_{y} \\ \frac{E_{z}}{c} \\ 0 \\ -\frac{E_{x}}{c} \\ \hline -B_{z} \\ -\frac{E_{y}}{c} \\ \frac{E_{x}}{c} \\ 0 \end{array}\right)$

tF = dual(coF);

Lorentz transformation tensor

Moving frame with velocity $v$ in the $x$ direction.

L = [
    gamma      -gamma*beta 0 0
   -gamma*beta  gamma      0 0
    0           0          1 0
    0           0          0 1
    ];

Electromagnetic field tensor in the moving frame

$F'_{\mu\nu} = \Lambda^{\mu}_{\ \varrho} \, \Lambda^{\nu}_{\ \sigma} \, F^{\varrho\sigma} = \left(\begin{array}{cccc} 0 & -\frac{E_{x}}{c} & -\frac{E_{y}-B_{z}\beta c}{c\sqrt{1-\beta ^2}} & -\frac{E_{z}+B_{y}\beta c}{c\sqrt{1-\beta ^2}} \\ \frac{E_{x}}{c} & 0 & -\frac{B_{z}c-E_{y}\beta }{c\sqrt{1-\beta ^2}} & \frac{B_{y}c+E_{z}\beta }{c\sqrt{1-\beta ^2}} \\ \frac{E_{y}-B_{z}\beta c}{c\sqrt{1-\beta ^2}} & \frac{B_{z}c-E_{y}\beta }{c\sqrt{1-\beta ^2}} & 0 & -B_{x} \\ \frac{E_{z}+B_{y}\beta c}{c\sqrt{1-\beta ^2}} & -\frac{B_{y}c+E_{z}\beta }{c\sqrt{1-\beta ^2}} & B_{x} & 0 \end{array}\right)$

F_   = simplify(L*F*L);

and the corresponding contravariant $F'^{\mu \eta}$ and dual tensors $\tilde F'_{\mu \eta}$

coF_ = g*F_*g;
tF_  = dual(coF_);

Lorentz transformations of the fields

$\vec E = \left(\begin{array}{c} E_{x} \\ E_{y} \\ E_{z} \end{array}\right) ~~\leadsto~~ \vec E' = \left(\begin{array}{c} \color{red}{E_{x}} \\ \frac{E_{y}-B_{z}\beta c}{\sqrt{1-\beta ^2}} \\ \frac{E_{z}+B_{y}\beta c}{\sqrt{1-\beta ^2}} \end{array}\right)$

$\vec B = \left(\begin{array}{c} B_{x} \\ B_{y} \\ B_{z} \end{array}\right) ~~\leadsto~~ \vec B' = \left(\begin{array}{c} \color{red}{B_{x}} \\ \frac{B_{y}c+E_{z}\beta }{c\sqrt{1-\beta ^2}} \\ \frac{B_{z}c-E_{y}\beta }{c\sqrt{1-\beta ^2}} \end{array}\right)$

E_observer = E(F);
E_moving   = E(F_);
B_observer = B(F);
B_moving   = B(F_);

Lorentz invariants

As a consequence of the Contraction Theorem [1, Section 6.6.3] the following quantities evaluates to the same value in all coordinate systems:

Recommended literature: [1, Section 6.3.2. Tensors Are the Key to Invariance]

1. Check the following identity:

$I_L = F_{\mu\nu} F^{\mu\nu} = 2 \vec B^2 - \frac{2}{c^2} \vec B^2 = 2B_{x}^2+2B_{y}^2+2B_{z}^2-\frac{2E_{x}^2}{c^2}-\frac{2E_{y}^2}{c^2}-\frac{2E_{z}^2}{c^2},$

where $F_{\mu \nu} = g_{\mu \varrho} \, g_{\nu \sigma} \, F^{\varrho \sigma}$.

I_L_lhs = trace(F.' * (g*F*g));
I_L_rhs = 2*B(F).'*B(F) - 2*E(F).'*E(F)/c^2;

ZERO = simplify(   I_L_lhs - I_L_rhs   )

Output:

 
ZERO =
 
0
 

2. Check that $I_L$ is invariant, namely, $I_L = F_{\mu\nu} F^{\mu\nu} = F'_{\mu\nu} F'^{\mu \nu}$.

ZERO = simplify(  trace(F.' * (g*F*g)) - trace(F_.' * (g*F_*g))  )

Output:

 
ZERO =
 
0
 

3. Check that

$I_D = \frac{c}{4} \tilde F^{\mu \nu} F_{\mu \nu} = \frac{c}{4} \tilde F'_{\mu \nu} F'^{\mu \nu} = \vec E \cdot \vec B = \vec E' \cdot \vec B'$

I_D_1 = simplify( -c/4 * trace(tF.' * coF) );
I_D_2 = simplify( -c/4 * trace(tF_.' * coF_) );
I_D_3 = E(F)' * B(F);
I_D_4 = E(F_).' * B(F_);

ZERO = simplify(  I_D_1 - I_D_2  )
ZERO = simplify(  I_D_2 - I_D_3  )
ZERO = simplify(  I_D_3 - I_D_4  )

Output:

 
ZERO =
 
0
 
 
ZERO =
 
0
 
 
ZERO =
 
0
 

References

[1] Pavel Grinfeld. Introduction to tensor analysis and the calculus of moving surfaces. Springer-Verlag New York, 2013, eISBN 978-1-4614-7867-6, DOI 10.1007/978-1-4614-7867-6, URL https://www.springer.com/gp/book/9781461478669.

Helper functions

A function, which generates the Levi-Civita symbol.

function [e_cell,e_mat,e_blkmat] = levicivita_4D

    e_mat = zeros(4,4,4,4);

    [i,j,k,l] = ndgrid(1:4,1:4,1:4,1:4);

    I = num2cell(num2cell([i(:) j(:) k(:) l(:)]),2).';

    for ijkl = I
        e_mat(ijkl{:}{:}) = parity([ijkl{:}{:}]);
    end

    e_cell = squeeze(num2cell(e_mat,[1 2]));

    e_blkmat = cell2mat(e_cell.');

end

function par = parity(per)

    I = eye(length(per));
    par = det(I(:,per));

end

function tF = dual(coF)

    % Levi-Civita symbol (alternating tensor)
    [~,~,e_blkmat] = levicivita_4D;

    % dual electromagnetic field tensor
    tF = 2 \ reshape(e_blkmat*coF(:),[4 4]);

end

Published with MATLAB® R2020b, and sanitarized by BeautifulSoup (Python script using BSoup is written by Polcz).