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| <!DOCTYPE html> | |
| <html lang="en"> | |
| <head> | |
| <meta charset="utf-8" /> | |
| <meta name="viewport" content="width=device-width, initial-scale=1.0"> | |
| <title>0.9.0</title> | |
| <link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/KaTeX/0.9.0/katex.min.css"> | |
| <script src="https://cdnjs.cloudflare.com/ajax/libs/KaTeX/0.9.0/katex.min.js"></script> | |
| <script src="https://cdnjs.cloudflare.com/ajax/libs/KaTeX/0.9.0/contrib/auto-render.min.js"></script> | |
| </head> | |
| <body> | |
| <p>The tricky thing is to pick the \(θ^μ\) without assuming that \(H\) is | |
| non-zero. The simplest way to do that is to assume that none of the | |
| coefficients of \(H\) vanish, and, since we have four unknowns (not | |
| counting \(ε_0\) and \(ε_1\)) and three equations, to set \(B = 1\). | |
| Then the first equation gives \(A = (ε_0 + H)/2\text{,}\) the second equation | |
| gives \(C = D(H - A)\), and plugging everything into the third equation | |
| gives \(D^2 = -ε_1 / ε_0\), which implies that \(ε_1 = -ε_0\) and | |
| \(D = ±1\). Set \(ε_0 = -1\) to make the frame have a Lorentzian signature | |
| \(({-} \; {+} \; {+} \; {+})\), and let \(D = ε\).</p> | |
| <script> | |
| renderMathInElement(document.body, { | |
| delimiters: [ | |
| {left: "\\[", right: "\\]", display: true}, | |
| {left: "\\(", right: "\\)", display: false} | |
| ], | |
| ignoredTags: [ | |
| "script", "noscript", "style", "textarea", "pre", "code", | |
| // We sometimes need a dedicated tag. | |
| "nokatex" | |
| ] | |
| }); | |
| </script> | |
| </body> | |
| </html> |
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
| <!DOCTYPE html> | |
| <html lang="en"> | |
| <head> | |
| <meta charset="utf-8" /> | |
| <meta name="viewport" content="width=device-width, initial-scale=1.0"> | |
| <title>0.9.0</title> | |
| <link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/KaTeX/0.9.0/katex.min.css"> | |
| <script src="https://cdnjs.cloudflare.com/ajax/libs/KaTeX/0.9.0/katex.min.js"></script> | |
| <script src="https://cdnjs.cloudflare.com/ajax/libs/KaTeX/0.9.0/contrib/auto-render.min.js"></script> | |
| </head> | |
| <body> | |
| <p>The tricky thing is to pick the \(θ^μ\) without assuming that \(H\) is | |
| non-zero. The simplest way to do that is to assume that none of the | |
| coefficients of \(H\) vanish, and, since we have four unknowns (not | |
| counting \(ε_0\) and \(ε_1\)) and three equations, to set \(B = 1\). | |
| Then the first equation gives \(A = (ε_0 + H)/2\), the second equation | |
| gives \(C = D(H - A)\), and plugging everything into the third equation | |
| gives \(D^2 = -ε_1 / ε_0\), which implies that \(ε_1 = -ε_0\) and | |
| \(D = ±1\). Set \(ε_0 = -1\) to make the frame have a Lorentzian signature | |
| \(({-} \; {+} \; {+} \; {+})\), and let \(D = ε\).</p> | |
| <script> | |
| renderMathInElement(document.body, { | |
| delimiters: [ | |
| {left: "\\[", right: "\\]", display: true}, | |
| {left: "\\(", right: "\\)", display: false} | |
| ], | |
| ignoredTags: [ | |
| "script", "noscript", "style", "textarea", "pre", "code", | |
| // We sometimes need a dedicated tag. | |
| "nokatex" | |
| ] | |
| }); | |
| </script> | |
| </body> | |
| </html> |
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