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akalin/test-0.9.0-with-text.html

Last active March 21, 2018 07:02
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line-breaking bug
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test-0.9.0-with-text.html
<!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>
Raw
test-0.9.0.html
<!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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