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# akalin/test-0.9.0-with-text.html

Last active March 21, 2018 07:02
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 0.9.0

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 $$D = ε$$.

This file contains 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
 0.9.0

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 $$D = ε$$.