Monday, August 17, 2015

Does Not Look Like What?

From the post "A Bloom Crosses Over" dated 10 Aug 2015,

\(\cfrac{ tan(\theta_1)}{tan(\theta_2)}=\cfrac{\varepsilon_2}{\varepsilon_1}\)

\(\varepsilon_2\gt\varepsilon_1\)

then,

\(\theta_2\lt\theta_1\)

\(\Delta\theta=\theta_2-\theta_1\lt0\)

In all the previous diagrams \(\Delta\theta\gt0\).

If \(\Delta\theta^{\,'}=-\Delta\theta\)


where \(\varepsilon_2\gt\varepsilon_1\).