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Ok, I must be missing something very obvious here. After applying the boundary conditions, we can write:

$$ A_R e^{i \delta_R} = (\frac{v_2 - v_1}{v_1 + v_2}) A_I e^{i \delta_I} $$

and

$$ A_T e^{i \delta_T} = (\frac{2v_2}{v_1 + v_2}) A_I e^{i \delta_I} .$$

Then, my book says if the second string is lighter, we have $v_2 > v_1$, so $ \delta_T = \delta_R = \delta_I $. I am really not seeing how we can deduce $ \delta_T = \delta_R = \delta_I $ from $v_2 > v_1$.

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It's due to the sign of $v_2 - v_1$ when $v_2 \gt v_1$. Note that $A_I$, $A_R$ and $A_T$ are all defined to be positive. Therefore a sign difference (if there is any) is subsumed in the phase. So when $v_2 - v_1 \lt 0$, $\delta_R = \delta_I + \pi$ because $\exp{i\pi} = -1$. Conversely, when $v_2 - v_1 \gt 0$, there is no $-1$ to account for, and the two phases are the same. Note that the phases must be equal up to a factor of $\pi$ due to the continuity condition for the string.

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  • $\begingroup$ Ahh, the A's defined as positive was the key. Thanks! $\endgroup$ – pmac Aug 18 at 4:03

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