# 1D string reflection and transmission phase

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 $$v_2 > v_1$$.

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.