For three-phase systems, in p-q theory instantaneous reactive power is
$$ q = \frac{(v_a-v_b)i_c + (v_b-v_c)i_a + (v_c-v_a)i_b}{\sqrt{3}} $$
(see, e.g., [1] Eq. (5))
If we substitute in expressions for a balanced inductive load, in which the current lags the voltage by 90 degrees: $$ \begin{align} v_a &= V \sin(\omega t)\\ v_b &= V \sin(\omega t - 2\pi/3)\\ v_c &= V \sin(\omega t + 2\pi/3)\\ i_a &= I \sin(\omega t - \pi/2)\\ i_b &= I \sin(\omega t - \pi/2 - 2\pi/3)\\ i_c &= I \sin(\omega t - \pi/2 + 2\pi/3)\\ \end{align} $$
we find (I used Sympy to verify this): $$ q = \frac{3VI}{2} > 0 \quad\text{(if V, I > 0)} $$ which matches up with, say, a phasor-based computation of total reactive power for a pure inductive load.
However, if we reverse the phase sequence by swapping phases b and c, this equation becomes $$ \begin{align} q' &= \frac{(v_a-v_c)i_b + (v_c-v_b)i_a + (v_b-v_a)i_c}{\sqrt{3}}\\ &=\frac{-(v_c-v_a)i_b - (v_b-v_c)i_a - (v_a-v_b)i_c}{\sqrt{3}}\\ &=\frac{-(v_a-v_b)i_c -(v_b-v_c)i_a -(v_c-v_a)i_b}{\sqrt{3}}\\ &=-q\\ \end{align} $$
and so the instantaneous reactive power changes sign.
My question: to relate p-q theory 3-phase instantaneous reactive power to "conventional" reactive power, must we take the phase sequence into account?
[1] "p-q Theory Power Components Calculations", Afonso, Freitas, and Martins. DOI 10.1109/ISIE.2003.1267279, available here
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