So we know that if $P$ is a conservative field, and $\tau$ is the unit tangent to closed curve $\Gamma$, then we have
$\int_\Gamma P \cdot \tau ds = 0$
So my question is if
$\int_\Gamma ( P \cdot \tau )^2 ds = 0$?
I feel like it probably does... but I am not sure how to prove (or disprove) it. Any help would be appreciated.
Just as a bit more info, in the work I am doing $P = \nabla Q$ for some $C^1$ function (maybe more than $C^1$, I am not 100% sure).
For completeness, as stated in the comments:
Note that $(P\cdot \tau)^2$ is always nonnegative. So the integral will only be zero if $P\cdot \tau=0$, almost everywhere. In other words, the answer is no.
The problem with the estimate $$\int_\Gamma (P\cdot \tau)^2 ds \leq \sup_\Gamma (P\cdot \tau) \int_\Gamma (P\cdot \tau)ds = 0$$ is the sign. While $(P\cdot \tau)$ is negative (which it is like half of the time), you cannot estimate against the supremum. Remember back from school, if you multiply with something negative, the inequality flips direction, so while $(P\cdot \tau)≤\sup_\Gamma(P\cdot \tau)$ holds everywhere, multiplying with $(P\cdot \tau)$ where it is negative yields $(P\cdot \tau)^2 \geq (\sup_\Gamma (P\cdot \tau)) (P\cdot \tau)$ instead.