TITLE: A question on motivic zeta-function
QUESTION [5 upvotes]: It's well-known that over $\mathbb F_q$ every smooth projective conic $C$ is isomorphic to a projective line. So the formula for the motivic zeta-function $Z_{mot}(C)$ is evident since $S^n\mathbb P^1 \simeq \mathbb P^n$.
But what can one say when the finite field in the formulation of this question is replaced by an arbitrary field? Are there any references which can help me to determine the answer?

REPLY [6 votes]: $S^nC$ will equal $\mathbb P^n$ for $n$ even and a Severi-Brauer variety with the same Brauer class as $C$ for $n$ odd. In the $n$ odd case, its class in the Grothendieck group will be $[C] ( 1+ L^2 + \dots + L^{n-1})$.
Combining the even and odd cases, you get a formula for the motivic zeta function.
For details, and a much more general result, see Zeta Functions of Curves with No Rational Points by Daniel Litt.