# What is local about the hidden variable in Bell's theorem?

A pion decays to a singlet electron/positron state. We will measure the component of the electron's spin in the $$\vec a$$ direction and positron's spin in the $$\vec b$$ direction. If there exists a local hidden variable(s) $$\lambda$$, then outcomes of the measurements must be given by functions $$A(\vec a,\lambda)$$ and $$B(\vec b,\lambda)$$ respectively. these functions can only output the values $$\pm1$$ (in units of $$\hbar/2$$.) Due to entanglement and the case of $$\vec a=\vec b$$, we may write $$B(\vec b,\lambda)=-A(\vec b,\lambda)$$. We also assume a non-negative probability density $$\rho(\lambda)$$ which integrates to unity over all of space. Then Bell's inequality follows in the usual way.

This is my question: What is it in the above definitions for $$\lambda$$ that makes it "local"?

Griffiths writes the following: Is the locality condition only that $$A$$ is strictly not a function of $$\vec b$$ (and $$B$$ not of $$\vec a$$), which could not be ruled out for FTL signals?

• A great resource for understanding Bell's theorem and the notion of locality it assumes is Travis Noreen's articles on the subject (you can search for them on arxiv); this one in particular is relevant: arxiv.org/abs/0707.0401 Jul 1, 2022 at 13:54

In addition to $$A$$ strictly not being a function of $$b$$ (and vice versa), the locality condition is ensures that $$\vec{a}$$, $$\vec{b}$$, and $$\lambda$$ are completely independent of each other. In particular, $$\vec{a}$$, $$\vec{b}$$ are not functions of $$\lambda$$ (or correlated with $$\lambda$$ in any way). Any such correlation would spoil the proof of Bell's theorem.
What makes the functions $$A(a,\lambda)$$ and $$B(b,\lambda)$$ local are two factors
1. As you said before the fact that $$A(a,\lambda)$$ does not depend on b.
2. The $$\lambda$$ variables are fixed at the source at the moment the particles are created. So their behavior is "locally" explained or predetermined by these variables.
• For a given direction in $A(a,\lambda)$ we must put $a=\theta$ and for the opposite direction $a=\theta+\pi$. So the function is well defined. Jul 20, 2022 at 20:05
• Suppose we consider variable 1 in a and 2 in b, if we look at thd setup in a spacetime diagram, then after the meadurement 1 can "know" the direction a. So looking at it through time the lambda1 can depend on a after ghe measurement and be transmitted to the averager, like in $cov(a,b,l1,l2)=1/(l1*l2)\int_0^{l1}Cos(A-S)ds\int_0^{l2}cos(b-t)dt$ Aug 7 at 10:36