Question
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Let's say I have a boolean field Es at each point S of a Lorentzian manifold:
d˜s2=−c2dt2+dx2
The mean probability is given by:
P(Es)=limn→∞∑sEs1+Es2+Es3+…+Esnn
The equations of motion are given by:
d˜s=Es(cdtˆet+dxˆex)+(1−Es)(cdtˆet−dxˆex)
Does this reduce to 1-dimensional brownian motion in the limit c→∞? Also is there some nice way to calculate the stress energy tensor of the boolean field?
Background
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Let's say I'm 1+1 Minkowski spacetime and want to model a random walk. Let Δs be the 2-displacement and Δx/Δt=v be a constant. We expect after Δs′=κΔs and event will happen which will either flip Δx or remain the same. Let us assume there are only 2 possibilities of Δs:
Δs0=(cΔtˆet+Δxˆex)
Δs1=(cΔtˆet−Δxˆex)
This can be expressed as:
Δs′κ=Ei(cΔtˆet+Δxˆex)+(1−Ei)(cΔtˆet−Δxˆex)
where Eκs is a boolean variable and denotes at the possibility of Δs0 or Δs1 happening. As an example to show what I mean:
∑Δs′κ=(cΔtˆet+Δxˆex)+(cΔtˆet+Δxˆex)
where E1=1 and E2=1
Going to the contiuum limit Δs′/κ→0 and Ei→Es where Es is a boolean field as point s:
d˜s=Es(cdtˆet+dxˆex)+(1−Es)(cdtˆet−dxˆex)
To integrate (2Es−1)dxdsds we use this:
∫˜sd˜s=∫˜scdtˆet+(2P(Es)−1)∫˜sdxˆex
More specifically:
∫˜s2Esdxdsds=2∑slimn→∞Es1+Es2+Es3+…+Esnn∫˜sdx
Making P(Es) the probability of a collision and ˜s the path.
Let, us consider the 2-distance:
d˜s2=c2dt2−E2sdx2−(1−Es)2dx2+2Es(1−Es)dx2=c2dτ2
Since, Es(1−Es)=0, E2s=Es and (1−Es)2=(1−Es) we have:
d˜s2=c2dt2−Esdx2−(1−Es)dx2=c2dt2−dx2