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  What is the stress energy tensor of this Boolean Field?

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Question
---


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)=limnsEs1+Es2+Es3++Esnn
The equations of motion are given by:

d˜s=Es(cdtˆet+dxˆex)+(1Es)(cdtˆetdxˆex)  

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
---

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=(cΔtˆetΔxˆex)

This can be expressed as:

Δsκ=Ei(cΔtˆet+Δxˆex)+(1Ei)(cΔtˆetΔxˆex)  

where Eκs is a boolean variable and denotes at the possibility of Δs0 or Δs1 happening. As an example to show what I mean:

Δsκ=(cΔtˆet+Δxˆex)+(cΔtˆet+Δxˆex)

where E1=1 and E2=1

Going to the contiuum limit Δs/κ0 and EiEs where Es is a boolean field as point s:

d˜s=Es(cdtˆet+dxˆex)+(1Es)(cdtˆetdxˆex)  

To integrate (2Es1)dxdsds we use this:

˜sd˜s=˜scdtˆet+(2P(Es)1)˜sdxˆex

More specifically:
˜s2Esdxdsds=2slimnEs1+Es2+Es3++Esnn˜sdx

Making P(Es) the probability of a collision and ˜s the path.

Let, us consider the 2-distance:

d˜s2=c2dt2E2sdx2(1Es)2dx2+2Es(1Es)dx2=c2dτ2 

Since, Es(1Es)=0, E2s=Es and (1Es)2=(1Es) we have:

d˜s2=c2dt2Esdx2(1Es)dx2=c2dt2dx2

asked May 7, 2020 in Theoretical Physics by Asaint (90 points) [ revision history ]
edited May 7, 2020 by Asaint

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