# What is the stress energy tensor of this Boolean Field?

+ 0 like - 0 dislike
162 views

## Question ---

Let's say I have a boolean field $E_s$ at each point $S$ of a Lorentzian manifold:

$$d \tilde s^2 = - c^2 dt^2 + dx^2$$

The mean probability is given by:

$$P(E_s) = \lim_{n \to \infty} \sum_{s}\frac{E_{s_1} + E_{s_2} + E_{s_3} + \dots _+ E_{s_n}}{n}$$
The equations of motion are given by:

$$d \tilde s = E_{ s} (c d t \hat e_t+ d x \hat e_x) +(1-E_{ s})(c d t \hat e_t - d x \hat e_x)$$

Does this reduce to $1$-dimensional brownian motion in the limit $c \to \infty$? 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 $\Delta s$ be the $2$-displacement and $\Delta x/ \Delta t =v$ be a constant. We expect after $\Delta s' = \kappa \Delta s$ and event will happen which will either flip $\Delta x$ or remain the same. Let us assume there are only $2$ possibilities of $\Delta s$:

$$\Delta s_0 = (c\Delta t \hat e_t+ \Delta x \hat e_x)$$

$$\Delta s_1 = (c\Delta t \hat e_t - \Delta x \hat e_x)$$

This can be expressed as:

$$\frac{\Delta s '}{\kappa} = E_{ i} (c\Delta t \hat e_t+ \Delta x \hat e_x) +(1-E_{ i})(c\Delta t \hat e_t - \Delta x \hat e_x)$$

where $E_{ \kappa s}$ is a boolean variable and denotes at the possibility of $\Delta s_0$ or $\Delta s_1$ happening. As an example to show what I mean:

$$\sum \frac{\Delta s '}{\kappa} = (c\Delta t \hat e_t+ \Delta x \hat e_x) + (c\Delta t \hat e_t+ \Delta x \hat e_x)$$

where $E_{1} = 1$ and $E_{2} = 1$

Going to the contiuum limit $\Delta s ' / \kappa \to 0$ and $E_i \to E_s$ where $E_s$ is a boolean field as point $s$:

$$d \tilde s = E_{ s} (c d t \hat e_t+ d x \hat e_x) +(1-E_{ s})(c d t \hat e_t - d x \hat e_x)$$

To integrate $(2E_s -1) \frac{dx}{ds} ds$ we use this:

$$\int_{\tilde s} d \tilde s = \int_{\tilde s} c dt \hat e_t + (2P(E_s) -1)\int_{\tilde s} dx \hat e_{x}$$

More specifically:
$$\int_{\tilde s} 2E_s \frac{dx}{ds} ds = 2 \sum_{s} \lim_{n \to \infty} \frac{E_{s_1} + E_{s_2} + E_{s_3} + \dots _+ E_{s_n}}{n} \int_{\tilde s} dx$$

Making $P(E_s)$ the probability of a collision and $\tilde s$ the path.

Let, us consider the $2$-distance:

$$d\tilde s^2 = c^2 dt^2 - E_s^2 dx^2 - (1-E_s)^2 d x^2 +2 E_s(1-E_s) dx^2 = c^2 d \tau^2$$

Since, $E_s(1-E_s) =0$, $E_s^2 = E_s$ and $(1-E_s)^2= (1-E_s)$ we have:

$$d\tilde s^2 = c^2 dt^2 - E_s dx^2 - (1-E_s) d x^2 = c^2 dt^2 - d x^2$$

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:$\varnothing\hbar$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.