knitr::opts_chunk$set(echo = TRUE) library(hadron)
For fitting thermal pollutions one wants to try the following model:
$$
A_1 \exp(E t) + A_2 \exp(E \cdot (T-t)) \,.
$$
It has two amplitudes but only one energy. I have implemented this as TwoAmplitudesModel
and restricted it to a single correlator. One could generalize this to fit a whole correlator matrix, but I cut the corners for now.
What we actually implement is the following to be consistent with the SingleModel
:
$$
\frac12 \left( A_1^2 \exp(E t) + A_2^2 \exp(E \cdot (T-t)) \right) \,.
$$
The samplecf correlation function does not have thermal pollutions. Therefore we expect the model to recover the same amplitude for forward and backward part.
scf <- bootstrap.cf(samplecf) plot(scf, log = 'y')
fit_sample <- new_matrixfit(scf, 8, 22, model = 'single') plot(fit_sample, log = 'y') residual_plot(fit_sample, ylim = c(1/1.05, 1.05))
fit_sample_2 <- new_matrixfit(scf, 8, 22, model = 'two_amplitudes') plot(fit_sample_2, log = 'y') residual_plot(fit_sample_2, ylim = c(1/1.05, 1.05))
Looking at the results from both fits, we see that the first fit produces $(E, A)$ which is reproduced by the second as $(E, A_1, A_2)$ pretty well:
mapply(tex.catwitherror, fit_sample$t0, fit_sample$se, with.dollar = FALSE) mapply(tex.catwitherror, fit_sample_2$t0, fit_sample_2$se, with.dollar = FALSE)
We can make up an example which has different forward and backward amplitudes and constant noise.
extent_time <- 48 time <- seq(0, extent_time - 1, by = 1) model_E <- 0.015 model_A1 <- 0.35 model_A2 <- 0.4 val <- 0.5 * model_A1^2 * exp(-model_E * time) + 0.5 * model_A2^2 * exp(-model_E * (extent_time - time))
plot(time, val, main = 'Model data', xlab = 't', ylab = 'C(t)')
measurements <- do.call(cbind, lapply(val, function (v) rnorm(400, v, 0.01))) cf <- cf_orig(cf_meta(Time = extent_time), cf = measurements) cf <- symmetrise.cf(cf) cf_boot <- bootstrap.cf(cf) plot(cf_boot, log = 'y')
We fit that using the new model and
fit <- new_matrixfit(cf_boot, 2, 23, model = 'two_amplitudes') plot(fit, log = 'y') residual_plot(fit)
Comparing with the input from the model gives a reasonable result:
print(c(model_E, model_A1, model_A2)) mapply(tex.catwitherror, fit$t0, fit$se, with.dollar = FALSE)
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