31 ms·

See my post here: https://news.ycombinator.com/item?id=16702599 https://news.ycombinator.com/item?id=16702599 From what I understand of MOND, the less attracti

by nonbel 9y ago

See my post here: https://news.ycombinator.com/item?id=16702599 https://news.ycombinator.com/item?id=16702599 From what I understand of MOND, the less attraction to visible mass (as you would expect from a "ultra diffuse" galaxy like this), the smaller the effect should be. So MOND predicts the effect should be relatively small for this galaxy. Further, in the paper that the OP references they do a calculation and get odds of something around 1/20 that a galaxy like this will exist if MOND is correct. The interesting part is that for some reason they interpret that as ruling out MOND, while meanwhile saying "no one has any idea how this thing can exist if the dark matter theory is right". I would think "no idea how it can exist" should correspond to odds less than 1/20.

sulam 9y ago
This galaxy is down in the 1 in 100_000 parts, not 1 in 20 fwiw. I am not going to debate MOND, I am just explaining the reasoning.
- nonbel 9y ago
  I drew the blue vertical line at 20 km/s: https://image.ibb.co/m38AH7/galaxy.png https://image.ibb.co/m38AH7/galaxy.png Are you claiming this is cutting off 1e-5 of the area of that Gaussian curve they fit? It looks very close to 1/20 to me. I don't care about .05, just to get an order of magnitude and compare it to the "this is an unbelievable result" quotes.
- nonbel 9y ago
  Ok, I see I read the figure 3b wrong (only glanced at it before). I extracted the data from their extended figure 5a: dV sigma R_major -4.16 16.41 -3.98 10.55 3.41 -1.97 15.03 7.25 -1.85 -2.24 5.12 -0.74 1.17 5.75 0.29 -14.39 5.97 1.16 -40.39 11.3 1.52 2.24 6.61 1.58 -3.52 13.43 4.49 -1.39 10.23 6.77 Then ran this simulation in R: # Used for biweight midvariance calculation (r.bw) require(asbio) # Utils: ## Simulation function: ### "For a given value of sigma_test we generated 1000 samples of 10 velocities, ### distributed according to a Gaussian of width sigma_test. The ten velocities ### in each sample were then perturbed with errors, drawn from Gaussians with ### widths equal to the empirically-determined uncertainties in the measured ### dispersions. run_sim = function(nSim, n, sigma_test, err){ replicate(nSim, rnorm(n, 0, sigma_test) + rnorm(n, 0, err)) } ## Clean up the biweight midvariance output r.bw2 = function(x){ as.numeric(sqrt(r.bw(x))) } # Scraped data roughly matches what is reported in paper ## MAD (~4.7) s_obs_nmad = mad(dat$dV) ## biweight (~8.4) s_obs_bi = r.bw2(dat$dV) ## RMS (~14.3) s_obs_rms = sqrt(mean(dat$dV^2)) # Run simulation to test sigma values in .1 km/s intervals ## "Using the biweight estimator, “measured” dispersions sigma_obs,test were calculated ## for all samples. If the value 8.4 is within the 16th – 84th percentile of the ## distribution of sigma_obs,test then sigma_test is within the + /-1 sigma uncertainty ## on sigma_intr. This method gives sigma_intr = 3.2 (+5.5, -3.2) kms^-1." sigmas = seq(.1, 30, by = .1) nSim = 1e4 out = as.data.frame(matrix(nrow = length(sigmas), ncol = 2)) colnames(out) = c("sigma_test", "s_obs_bi_percentile") for(i in 1:length(sigmas)){ # Calculate biweights for each sim result sim_res = run_sim(nSim, 10, sigmas[i], dat$sigma) sigma_obs_test = apply(sim_res, 2, function(x) r.bw2(x)) # Output sigma_test and the percentile of 8.4 out[i, ] = c(sigmas[i], ecdf(sigma_obs_test)(8.4)) # Print progress print(out[i, ]) } # Check which sigma_test values are within the accepted percentile range out$check = out$s_obs_bi_percentile > .16 & out$s_obs_bi_percentile < .84 # Plot the results png("results.png", width = 960, height = 960) plot(out[, 1], out[, 2], xlab = "sigma_test", ylab = "Percentile(8.4)", panel.first = grid()) dev.off() I get that their observed sigma_biweight = 8.4 has percentile of p = .0146 for sigma_test = 20 (https://image.ibb.co/ncRF27/results.png https://image.ibb.co/ncRF27/results.png). In their terminology that would mean it is just within the 99% CI, ie had odds of happening of ~ 1/70 if MOND were correct. I believe I replicated their results since it had percentile of p = .6336 for sigma_test = 0.1 and p = .1607 for sigma_test = 8.6, which matches to their sigma_int = 3.2 (+5.5, -3.2) using 16th - 84th percentiles result. Where did you get 1/100,000?

by nonbel 9y ago

sulam 9y ago

This galaxy is down in the 1 in 100_000 parts, not 1 in 20 fwiw. I am not going to debate MOND, I am just explaining the reasoning.

nonbel 9y ago
I drew the blue vertical line at 20 km/s: https://image.ibb.co/m38AH7/galaxy.png https://image.ibb.co/m38AH7/galaxy.png Are you claiming this is cutting off 1e-5 of the area of that Gaussian curve they fit? It looks very close to 1/20 to me. I don't care about .05, just to get an order of magnitude and compare it to the "this is an unbelievable result" quotes.
nonbel 9y ago
Ok, I see I read the figure 3b wrong (only glanced at it before). I extracted the data from their extended figure 5a: dV sigma R_major -4.16 16.41 -3.98 10.55 3.41 -1.97 15.03 7.25 -1.85 -2.24 5.12 -0.74 1.17 5.75 0.29 -14.39 5.97 1.16 -40.39 11.3 1.52 2.24 6.61 1.58 -3.52 13.43 4.49 -1.39 10.23 6.77 Then ran this simulation in R: # Used for biweight midvariance calculation (r.bw) require(asbio) # Utils: ## Simulation function: ### "For a given value of sigma_test we generated 1000 samples of 10 velocities, ### distributed according to a Gaussian of width sigma_test. The ten velocities ### in each sample were then perturbed with errors, drawn from Gaussians with ### widths equal to the empirically-determined uncertainties in the measured ### dispersions. run_sim = function(nSim, n, sigma_test, err){ replicate(nSim, rnorm(n, 0, sigma_test) + rnorm(n, 0, err)) } ## Clean up the biweight midvariance output r.bw2 = function(x){ as.numeric(sqrt(r.bw(x))) } # Scraped data roughly matches what is reported in paper ## MAD (~4.7) s_obs_nmad = mad(dat$dV) ## biweight (~8.4) s_obs_bi = r.bw2(dat$dV) ## RMS (~14.3) s_obs_rms = sqrt(mean(dat$dV^2)) # Run simulation to test sigma values in .1 km/s intervals ## "Using the biweight estimator, “measured” dispersions sigma_obs,test were calculated ## for all samples. If the value 8.4 is within the 16th – 84th percentile of the ## distribution of sigma_obs,test then sigma_test is within the + /-1 sigma uncertainty ## on sigma_intr. This method gives sigma_intr = 3.2 (+5.5, -3.2) kms^-1." sigmas = seq(.1, 30, by = .1) nSim = 1e4 out = as.data.frame(matrix(nrow = length(sigmas), ncol = 2)) colnames(out) = c("sigma_test", "s_obs_bi_percentile") for(i in 1:length(sigmas)){ # Calculate biweights for each sim result sim_res = run_sim(nSim, 10, sigmas[i], dat$sigma) sigma_obs_test = apply(sim_res, 2, function(x) r.bw2(x)) # Output sigma_test and the percentile of 8.4 out[i, ] = c(sigmas[i], ecdf(sigma_obs_test)(8.4)) # Print progress print(out[i, ]) } # Check which sigma_test values are within the accepted percentile range out$check = out$s_obs_bi_percentile > .16 & out$s_obs_bi_percentile < .84 # Plot the results png("results.png", width = 960, height = 960) plot(out[, 1], out[, 2], xlab = "sigma_test", ylab = "Percentile(8.4)", panel.first = grid()) dev.off() I get that their observed sigma_biweight = 8.4 has percentile of p = .0146 for sigma_test = 20 (https://image.ibb.co/ncRF27/results.png https://image.ibb.co/ncRF27/results.png). In their terminology that would mean it is just within the 99% CI, ie had odds of happening of ~ 1/70 if MOND were correct. I believe I replicated their results since it had percentile of p = .6336 for sigma_test = 0.1 and p = .1607 for sigma_test = 8.6, which matches to their sigma_int = 3.2 (+5.5, -3.2) using 16th - 84th percentiles result. Where did you get 1/100,000?