H0 measurements over time

Hubble, 1929

75

stochastic or random errors

1) As the size of a sample tends to infinity the mean of the sample tends to the mean of the population

2) Count statistics follow a Poisson distribution with mean λ = N number of counts

if λ is large (>30) Poisson ~ Gaussian

\hat{X} \sim N(\lambda, \sqrt{\lambda})

\hat{X} \sim N(\lambda, \sqrt{\lambda})

For inherently random phenomena that involve counting individual events or occurrences, we measure only a single number N. This kind of measurement is relevant to counting the number of radioactive decays in a specific time interval from a sample of material. It is also relevant to counting the number of Lutherans in a random sample of the population. The (absolute) uncertainty of such a single measurement, N, is estimated as the square root of N.

As example, if we measure 50 radioactive decays in 1 second we should present the result as 50±7 decays per second.

https://www.edouniversity.edu.ng/oerrepository/articles/phy_119_general_physics_practical_20182019.pdf

Statistical	Systematic
No preferred direction	Biases the measurement in one direction
Shrinks with the sample size (typically as )	Affects the sample regardless of the size
Typically Gaussian or Poisson	Any distribution, including constant

SN are enable life in the Universe

SN are natural extreme physics laboratories

SN trace the evolution of the Universe

Reason to study Supernovae

high initial mass

8-100 MSun

live fast and die in spectacular explosions

t_\mathrm{MS; Sun} = \mathrm{10 ~Billion~ years}

t_\mathrm{MS; Sun} = \mathrm{10 ~Billion~ years}

lives of stars

\frac{t_\mathrm{MS}}{t_\mathrm{Sun}} = \left(\frac{M_\mathrm{Sun}}{M}\right)^{2.5}

\frac{t_\mathrm{MS}}{t_\mathrm{Sun}} = \left(\frac{M_\mathrm{Sun}}{M}\right)^{2.5}

….. live to be old and die peacefully by slowly cooling down

low initial mass

0.1-8 MSun

t_\mathrm{MS; Sun} = \mathrm{10 ~Billion~ years}

t_\mathrm{MS; Sun} = \mathrm{10 ~Billion~ years}

lives of stars

\frac{t_\mathrm{MS}}{t_\mathrm{Sun}} = \left(\frac{M_\mathrm{Sun}}{M}\right)^{2.5}

\frac{t_\mathrm{MS}}{t_\mathrm{Sun}} = \left(\frac{M_\mathrm{Sun}}{M}\right)^{2.5}

Nobel Prize winning - SN Ia

\mu_0 = m_0 - M_0 = 5 \log{D} + 25 [Mpc]\\

\mu_0 = m_0 - M_0 = 5 \log{D} + 25 [Mpc]\\

\mu_0 = 43.17 - 5\log_{10}(\frac{H_0}{70 \frac{m}{s~Mpc}}) + 5\log_{10}(z)+1.086(1-q_0)z

\mu_0 = 43.17 - 5\log_{10}(\frac{H_0}{70 \frac{m}{s~Mpc}}) + 5\log_{10}(z)+1.086(1-q_0)z

Hubble 1929

Nobel Prize winning - SN Ia

\mu_0 = m_0 - M_0 = 5 \log{D} + 25 [Mpc]\\

\mu_0 = m_0 - M_0 = 5 \log{D} + 25 [Mpc]\\

\mu_0 = 43.17 - 5\log_{10}(\frac{H_0}{70 \frac{m}{s~Mpc}}) + 5\log_{10}(z)+1.086(1-q_0)z

\mu_0 = 43.17 - 5\log_{10}(\frac{H_0}{70 \frac{m}{s~Mpc}}) + 5\log_{10}(z)+1.086(1-q_0)z

\mu_{i}^{model}(z_{i};\Omega_{m},w_{0}) \propto 5log_{10}(\frac{c(1+z)}{h_{0}})\int_{0}^{z}\frac{dz'}{E(z')}\\ E(z) = \sqrt{\Omega_{m} (1+z)^{3} + (1-\Omega_{m})e^{3\int_{0}^{z} dln(1+z')[1+w(z')]}}

\mu_{i}^{model}(z_{i};\Omega_{m},w_{0}) \propto 5log_{10}(\frac{c(1+z)}{h_{0}})\int_{0}^{z}\frac{dz'}{E(z')}\\ E(z) = \sqrt{\Omega_{m} (1+z)^{3} + (1-\Omega_{m})e^{3\int_{0}^{z} dln(1+z')[1+w(z')]}}