# A Deeper Dive into Core Particles To understand our sun, we need to look deep into its internals. There, many factors combine to turn our angry plasma sphere into the one that we know today! One of these factors is the core hydrogen density of our sun. Our sun fuses hydrogen into helium to generate energy - I hope that's been ingrained in you, but by understanding the densities within our star we can truly dissect and start to see this hidden symmetry within our star. Besides, we can also extrapolate our findings to most other stars out there - although high-mass and low mass stars are different thanks to their interiors. we can generalise this into an equation: $n = \frac{P}{m_{n}} \cdot \left( \frac{m_{H}}{m_{tot}} \right)$ where: $m_{H}$ is the total mass of the hydrogen within the sun $m_tot$ is the total mass of the sun $n$ is the number particle density of hydrogen within the sun $P$ is the internal pressure within the sun $m_n$ is the pressure exerted by each hydrogen molecule This is also given by: $n = \frac{N}{V}$ Where: - $N$ is the number of particles - V is one unit volume (the base ) ## Core Particle Velocity Now it's time to move to finding the velocities of the core particles. Remember that the kinetic and thermal energies of the particles within a star are equal - or at least proportional (i need to check this, remember!) We can use the equations for thermal and kinetic energy: $E_{T} = \frac{3}{2}k_{B}T$ $E_{k} = \frac{1}{2}mv^2$ where $e_t$ is the thermal energy and $e_{k}$ is the kinetic energy. The temperature of the sun's core is given as $T_{C} = 10^7K$ the mass of a hydrogen atom is given by $1.67 \times 10^{-24}$ We can write that $E_{T} = E_{K}$, as the temperature is a direct measure for the velocity. This means we can rewrite the equation: $\frac{3}{2}k_{B}T = \frac{1}{2}mv^2$ isolate the $v$ to get your answer! $v = \sqrt{ \frac{3k_{B}T}{m} }$ >[!Tip]- Boltzmann Constant Units >Remember that the Boltzmann constant is given with the units cm^2. This means we'll have to knock off two decimal places for each calculation you make! units, units. Most things are based off metres after all, not centimetres. # Stellar Lifetimes A star only has a certain amount of fuel. Once this (initial) fuel runs out, the star expands into a red giant. However, that leaves the question - how can we model this initial hydrogen burning phase? The expression below shows us the $\tau_{MS} \approx 10^{10}years\left[ \frac{m}{m_{\odot}} \right]^{-2.5}$ where: - $\tau_{MS}$ is the time a star will spend on the main sequence - $\frac{m}{m_{\odot}}$ is the ratio between the subject star's solar mass and the mass of our sun - $-2.5$ is the **mass-luminosity coefficient** >[!Abstract]- Deriving the Equation >Remember that luminosity, $L$ is given by the energy produced by hydrogen burning over the total time for burning, or: >$L = \frac{E}{t}$ >The energy $E$ can be given by the mass-energy equivalence equation: >$E = fMc^2$ >Where $f$ is the fraction of the stellar mass being converted into stellar energy. >We can therefore combine both equations to isolate $t$, giving us a total value for the time on the hydrogen burning phase: >$t_{MS} \approx \frac{M}{L}$ >Now we must substitute something we haven't learned - the mass-luminosity relationship. For the purposes of this, we can derive the relationship as shown: >$L \propto M^{3.5}$ >Now we substitute for $L$, given the equation at the very start $\frac{E}{t}$ >$L \approx M^{-2.5}$ >Now we just have to match this with the ratio of the star's mass and luminosity relative to the sun. This gives us: >$\frac{t_{MS}}{t_{\odot}} \approx (\frac{M}{M_{\odot}})^{-2.5}$ >Where $t_{\odot}$ is the predicted solar lifetime, 10^10 years. # Example Problems - Carroll, Ostlie >[!Example]