What is a star? How do we model them and what properties do we care about these objects? In the first lecture we will explore these properties.
By definition a star is : A body that is bounded by self-gravity and radies energy supplied by an internal source. (grav. or nuclear). This means that evolution (birth + death)
Things we care about in a star
Here we often normalise by units related to our own sun called \(\odot\) symbol. With some numbers like:
Map that relates Luminosity \(L\)to effective surface temperature \(T\). But these are not directly plotted but instead it plots Magnitudes and Color. We define color the difference in brightness between two specific wavelengths of light observed from a celestial object, and this difference is directly related to the object's temperature, with hotter objects appearing bluer and cooler objects appearing redder. We define magnitude (absolute) as the log scale of Luminosity, meaning that a difference of one magnitude corresponds to a roughly 2.5 times change in brightness.
Red Giant \[T = 3000K,\quad L = 10^4 L_\odot \Rightarrow \frac{R_{giant}}{R_\odot} = (\frac{L_{giant}}{L_\odot})^{1/2} \cdot (\frac{T_{giant}}{T\odot})^{-2} \approx (10^4)^{1/2}(1/2)^{-2} \approx 100 \times 4 \] Red giant is hundreds of times larger
White Dwarf \[T = 24000K,\quad L = 10^{-3} L_\odot \Rightarrow \frac{R_{dwarf}}{R_\odot} = (\frac{L_{dwarf}}{L_\odot})^{1/2} \cdot (\frac{T_{dwarf}}{T\odot})^{-2} \approx (10^4)^{-3/2}/36 \approx 10^{-3} \] White Dwarf is thousand of times smaller. And thus compact object.
Okay but why did we choose luminosity and the effective temperature? Well effective temperature is in the blackbody limit that produces Bolometric luminosity.
Stellar energy has a hidden variable which is age. We see that the luminosity of main sequence stars correlate with mass of \(L \propto M^{\alpha},\quad \alpha \approx 3-3.5\). Large stars loose energy FAST.
Takeaway: older populations have fewer high luminosity objects because they burn away quicker. And their fuel resvoir is proportional to mass and thus for a fixed mass we expect the life on a main sequence to be \(1/M^2\).
We can approximate time of evolution to be \(t = t_\odot (\frac{M}{M_\odot})^{-2.5}\). This is the ratio of time relative to our sun.
We have a few equations that we care about (1) Continuity of mass (2) hydrostatic equilibrium (3) energy equation (4) energy transport
We assume the setting: (1) isolated (2) spherical (3) uniform composition (4) its gas
Mass conservation: \(dm = 4\pi r^2 \rho(r) dr\) since we are spherically symmetrical. \[m(r) = \int_0^r 4\pi (r`)^2\rho(r`) dr`\] With total mass being \(M= m(R)\) where its the total radius of the star.
hydrostatic Equillibrium: this is force balance. \[\Delta m \ddot r = \overbrace{\frac{-Gm\Delta m}{r^2}}^{grav}+ \underbrace{[P(r) - P(r+dr)] dS}_{pressure force}\] This becomes when we divide by the mass and volume on everything \[\frac{d^2r}{dt^2} = \overbrace{\frac{-Gm}{r^2}}^{grav}+ \underbrace{-\frac{1}{\rho}\frac{dP}{dr}}_{pressure force}\]
Add (1)+(2) If its in equilibrium then we have the net force is zero \[\frac{-Gm}{r^2} = \frac{1}{\rho}\frac{dP}{dr}\] We can then mass conservation equation \(dr = \frac{dm}{4\pi r^2 \rho}\) we can then replace \[\frac{-Gm}{r^2} = \frac{1}{\rho}\frac{dP}{\frac{dm}{4\pi r^2 \rho}}\] \[\frac{-Gm}{r^2} = \frac{1}{\rho} (4\pi r^2 \rho)\frac{dP}{dm}\] \[\boxed{\frac{-Gm}{4\pi r^4} = \frac{dP}{dm}}\]
We can simply integrate the equation \[\int_0^R \frac{-Gm}{4\pi r^4}dm = \int dP\] \[\frac{-GM^2}{4\pi r^4}dm = P(R)-P(0)\] We se the pressure outside to be 0. \[\frac{GM^2}{4\pi r^4}dm = P(0)\]
Free-fall Time: Gravitation contraction occurs when there is no pressure to counter act the force of the gravitational pull. Then we have the following PDE (no pressure then becomes): \[\frac{\partial^2 r}{\partial t^2} = \frac{-Gm}{r^2} - \frac{\partial P}{\partial r} \frac{1}{\rho}\Rightarrow a = \frac{Gm(r)}{r^2}\] The time to collapse we can just do simple kinematics \[R = \frac{1}{2}at^2\] \[R = \frac{1}{2}\frac{GM}{R^2}t^2\] Rearrange \[t_{eff} = (\frac{2R^3}{GM})^{1/2}\] We can rearrange by multiplying up and down by the surface area \[t_{eff} = (\frac{2}{G (4/3\pi)}\frac{(4/3\pi) R^3}{M})^{1/2}\] We can then get the average density \[t_{eff} = (\frac{2}{G (4/3\pi)}\ \frac{1}{bar \rho})^{1/2}\] \[\boxed{t_{eff} = \sqrt{\frac{3}{2 \pi G \bar \rho}}}\]
Another way of deriving free fall: The other way is going through energy conservation. We match the potential energy with the kinetic energy. And then integrate to get out the velocities. \[\boxed{t_{eff} \approx \frac{1}{\sqrt{G \bar\rho}}}\]
If we compared this to the sun which has a density of water... we get a free fall of 1000seconds while a neutron star is \(10^{-4}sec\)...
Gravitational collapse is never UNOPPOSED. And so the change in gravitational potential introduces thermal motion of particles giving increase pressure that counters the collapse. The star can reach a new equilibrium and evolve quasi-statically...
Dynamical Timescale: is the free fall time. If dynamic time much smaller than the age, then its fast variation in the star can be related to dynamical process being in action, sometimes not involving the entire star..
If the star is in equilibrium then the dynamic time scale is the radial oscillation. If the star cant be in equilibirum and recover then this is the timescale for supernovae. The density can be calculate and thus we can compute the radial oscillation of stars. So denser stars has faster oscillation.
We can integrate the hydrostatic equilibrium equation \[\frac{dP}{dr} = - \frac{G \rho(r) m(r)}{r^2}\] Mutliply both sides \[\frac{dP}{dr} 4\pi r^3 dr= - \frac{G \rho(r) m(r)}{r^2}4\pi r^3 dr \] Integrate by parts the left handside \(dv = dP\quad u = 4\pi r^3 \Rightarrow du = 12 \pi r^2 dr, v = P\) \[\int_0^R dP 4\pi r^3 = uv - \int vdu = P(r) 4\pi r^3 - \int 12 \pi P r^2 dr = -3 \int P4\pi r^2 dr = -3 \bar{P}V \] Here we get an average pressure. \[\] On the right hand side we get gravitational potential \[\boxed{\Omega = -3 \bar P V}\] So in equilibrium the pressure is supported by one third of the gravitational force to collapse inwards.
What we should note is, there is not specified source yet... This internal pressure can be a gas or raditaion. If its a gas it can be classical or quantised gass and those can be relativistic or non-relativistic. And so this will drive the way we discuss the structures within the stars.
In this case assume ideal gass so we have \(PV = N kT \Rightarrow P=nkT\) where small n is the number density.
Kinetic energy is \(E_k = \frac{3}{2}kT \Rightarrow u = \frac{3}{2}nkT = \frac{3}{2} \rho/m_g kT \)
The pressure is thus \(P = \frac{3}{2}u\) We can then plug this into the virial theorem:
\[\Omega = -3 \bar P V\] \[\Omega = -3 \int_0^R 4\pi r^2 P(r) dr = -3 \int_0^R 4\pi r^2 \frac{2}{3}u(r) dr \] \[\Omega = -2\int_0^R 4\pi r^2 u(r) dr \] \[\Omega = -2U \] Here \(U\) is the total internal kinetic energy. \[\] Let us try to get the potential energy \[\Omega = -\int \frac{Gm(r)}{r}dm\] if we normalise by the total mass we get unitless scaling \(m` = m/M\) and \(r` = r/R\) \[\Omega = - \frac{GM^2}{R} \int_0^1 m'/r'dm\] \[\Omega = - \frac{GM^2}{R} \alpha\] Let us try to rewrite the internal kinetic energy in terms of temperature: \[U =\int_0^R 4\pi r^2 u(r) dr\] \[U =\int_0^R 4\pi r^2 \frac{3}{2}nkT dr\] \[U =\frac{3k}{2} \int_0^R 4\pi r^2 nT dr\] \[U =\frac{3k}{2} \int_0^R 4\pi r^2 \frac{\rho}{m_g}T dr\] We notice that \(\rho 4\pi r^2 dr = dm\) \[U =\frac{3k}{2 m_g} \int_0^M T dm\] And so we see that this is the average temperature over mass. \[U =\frac{3k}{2 m_g} \bar{T} M\] We can sub into the Virial theorem as \[\Omega = -2 U \Rightarrow \frac{\alpha GM^2}{R} = 2 \cdot 3/2 K/m_g \bar{T}M\] \[\Omega = -2 U \Rightarrow \boxed{\bar{T} =\frac{\alpha}{3}\frac{GM}{R}} \] This simply means that the scaling is \[\boxed{\bar{T} \approx M^{2/3}\bar{\rho}^{1/3}}\] The temperature of the star is the following.
We can derive the general relation between pressure and internal energy. Consider a box of size \(L_x = L_y = L_z\) we get the following: Momentum and velocity \(v_x,v_y,v_z\) and \(p_x,p_y,p_z\). We can see that the number of collisions per tme is \[\frac{v_z}{2L}\] The momentum transfer is thus \(\frac{v_z}{2L} \cdot 2 P_z = \frac{\Delta p}{\Delta t}\). This means the number of total gas particles in the box \(N\) and the pressure on the area of \(L^2\) is \[P = \frac{N \langle \frac{P_z v_z}{L} \rangle }{L^2} = \frac{N}{L^3} \langle p_z v_z \rangle\] In general though we have some isotropic fluid \(\langle p\cdot v \rangle =3 \langle p_z v_z \rangle\). And thus \[\boxed{P = \frac{n}{3} \langle p\cdot v \rangle}\] This also works for quantum like particle. \[\] We reconzie that for classical non-relativistic the momentum is just \[p = mv \Rightarrow \langle p \cdot v \rangle = \langle m v^2 \rangle\] And so we ahev that \[P = \frac{n}{3} \langle mv^2 \rangle = \frac{2}{3} n \langle \frac{1}{2} mv^2 \rangle = \frac{2}{3} [E_k/V]\] This is case of perfect monatomic gas.
\[\boxed{P = \frac{2}{3} [E_k /V]}\] \[\langle P \rangle = -3 \Omega/V\] \[\frac{2}{3} [E_k /V] \rangle = -3 \Omega/V\] \[\boxed{2 E_k + \Omega = 0}\] This is for non-relativistic gas.
We see that the total energy is \[E = E_k + \Omega\] if the gas is in hydrostatic equilibirum. \[\] if the total energy in the system is less than zero then the system is bound. Tightly bounded systems are hot as they have large \(E_K\).
For the sun we can define a thermal timescale. This means trying to find the time scale where the interal energy changes. \[\boxed{\frac{U}{L} = \frac{\alpha GM^2}{2RL}}\] Is the time scale to radiate away the internal energy. This is also called the Kelvin Kelvin-Helmholtz time scale. Here \(L\) is the luminosity.
During collapse the radius decreases which increase the potential energy which increaes timeperature which increases fusion. If there is no fusion then star goes into a less bound state and radius increases and temperature drops.
The internal energy becomes the following: \[\epsilon^2 = p^2c^2+m^2c^4\] Relation between energy of a particle \(\epsilon\) and momentum is the relatistic limit. \(pc >> mc^2\). Here we have that \[\boxed{P = \frac{n}{3}\langle pc \rangle = \frac{1}{3} [E_k/V]}\] Virial theorem \[\boxed{E_k +\Omega = 0}\] This means zero binding energy and thus is on the cusp of about to be destroyed and go unstable.
Baryon number is the sum of protons and electrons which is \(p+e^- =A\) where as the atomic number \(p= Z\).
Single Atom gas has multiple species, thermal equilibirum with \(N\) species. \[p = \sum_i P_i = \sum_i (n_i)kT = \frac{\rho}{m_p}kT \sum \frac{x_i}{A_i} = \left(\frac{1}{\mu_I m_p} \right)kT\]