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--- un:virial-theorem [2025/10/25 13:58] – asad
+++ un:virial-theorem [2025/10/25 14:17] (current) – asad
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 ====== Virial theorem ======
-If a bound system of particles is in steady (time-averaged) equilibrium, the **virial theorem** relates its total kinetic energy \(K\) and the force field acting on the particles. For inverse–square gravity it reduces to a remarkably simple constraint between kinetic and potential energies.
+The **virial theorem** provides a fundamental link between the internal pressure (or kinetic energy) and the gravitational binding energy of a self-gravitating system.
+For a star, galaxy, or molecular cloud in steady equilibrium, it can be derived directly from the condition of **hydrostatic equilibrium**.
-We start from the scalar virial \(G \equiv \sum_i \mathbf{r}_i \!\cdot\! \mathbf{p}_i\). Differentiating twice and time-averaging over many dynamical times for a system whose size does not grow without bound (\(\langle \ddot{G}\rangle = 0\)) gives
+{{:courses:ast301:virial.webp?nolink&650|}}
+Starting from hydrostatic equilibrium,
 $$
-\left\langle \sum_i \mathbf{r}_i \!\cdot\! \mathbf{F}_i \right\rangle + 2\langle K\rangle = 0.
+\frac{dP}{dr}=-\rho(r)\frac{G\,M(r)}{r^{2}},
+$$
+multiply by the spherical volume element \(4\pi r^{3}dr\) and integrate from center (\(r=0\)) to surface (\(r=R\)):
+$$
+\int_{0}^{R}4\pi r^{3}\frac{dP}{dr}\,dr=-\int_{0}^{R}4\pi r^{3}\rho(r)\frac{G\,M(r)}{r^{2}}\,dr.
 $$
-\begin{equation}
+**Left-hand side (pressure term).** Integrate by parts:
-\boxed{\,2\langle K\rangle + \left\langle \sum_i \mathbf{r}_i \!\cdot\! \mathbf{F}_i \right\rangle = 0\,} \tag*{}
+$$
-\end{equation}
+\int_{0}^{R}4\pi r^{3}dP=\Big[4\pi r^{3}P\Big]_{0}^{R}-\int_{0}^{R}12\pi r^{2}P\,dr
+\approx-3\!\int_{0}^{R}4\pi r^{2}P\,dr=-3\!\int P\,dV,
+$$
+where \(P(R)\simeq 0\).
-For Newtonian gravity between pairs \(i,j\), each term \(\mathbf{r}_i\!\cdot\!\mathbf{F}_{ij}= -G m_i m_j/r_{ij}\). Summing over all pairs yields the **gravitational potential energy**
+**Right-hand side (gravity term).** Using \(dM=4\pi r^{2}\rho\,dr\),
 $$
-\Omega \equiv -\sum_{i<j}\frac{G m_i m_j}{r_{ij}}.
+-\int_{0}^{R}4\pi r^{3}\rho\,\frac{G\,M}{r^{2}}\,dr
+= -\int_{0}^{R}\frac{G\,M(r)}{r}\,dM(r)
+\equiv \Omega ,
 $$
-Hence, for self-gravitating systems,
+which is the (negative) **gravitational potential energy**.
+Equating both sides gives the pressure–gravity balance:
 \begin{equation}
-\boxed{\,2\langle K\rangle + \langle \Omega\rangle = 0\,} \tag*{}
+\!\int P\,dV+\Omega=0.
 \end{equation}
-—a result that holds for any bound, steady system interacting via \(1/r^2\) gravity (angles denote time averages).
+For an **ideal gas** the thermal kinetic energy is \(K=\tfrac{3}{2}\!\int P\,dV\), hence
+\begin{equation}
+K+\Omega=0,
+\end{equation}
+the virial theorem in its standard energy form.
+===== Satellites and stars =====
-**Example (circular orbit):** a satellite of mass \(m\) around a mass \(M\) at radius \(r\) has \(v^2=GM/r\). Then
+A satellite of mass \(m\) in a circular orbit around a mass \(M\) at radius \(r\) obeys
 $$
-E_k=\tfrac12 m v^2=\tfrac12 \frac{G M m}{r},\qquad
+K=\tfrac12 m v^{2}=\tfrac12\,\frac{G M m}{r},\qquad
-E_p=-\frac{G M m}{r},
+\Omega=-\frac{G M m}{r},
 $$
-so
+so that
 \begin{equation}
-\boxed{\,E_k = -\tfrac12 E_p\,} \tag*{}
+K=-\tfrac12\,\Omega ,
 \end{equation}
-in agreement with \(2K+\Omega=0\).
+consistent with \(2K+\Omega=0\).
----
+{{https://upload.wikimedia.org/wikipedia/commons/thumb/8/83/The_Sun_in_white_light.jpg/1024px-The_Sun_in_white_light.jpg?nolink&500}}
-**Derivation from hydrostatic equilibrium (stellar virial theorem).**
+A **star without an internal nuclear source** behaves analogously: as it radiates, its total energy \(E=K+\Omega\) decreases; to re-establish virial balance the star **contracts**, making \(\Omega\) more negative and increasing \(K\) (heating the gas).
+As the star shrinks:
+  * \(\Omega\) becomes more negative (stronger binding),
+  * \(K\) increases (higher temperature).
-For a spherical star in hydrostatic balance,
+Thus a self-gravitating gas has **negative specific heat**: losing energy can raise its temperature.
+During stellar birth the contracting cloud spins up and heats until pressure and rotation counter gravity; during late stages, core contraction again heats the gas and ignites progressively heavier fuels.
+When nuclear power equals surface losses, the star reaches **true equilibrium**—hydrostatic balance sustained by the virial relation.
+===== Clusters of galaxies =====
+{{https://upload.wikimedia.org/wikipedia/commons/thumb/8/8f/Webb%27s_First_Deep_Field_%28adjusted%29.jpg/1007px-Webb%27s_First_Deep_Field_%28adjusted%29.jpg?nolink&600}}
+The virial theorem also yields the **total mass** of a galaxy cluster from member velocities.
+For \(N\) identical galaxies (mass \(m\)),
 $$
-\frac{dP}{dr}=-\rho(r)\frac{G M(r)}{r^2}.
+\sum_i \tfrac12 m v_i^{2}-\sum_{i\neq j}\frac{G m^{2}}{r_{ij}}=0.
 $$
-Multiply by the volume element \(4\pi r^3\,dr\) and integrate from \(0\) to \(R\):
+Rearranging,
 $$
-\int_0^R 4\pi r^3 \frac{dP}{dr}\,dr
+m\sum_i v_i^{2}-G m^{2}\sum_{i\neq j}\frac{1}{r_{ij}}=0.
-= -\int_0^R 4\pi r^3 \rho(r)\frac{G M(r)}{r^2}\,dr .
 $$
-Integrate the **left** side by parts:
+Multiplying the first term by \(N/N\) and the second by \(N^{2}/N^{2}\),
 $$
-\int_0^R 4\pi r^3 dP
+N m\!\left[\frac{1}{N}\sum_i v_i^{2}\right]
-=\Big[4\pi r^3 P\Big]_0^R - \int_0^R 12\pi r^2 P\,dr
+-\frac{G(Nm)^{2}}{2}\!\left[\frac{1}{N(N-1)/2}\sum_{i\neq j}\frac{1}{r_{ij}}\right]=0.
-\approx -3\!\int_0^R 4\pi r^2 P\,dr
-= -3\!\int P\,dV ,
 $$
-where the surface term \((4\pi R^3 P(R))\) vanishes for negligible external pressure.
+With \(M\equiv Nm\) (total mass) and defining \(\langle \cdot \rangle\) as the average over members, we obtain
+\begin{equation}
-Rewrite the **right** side using \(dM=4\pi r^2\rho\,dr\):
+M\,\langle v^{2}\rangle-\frac{G M^{2}}{2}\,\langle r_{ij}^{-1}\rangle=0,
-$$
+\end{equation}
--\int_0^R 4\pi r^3 \rho\frac{G M}{r^2}\,dr
+hence the **virial mass**
-= -\int_0^R \frac{G M(r)}{r}\,dM(r)
-= \Omega ,
-$$
-because \( \Omega = -\int_0^R \frac{G M(r)}{r}\,dM(r) \) is the gravitational binding energy of a sphere.
-Equating both sides gives the **pressure–gravity identity**
 \begin{equation}
-\boxed{\,3\!\int P\,dV + \Omega = 0\,} \tag*{}
+M=\frac{2\,\langle v^{2}\rangle}{G\,\langle r_{ij}^{-1}\rangle}.
 \end{equation}
-For an ideal gas, the internal (thermal) kinetic energy is
+Only the **line-of-sight** component \(v_{\text{los}}\) is observed.
+Assuming isotropy,
 $$
-K = \tfrac{3}{2}\!\int P\,dV .
+\langle v^{2}\rangle=3\,\langle v_{\text{los}}^{2}\rangle .
 $$
-Insert this in the previous box:
+Thus the virial theorem converts measured velocity dispersions and projected separations into the **total (including dark) mass** of a cluster.
-\begin{equation}
-\boxed{\,2K + \Omega = 0\,} \tag*{}
-\end{equation}
-— the virial theorem recovered **directly from hydrostatic equilibrium**.
 ===== Insights =====
-The virial theorem ties global energetics to local balance:
+The virial theorem states that for any **self-gravitating system in equilibrium**, the total kinetic energy equals half the magnitude of the gravitational binding energy.
-  * In any bound, steady, self-gravitating system, kinetic energy is **half** the magnitude of the (negative) potential energy.
+  * In stars: \(3\int P\,dV+\Omega=0\) links pressure support to binding energy.
-  * From hydrostatic equilibrium, the identity \(3\int P dV + \Omega=0\) shows that **pressure support** exactly balances **gravity** in aggregate; with an ideal gas this becomes \(2K+\Omega=0\).
+  * In clusters: velocity dispersion \(\langle v^{2}\rangle\) maps to total mass through \(M=2\langle v^{2}\rangle/(G\langle r_{ij}^{-1}\rangle)\).
-  * Heating a star (increasing \(K\)) makes \(\Omega\) more negative; the star expands or contracts until the virial relation is restored—central to stellar stability and pre-main-sequence contraction.
+  * The negative specific heat of bound gravitating systems explains why contraction heats stars and clouds.
 ===== Inquiries =====
-  - Starting from \( \frac{dP}{dr}=-\rho GM(r)/r^2 \), reproduce the step \( 3\int P\,dV + \Omega=0 \) (show the integration by parts and the \(dM\) substitution).
+  - Starting from \( \frac{dP}{dr}=-\rho\,G M(r)/r^{2} \), derive \(3\int P\,dV+\Omega=0\) (show the parts integration and the \(dM\) substitution).
-  - Use \(2K+\Omega=0\) to explain why removing energy from a self-gravitating gas can make it **hotter** (negative heat capacity).
+  - Explain physically how \(2K+\Omega=0\) implies **negative heat capacity** for self-gravitating gas.
-  - For a cluster of \(N\) equal-mass stars with 1D velocity dispersion \(\sigma\), show that \( M \approx \frac{\eta\, \sigma^2 R}{G} \) (state a reasonable \(\eta\) from geometry) and discuss observational uses of this **virial mass**.
+  - Using the cluster derivation, show how \( \langle v_{\text{los}}^{2}\rangle \) and projected \( r_{ij} \) yield \( M=2\langle v^{2}\rangle/(G\langle r_{ij}^{-1}\rangle) \); state assumptions and corrections (isotropy, projection).