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--- un:virial-theorem [2025/10/25 14:04] – asad
+++ un:virial-theorem [2025/10/25 14:17] (current) – asad
@@ Line 7: / Line 7: @@
 Starting from hydrostatic equilibrium,
 $$
-\frac{dP}{dr} = -\rho(r)\frac{G M(r)}{r^2},
+\frac{dP}{dr}=-\rho(r)\frac{G\,M(r)}{r^{2}},
 $$
+multiply by the spherical volume element \(4\pi r^{3}dr\) and integrate from center (\(r=0\)) to surface (\(r=R\)):
-we multiply both sides by the spherical volume element \(4\pi r^3 dr\) and integrate from the center (\(r=0\)) to the surface (\(r=R\)):
 $$
-\int_0^R 4\pi r^3 \frac{dP}{dr} \, dr = -\int_0^R 4\pi r^3 \rho(r)\frac{G M(r)}{r^2} \, dr.
+\int_{0}^{R}4\pi r^{3}\frac{dP}{dr}\,dr=-\int_{0}^{R}4\pi r^{3}\rho(r)\frac{G\,M(r)}{r^{2}}\,dr.
 $$
----
+**Left-hand side (pressure term).** Integrate by parts:
-**Left-hand side (pressure term):**
-Integrate by parts:
 $$
-\int_0^R 4\pi r^3 dP = \Big[4\pi r^3 P\Big]_0^R - \int_0^R 12\pi r^2 P\, dr.
+\int_{0}^{R}4\pi r^{3}dP=\Big[4\pi r^{3}P\Big]_{0}^{R}-\int_{0}^{R}12\pi r^{2}P\,dr
+\approx-3\!\int_{0}^{R}4\pi r^{2}P\,dr=-3\!\int P\,dV,
 $$
+where \(P(R)\simeq 0\).
-At the surface \(P(R)\approx 0\), and at the center \(r=0\), the first term vanishes, leaving:
+**Right-hand side (gravity term).** Using \(dM=4\pi r^{2}\rho\,dr\),
 $$
-\int_0^R 4\pi r^3 dP \approx -3\!\int_0^R 4\pi r^2 P\, dr = -3\!\int P\, dV.
+-\int_{0}^{R}4\pi r^{3}\rho\,\frac{G\,M}{r^{2}}\,dr
+= -\int_{0}^{R}\frac{G\,M(r)}{r}\,dM(r)
+\equiv \Omega ,
 $$
+which is the (negative) **gravitational potential energy**.
----
+Equating both sides gives the pressure–gravity balance:
-**Right-hand side (gravitational term):**
-Using \(dM = 4\pi r^2 \rho\, dr\),
-$$
--\int_0^R 4\pi r^3 \rho \frac{G M}{r^2} dr = -\int_0^R \frac{G M(r)}{r}\, dM(r).
-$$
-But this is precisely the **gravitational potential energy** of the system,
-$$
-\Omega = -\int_0^R \frac{G M(r)}{r}\, dM(r).
-$$
----
-Equating both sides gives the **pressure–gravity balance**:
 \begin{equation}
-\!\int P\, dV + \Omega = 0.
+\!\int P\,dV+\Omega=0.
 \end{equation}
-For an **ideal gas**, where the internal (thermal) kinetic energy is related to pressure by
+For an **ideal gas** the thermal kinetic energy is \(K=\tfrac{3}{2}\!\int P\,dV\), hence
-\( K = \tfrac{3}{2}\!\int P\, dV \),
-this can be written as:
 \begin{equation}
-K + \Omega = 0.
+K+\Omega=0,
 \end{equation}
+the virial theorem in its standard energy form.
-This is the **virial theorem** — it expresses the global equilibrium between the total kinetic energy and the gravitational potential energy of a bound system.
+===== Satellites and stars =====
----
-===== Satellite and Stellar Contraction =====
-A simple example of the virial theorem in action is a satellite orbiting a planet of mass \(M\) at radius \(r\). Its kinetic and potential energies are:
+A satellite of mass \(m\) in a circular orbit around a mass \(M\) at radius \(r\) obeys
 $$
-E_k = \tfrac{1}{2} m v^2 = \tfrac{1}{2}\frac{G M m}{r}, \qquad
+K=\tfrac12 m v^{2}=\tfrac12\,\frac{G M m}{r},\qquad
-E_p = -\frac{G M m}{r}.
+\Omega=-\frac{G M m}{r},
 $$
+so that
-Hence,
 \begin{equation}
-E_k = -\tfrac{1}{2} E_p,
+K=-\tfrac12\,\Omega ,
 \end{equation}
+consistent with \(2K+\Omega=0\).
-which satisfies \(2K + \Omega = 0\).
+{{https://upload.wikimedia.org/wikipedia/commons/thumb/8/83/The_Sun_in_white_light.jpg/1024px-The_Sun_in_white_light.jpg?nolink&500}}
----
-Now imagine a **star with no internal nuclear energy source**. It behaves similarly:
-as it radiates energy from its surface, its total energy \(E = K + \Omega\) decreases.
-To restore equilibrium, the star **contracts**, making \(\Omega\) more negative and \(K\) larger, because energy conservation requires that the potential energy released during contraction appears as heat.
+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:
-  * \(E_p\) becomes more negative (stronger gravity),
+  * \(\Omega\) becomes more negative (stronger binding),
-  * \(E_k\) increases (particles move faster),
+  * \(K\) increases (higher temperature).
-  * and the temperature rises.
-Thus, when the star loses energy, it actually gets hotter — a hallmark of **negative specific heat**.
+Thus a self-gravitating gas has **negative specific heat**: losing energy can raise its temperature.
-Removing energy makes it hotter because more gravitational binding energy is converted into kinetic energy.
+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.
-This leads to a **quasi-stable equilibrium** during stellar formation or collapse.
+When nuclear power equals surface losses, the star reaches **true equilibrium**—hydrostatic balance sustained by the virial relation.
-During stellar birth, the collapsing gas cloud spins faster and heats up until internal pressure and rotation counteract gravity.
+===== Clusters of galaxies =====
-During stellar death, as nuclear fuel runs out, the core again contracts and heats, triggering fusion of heavier elements until no further fusion is possible.
+{{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}}
-When nuclear reactions in the core produce energy at exactly the rate it is lost from the surface, the star remains in **true equilibrium** — this is **hydrostatic equilibrium**, maintained by the virial balance.
----
-===== Measuring Galaxy Cluster Mass =====
-The virial theorem can also be used to estimate the total mass of a galaxy cluster from the observed velocities of its member galaxies.
-For a system of \(N\) galaxies, each of mass \(m\), the virial condition is:
+The virial theorem also yields the **total mass** of a galaxy cluster from member velocities.
+For \(N\) identical galaxies (mass \(m\)),
 $$
-\sum_i \frac{1}{2} m v_i^2 - \sum_{i\neq j} \frac{G m^2}{r_{ij}} = 0.
+\sum_i \tfrac12 m v_i^{2}-\sum_{i\neq j}\frac{G m^{2}}{r_{ij}}=0.
 $$
+Rearranging,
-Rewriting and simplifying for identical masses:
 $$
-m \sum_i v_i^2 - G m^2 \sum_{i\neq j} \frac{1}{r_{ij}} = 0.
+m\sum_i v_i^{2}-G m^{2}\sum_{i\neq j}\frac{1}{r_{ij}}=0.
 $$
+Multiplying the first term by \(N/N\) and the second by \(N^{2}/N^{2}\),
-Multiply the first term by \(N/N\) and the second term by \(N^2/N^2\):
 $$
-N m \left[\frac{1}{N}\sum_i v_i^2 \right]
+N m\!\left[\frac{1}{N}\sum_i v_i^{2}\right]
-- G \frac{(N m)^2}{2} \left[\frac{1}{N(N-1)/2} \sum_{i\neq j} \frac{1}{r_{ij}}\right] = 0.
+-\frac{G(Nm)^{2}}{2}\!\left[\frac{1}{N(N-1)/2}\sum_{i\neq j}\frac{1}{r_{ij}}\right]=0.
 $$
+With \(M\equiv Nm\) (total mass) and defining \(\langle \cdot \rangle\) as the average over members, we obtain
-If \(N(N-1)\approx N^2\) and \(M = N m\) is the total mass, we obtain:
 \begin{equation}
-M \langle v_i^2 \rangle_{av} - \frac{G M^2}{2} \langle r_{ij}^{-1} \rangle_{av} = 0.
+M\,\langle v^{2}\rangle-\frac{G M^{2}}{2}\,\langle r_{ij}^{-1}\rangle=0,
 \end{equation}
+hence the **virial mass**
-Hence, the **virial mass** of the cluster is:
 \begin{equation}
-M = \frac{2 \langle v_i^2 \rangle_{av}}{G \langle r_{ij}^{-1} \rangle_{av}}.
+M=\frac{2\,\langle v^{2}\rangle}{G\,\langle r_{ij}^{-1}\rangle}.
 \end{equation}
----
+Only the **line-of-sight** component \(v_{\text{los}}\) is observed.
+Assuming isotropy,
-In practice, we observe only the **line-of-sight velocity** component \(v_{i,\text{los}}\) via Doppler shift.
-Assuming the velocity distribution is isotropic:
 $$
-\langle v_i^2 \rangle_{av} = 3 \langle v_{i,\text{los}}^2 \rangle_{av}.
+\langle v^{2}\rangle=3\,\langle v_{\text{los}}^{2}\rangle .
 $$
+Thus the virial theorem converts measured velocity dispersions and projected separations into the **total (including dark) mass** of a cluster.
-Thus, using the virial theorem, one can infer the **total (including dark) mass** of a cluster from measured velocity dispersions and projected separations between galaxies.
 ===== Insights =====
-The virial theorem states that in any **self-gravitating system in equilibrium**, the total kinetic energy equals half the magnitude of the (negative) potential energy.
+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.
-It connects local pressure balance to global energetics.
+  * In stars: \(3\int P\,dV+\Omega=0\) links pressure support to binding energy.
+  * In clusters: velocity dispersion \(\langle v^{2}\rangle\) maps to total mass through \(M=2\langle v^{2}\rangle/(G\langle r_{ij}^{-1}\rangle)\).
-  * In stars, it links thermal pressure to gravitational binding energy: \(3\int P\,dV + \Omega = 0\).
+  * The negative specific heat of bound gravitating systems explains why contraction heats stars and clouds.
-  * In galaxy clusters, it relates velocity dispersion to total mass, including unseen dark matter.
-  * It explains why stars heat up as they lose energy — a consequence of negative specific heat.
-  * It underlies stellar stability, cluster dynamics, and the estimation of galaxy masses through velocity dispersion.
 ===== Inquiries =====
-  - Derive the step \(3\int P\, dV + \Omega = 0\) from the hydrostatic equilibrium equation, showing the integration by parts.
+  - 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).
-  - Explain why a system obeying \(2K + \Omega = 0\) can have a **negative heat capacity** (losing energy makes it hotter).
+  - Explain physically how \(2K+\Omega=0\) implies **negative heat capacity** for self-gravitating gas.
-  - Using the virial theorem, derive the expression for the **virial mass** \(M = \frac{2\langle v^2 \rangle}{G \langle r^{-1}\rangle}\) and describe how \(v_{\text{los}}\) measurements yield total cluster 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).