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In the center of mass frame, the velocities are ${\mathbf{v}}_1=(M_2/M)\mathbf{v}$ and ${\mathbf{v}}_2=-(M_1/M)\mathbf{v}$. Hence the kinetic energy is $T=\frac{1}{2}\mu v^2$, while the Newtonian gravitational potential energy is $U=-G{M}_1{M}_2/r=-GM\mu /r$. The two-body orbital energy is thereforeThe relative equation of Newtonian gravity is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. It is a central force, so conservation of angular momentum confines the motion to a plane and preserves the specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|=r^2\dot{\theta }$.

For completeness, derive the polar equation of a Kepler orbit. Let $w=1/r$ and use primes for $\theta$ derivatives. Then $\dot{r}=-h{w}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{w}^2(w^{\prime \prime }+w)$. The radial equation gives the Binet equationChoose the angular origin at closest approach. Its solution is $w=(GM/h^2)(1+e\cos \theta )$, soFor the ellipse, the orbital eccentricity satisfies $0\le e<1$. Its extreme separations are $r_p=l/(1+e)$ and $r_a=l/(1-e)$. The semi-major axis is half their sum, giving $a=l/(1-e^2)$ and thus $l=a(1-e^2)$. The true anomaly $\theta$ runs through a full $2\pi$; the endpoints are the same position, and for a circular Kepler orbit the angular origin is arbitrary.

The radial and transverse relative speeds are $\dot{r}=(GM/h)e\sin \theta$ and $r\dot{\theta }=h/r$. Substituting these into the specific orbital energy givesTherefore the conserved orbital energy isThe negative specific orbital energy and fixed angular momentum characterize the bound Kepler orbit; neither should be assumed unchanged through a mass-ejecting explosion.

Initially the circular Kepler orbit has $r=a$ and $v^2=GM/a$. Treat the supernova kick in a binary star as impulsive: the relative position does not change, the companion's velocity is unchanged during the impulse, and the new neutron star receives $\mathbf{u}$ with $|\mathbf{u}|=\alpha v$. Thus ${\mathbf{v}}^{\prime }=\mathbf{v}+\mathbf{u}$, and with $\psi$ the angle between these vectors before the kick,Write $M^{\prime }=M_1^{\prime }+M_2$, $\beta =M^{\prime }/M$ and ${\mu }^{\prime }=M_1^{\prime }{M}_2/M^{\prime }$. The post-explosion specific orbital energy isUsing ${\epsilon }^{\prime }=-G{M}^{\prime }/(2a^{\prime })$ givesHere $a^{\prime }>0$ for a bound Kepler orbit. For a hyperbolic Kepler orbit, this formula uses the signed energy parameter $a^{\prime }<0$, rather than the positive geometric magnitude of the hyperbola's semi-major axis. At the parabolic Kepler orbit boundary, $1/a^{\prime }=0$.

The kick-direction binary survival criterion is bound if $S<2\beta$ and unbound with positive asymptotic speed if $S>2\beta$. To prove the printed sufficient disruption condition, choose a perpendicular kick, $\cos \psi =0$: then $S=1+{\alpha }^2$, so $M^{\prime }<\frac{1}{2}(1+{\alpha }^2)M$ produces positive specific orbital energy. This is sufficient, not the sharp existence threshold. Sincethe complete kick-direction binary survival criterion isThe second inequality follows by taking a kick directly opposite to $\mathbf{v}$. All directions remain bound if $2\beta >(1+\alpha {)}^2$, while every direction has positive escape energy if $2\beta <(1-\alpha {)}^2$. For $\alpha >0$, the marginal direction obeys $\cos \psi =(2\beta -1-{\alpha }^2)/(2\alpha )$ when the right-hand side lies in $[-1,1]$. Equalities give marginal parabolic Kepler orbits for the relevant extreme direction. With zero kick, losing more than half the original total mass unbinds the circular Kepler orbit.

For an escaping pair, the Newtonian gravitational potential energy approaches zero at infinite separation. Conservation of the post-explosion two-body orbital energy then gives $\frac{1}{2}{\mu }^{\prime }{V}^2={\mu }^{\prime }{\epsilon }^{\prime }$, or the asymptotic relative speed of a disrupted binaryThis is the relative recession speed, not the velocity of the new center of mass or either star's individual velocity in the original inertial frame. A marginal parabolic Kepler orbit separates with $V=0$ at infinity.

The magnetic torque transfers angular momentum from the faster rotator to the slower one. When ${\Omega }_m>{\Omega }_K(r_m)$, the neutron star supplies angular momentum to the inner accretion disk and possible propeller flow, so it tends to spin down. When the inner accretion disk rotates faster than the star, the magnetic connection and accreted angular momentum tend to spin it up.

For slowly varying mass accretion rate and magnetic dipole moment, these tendencies favour a magnetic spin equilibrium with the corotation radius near the magnetospheric truncation radius:The precise balance depends on accreted and ejected angular momentum and on the extent of magnetic coupling across the accretion disk; corotation is a scaling estimate, not an exact zero-torque theorem.

The exact initial-mass boundaries depend on metallicity, convective overshooting, rotation, and mass loss, but stars near the upper intermediate-mass range, roughly $6.5$--$9M_{\odot }$, can become Super-AGB stars. They ignite carbon but do not immediately ignite neon hydrostatically throughout the core.

Their final fate is set by competition between shell-driven core growth and envelope loss. If a stellar wind removes the envelope first, the remnant is an oxygen-neon, or oxygen-neon-magnesium, white dwarf. If the degenerate core grows toward the Chandrasekhar mass, electron captures on magnesium and neon reduce the electron pressure and can trigger an electron-capture supernova, leaving a neutron star. Slightly higher-mass stars can ignite further fuels and proceed to ordinary iron-core collapse.

Russell Alan Hulse and Joseph Hooton Taylor Jr. discovered this binary pulsar at Arecibo in 1974; their January 1975 paper reports a roughly $59\mathrm{ms}$ pulse period and $7.75\mathrm{h}$ orbital period. Its unseen companion was not detected as a second radio pulsar. Discovery of the orbit preceded the later binary-pulsar orbital decay test.^[ref]

The 1993 Nobel Prize in Physics recognized the discovery.^[ref]

By 1993, the system was interpreted as a pair of neutron stars.^[ref]

A pulsar is a rotating magnetized neutron star whose directed emission is observed as periodic pulses. Rotation supplies the energy of an ordinary radio pulsar.

A pulsar turns stellar rotation into an observable clock: the neutron star's emission can continue while its beam sweeps past the observer only briefly. The pulse spacing measures a repetition period; identifying that period with stellar rotation requires a physical model. The discovery of the first pulsar established a remarkably regular astronomical signal before its engine was understood. Today, pulsar dispersion delay also makes these clocks probes of the intervening interstellar medium.

A stellar remnant is the object left after a star finishes its principal nuclear-burning evolution, such as a white dwarf, neutron star or black hole. Stellar remnants retain mass while often contributing little optical light, affecting an old stellar population's mass-to-light ratio.

A Type I X-ray burst is a transient thermonuclear runaway in accreted hydrogen or helium on a neutron star surface. It rises rapidly and cools after the fuel burns. A black hole has no material surface on which the same accumulated surface-fuel event can occur. This is distinct from an accretion-powered X-ray pulse or an accretion instability.