IPhO 2001 · Theory problem 2

Binary Star System

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a) It is well known that most stars form binary systems. One type of binary system consists of an ordinary star with mass \(m_0\) and radius \(R\), and a more massive, compact neutron star with mass \(M\), rotating around each other. In all the following ignore the motion of the earth. Observations of such a binary system reveal the following information:

  • The maximum angular displacement of the ordinary star is \(\Delta\theta\), whereas that of the neutron star is \(\Delta\phi\) (see Fig. 1).
  • The time it takes for these maximum displacements is \(\tau\).
  • The radiation characteristics of the ordinary star indicate that its surface temperature is \(T\) and the radiated energy incident on a unit area on earth’s surface per unit time is \(P\).
  • The calcium line in this radiation differs from its normal wavelength \(\lambda_0\) by an amount \(\Delta\lambda\), due only to the gravitational field of the ordinary star. (For this calculation the photon can be considered to have an effective mass of \(h/c\lambda\).)
Figure 1

Fig. 1

Find an expression for the distance \(\ell\) from earth to this system, only in terms of the observed quantities and universal constants. Copy your result onto the answer form. [7 pts]

b) Assume that \(M\gg m_0\), so that the ordinary star is basically rotating around the neutron star in a circular orbit of radius \(r_0\). Assume that the ordinary star starts emitting gas toward the neutron star with a speed \(v_0\), relative to the ordinary star (see Fig. 2). Assuming that the neutron star is the dominant gravitational force in this problem and neglecting the orbital changes of the ordinary star find the distance of closest approach \(r_f\) shown in Fig. 2. Copy your result onto the answer form. [3pts]

Figure 2

Fig. 2

Official English text of the problem.

Answers

  • a) With the angular displacements in radians, \[ \boxed{ \ell= \frac{2\sqrt{2}\,c\tau}{\pi(\Delta\theta+\Delta\phi)} \left(\frac{\Delta\lambda}{\lambda_0\Delta\phi}\right)^{1/2} \left(\frac{P}{\sigma T^4}\right)^{1/4} } \] where \(\sigma\) is the Stefan–Boltzmann constant. The orbital interpretation is specified below.

  • b) \[ \boxed{ r_f=\frac{r_0}{1+v_0\sqrt{\dfrac{r_0}{GM}}} }. \]

Solution

a)

We use Newtonian orbital dynamics, the small-angle approximation, blackbody radiation from the ordinary star, and weak-field gravitational redshift. The system’s dimensions are negligible compared with its distance from earth.

Interpret Fig. 1 as a circular binary orbit whose observed angular excursions span opposite extrema. Thus \(\tau\) is half the orbital period: \[ \mathcal T=2\tau. \]

Let \(a_0\) and \(a_N\) be the ordinary and neutron stars’ orbital radii about their common centre of mass. Since the angular excursions are full diameters, \[ a_0=\frac{\ell\Delta\theta}{2}, \qquad a_N=\frac{\ell\Delta\phi}{2}. \] The separation is therefore \[ a=a_0+a_N=\frac{\ell(\Delta\theta+\Delta\phi)}{2}. \]

The centre-of-mass relation gives \[ m_0a_0=Ma_N, \qquad m_0=(m_0+M)\frac{a_N}{a}. \] Kepler’s third law for the relative orbit is \[ G(m_0+M)=\frac{4\pi^2a^3}{\mathcal T^2}. \] Combining these equations, \[ Gm_0=\frac{4\pi^2a^2a_N}{\mathcal T^2} =\frac{\pi^2\ell^3}{8\tau^2} \Delta\phi(\Delta\theta+\Delta\phi)^2. \tag{1} \]

Next, the ordinary star’s luminosity is \[ L=4\pi R^2\sigma T^4. \] Assuming isotropic emission and negligible absorption between the star and earth, \[ P=\frac{L}{4\pi\ell^2} =\sigma T^4\frac{R^2}{\ell^2}, \] so \[ R=\ell\left(\frac{P}{\sigma T^4}\right)^{1/2}. \tag{2} \]

Finally, consider the photon’s gravitational energy loss. Its emitted energy and approximate effective mass are \[ E_0=\frac{hc}{\lambda_0}, \qquad m_\gamma\simeq\frac{E_0}{c^2}. \] On travelling from the stellar surface to earth, effectively at infinity relative to \(R\), it loses energy \[ E_0-E_{\mathrm{obs}}\simeq\frac{Gm_0m_\gamma}{R}. \] Consequently, \[ \frac{E_0-E_{\mathrm{obs}}}{E_0} \simeq\frac{Gm_0}{Rc^2}. \] For \(\Delta\lambda\ll\lambda_0\), \[ \frac{E_0-E_{\mathrm{obs}}}{E_0} =1-\frac{\lambda_0}{\lambda_0+\Delta\lambda} \simeq\frac{\Delta\lambda}{\lambda_0}. \] Thus the positive gravitational redshift determines \[ Gm_0=c^2R\frac{\Delta\lambda}{\lambda_0} =c^2\ell\left(\frac{P}{\sigma T^4}\right)^{1/2} \frac{\Delta\lambda}{\lambda_0}. \tag{3} \]

Equating (1) and (3) and cancelling one factor of \(\ell\), \[ \ell^2= \frac{8c^2\tau^2} {\pi^2\Delta\phi(\Delta\theta+\Delta\phi)^2} \frac{\Delta\lambda}{\lambda_0} \left(\frac{P}{\sigma T^4}\right)^{1/2}. \] Taking the positive square root, \[ \boxed{ \ell= \frac{2\sqrt{2}\,c\tau}{\pi(\Delta\theta+\Delta\phi)} \left(\frac{\Delta\lambda}{\lambda_0\Delta\phi}\right)^{1/2} \left(\frac{P}{\sigma T^4}\right)^{1/4} }. \]

The gravitational constant cancels because both the orbital dynamics and the redshift measure \(Gm_0\). All factors except \(c\tau\) are dimensionless.

b)

Since \(M\gg m_0\), treat the neutron star as stationary. The ordinary star’s circular orbital speed follows from \[ \frac{v_c^2}{r_0}=\frac{GM}{r_0^2}, \qquad v_c=\sqrt{\frac{GM}{r_0}}. \]

A gas element inherits this tangential speed and receives an additional inward radial speed \(v_0\). Its initial velocity components are therefore \[ v_r=-v_0,\qquad v_t=v_c. \]

Neglecting the ordinary star’s gravity, the gas moves in the neutron star’s central gravitational field. Its specific angular momentum \(j\) and specific mechanical energy \(\varepsilon\) are conserved: \[ j=r_0v_c=\sqrt{GMr_0}, \] \[ \varepsilon =\frac12(v_c^2+v_0^2)-\frac{GM}{r_0} =\frac12v_0^2-\frac{GM}{2r_0}. \]

At closest approach, the radial velocity vanishes. If the tangential speed there is \(v_f\), angular-momentum conservation gives \[ r_fv_f=r_0v_c, \qquad v_f=\frac{r_0v_c}{r_f}. \] Energy conservation then gives \[ \frac12v_0^2-\frac12v_c^2 = \frac12v_c^2\frac{r_0^2}{r_f^2} -v_c^2\frac{r_0}{r_f}. \] Rearranging, \[ v_0^2=v_c^2\left(\frac{r_0}{r_f}-1\right)^2. \]

The gas initially moves inward, so its closest approach satisfies \(r_f\le r_0\). We therefore choose \[ \frac{r_0}{r_f}-1=\frac{v_0}{v_c}, \] which yields \[ \boxed{ r_f=\frac{r_0}{1+v_0/v_c} =\frac{r_0}{1+v_0\sqrt{\dfrac{r_0}{GM}}} }. \]

For \(v_0=0\), this correctly gives the original circular orbit, \(r_f=r_0\). The gas trajectory is elliptical as drawn when \(v_0<v_c\), parabolic when \(v_0=v_c\), and hyperbolic when \(v_0>v_c\); the closest-approach formula holds in all three cases.

The solution was checked against the official solution.