IPhO 1995 · Theory problem 1

Gravitational Red Shift and the Measurement of Stellar Mass

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(a) (3 marks)

A photon of frequency \(f\) possesses an effective inertial mass \(m\) determined by its energy. Assume that it has a gravitational mass equal to this inertial mass. Accordingly, a photon emitted at the surface of a star will lose energy when it escapes from the star’s gravitational field. Show that the frequency shift \(\Delta f\) of the photon when it escapes from the surface of the star to infinity is given by

\[ \frac{\Delta f}{f}\simeq-\frac{GM}{Rc^2} \]

for \(\Delta f\ll f\) where:

  • \(G=\) gravitational constant
  • \(R=\) radius of the star
  • \(c=\) velocity of light
  • \(M=\) mass of the star.

Thus, the red-shift of a known spectral line measured a long way from the star can be used to measure the ratio \(M/R\). Knowledge of \(R\) will allow the mass of the star to be determined.

(b) (12 marks)

An unmanned spacecraft is launched in an experiment to measure both the mass \(M\) and radius \(R\) of a star in our galaxy. Photons are emitted from \(\mathrm{He}^{+}\) ions on the surface of the star. These photons can be monitored through resonant absorption by \(\mathrm{He}^{+}\) ions contained in a test chamber in the spacecraft. Resonant absorption occurs only if the \(\mathrm{He}^{+}\) ions are given a velocity towards the star to allow exactly for the red shifts.

As the spacecraft approaches the star radially, the velocity relative to the star \((v=\beta c)\) of the \(\mathrm{He}^{+}\) ions in the test chamber at absorption resonance is measured as a function of the distance \(d\) from the (nearest) surface of the star. The experimental data are displayed in the accompanying table.

Fully utilize the data to determine graphically the mass \(M\) and radius \(R\) of the star. There is no need to estimate the uncertainties in your answer.

Data for Resonance Condition

Velocity parameter \(\beta=v/c\quad(\times10^{-5})\) 3.352 3.279 3.195 3.077 2.955
Distance from surface of star \(d\quad(\times10^8\mathrm{m})\) 38.90 19.98 13.32 8.99 6.67

(c) (5 marks)

In order to determine \(R\) and \(M\) in such an experiment, it is usual to consider the frequency correction due to the recoil of the emitting atom. [Thermal motion causes emission lines to be broadened without displacing emission maxima, and we may therefore assume that all thermal effects have been taken into account.]

(i) (4 marks)

Assume that the atom decays at rest, producing a photon and a recoiling atom. Obtain the relativistic expression for the energy \(hf\) of a photon emitted in terms of \(\Delta E\) (the difference in rest energy between the two atomic levels) and the initial rest mass \(m_0\) of the atom.

(ii) (1 mark)

Hence make a numerical estimate of the relativistic frequency shift \(\left(\dfrac{\Delta f}{f}\right)_{\mathrm{recoil}}\) for the case of \(\mathrm{He}^{+}\) ions.

Your answer should turn out to be much smaller than the gravitational red shift obtained in part (b).

Data:

Velocity of light \(c\) \(=\) \(3.0\times10^8\mathrm{m\,s^{-1}}\)
Rest energy of He \(m_0c^2\) \(=\) \(4\times938(\mathrm{MeV})\)
Bohr energy \(E_n\) \(=\) \(-\dfrac{13.6Z^2}{n^2}(\mathrm{eV})\)
Gravitational constant \(G\) \(=\) \(6.7\times10^{-11}\mathrm{N\,m^2\,kg^{-2}}\)

Official English text of the problem.

Answers

  • (a) With \(\Delta f=f_\infty-f\), \[ \boxed{\frac{\Delta f}{f}\simeq-\frac{GM}{Rc^2}}. \]

  • (b) A graph of \(1/\beta\) against \(1/d\) is approximately a straight line. Using all five measurements gives \[ \boxed{R\approx1.1\times10^8\ \mathrm m,\qquad M\approx5.2\times10^{30}\ \mathrm{kg}}. \] The corresponding surface-to-infinity fractional frequency shift is approximately \(-3.46\times10^{-5}\).

  • (c)(i) The exact photon energy is \[ \boxed{hf=\Delta E\left(1-\frac{\Delta E}{2m_0c^2}\right)}. \]

  • (c)(ii) For a transition \(n_i\to n_f\) in \(\mathrm{He}^{+}\), \[ \boxed{\left(\frac{\Delta f}{f}\right)_{\mathrm{recoil}} \simeq-7.25\times10^{-9} \left(\frac1{n_f^2}-\frac1{n_i^2}\right)}. \] Its magnitude is below \(7.3\times10^{-9}\) for bound-bound transitions. For example, \(2\to1\) gives \(-5.4\times10^{-9}\), several thousand times smaller than the gravitational shift.

Solution

(a)

The photon has energy \(E=hf\), so its effective inertial and gravitational mass, under the assumptions of the question, is

\[ m=\frac{hf}{c^2}. \]

The gravitational potential outside the star is

\[ \Phi(r)=-\frac{GM}{r}. \]

As the photon travels from the surface to infinity, its gravitational potential energy increases by \(GMm/R\). Conservation of energy therefore gives

\[ h\Delta f\simeq-\frac{GMm}{R}. \]

Here \(m\) can be treated as constant to first order because the fractional change in photon energy is small. Substituting \(m=hf/c^2\),

\[ h\Delta f\simeq-\frac{GM}{R}\frac{hf}{c^2}, \]

and hence

\[ \boxed{\frac{\Delta f}{f}\simeq-\frac{GM}{Rc^2}}. \]

The negative sign represents a red shift. The approximation requires \(GM/(Rc^2)\ll1\).

(b)

At the spacecraft, the distance from the star’s centre is \(r=R+d\). Applying the same energy argument between \(R\) and \(R+d\), the frequency \(f_r\) measured by an observer stationary relative to the star is

\[ f_r\simeq f_0\left[ 1-\frac{GM}{c^2} \left(\frac1R-\frac1{R+d}\right) \right], \]

where \(f_0\) is the emitted transition frequency. Recoil corrections are neglected here and assessed in part (c).

Define the positive fractional gravitational frequency loss

\[ \epsilon(d)=\frac{GM}{c^2} \left(\frac1R-\frac1{R+d}\right). \]

The photons travel outwards, while the absorbing ions move towards the star. Thus the ions see a Doppler blueshift:

\[ f_{\mathrm{ion}} =\gamma(1+\beta)f_r =f_r\sqrt{\frac{1+\beta}{1-\beta}}. \]

Since \(\beta\sim3\times10^{-5}\),

\[ f_{\mathrm{ion}}\simeq f_r(1+\beta). \]

Resonant absorption requires \(f_{\mathrm{ion}}=f_0\), so, to first order,

\[ (1-\epsilon)(1+\beta)\simeq1 \quad\Longrightarrow\quad \beta\simeq\epsilon. \]

Consequently,

\[ \boxed{ \beta=\frac{GM}{c^2} \left(\frac1R-\frac1{R+d}\right) =\frac{GM}{Rc^2}\frac{d}{R+d} } \]

to the accuracy required.

Taking reciprocals produces a linear relation:

\[ \frac1\beta =\frac{Rc^2}{GM} +\frac{R^2c^2}{GM}\frac1d. \]

Therefore, plot

\[ Y=\frac1\beta \qquad\text{against}\qquad X=\frac1d. \]

Writing the straight line as \(Y=A+BX\), its intercept and slope are

\[ A=\frac{Rc^2}{GM}, \qquad B=\frac{R^2c^2}{GM}. \]

These determine both unknowns:

\[ \boxed{R=\frac BA}, \qquad \boxed{M=\frac{c^2B}{GA^2}}. \]

The five points to plot, with convenient axis scales, are:

\(X/(10^{-9}\ \mathrm{m^{-1}})\) \(Y/10^4\)
0.25707 2.98329
0.50050 3.04971
0.75075 3.12989
1.11235 3.24992
1.49925 3.38409

The graph is an approximately straight, rising line. A best-fitting line using all five points is

\[ \frac{Y}{10^4}\simeq 2.8917+0.3249 \frac{X}{10^{-9}\ \mathrm{m^{-1}}}. \]

Thus the vertical intercept and slope in unscaled units are

\[ A\simeq2.8917\times10^4, \qquad B\simeq3.2489\times10^{12}\ \mathrm m. \]

The extrapolated horizontal intercept is \(-A/B\simeq-8.90\times10^{-9}\ \mathrm{m^{-1}}\).

It follows that

\[ R=\frac{3.2489\times10^{12}}{2.8917\times10^4} \simeq1.12\times10^8\ \mathrm m, \]

and

\[ M= \frac{(3.0\times10^8)^2(3.2489\times10^{12})} {(6.7\times10^{-11})(2.8917\times10^4)^2} \simeq5.2\times10^{30}\ \mathrm{kg}. \]

To sensible precision for a graphical determination,

\[ \boxed{R\approx1.1\times10^8\ \mathrm m,\qquad M\approx5.2\times10^{30}\ \mathrm{kg}}. \]

The surface-to-infinity red shift is

\[ \frac{\Delta f}{f}\simeq-\frac{GM}{Rc^2} =-\frac1A \simeq-3.46\times10^{-5}. \]

This also verifies the weak-field approximation used throughout.

(c)(i)

Let the initial atom have rest mass \(m_0\), and let the final atom have rest mass \(m_1\). The difference between the two internal rest energies is

\[ \Delta E=(m_0-m_1)c^2. \]

Write the emitted photon energy as \(E_\gamma=hf\). Since the initial atom is at rest, conservation of momentum requires the recoiling atom to have momentum of magnitude

\[ p=\frac{E_\gamma}{c}. \]

Using the relativistic energy–momentum relation for the final atom, conservation of energy gives

\[ m_0c^2 =E_\gamma+\sqrt{m_1^2c^4+p^2c^2} =E_\gamma+\sqrt{m_1^2c^4+E_\gamma^2}. \]

Rearranging and squaring,

\[ (m_0c^2-E_\gamma)^2=m_1^2c^4+E_\gamma^2, \]

so

\[ E_\gamma=\frac{(m_0^2-m_1^2)c^2}{2m_0}. \]

Substituting \(m_1=m_0-\Delta E/c^2\) yields

\[ \boxed{ hf=\Delta E-\frac{(\Delta E)^2}{2m_0c^2} =\Delta E\left(1-\frac{\Delta E}{2m_0c^2}\right) }. \]

This expression is exact for the stated two-body decay. The photon energy is smaller than the internal energy difference because some energy becomes kinetic energy of the recoiling atom.

(c)(ii)

Without recoil, the transition frequency would be \(f_0=\Delta E/h\). Therefore,

\[ \frac{f-f_0}{f_0} =-\frac{\Delta E}{2m_0c^2}. \]

Since the correction is very small, the fractional recoil shift is

\[ \left(\frac{\Delta f}{f}\right)_{\mathrm{recoil}} \simeq-\frac{\Delta E}{2m_0c^2}. \]

For \(\mathrm{He}^{+}\), \(Z=2\), and the given Bohr formula gives

\[ E_n=-\frac{54.4}{n^2}\ \mathrm{eV}. \]

Thus, for emission from \(n_i\) to \(n_f<n_i\),

\[ \Delta E=54.4 \left(\frac1{n_f^2}-\frac1{n_i^2}\right)\mathrm{eV}. \]

The given rest energy is

\[ m_0c^2=4\times938\ \mathrm{MeV} =3.752\times10^9\ \mathrm{eV}. \]

Hence

\[ \boxed{ \left(\frac{\Delta f}{f}\right)_{\mathrm{recoil}} \simeq-7.25\times10^{-9} \left(\frac1{n_f^2}-\frac1{n_i^2}\right) }. \]

For any bound-bound transition, \(\Delta E<54.4\ \mathrm{eV}\), giving the numerical upper estimate

\[ \boxed{ \left|\frac{\Delta f}{f}\right|_{\mathrm{recoil}} <7.3\times10^{-9} }. \]

For example, the \(2\to1\) transition has

\[ \Delta E=54.4\left(1-\frac14\right)=40.8\ \mathrm{eV}, \]

and therefore

\[ \boxed{ \left(\frac{\Delta f}{f}\right)_{\mathrm{recoil},\,2\to1} \simeq-5.4\times10^{-9} }. \]

Even the upper estimate is several thousand times smaller than the gravitational shifts of approximately \(3\times10^{-5}\) in part (b).

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.