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IPhO 1995, Теория — Задача 1. Gravitational Red Shift and the Measurement of Stellar Mass

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Всички материали от IPhO 1995

XXVI IPhO — Theoretical Questions · 20 т.

Условие

(a) 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. [3 т.]

(b) 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 He $^+$ ions on the surface of the star. These photons can be monitored through resonant absorption by He $^+$ ions contained in a test chamber in the spacecraft. Resonant absorption occurs only if the 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 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	$\beta = v/c\ \ (\times 10^{-5})$	3.352	3.279	3.195	3.077	2.955
Distance from surface of star	$d\ \ (\times 10^{8}\mathrm{m})$	38.90	19.98	13.32	8.99	6.67

[12 т.]

(c) 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 He $^+$ ions.

Your answer should turn out to be much smaller than the gravitational red shift obtained in part (b). [5 т.]

Data:

\begin{array}{llcl} \text{Velocity of light} & c & = & 3.0 \times 10^{8}\mathrm{ms}^{-1} \\ \text{Rest energy of He} & m_0c^2 & = & 4 \times 938(\mathrm{MeV}) \\ \text{Bohr energy} & E_n & = & -\dfrac{13.6Z^2}{n^2}(\mathrm{eV}) \\ \text{Gravitational constant} & G & = & 6.7 \times 10^{-11}\mathrm{Nm}^{2}\mathrm{kg}^{-2} \end{array}

Отговори

Решение

(a)

If a photon has an effective inertial mass $m$ determined by its energy then $mc^2 = hf$ or $m = \dfrac{hf}{c^2}$ . Now, assume that gravitational mass = inertial mass, and consider a photon of energy $hf$ (mass $m = \dfrac{hf}{c^2}$ ) emitted upwards at a distance $r$ from the centre of the star. It will lose energy on escape from the gravitational field of the star.

Apply the principle of conservation of energy:

Change in photon energy $(hf_i - hf_f)$ = change in gravitational energy, where subscript $i \rightarrow$ initial state and subscript $f \rightarrow$ final state.

\begin{aligned} hf_i - hf_f &= -\frac{GMm_f}{\infty} - \left[-\frac{GMm_i}{r}\right] \\ hf_f &= hf_i - \frac{GMm_i}{r} \\ hf_f &= hf_i - \frac{GM\frac{hf_i}{c^2}}{r} \\ hf_f &= hf_i\left[1 - \frac{GM}{rc^2}\right] \\ \frac{f_f}{f_i} &= \left[1 - \frac{GM}{rc^2}\right] \\ \frac{\Delta f}{f} &= \frac{f_f - f_i}{f_i} = -\frac{GM}{rc^2} \end{aligned}

The negative sign shows red-shift, i.e. a decrease in $f$ , and an increase in wavelength.
Thus, for a photon emitted from the surface of a star of radius $R$ , we have

\boxed{\frac{\Delta f}{f} = \frac{GM}{Rc^2}}

Since the change in photon energy is small, $(\delta f \ll f)$ ,

m_f \simeq m_i = \frac{hf_i}{c^2}\ .

(b)

The change in photon energy in ascending from $r_i$ to $r_f$ is given by

\begin{aligned} hf_i - hf_f &= -\frac{GMm_f}{r_f} + \frac{GMm_i}{r_i} \\ &\simeq \frac{GMhf_i}{c^2}\left[\frac{1}{r_i} - \frac{1}{r_f}\right] \\ \therefore \frac{f_f}{f_i} &= 1 - \frac{GM}{c^2}\left[\frac{1}{r_i} - \frac{1}{r_f}\right] \end{aligned}

In the experiment, $R$ is the radius of the star, $d$ is the distance from the surface of the star to the spacecraft and the above equation becomes:

\frac{f_f}{f_i} = 1 - \frac{GM}{c^2}\left[\frac{1}{R} - \frac{1}{R+d}\right] \qquad (1)

The frequency of the photon must be doppler shifted back from $f_f$ to $f_i$ in order to cause resonance excitation of the He $^+$ ions in the spacecraft.
Thus apply the relativistic Doppler principle to obtain:

\frac{f'}{f_f} = \sqrt{\frac{1+\beta}{1-\beta}}

where $f'$ is the frequency as received by He $^+$ ions in the spacecraft, and $\beta = v/c$ .
That is, the gravitationally reduced frequency $f_f$ has been increased to $f'$ because of the velocity of the ions on the spacecraft towards the star. Since $\beta \ll 1$ ,

\frac{f_f}{f'} = (1-\beta)^{\frac{1}{2}}(1+\beta)^{-\frac{1}{2}} \simeq 1 - \beta

Alternatively, since $\beta \ll 1$ , use the classical Doppler effect directly.
Thus

f' = \frac{f_f}{1-\beta}

\frac{f_f}{f'} = 1 - \beta

Since $f'$ must be equal to $f_i$ for resonance absorption, we have

\frac{f_f}{f_i} = 1 - \beta \qquad (2)

Substitution of 2 into 1 gives

\beta = \frac{GM}{c^2}\left(\frac{1}{R} - \frac{1}{R+d}\right) \qquad (3)

Given the experimental data, we look for an effective graphical solution. That is, we require a linear equation linking the experimental data in $\beta$ and $d$ .
Rewrite equation 3:

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

Inverting the equation gives:

\frac{1}{\beta} = \left(\frac{Rc^2}{GM}\right)\left[\frac{R}{d} + 1\right]

\boxed{\frac{1}{\beta} = \left(\frac{R^2c^2}{GM}\right)\frac{1}{d} + \frac{Rc^2}{GM}}

Graph of 1/β (vertical axis) against 1/d (horizontal axis): five data points (circles) on a straight line; arrows mark the intercept = Rc²/GM = α and the slope = R²c²/GM = Rα.

Graph of $\dfrac{1}{\beta}$ vs. $\dfrac{1}{d}$

The slope is $\left(\dfrac{Rc^2}{GM}\right)R = \alpha R$ (A)

The $\dfrac{1}{\beta}$ -intercept is $\left(\dfrac{Rc^2}{GM}\right) = \alpha$ (B)

and the $\dfrac{1}{d}$ -intercept is $-\dfrac{1}{R}$ (C)

$R$ and $M$ can be conveniently determined from (A) and (B). Equation (C) is redundant. However, it may be used as an (inaccurate) check if needed.
From the given data:

R = 1.11 \times 10^{8}\ \mathrm{m}

M = 5.2 \times 10^{3}0\ \mathrm{kg}

From the graph, the slope $\alpha R = 3.2 \times 10^{12}\ \mathrm{m}$ (A)

The $\dfrac{1}{\beta}$ -intercept $\alpha = \dfrac{Rc^2}{GM} = 0.29 \times 10^{5}$ (B)

Dividing (A) by (B)

R = \frac{3.2 \times 10^{12}\ \mathrm{m}}{0.29 \times 10^{5}} \simeq \boxed{1.104 \times 10^{8}\ \mathrm{m}}

Substituting this value of $R$ back into (B) gives:

M = \frac{Rc^2}{g\alpha} = \frac{(1.104 \times 10^{8}) \times (3.0 \times 10^{8})^2}{(6.7 \times 10^{-11}) \times (0.29 \times{}^{1}0^{5})}

or $M = 5.11 \times 10^{30}\ \mathrm{kg}$

Diagram headed 'Atom before the decay' and 'Atom and photon after the decay': an atom of mass m0 at rest, an arrow, then a recoiling atom of mass m0′ moving left plus a photon hf moving right.

(c)

(i)

For the photon, photon momentum is $p = \dfrac{hf}{c}$ and photon energy is $E = hf$ .
Use the mass-energy equivalence, $E = mc^2$ , to relate the internal energy change of the atom to the rest-mass change. Thus:

\Delta E = (m_0 = m_0')\,c^2 \qquad (1)

In the laboratory frame of reference the energy before emission is

E = m_0c^2 \qquad (2)

Recalling the relativistic relation

E^2 = p^2c^2 + m_0^2c^4

The energy after emission of a photon is

E = \sqrt{p^2c^2 + {m_0'}^2c^4} + hf \qquad (3)

where also $p = hf/c$ by conservation of momentum.
Conservation of energy requires that (2) = (3), so that:

\left(m_0c^2 - hf\right)^2 = (hf)^2 + m_0^2c^4

\left(m_0c^2\right)^2 - 2hfm_0c^2 = m_0^2c^4

Carrying out the algebra and using equation (1):

\begin{aligned} hf(2m_0c^2) &= (m_0^2 - {m_0'}^2)c^4 \\ &= (m_0 - m_0')c^2(m_0 + m_0')c^2 \\ &= \Delta E[2m_0 - (m_0 - m_0')]c^2 \\ &= \Delta E[2m_0c^2 - \Delta E] \end{aligned}

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

(ii)

For the emitted photon,

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

If relativistic effects are ignored, then

hf_0 = \Delta E\ .

Hence the relativistic frequency shift $\dfrac{\Delta f}{f_0}$ is given by

\boxed{\frac{\Delta f}{f_0} = \frac{\Delta E}{2m_0c^2}}

For He $^+$ transition $(n = 2 \rightarrow 1)$ , applying Bohr theory to the hydrogen-like helium ion gives:

\Delta E = 13.6 \times 2^2 \times \left[\frac{1}{1^2} - \frac{1}{2^2}\right] = 40.8\ \mathrm{ev}

Also, $m_0c^2 = 3.752 \times 10^{6}\ \mathrm{eV}$ . Therefore the frequency shift due to the recoil gives

\boxed{\frac{\Delta f}{f_0} \simeq 5.44 \times 10^{-12}}

This is very small compared to the gravitational red-shift of $\dfrac{\Delta f}{f} \sim 10^{-5}$ , and may be ignored in the gravitational red-shift experiment.

Оригинал в Архива: th.pdf · официални решения: th_sol.pdf