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Unformatted text preview: Solutions to Theoretical Question 1 Gravitational Red Shift and the Measurement of Stellar Mass (a) If a photon has an effective inertial mass m determined by its energy then mc 2 = hf or m = hf c 2 . Now, assume that gravitational mass = inertial mass, and consider a photon of energy hf (mass m = 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 → initial state and subscript f → final state. hf i hf f = GMm f ∞ • GMm i r ‚ hf f = hf i GMm i r hf f = hf i GM hf i c 2 r hf f = hf i • 1 GM rc 2 ‚ f f f i = • 1 GM rc 2 ‚ Δ f f = f f f i f i = GM rc 2 The negative sign shows redshift, 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 Δ f f = GM Rc 2 Since the change in photon energy is small, ( δf ¿ f ), m f ’ m i = hf i c 2 . (b) The change in photon energy in ascending from r i to r f is given by hf i hf f = GMm f r f + GMm i r i ’ GMhf i c 2 • 1 r i 1 r f ‚ ∴ f f f i = 1 GM c 2 • 1 r i 1 r f ‚ 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: f f f i = 1 GM c 2 • 1 R 1 R + d ‚ (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: f f f = s 1 + β 1 β where f is the frequency as received by He + ions in the spacecraft, and β = 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 β ¿ 1, f f f = (1 β ) 1 2 (1 + β ) 1 2 ’ 1 β Alternatively, since β ¿ 1, use the classical Doppler effect directly. Thus f = f f 1 β or f f f = 1 β Since f must be equal to f i for resonance absorption, we have f f f i = 1 β (2) Substitution of 2 into 1 gives β = GM c 2 µ 1 R 1 R + d ¶ (3) Given the experimental data, we look for an effective graphical solution. That is, we require a linear equation linking the experimental data in β and d . Rewrite equation 3: β = GM c 2 • R \ + d R \ ( R + d ) R ‚ Inverting the equation gives: 1 β = µ Rc 2 GM ¶• R d + 1 ‚ or 1 β = µ R 2 c 2 GM ¶ 1 d + Rc 2 GM Graph of 1 β vs. 1 d 1 1 intercept = = α Rc GM GM R c slope = = R α 2 2 2 β d The slope is µ Rc 2 GM ¶ R = αR (A) The 1 βintercept is µ Rc 2 GM ¶ = α (B) and the 1 dintercept is 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....
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 Spring '11
 NA
 Energy, Inertia, Mass, Photon, Sin, Cos

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