2.7: S07. Distances as Determined by Standard Candles - SOLUTIONS
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Exercise 7.1.1
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Constant t, r, and ϕ so we have
ds=a(t)rdθ
Exercise 7.1.2
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Similar to 7.1.1 above, we now have constant t, r, and θ, which gives
ds=a(t)rsinθdϕ
Exercise 7.1.3
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This is simply a2(t)r2sinθdθdϕ.
Exercise 7.1.4
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Using the result from 7.1.3 above, we now just integrate over θ and ϕ, so the area is
A=∫2π0∫π0a(t)rsinθdθdϕ=∫2π02a2(t)r2dϕ=4πa2(t)r2
Exercise 7.1.5
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We know that Luminosity = (Flux)×(Surface Area). Making the appropriate substitutions we do indeed find that F=L/(4πd2a2).
Exercise 7.2.1
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As we found in Chapter 6, the rate of arrival of the wave crests will be slower than the rate of emission by the factor of a(tr)/a(te). The same argument applies to the rate of arrival of photons. We also saw that wavelength would be stretched out by a factor 1+z≡λrλe=a(tr)/a(te). Therefore the rate of arrival of photons will be slowed down by a factor of 1+z.
Exercise 7.2.2
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The relationship between photon energy and wavelength is E=hcλ. Substituting this into the definition of redshift, we find that
Er/Ee=11+z
Exercise 7.3.1
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Solving Equation 7.4 for dlum we get
dlum=√L4πF
Exercise 7.3.2
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Substituting in Equation 7.3 to our result above we get
d2lum=L4π4πd2(1+z)2L⟹dlum=d(1+z)