Homework 6, due October 27: Problems in Measuring the Universe: 4.2. You want to observe a star at -10 degrees declination through no more than two air masses from a telescope at 20 degrees north latitude. On the most favorable nights, how long can you observe the star? ----------------- Ignoring the minor effects of atmospheric refraction, two air masses occurs at an elevation of 30 o (i.e., 1/sin(30 o ) = 2). The coordinate transformations in Section 4.6.3 can be used to determine the range of hour angle when the source will be at greater than 30 o altitude. Solving the one giving the sine of the altitude in terms of the latitude for the observation and the declination and hour angle of the source, we have Substituting the given values yields cos h = 0.604475, that is for angles < 52.81 o from the meridian, the source is above 2 air masses. Converting to hour angle, this condition is met for about seven hours on a night when the source can be observed fully before and after meridian passage. 4.4. The first moment can be used to determine the centroid of a distribution of measurements (e.g., the measurements of an image by the pixels of an array): where the I i are the signals at positions x i. Use this approach to evaluate the impact of interpixel gaps and sampling on centroiding accuracy. Assume a Gaussian image, and that the array pixels have uniform response over 80% of their widths, but the response is only 0.9 times the central value for the outer 10% at either side. Place the image half way between the center and edge of one of the pixels and calculate the centroiding error as a function of the pixel sampling. For 1 arcsec FWHM seeing, what level of sampling is required to keep the centroiding error below 10 mas? ------------------------- This problem requires constructing a one-dimensional model of the signal and response to it, and then making the calculations based on the model. The answer is that the pixels need to be no larger than 0.98 arcsec. Of course, other considerations (less than infinite signal to noise, seeing fluctuations, etc.) would make the required pixel size smaller in realistic situations.
5.1 The list below gives Johnson system photometry of a number of stars at J and K. Kidger and Martin- Luis (2003) report JHK observations of bright stars, including the ones listed below. Look up these stars in the paper and compute the transformations to put the Johnson J, K photometry on the same basis as the Kidger and Martin-Luis J, K photometry. Name Type J K Name type J K HR 33 F5V 3.94 3.65 HR 5107 A3V 3.20 3.11 HR 509 G8V 2.16 1.68 HR 5854 K2IIIb 0.76 0.06 HR 1256 K0III 2.63 1.97 HR 5947 K2III 2.09 1.30 HR 1286 K1II-III 3.36 2.49 HR 6623 G5IV 2.18 1.77 HR 1791 B7III 1.97 2.05 HR 6698 G9III 1.68 1.12 HR 1907 K0IIIb 2.35 1.69 HR 6705 K5III -0.39-1.34 HR 1963 K1III 2.89 2.11 HR 6707 F2II 3.55 3.23 HR 2077 K0III 2.09 1.46 HR 7236 B9Vn 3.64 3.67 HR 2427 K3Iab 2.79 2.05 HR 7525 K3II 0.30-0.59 HR 2560 G5III-IV 2.89 2.33 HR 7557 A7V 0.39 0.26 HR 3003 K4III 2.28 1.33 HR 7615 K0III 2.28 1.67 HR 4335 K1III 1.16 0.44 HR 7949 K0III 0.77 0.11 HR 4377 K3III 1.18 0.31 HR 8143 B9Iab 3.95 3.79 HR 4608 G8IIIa 2.48 1.90 HR 8632 K2III 2.36 1.51 HR 4737 K1III 2.55 1.90 HR 8905 F8IV 3.37 3.02 HR 4983 F9.5V 3.24 2.90 ------------------------------- The figure below is a plot of the difference in J (Johnson) -J(Kidger/Martin-Luis) vs. J-K. Immediately one sees two discordant points, which since one was high and the other low, I threw out (and assigned hollow symbols). They might be bad measurements, or maybe variable stars. The line is a linear regression fit to the remaining points. The rms scatter around the line is 0.022, so it is a good fit.
The same thing can be done at K, see next figure. Again, I got rid of the high and low outliers (in this case I did a preliminary fit to be sure I knew which one was the high outlier - the point to the upper left was also a candidate but the fit showed it was not so bad). The linear regression fit to the remaining points has rms scatter of 0.024, again showing the fit is good. What is really surprising is that the Johnson photometry is so accurate.
My results are J Johnson = JKM L + 0.0766 0.0045(J K)KM L KJohnson = KKM L + 0.0567 0.056(J K)KM L 5.3. Assume you have a somewhat idealized J-band system with spectral response of 0.8 from 1.15 through 1.35 m and zero outside this range. What is the effective (mean) wavelength? What is the nominal (IRAS definition) wavelength? Assume your standard star has a Rayleigh Jeans spectrum across the J filter. Using the mean wavelength, compute the bandpass correction relative to the standard for the ultraluminous galaxy spectrum of d with in m. --------------------------------------------- The mean (or effective) wavelength of the filter is The nominal wavelength is the same expression with an extra factor of λ in both integrals; 1.253μm. To compute the bandpass correction, assume that the standard is Rayleigh-Jeans, i.e., with a spectrum
going as λ -4, and that it is normalized to 1 at λ 0. Convolve with the filter and integrate. I get 0.16346. For the galaxy, I again normalize to 1 at 1.25 m and then convolve with the filter and integrate. I get 0.16007. Therefore, if the galaxy had the same flux density as the star at the filter effective wavelength, we would have gotten a smaller signal from it by the ratio 0.16007/0.16346 = 0.9792. Without the band pass correction, if we had gotten the same signals we would have thought that they had the same flux densities at the effective wavelength. To correct our error, we need to take the flux density we got for the galaxy under this assumption and increase it by the factor 1/0.9792 = 1.021.
5.8 You have the following set of measurements, obtained from a telescope at 32 degrees north latitude: Determine m V for star 3. Hint: if you use the appropriate units, you can determine the air mass correction from a linear fit. Star m V Dec HA DU 1 5.73 +55 o -3.5 hours 1554 1.3 hours 1605 5.1 hours 1333 2 7.86 +14 o -2.1 hours 220 3.7 hours 195 3-3 o 1.4 hours 487 ----------------------------------------------- We use equation (4.3) to compute the air masses for each observation: star 1; 1.369, 1.120, and 1.826; star 2; 1.205, 1.683; star 3; 1.311. We can plot log(du) vs. airmass (equation (5.10) suggests this approach) and fit the signals for stars 1 and 2 requiring the same slope for the two stars. This yields log(du) = -0.12485 * (air mass) + 3.3535 for star 1 and log(du) = -0.12485 * (air mass) +2.4965 for star 2. At one airmass, star 1 then would have a signal of log(du) = 3.22865, or DU = 1693. At one airmass, star 2 similarly would have 235.3 DU, and star 3 would have 532.5 DU at one airmass (for the latter, we take (am-1)*(-0.12485) and apply the resulting correction to the log of the observed DU). Relative to star 1, star 3 is then 2.5*log(1693/532.5) = 1.256 magnitudes fainter, or at magnitude 6.99. Relative to star 2, star 3 is 2.5*log(532.5/235.3) = 0.89 magnitudes brighter, or at magnitude 6.97. We can take the average as our best estimate, 6.98. You can also use the given magnitudes of the two stars to put them on the same scale to determine the airmass correction, although that depends on the magnitudes being accurate and is a bit more risky than the method used above.