Measuring a Source in an Observation

In class, you are given this information about a specific observation done with a telescope with the below properties in order to make a flux measurement of the source (in this case an unresolved X-ray binary star system).

1. First, we want to estimate the total number of source counts $S$ from the XRB. We know the number of counts inside the yellow circle, $T = 2108$, but that contains counts from the background, which we need to subtract off, i.e., $S^\prime = T-B$. We also know the number of counts inside the red circle; let's call that $B^\prime = 1438$. That's not the number of background counts $B$ inside the yellow circle though, because the red circle is bigger. However, if we assume that the number of background counts per pixel (equivalent to per solid angle, assuming the plate scale is constant everywhere) is a constant, we can estimate the number of background counts inside the yellow circle by accounting for the ratio of the areas of the two circles. The number of background counts inside the yellow circle is then $B = B^\prime \left( \frac{\pi r_y^2}{\pi r_r^2} \right) = B^\prime (r_y/r_r)^2$, where $r_y = 30^{\prime\prime}$ and $r_r = 2^\prime \cdot 60~{\rm arcsec/arcmin} = 120^{\prime\prime}$. So, the number of counts inside the yellow circle, $S^\prime$, is

   $$S^\prime = T-B = T - B^\prime \left(\frac{r_y}{r_r} \right)^2 = 2108 - 1438 \cdot \left(\frac{30}{120}\right)^2 = 2018.1~{\rm counts}\, .$$

   This is NOT the total number of source counts, however, just the number inside the yellow circle. We still need to correct for the number outside of the circle (after all, we could have chosen a circle of any size, which would yield a different number $S^\prime$ of counts). This correction depends on the PSF and typically depends on a detailed calibration of the PSF shape, which can be non-trivial. A common shortcut for this calibration is the half-power diameter, HPD, which is the diameter of a circle within which half the "power" (in this case, number of photons) is contained. Since the yellow circle above has the same size as the HPD, we know that it contains half the photons from the source, and finally we can say

   $$S = 2S^\prime = 2\cdot 2018.1 = \fbox{4036.2~{\rm counts}}\, .$$

2. To convert this number to a flux $F$, we need to account for the energy of the photons, the collecting area of the telescope, and exposure time of the observation:

   $$\begin{align} F = \frac{S\cdot E_\gamma}{A_{\rm coll}\cdot t_{\rm exp}} &= \frac{4036.2\cdot (10~{\rm keV}\cdot1.602\times10^{-16}~{\rm J/keV})}{(400~{\rm cm}^2\cdot(1~{\rm m}/100~{\rm cm})^2\cdot 37547~{\rm s}} \\ &= \boxed{4.31 \times 10^{-15}~{\rm J~m}^{-2}~{\rm s}^{-1}}\, . \end{align}$$

   This flux is really a rough estimate, because we've made the assumption that all the photons have an energy $E_\gamma = 10~{\rm keV}$, when in fact they are distributed in a spectrum from 4 keV up to 25 keV. What we need to know to estimate a more accurate flux is the spectrum of the source, which is information not provided on the slide. If the spectrum is flat, meaning there are roughly the same number of photons at all energies, then we could take the average photon energy to be something like $\left<E_\gamma\right> = (4~{\rm keV}+25~{\rm keV})/2 \approx 15~{\rm keV}$, but X-ray spectra are rarely if ever flat and have more photons with lower energies than higher ones, so I used $\left<E_\gamma\right> \approx 10~{\rm keV}$, but you wouldn't know that, so ~15 keV is perfectly reasonable.

3. The flux is the amount of energy per second produced by a source that passes through some unit area at our location. Assuming the source emits the same amount of light in all directions, the total amount of energy per second emitted by the star is then just the flux times the area of a sphere centered on the source with a radius equal to our distance from the source:

   $$L = F\cdot4\pi d^2 = 4.31\times10^{-15}~{\rm J~m}^{-2}~{\rm s}^{-1}\cdot 4\pi\cdot (780~{\rm kpc}\cdot3.086\times10^{19}~{\rm m/kpc})^2 = \boxed{3.13\times 10^{31}~{\rm J~s}^{-1}}\, .$$

4. The signal-to-noise ratio tells us how precise our measurement is, the ratio of our measurement to our estimate of the error (or better, uncertainty) in that measurement. As such, we need to start with what is actually measured, $S^\prime$:

   $$S/N = \frac{S^\prime}{\sigma} = \frac{T-B}{\sqrt{\sigma_S^2+\sigma_B^2}} = \frac{T-B}{\sigma_T} = \frac{2018.1}{\sqrt{2108}} = \boxed{44}\, .$$

   Uncertainty (or error) is contributed by both the photons from the source, $\sigma_S$, and the events from the background, $\sigma_B$, both inside the yellow circle. Based on our estimate of the background from the red circle, the yellow circle contains $\sim$2018 source counts and $2108-2018 = 90$ background counts, so the noise (denominator of the ratio) is then $\sqrt{\sqrt{2018}^2+\sqrt{90}^2} = \sqrt{2108}$.

5. Another way to think of the S/N is that it tells you how to derive the uncertainty when doing error propagation. The actual measurement done above yields $2018.1 \pm \sqrt{2108} = 2018.1 \pm 46$ counts. When converting this measurement to the total number of source counts $S = 2S^\prime$, as long as the HPD is precisely known (it is exactly 1 arcmin, or say 1.000 arcmin with all those digits being significant), the propagation of the uncertainty on the total counts $S$ won't significantly inflate the uncertainty. In other words, S/N stays the same:

   $$S = 2S^\prime = 2(2018.1\pm46) = \boxed{4036 \pm 92}\, .$$

   Similarly, to get the error on the flux, you can replace 4036.2 in equation (1) above with 92, assuming again that the collecting area and exposure time are known with high precision. Equivalently, the uncertainty on the flux, $\sigma_F$, can be computed directly from the signal-to-noise, $\sigma_F = F / (S/N) = 4.31\times10^{-15}~{\rm J~m}^{-2}~{\rm s}^{-1} / 44 = 9.8\times10^{-17}~{\rm J~m}^{-2}~{\rm s}^{-1}$, which can also be written as

   $$\boxed{F = (4.31\pm 0.10)\times10^{-15}~{\rm J~m}^{-2}~{\rm s}^{-1}}\, .$$

   Given that we assumed all the photons had an energy of 10 keV, when in fact they have a range of energies that depend on the spectrum, this flux could be systematically biased by this assumption; for example, if the true average energy of the photons is 8 keV, then our flux should be $F = F_{\rm 10~keV} (8~{\rm keV}/10~{\rm keV}) = 3.45\times10^{-15}~{\rm J~m}^{-2}~{\rm s}^{-1}$, or $(3.45\pm0.08)\times10^{-15}~{\rm J~m}^{-2}~{\rm s}^{-1}$. Similarly, perhaps the average photon energy is closer to 12 keV, and we'd have $F = F_{\rm 10~keV} (12~{\rm keV}/10~{\rm keV}) = (5.17\pm0.12)\times10^{-15}~{\rm J~m}^{-2}~{\rm s}^{-1}$. If we had reason to think that there was a roughly 68% chance that the average energy of a photons was between 8 keV and 12 keV, we could use these calculations to estimate the systematic error on the flux, which is an uncertainty that, if we had more information, could be eliminated entirely (if, say, we measured the spectrum directly and found $\left<E_\gamma\right> = 10.146\pm0.001~{\rm keV}$). In the above example, since $5.17-4.31\approx4.31-3.45 \approx 0.8$, our final measurement of the flux could be reported as

   $$F \pm\sigma_{\rm F, stat}\pm \sigma_{\rm F, sys} = (4.31\pm0.10\pm 0.80)\times10^{-15}~{\rm J~m}^{-2}~{\rm s}^{-1}\, .$$

   It is also reasonable, especially when the systematic uncertainty is lower than the statistical uncertainty, to combine the two errors together in a total error, again summing in quadrature so that $\sigma_{\rm tot}^2 = \sigma_{\rm stat}^2+\sigma_{\rm sys}^2$. Showing both uncertainties separately makes it more clear what the sources of uncertainty are and how important they are relatively to each other. The statistical uncertainty can be reduced in size with more data, but the systematic uncertainty cannot — it generally comes from a lack of knowledge, usually about some aspect regarding how the data were collected, or the conditions under which they were collected.