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## Hot answers tagged flux

5

The R in that equation is the distance from the star to observer, not the star radius. The light emitted from the star is distributed uniformly on a sphere of radius R, and when the light arrives to the Earth, that sphere will have a radius equal to the distance Earth-star. Therefore, the second relation for the two fluxes is about the apparent magnitudes (...

4

This expression is valid for low frequencies, including the case of the 1420 MHz hydrogen line. It arises from treating the source as a black body with temperature $T(\theta,\phi)$$^{\dagger}, and assuming h\nu\ll k_BT:$$I_{\nu}=B_{\nu}=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT}-1}\approx\frac{2h\nu^3}{c^2}\frac{1}{h\nu/k_BT}=\frac{2k_BT\nu^2}{c^2}$$... 3 You need to do what is called “propagation of uncertainties”. You can search to get more information on that, but briefly if you have some function f(x) that depends on variable x, then the uncertainty \sigma_x on the quantity x is related to the uncertainty on f by$$ \sigma_f = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 \sigma_x^2}$$... 2 Imagine your line as a rectangle of width$w$and depth$d$relative to a normalised continuum. Without scattered light, the area blocked off by the line is$wd$and if the continuum level is normalised to 1, then the equivalent width is$wd\$. Now add 5% scattered light. The height of the continuum is 1.05 (but we're going to renormalise it) and the depth ...

1

From what I can see, you already found what you were looking for. The table you linked is the description of the data available in the Catalog Archive Server (CAS) database. All you have to do now is use those table names (it seems that masses are on different tables, here and here) and then search whatever you want. If you do not have constraints, you could ...

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