# Tag Info

## Hot answers tagged python

21

Having now looked at the paper by Aiola et al. (2020), it emerges that for that map, they filtered the data to exclude low frequency multipoles with $|l|<150$, corresponding to about 1 degree. This filtering was done to all the maps in the paper and will be responsible for the dramatic "hole" in your Fourier transform. As for the high frequency ...

19

For that specific E-mode map we have applied a Wiener filter to highlight the high SN modes (those "rings"). I also further apply the following filter: $((1 + (kx/5)^{-4})^{-1}) * ((1 + (k/150)^{-4})^{-1})$. This second filter gives the "hole" and a "thin" vertical line in your 2D PS. The image above is just for PR purposes. In ...

8

The fringing pattern is caused by thin-film interference within the CCD. The signal received in a pixel will be proportional to the light falling on it, multiplied by a sensitivity, but then some extra signal is added or subtracted which depends on how much of the incoming light is at particular wavelengths that are affected by the interference (i.e. the ...

7

Your approach is completely correct, just note three things: Logarithmic distribution First, since the distribution of masses is logarithmic in nature (as is most other things), be sure to bin them logarithmically. Otherwise you will oversample (undersample) the bins at the low-(high-)mass end. Comoving densities Second, to be able to compare mass ...

6

Random points on the surface of a sphere can be generated by allowing the azimuthal angle $\phi$ to take a uniformly distributed random value between 0 and $2\pi$. To convert this to RA in degrees you multiply by $180/\pi$. To convert to hours, minutes and seconds you divide the $\phi$ in degrees by 15, which gives the hours, divide the remainder by 60 which ...

5

While I'm not familiar with the package, a very quick look at the documentation suggests that you want In [90]: c.M_sun.uncertainty instead. I've just checked and this appears to be correct. > python -c "from astropy import constants as c ; print c.M_sun.uncertainty" 5e+25

5

There is no obvious astronomy error, but your use of math.pow is wrong You write constgrav = math.pow(6.67,-11). That means $G = 6.67^{-11}$. You mean constgrav = 6.67e-11 or $G=6.67\times 10^{-11}$. With completely different initial conditions, it is not surprising that the Earth flies off into space. As a suggestion, try not using SI units for this, ...

5

This code reads coordinates as equatorial (ra, dec) and transforms them to galactic (l, b): eq = SkyCoord(xarr[:], yarr[:], unit=u.deg) gal = eq.galactic The contents of 'galacticwperiod.csv' are already in galactic coordinates and should not be transformed. Something like this may give better results: gal = SkyCoord(xarr[:], yarr[:], frame='galactic', ...

5

The second light curve you show has no obvious periodic behaviour and I cannot see any sign of a planetary transit. The period-finding algorithm appears to be working correctly. The planet (if it exists) is supposed to be one of the smallest planetary candidates found by Kepler and will have a barely detectable transit (depth of order 0.004%). The small ...

5

The visual appearance of fringing is caused by the CCD (thickness) being comparable to the size of the wavelength (thin-film interference). An everyday example (same physics except with more colors) is an oilslick one sees in a puddle. The wavelengths of visible light are similar in size to the layer of oil on top of the water. The slight variation in ...

5

You are doing many things wrong. You are computing the eccentricity of one body with respect to the center of mass. You need to compute the eccentricity of one body with respect to the other. You are using reduced mass in np.cross(Ve, Le, axis=0) / mred - Xe / np.sqrt(np.sum(np.square(Xe), axis=0)) This is wrong for multiple reasons. First off, look at the ...

4

I want to say that's a Mollweide projection. I know that they're pretty common in astronomy; many images of the CMB use them. I was actually working with one recently. Given latitude $\varphi$ and longitude $\lambda$, the $x$ and $y$ coordinates of an object are $$x=R\frac{2\sqrt{2}}{\pi}(\lambda-\lambda_0)\cos\theta,\quad y=R\sqrt{2}\sin\theta$$ for ...

4

Heres more python than you can shake a telescope at. I just used @RobJeffries' algorithm. This is just a python script, the real answer to the question is @RobJeffries' answer and I've just scripted it. The mathematics behind generating statistically uniform distributions is explained very nicely there as well. Python is below the plots. You can see on an X-...

4

I personally use Astropy, specifically astropy.io.fits, although I'm not a seasoned user of FITS files and I don't really know their layout. As an example snippet of code, I often load data from FITS files using from astropy.io import fits data = fits.open('data_file.fits')[0].data You'll find more information in the documentation on the FITS module.

4

Your feeling is right: You shouldn't convolve the spectrum and the filter, you should only multiply so that flux outside the bandpass is suppressed. Subsequently you integrate the resulting function over wavelength, so that flux density (in energy/time/area/wavelength) becomes flux (in energy/time/area). Simply setting the flux to 0 outside $\lambda_1$ and $... 4 I haven't done much astronomical image processing before, but as this question is unanswered I'll give it a shot - hopefully to some avail. If the problem is more specific, a code sample/image sample would probably be useful for further diagnosis, but otherwise this example may help. It discusses the process of writing a 3-channel image to separate FITS ... 4 Given you haven't specified which kind of DSOs you want or fully given the list of properties, it's difficult to recommend a specific source. I would suggest searching for catalogues on VizieR, which allows you to search with keywords and/or various predefined categories. Bear in mind that you may need to use several different catalogues for the various ... 3 From eq. 10 in Hogg's classic paper, assuming that the peculiar velocity$v_\mathrm{pec} \ll c$: $$v_\mathrm{pec} = c \frac{z_\mathrm{obs} - z_\mathrm{cos}}{1 + z_\mathrm{cos}},$$ where$z_\mathrm{obs}$is the observed redshift, and$z_\mathrm{cos}$is the redshift from cosmological expansion only. Let me invert that and wrap it up in Python for ya: def ... 3 If you were working with a rectangular coordinate system as shown in the figure below (that also had no bounds), then it is correct that the right ascension would be$RA=RA_p\pm d$(points 1 and 3) and the declination would be$Dec=Dec_p \pm d\$ (points 2 and 4) where d is the radius of the circle. The sky is spherical, so the lines of RA are not a constant ...

3

Not sure what you are expecting to see from the two datasets. Both datasets are examples of light curves, flux against time with different arbitrary origins for the time axis; SuperWASP TMID is integer seconds from Julian Date 2453005.5, Kepler uses BKJD = Barycentric Kepler Julian Date, but offset by 2454833.0. i.e., BKJD = BJD - 2454833.0. The SuperWASP ...

3

Assuming x, y = rrl_pm.l, rrl_pm.b The transformation you want is xx = [(q+180)%360 - 180 for q in x] Add 180, do modulo 360, then subtract 180. Then set your limits ax.set_xlim(180, -180) import matplotlib.pyplot as plt x = (0, 10, 20, 40, 80, 160, 200, 280, 320, 340, 350) y = (0, 10, 20, 30, 40, 50, -50, -40, -30, -20, -10) fig = plt.figure(...

3

The La2010 long-term ephemeris Laskar et al. (2011), which is based on the INPOP numerical ephemeris integrated for 1 Myr, is valid for 250 million years before the present day and a unknown distance into the future. The authors note that due to chaotic motion in the Solar System, the accuracy degrades significantly beyond -50 Myr and likely a similar time ...

3

I think you may be seeing the planet in the periodogram! But also another signal - higher harmonics of other periodic signals example - various periodic signals in our sun - with a characteristic period of 11 years (in the bottom panel you also see Earth's, as this is a local measure, and thus is mostly affected by our distance from the sun) Patterns of ...

3

The accepted string formats for date-hms are: Format Class Example argument fits TimeFITS ‘2000-01-01T00:00:00.000’ iso TimeISO ‘2000-01-01 00:00:00.000’ isot TimeISOT ‘2000-01-01T00:00:00.000’ https://docs.astropy.org/en/stable/time/index.html You can either change one of these, or you can define your own format by deriving a ...

3

An Earth-centered observer looking at that point on Earth’s surface would also be looking toward the point that is the zenith for that location on Earth. Thus, you just need to find the local sidereal time (LST) for that Earth location. That gives you the right ascension on the meridian for that place and time. The meridian passes through the zenith, so ...

3

While this is not exactly how folding works, one-dimensional data encourages trying some simple statistical tallying first. What very easily reveals those periodicities is just tallying up the deltas, with an appropriate fuzziness for what periodicities are relevant: Deltas [0, 9308, 28857, 36709, 28857, 9308, 28858, 36708, 28858, 9308, 28857, 36708, 9309,...

2

From the documentation >>> c1 = SkyCoord(ra=10*u.degree, dec=9*u.degree, distance=10*u.pc, frame='icrs') >>> c2 = SkyCoord(ra=11*u.degree, dec=10*u.degree, distance=11.5*u.pc, frame='icrs') >>> c1.separation_3d(c2) <Distance 1.5228602415117989 pc> The rest of the code is just reading the excel files and printing the ...

2

Generating random galaxy catalogs for correlation functions ... the random catalog can be just be a uniform distribution of points in the same volume as the real data. However, real-world galaxy catalogs are not homogenous boxes, and have very irregular shapes and observationally induced redshift and angular distributions. Even some mock-catalogs are endowed ...

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