Wednesday, 25 June 2014

galaxy - What methods exist to calculate the ellipticity of galaxies

What methods exist to calculate the ellipticity of galaxies and what are their drawbacks? I have asked this question about ellipticity in the SDSS but I want to know about general methods for cases where I have just an image of a galaxy.



I do so far know these methods:



  • ellipticity from stoke parameters or the flux-weighted second moment as described here.

  • from adaptive moments as described here (which would allow me to do shear calibration).

  • fit an ellipse on a particular isophote as decribed here.

My question is whether there are further methods I am not aware of. Does there also exist a method when I fit an ellipse to all isophotes? And how would I calculate the 'average' from it (such that the spherical center does not contribute to much)?



I would like to learn about the pro and cons of each method. I am especially interessted in methods that get close to what people would estimate by eye. I would also be glad to hear about references (textbooks or papers) for the different methods.

Monday, 23 June 2014

amateur observing - Stars in the sky

The resolved stars (those that can be seen as individuals) are all part of the Milky Way Galaxy (unless there are any interlopers that have been captured!).



The distances to the next nearest galaxies of any size are more than 100,000 light years. Andromeda is 2 million light years away. Unless one goes supernova, there basically aren't any types of star that are intrinsically bright enough to be seen with naked eye at these distances. Most of the stars we do see are at distances of one AU (the Sun) to a few thousand light years, with a median of a few hundred light years (See Aren't there more naked-eye-visible stars in the Milky Way plane? for more detail).

Sunday, 22 June 2014

light - Gravitational Waves and the Big bang

Firstly, Big Bang didn't happen at a point in space, away from which we are traveling. Big Bang was the creation of space. This space has been expanding ever since, so that the distances between everything increases, but Big Bang happened right where you are, where the Andromeda galaxy is, where GN-z11 is, and so on.



The Universe has evolved ever since. If we want to know how galaxies look 13.8 billion years (Gyr) after Big Bang, we can just look around in our neighborhood. If we look too far away, we don't see 13.8 Gyr old galaxies, because the light has taken some time to reach us. Thus, if we want to see galaxies that are 12.8 Gyr old, we simply look 1 billion lightyears away; if we want to se 10 Gyr old galaxies, we look (roughly) 3.8 billion lightyears away, and so on.$^dagger$ In this way, we look back in time.



If we look away far enough, we would in principle be able to look 13.8 Gyr back in time. However, we face a problem in that until the Universe was 380,000 years old, it was opaque to light (why this is so is another story). It wasn't opaque to gravitational waves, however. And since lots of GWs are thought to have originated during the epoch called inflation, which were responsible for the expansion of the Universe, and which took place a fraction of a second after the creation of space, we say that GWs offer the possibility of looking all the way back to Big Bang.




$^dagger$This is somewhat imprecise, since galaxies weren't created instantly after Big Bang, and since all galaxies weren't created at the same time. But for the sake of the arument, let's pretend they were.

Saturday, 21 June 2014

planet - Are there any works discussing planetary bodies being forms of life?

I'm searching for works or individuals that consider and discuss the idea of planetary bodies being considered stand alone living creatures, rather than objects either capable or incapable of supporting life. Anyone familiar with something along those lines?

Thursday, 19 June 2014

planet - What is the fifth moon-like object I saw around jupiter through my telescope?

Tonight while observing Jupiter I was able to see a five (what looked like) moons. I have observed Jupiter often before and only been able to make out the four galilean moons. It's my understanding that through my 8-inch scope it would be impossible for me to see any others. Switching to a widefield lens I was able to see Jupiter along with some nearby stars however the 5 moon-like points around it all still appeared... well, moon-like in comparison to the bright stars.



I observed Jupiter from about 2016-04-15 05:20 to 05:40 UTC and all five of the moon-like objects were present the whole time. What was I probably looking at?

Sunday, 15 June 2014

orbit - Does Sun have a reflection on Earth?

The new Google Maps1 presents an actual view of Earth, with the current position of Sun illuminating half part of Earth in real time. It is quite an exquisite view.



My question is based on the following image:



Earth



As you can see, Sun's reflection from southern Atlantic Ocean looks very charming. But does it really happen like that? If we travel far2 from Earth, can we actually see Sun's reflection or it is just something Google added for aesthetics?




1: Yes, I don't like most of its new features too, especially when they broke several of the old ones. But that's a separate discussion.



2: Although, the camera's supposed position would be close to geostationary orbit, I guess the reflection can also been seen as close as ISS or Hubble.

Tuesday, 10 June 2014

mars - Can we find rocks from Venus or Mercury on Earth?

You can think of it in terms of Hohmann transfer orbits, which define the minimum $Delta v$ that needs to be applied to bring something from one orbital radius to another orbital radius when orbiting a massive body. This calculation takes into account that the two objects have Keplerian orbits where the objects begins with at least the orbital speed of the initial orbital radius.



The Hohmann $Delta v$ is given by
$$Delta v = sqrt{frac{GM_{odot}}{r_1}} left(sqrt{frac{2r_2}{r_1 +r_2}} -1 right),$$
where $r_1$ and $r_2$ are the initial and final radius from the Sun.



Conceptually what you have to do is give a rock enough energy to escape from the planet and then give it an additional $(m/2)(Delta v)^2$ of kinetic energy to get it to transfer into the other orbit. If the ejection speed is $v_{ej}$ then
$ v_{eg} > sqrt{(Delta v)^2 + v_{esc}^2},$
where $v_{esc}$ is the escape velocity.



The numbers for Mercury $rightarrow$ Earth are $Delta v = 9.2$ km/s and for Mars $rightarrow$ Earth $Delta v = -2.6$ km/s (you have to slow it down to allow it to fall inwards).



The escape velocities for these planets are 4.35 km/s and 5 km/s respectively (so almost the same).



This means you need to give a rock more kinetic energy to get it to Earth from Mercury as from Mars. In the case of Mars, the transfer kinetic energy is almost negligible once the rock can escape Mars' gravity. In the case of Mercury, the rock needs to be given an initial ejection velocity of $> sqrt{9.2^2 + 4.3^2}= 10.1 $ km/s. This compares with $> sqrt{2.4^2 + 5^2}= 6.5$ km/s for Mars. At lower ejection speeds most of the ejected objects will be reaccreted by the planet.



Against this, the leading theory to explain migration of rocks between planets is high velocity impacts. Objects falling from much further out will hit Mercury with greater speeds than Mars and impart greater energies to the ejecta.



A paper addressing the possibility of Mecurean meteorites was presented by Gladman & Coffey (2008). They concluded that once ejection speeds are large enough ($sim 10$ km/s) to produce Earth-crossing ejecta, that significant accretion of meteorites should take place. Several per cent of high speed ejecta should impact the Earth (or its atmosphere at least) within 30 million years. This compares with an efficiency a factor of 2-3 higher for Mars.



There are various reports and speculations that at least one meteorite in existing collections (NWA7325, pictured) may have come from Mercury. See here for example. It appears that the main problem is getting agreement on what the chemical signatures of such meteorites are.



NWA7325



Accretion of material from Venus is a different matter. The required ejection velocities are higher because the escape velocity for Venus 10.4 km/s. But more importantly, drag in the dense Venusian atmosphere would prevent anything emerging from the planet with anything like these speeds.

the sun - Why would the Earth's solar analemma would be still a figue eight even if Earth's orbit was circular?

It is correct that the Earth's axial tilt alone would result in an analemma of a figure of eight with equally sized lobes. The axial tilt contributes to the equation of time.



Circular Orbit, no Axial Tilt



Let's look at an example case with a perfectly circular orbit and no axial tilt, to observe what happens with no equation of time. In this example, it's noon at the equator and prime meridian. The Sun is directly overhead. Next, the Earth rotates exactly once. This isn't a normal solar day, but a sidereal day, about 23 hours, 56 minutes. The Earth has rotated exactly once with respect to the stars. Where is the Sun directly overhead now? The Earth has moved a little bit in it's orbit around the Sun, so that our observer at the equator and 0 degrees longitude has observed that the Sun has moved in the sky with respect to the stars. The Sun is now directly overhead at a location to the east, a little less than 1 degree longitude away (360 degrees in a circle, usually 365 days in a year). The Earth must rotate a little more so that the Sun is directly overhead again (which completes the 24 hours in a day).



After each rotation, the Earth must rotate a little more each day to bring the Sun back overhead at 0 degrees latitude, 0 degrees longitude. After a quarter of a year, the Earth must rotate another quarter of a rotation (90 degrees) for the Sun to be overhead at that spot. Without that extra quarter of a rotation, the Sun is overhead at 90 degrees east longitude. That makes sense; the Earth has moved 90 degrees in its orbit. With a perfectly circular orbit, the rate at which this point moves eastward per day is constant, a little under a degree per day.



These points define a great circle around the Earth that coincides with the equator. A great circle is a circle on the surface of a sphere whose center is also the center of the sphere; it's the biggest possible circle that can exist on the surface of a sphere.



A Large Axial Tilt



Let's keep Earth's perfectly circular orbit, but let's give it a very large axial tilt, 80 degrees, for the purpose of this explanation. This Earth now begins at noon on the northward equinox at 0 degrees latitude, 0 degrees longitude. After one Earth rotation, where on Earth is the Sun overhead now? It's still a little less than one degree of angle away, but the direction of the displacement has changed. Instead of this point moving due east, it is mostly northward and only a little bit east. The Earth needs to rotate far less than 4 more minutes for the Sun to reach local noon for our observer at 0 degrees latitude, 0 degrees longitude. These points are west of where they would be without axial tilt. Solar noon has occurred before 24 hours have passed. As days continue to pass after the equinox, solar noons continue to occur earlier each solar day, as these points continue mostly northward and only a little bit eastward. The analemma shape as seen in the Northern hemisphere sky is moving up and to the right (west is to the right when looking south towards the Sun in the Northern hemisphere).



A quarter of a year later, it's the northern solstice, so let's see what has happened to our sun-overhead plots as the sidereal days have piled up. These points have continued to move mostly northward and a little eastward, but now they are moving purely eastward, because at the solstice the sun has stopped moving northward in the sky. It's still moving at a little less than 1 degree per day along that great circle, but because longitude lines are spaced much more closely together at 80 degrees latitude, one degree along a great circle covers many degrees of longitude. In other words, the eastward movement of these points has been "catching up" with the longitude of the corresponding no-axial tilt points. At the solstice, both the no-axial tilt point and the axial tilt point have the same longitude, 90 degrees east. The analemma shape as seen in the Northern hemisphere sky is moving up and to the left (eastward), where it reaches its highest point.



These points will continue to move through longitude lines rapidly while the Sun's overhead point remains far in the north. Now, the sun's overhead point is further east than it would be with no axial tilt. The analemma shape as seen in the Northern hemisphere sky is moving down and to the left (eastward).



As the southward equinox approaches, the sun-overhead points are moving mostly southward and only a little eastward. The analemma shape as seen in the Northern hemisphere sky is moving down and to the right (westward). This allows the non-axial tilt points to "catch up", and the longitude differences begin to decrease again, until the time of the southward equinox comes, when the two points coincide again.



Here is a crude ASCII drawing of these points so far, with longitude along the horizontal direction, and latitude along the vertical direction.



                                   (2)
K L M N O
J P
I Q
H R
G S
(1)F T(3)
E U
D V
C W
B X
A B C D E F G H I J K L M N O P Q R S T U V W X Y

A = northward equinox
M = northern solstice
Y = southward equinox


The points around (1) are further west than their corresponding points on the equator. At (2), the points have caught up in longitude. At (3), they are further east than their corresponding points on the equator, but the equator points are catching up.



After the southward equinox, the longitudes of the sun-overhead point diverge once again, with the same mechanism as described above for the northward equinox. Just reverse north and south, and the eastward movements are the same. The analemma shape as seen in the Northern hemisphere sky first moves down and to the right, then down and to the left where it reaches its lowest point at the southern solstice, then up and to the left, then up and to the right where it reaches its starting point at the northward equinox.



Back to Reality



With an axial tilt of 80 degrees, the equation of time would show some extreme values, approaching almost 6 hours of divergence from the no-axial tilt case. With the true axial tilt of about 23.5 degrees, our equation of time difference values are far less substantial, but the effect is real.



The true analemma shape we see on Earth is the combined effect of the axial tilt that we have seen plus the fact that Earth's orbit is elliptical and it slows down during the parts of its orbit when it's farther from the Sun and it speeds up during the parts of its orbit when it's closer to the Sun.



The site Analemma has good explanations about the individual effects of the elliptical orbit and the axial tilt and how they combine to create our analemma.

Friday, 6 June 2014

amateur observing - How to orientate a telescope using an accelerometer and magnetometer?

For a project with my astronomy professor, we are hooking up an array of 4 smaller telescopes to observe objects in space together.



To do this, we wanted to put an accelerometer and magnetometer on each telescope and have it read data from the gravitational pull on the accelerometer and the pull from the north magnetic pole on the magnetometer.



The accelerometer would measure the altitude, or the absolute height of measure point where accelerometer is at on the telescope. This would be converted to the angle of the telescope from its initial plane (when it makes a 90 degree angle with its stand and Cartesian z is zero).



The magnetometer would measure the pull of the of the magnetic north pole, and be able to give the azimuth angle, or the angle (from 0-360) that the telescope is pointing away from the magnetic north pole (0 degrees in the x direction would be pointing at the magnetic north pole).



Both readings are recorded in microteslas.



My professor said he was able to solve the altitude equation, but doesn't have time for the azimuth. He suggested treating the telescope itself as a 3D vector, normalizing it, and the projecting it down onto a flat plane, i.e. where it makes a 90 degree angle with the stand.



Would I use a rotational vector for this? And should I treat the magnetometer and accelerometer vectors as two separate matrix vectors and use one to solve the other? What would the end equation look like?

Sunday, 1 June 2014

cosmological inflation - Problem with denoising the BICEP2 data?

This question pertains to this article which talks about why the BICEP2 measurements of B-mode polarization in Cosmic microwave background radiation turned out to be noise from galactic stardust. They go on to add that the Plank data of the noise does not lend to getting a good B-mode estimate from the BICEP2 data.




However, the results of the joint assessment would suggest that
whatever signal BICEP2 detected, it cannot be separated at any
significant level from the spoiling effects. In other words, the
original observations are equally compatible with there being no
primordial gravitational waves. "This joint work has shown that the
detection of primordial B-modes is no longer robust once the emission
from galactic dust is removed," Jean-Loup Puget, principal
investigator of Panck's HFI instrument, said in the Esa statement.
"So, unfortunately, we have not been able to confirm that the signal
is an imprint of cosmic inflation."




Can someone explain why(from a signal processing perspective) exactly the detection of primordial B-modes is not robust even after the emission from galactic dust is removed? And what are the proposed workarounds to tackle this problem?