Thursday, 28 October 2010

special relativity - What is the bulk Lorentz factor?

I think it is referring to the speed and Lorentz factor $(beta = v/c$ and $gamma = [1-beta^2]^{-1/2})$ of the gas as a whole. Within the gas, there could be particles moving with a variety of velocities.



So if you pick up a ball of gas at 10,000 K (ouch) and throw it at 100 m/s then the bulk speed is 100 m/s, but obviously the particles in the gas have their own individual velocities.

mycology - How do fungi react to being grown in a tissue culture?

Generally, fungi are cultured on agar with a food source such as malt extract. One way to do this is to clone a mushroom from the fruitbody of a mushroom or from colonized substrate (e.g. rotting wood). Another approach is to germinate spores.



The fungus will grow as vegetative mycelia until it runs out of food. At that time, some species will grow tiny fruitbodies or sporulate before entering a more dormant phase, dying out. Unless refrigerated, standard agar plates dehydrate. Also, fungi may take on different textures: wispy, bumpy, root-like (rhizomorphic), leathery, powdery, and may darken or change color as they age.



Some cultures start to adapt their metabolism to the media after a large number of transfers (based on observation and discussed in Stamets' "The Mushroom Cultivator".) at this point, cultures may also start to grow more slowly.



The mycelia can be transferred from one plate to another or to another media (substrate) such as a log, 'fortified sawdust' (sawdust with ~30% w/w wheat bran), or pasteurized compost. With the appropriate nutrition and environmental stimuli (CO2 levels, temperature, moisture, light, etc), the mycelia will fruit, creating mushrooms - generally after fully colonizing the substrate.



Here is a figure of 35 agar plates with 16 different fungal cultures that I isolated from the wild (soil, wood, mushrooms, as described in Allison et. al, 2009; LeBauer, 2010), and have transferred from a mature culture (about a 2 months old) to a new plate, which are about two weeks old in this picture. Parent cultures are in the first and third row, with children below (they are genetic clones; instead of counting generations, I count number of transfers.



enter image description here



You can also find cultures of fungi (Yeast) in unfiltered beer.



I have used many techniques to isolate and culture fungi (Isikhuemhen and LeBauer, 2004; LeBauer, 2010) but I would recommend the books "The Mushroom Cultivator" and "Growing Gourmet and Medicinal Mushrooms" by Paul Stamets for more information on sterile culture techniques used in mushroom cultivation.

positional astronomy - How/where to check where the Sun is (constellation)?

In this case, Wolfram|Alpha has you covered. For example, look at the result for "which constellation was the Sun in Jan 1 1400". That said, I'm not sure this calculate includes the precession of the Earth's rotation axis, since I would've expected it to get my astrological birth sign right in 500 BC.

Wednesday, 27 October 2010

Would it be possible to equip an asteroid to collect space junk in Earth's orbit?

This is plausible, and might even be a good idea if used right.



First off, NASA has been working on plans for an asteroid redirect mission, called ARM. While it remains to be seen whether this will be approved, and they plan to put it into lunar orbit, this is a hypothetical scenario anyways so I'll ignore that.



Now, putting an asteroid into earth orbit is a little bit of a difficult subject. Sure, you could do it, but messing up would net you a whole lot of problems. I presume it would have to go into high earth orbit if it goes into orbit at all.



This rules out using the asteroid directly like a satellite, but that would have been impractical even if it was sanctioned by everyone with enough influence to have an opinion on earth. From there the satellite would be best used as a base of operations, from which you could send drones to collect junk and keep refueling stations. I'd also include a telescope or two, but that's personal preference.



If you want a list of targets though, here would be a good place to look.

Tuesday, 26 October 2010

general relativity - Does mass create space?

This is a tough question to answer because the dimensions of your proverbial cube would be affected by the mass inside (as dimensions can only exist in space, so anything that affects space will also bend your cube), and mass does not bend space, it bends spacetime. You have to keep in mind that when talking about special and general relativity, you are talking about four-dimensional spacetime. This is what is curved by mass. For example, due to the curvature of spacetime due to Earth's gravity, you feel the same effects on its surface that you would if you were accelerating upwards at 9.8 m/s^2 through deep space (where we assume there is no gravity). This doesn't mean the physical space is curved - if you draw a straight line on a piece of paper in space and then travel down to Earth's surface, it will remain straight. Instead, spacetime is curved, which can affect objects' paths of motion, but not the objects themselves. Now to your questions:



  • The volume inside of the cubes would be exactly the same, but two identical objects flying through each cube would follow two different paths, and if either looked at the other, it would see that the other's clock is moving at a different speed than its own due to time dilation (due to the curvature of spacetime towards mass, time moves more slowly closer to massive objects). We can see evidence of the effect of gravity on paths with gravitational lensing. Light travels in a straight line through spacetime, but when looking at the space around a star, you will actually see objects that should be directly behind the star and blocked from sight. This is because light rays that, in the absence of gravity, would pass next to the star and continue off at an angle are pulled towards the star by it's gravitational field (in reality they're just following the curvature of spacetime around the massive star), so they curve around the star, becoming visible to us, even though, in space alone, they should be blocked by the star.

  • The distribution of the mass has no effect on the gravitational field around it (same with electromagnetic fields). If you look at any equation that has to do with gravity (gravitational potential energy, gravitational force, etc.), you will find a term for the total mass of the object, but unless you are within the bounds of the control volume containing the mass, the distribution does not matter. This is one of the reasons why black holes are so hard to study - we can tell how much mass is inside of them, but without the ability to see inside, we have no other way of detecting their properties.

  • Like I said above, the volume is not affected because space is not warped by gravity, only spacetime. If you made a physical box (or we'll say a cube frame) around empty space and then moved it to a space that contained a star, it would remain the same size and shape (assuming that it was strong enough to withstand the force of gravity pulling it towards the massive star).

  • Neither, it simply curves spacetime.

Saturday, 23 October 2010

cosmology - Three-torus model of the universe

A three-torus is the boundary of the solid three torus, just like the two-torus is the surface of a solid donut. You can imagine it as a cubical room where each wall/ceiling/floor is a portal to the opposite-facing wall (i.e. the wall to your right is a portal that sends you to your left), but preserves orientation (when you walk out of the portal, your heart is still on your left).



You can also think of it as a world where your position is described by three coordinates (like x, y, and z in Euclidean space), but each coordinate corresponds to an angle on the unit circle. If you go far enough in one coordinate, you loop back to where you started (360 degrees is the same as 0 on a circle).

history - Which stars have been named after astronomers?

You probably need someone better versed in the history of astronomy than me, but I'll give it a stab.



As you've already noticed, the Wikipedia article on stars named after people has some entries, and that's probably about it. In terms of modern naming procedures, we tend to use catalogue names. Unlike, say, asteroids, there's no systematic way of naming stars after people or anything else, so stars usually continue to be referred to only by catalogue number. Those rare stars that are named after people (or something else) are usually because they are now somehow historically associated with that system. Unless, of course, the catalogue is named after someone! Another exception are those very bright stars with pre-telescope names, but those aren't named after people (e.g. many scientific papers refer to Betelgeuse as such).



I guess in short, there's just no standard modern naming procedure by which a star can end up with a person's name, so few exist, and probably most are listed in the Wikipedia article.

Wednesday, 20 October 2010

asteroids - Dynamic Method-Please explain

I have, for years, been an Astronomy and physics nerd. For the first while, I was a total astronomy nerd. Then I slowly transitioned into physics and have been, for less than a year, a physics nerd. In this shirt while, I have taught my self calculus up to Calculus II (over summer and early this school year; I was 14 then). I know how to do geodesics, Schrodinger's equation, classical physics, etc and I'm teaching myself QFT and even planning to make one on quintessence. When I was an astronomy nerd, I didn't know how to do a lot of this math, and so I quit. Now, I am in an "astronomy comeback" phase. There's still one particular thing involving classical physics and astronomy that I still can't quite get-



The question-
How exactly, with steps and math shown, do you derive the masses of asteroids via perturbations? please, no links, because I have searched and searched for years and can't find anything. Thank you!



P.s. An extra thank you if you read the whole thing(including the story)! 😄

Tuesday, 19 October 2010

star cluster - What exactly is a stellar association?

A stellar association is a loose cluster of stars, that formed at the same time from the same molecular cloud, and so have the same proper motion. Unlike open clusters, they are not gravitationally bound together, so the stars in a stellar association will gradually separate, forming a moving group of stars.



An example is the Scorpius–Centaurus Association

Monday, 18 October 2010

human biology - What is the relationship between migraines and histamine?

Yes, this is true.



Histamine is thought to induce the enzyme Nitric Oxide (NO) Synthase. NO is then available to act locally on the vasculature as a vasodilator.



NO binds to guanylyl cyclase in vascular smooth muscle cells, which leads to the production of cyclic GMP, which in turn forms phosphorylated protein kinase G. PKG phosphorylates Ca2+ channels, slowing the influx of calcium into the cell, which leads to smooth muscle relaxation, and vasodilation, which leads to migraine.



The only silver lining is that there is a check in place: with the binding of histamine to H3 receptors on c-fibers in the central nervous system, feedback inhibition prevents the further release of histamine from these sites.



References:




Akerman S, Williamson DJ, Kaube H, Goadsby PJ. (2002). The role of histamine in dural vessel dilation. Brain Res. 956(1):96-102.



Gupta, S., Nahas, S.J.,Peterlin, B.L. (2011) Chemical Mediators of Migraine:
Preclinical and Clinical Observations. Headache: The Journal of Head and Face Pain, 51(6): 1029–1045.


Sunday, 17 October 2010

apparent magnitude - Why do dark objects look white from a distance? (Moon, Ceres, but not Earth!)

The photo in your question is -- well, not exactly fake, but a composite. The biggest clue is that Earth is too close to the horizon; it would have had to be taken from within a few degrees of the boundary between the near and far sides of the Moon, and none of the Apollo missions landed there.



Furthermore, take a close look at the cloud patterns. The view of Earth appears to be identical to that in famous Earthrise photo taken from lunar orbit by Apollo 8 (click to see a larger version):



Earthrise



As you can see in that photo, the surface of the Moon is considerably darker than the Earth.



In the photo in the question, ignoring the inserted image of the Earth, the sunlit surface of the Moon is the brightest thing in the photo. The light balance must have been adjusted to make everything in the image easy to see.



As for why the Moon looks white in the night sky, it has an albedo of about 0.37, whereas the night sky has an albedo of about 0.00. Human eyes are very good at adapting to varying lighting conditions. When you look at a dark object against an even darker background, it's going to look white or light gray, even if it's intrinsically dark gray.

Why did time progress past the singularity?

I hope this doesn't sound stupid but I've always wondered:
Why didn't time stop with the singularity before the big bang?
I might be mistaken but doesn't time slow in proportion to the distortion of space-time caused by a massive object? And,if that's so, and if the singularity was infinity massive then wouldn't time theoretically stop?

human anatomy - Foveal ganglion cell density (Tay-Sachs Disease)

As you've already mentioned, cells near the primate macula tend to make one to one connections. Due to this lack of convergence they can be somewhat smaller than normal cells (particularly in the size of their dendritic arbors) earning them the moniker "midget" ganglion and bipolar cells (also P cells). By reducing the magnitude of photoreceptor convergence onto ganglion cells, the retina sacrifices luminal sensitivity for spatial acuity. Or put another way, a downstream neuron in the LGN/V1 can be more "certain" about which photoreceptor is sending information when light is bright, but less certain that any information is being sent when light is dim.



The other side of the question you bring up is how the retinal ganglion cells themselves alter the image formed onto the photoreceptor layer. For those that don't already know, photoreceptors are located at the very back of the retina. Since they're the major light sensitive part of the retina, that means light has to go through the ganglion cells, bipolar cells, amacrine cells, and all of their associated processes before reaching the actual photosensitive part of the eye. So, why not put photoreceptors at the front of the eye? One major reason is metabolic, photoreceptors are tightly coupled with the retinal pigment epithelium which helps to re-process used photopigment as part of the retina's vitamin A cycle.



How can those two factors, a one to one correspondence and a decrease in thickness be present at the same time? For reference, let's locate a cross section of a primate macula



monkey retina h&e



Dark areas are cell nuclei. The innermost layer of cells (bottom of the image) are ganglion cells, the next layer of cells up is bipolar cells (the inner nuclear layer) and the next layer up is photoreceptor cell bodies. you can make out the cilial stalks of the photoreceptors pointing upwards toward the RPE.



From this image we can see a few transformations happening in the macula:



  1. The photoreceptor layer is thicker, up to 7 or 8 cell bodies deep. This means higher photoreceptor density.

  2. The outer plexiform layer (photoreceptors to bipolar cell connections) is considerably thicker. A single cone should connect to fewer bipolar cells in the retina, so how does this make sense? We'll find the answer in #3.

  3. The outer nuclear layer is only one cell thick. This is the first layer that is actually thinner in the foveal region. The cell bodies for photoreceptors from tis region have been displaced laterally, and the outer plexiform layer from #2 had to be larger to accommodate this lateral shift

  4. The inner plexiform layer, connections between bipolar cells and ganglion cells, is nonexistent. Also absent is the ganglion cell layer itself.

So how do these observations resolve your question? Quite simply, the lateral movement of information from the foveal/macular photoreceptors happens predominantly in the outer plexiform layer. That buys the inner plexiform layer considerably more space to further spread those signals laterally to ganglion cells, and avoids the problematic situation you suggested of having an extra-thick ring of ganglion cells immediately around the fovea. The end result is that ganglion cells that respond to photoreceptors in the foveal region of space are even further out from the retina than the bipolar cells they connect to, which are themselves displaced.



As far as why a cherry red spot looks so large, that honestly depends on two factors: the disease or disorder in question and the zoom used. Bear in mind that the red color comes from, as far as I know, the blood vessels in the choroid itself. It would therefore not necessarily correspond exactly to the size of the macula, but instead to any area which is thinned in that vicinity. I think you'll become more comfortable with that idea if you look back at the thickness of the total retina in the above figure and notice that even outside of the foveal region it is still growing thicker.



I think your question shows good instincts, because I believe that if the eye evolved to have a larger "true" fovea then you would get exactly what you describe: a large area of super high density vision surrounded by a region with non-linearly worse vision designed to support the central region. Instead it appears that the primate macula is just large enough that we have a moderate sized central acute region and a (relatively) linear drop in spatial acuity moving out from it.

stellar evolution - Description of Henyey tracks on wikipedia incorrect?

So, if you search for Henyey tracks on wikipedia (I know, the shame of it!) you will come across this statement:




The Henyey track is characterized by a slow collapse in near hydrostatic equilibrium. They are approaching the main sequence almost horizontally in the Hertzsprung–Russell diagram (i.e. the luminosity remains almost constant).




Ignoring the bad grammar, I take umbrage with the following:




the luminosity remains almost constant.




I would say that the apparent horizontal track associated with the Henyey track is more due to the fact that in a typical HR diagram the temperature axis ranges over about one order of magnitude, while the luminosity axis may vary over five or more orders of magnitude.



So, to frame this as a question, would you agree or disagree with the statement on the site or with what I have stated (i.e., that the Henyey track is an apparent horizontal track caused by the nuance in the magnitude scales of luminosity in comparison with the temperature)?

Saturday, 16 October 2010

molecular biology - How long can E. coli stocks be stored at -20°C?

I'm volunteering for a biohacker lab - biocurious in Sunnyvale. The have a pretty good set of equipment - gel boxes, incubators, but they don't have a -80°C freezer yet.



I'd like to set up some glycerol stab stocks of some E. coli strains, but i'm told even untransformed (i.e. no plasmid) bacteria will get funny after a while at -20°C. It would also be good to hear how long you can keep a plasmid transformed strain at -20°C.



Can anyone be specific about what happens and how long it takes when you store them at -20°C for long periods of time?



I have done a little reading for competent cells and I guess they stay active for about 4 days without a -80°C freezer.

Wednesday, 13 October 2010

space - Is it possible for a virus to come from a meteorite?

I think it's a fun question and I can kind of give a layman's answer. (though reading the above answer, there is some overlap).



It seems likely that "tiny", life or viruses could survive/hibernate for a long time if blown into space on space rocks, so it follows that life could travel to earth that way, but I think we could apply a modified version of the drake equation.



The Earth gets hit by comets or meteors every day but most of those burn up in the atmosphere. There is a temperature beyond which even bacteria and viruses have a hard time surviving. So, many of those meteors or bits of comet, I suspect, would lose any life they might have. Only a percentage could be the right size to both slow down enough while breaking apart to not impact at too high a speed and successfully deliver life to the surface or into the ocean. Like the real drake equation, we get to plug in our own numbers, but I suspect most impacts wouldn't successfully deliver life to the planet they impact. How many would? 1 in 10? 1 in 100? I'm not sure.



Now, what percentage of comets or meteors have life on them? This is a tough one, I'm pretty sure life can only form on a planet because you need (I would think), liquid water, atmosphere to maintain surface water and reduce evaporation and some kind of heat to drive the mixing of chemicals, perhaps lightning or undersea volcanic vents. Life likely cannot be formed in space or in a nebula. It can be blown into space off a planet with life, but it can't form in space. That likely reduces the number of meteors or objects in space that even have the possibility of carrying life. I hate to guess this one, cause who knows, but 1 in 1000? 1 in 10,000? Most of the meteors, asteroids and comets in space have probably formed without the possibility of carrying life.



There are some simple organic molecules in space, alcohol for example, but there's an enormous chasm between organic molecules and even the simplest life.



Some planets are very difficult to blow debris off of. Earth with an atmosphere and an escape velocity of some 25,000 miles per hour needs a very big impact to lose material into space. Mars, with a very thin atmosphere and lower escape velocity is much easier. Of some 60,000 recognized meteorites, 124 have been identified from Mars. (Source) and Mars is the easiest one. Venus, with it's thick atmosphere and Mercury, being as close to the sun it is, adding the suns escape velocity to it's own, are much harder. Material being blown off the gas giants is even less likely, so Mars is the easiest plant to potentially lose material where it could land on Earth or other planets. Even if we use the Mars Meteor ratio of 400 to 1, which is, I think, too high, it's still a very low percentage of objects in space that potentially came off a planet or moon and carry life with them.



A 3rd point worth bringing up is repetition. Lets say, just for fun, that Europa and Enceladus both have life in their under the surface oceans and both planets regularly eject jets of life filled water into space, that freezes and carries basic life into orbits around Jupiter / Saturn and some of those life carrying bits of ice make their way on occasion to other planets. If this happens it probably happens more than once and any bits of life that may come to earth from Europa or Enceladus, if it happens at all, probably happens more than once. The first introduction of invasive species, even simple ones, can be problematic, but after the first, the planet should have adapted. That's one of the mathematical quirks. One Andromeda strain can in theory be a huge problem, but repeated Andromeda strains from the same source, the risk drops with each impact. What you need, for a dangerous situation to happen is a new introduction of unfamiliar that's never happened before. That's a pretty specific requirement. Mars may have once had life, and if it had, life on Mars probably landed on earth many times over 4 plus billion years.



a 4th point, space is full of radiation. Even viruses will eventually die exposed to the radiation of space. A space rock, carrying life from another solar-system light-years from earth has a very long journey and a very small target. Bits of frozen ice or rock probably travel from other solar systems into ours quite rarely over very long periods of time, and, as I pointed out above, most of those probably won't have life.



This is just a kind of "shoot from the hip" sort of answer, but I think you would need a perfect storm of events for your scenario to actually happen. statistically, I suspect it's a very rare occurrence and not something we need to worry about on a human time frame.

Monday, 11 October 2010

black hole - Why do certain massive stars leave no remnants?

The gap appears because of pair instability supernovae. In short, as one looks at such massive stellar cores at increasing temperatures, an ever-larger fraction of the photons are sufficiently energetic to spontaneously form electron-positron pairs. True, they soon recombine, but there is nevertheless a loss in (radiation) pressure, which causes contraction, and thus a further increase of temperature. If unstable, as appears to be the case in such massive stars, this leads to more high-energy photons, more pairs, and a further loss of pressure.



Models of such objects suggest that the collapse leads to a sudden, rapid ignition of oxygen- and silicon-burning reactions. In a mass range between about 150 and 250 solar masses (corresponding to core masses of very roughly half of that), the thermonuclear explosion is enough to rip the core apart, and there's nothing left to collapse. This is a bit like how the sudden onset of carbon burning destroys white dwarfs in type Ia supernovae. More massive cores are more strongly bound by gravity, and models suggest that the nuclear explosion isn't enough to unbind the core.



So below about 150 solar masses (and at higher metallicities, I think), the core doesn't reach pair-unstable conditions. Above about 250 solar masses, the induced nuclear explosion isn't powerful enough to destroy the core.

general relativity - View from inside the black hole


If an observer is falling toward a black hole with his face away from singularity then what will he observe after crossing the event horizon?




Nothing.




The reason that why I am asking this question because as far as I know for an outside observer, the falling observer appears to freeze at the event horizon i.e. time appears to stop for the falling observer.




Correct. Gravitational time dilation goes infinite, and the distant observer says the "coordinate" speed of light at the event horizon is zero.




So if the falling observer is able to look outward after crossing the event horizon then he will be able to see an infinite amount of time which is impossible. So what will the observer see after crossing the event horizon?




Nothing. The coordinate speed of light is zero at that location, which means that by our clocks, it takes forever to see anything. So the falling observer hasn't seen anything yet, and he never ever will.



IMHO it's worth reading the mathspages Formation and Growth of Black Holes and paying attention to the frozen star interpretation:



"Incidentally, we should perhaps qualify our dismissal of the 'frozen star' interpretation, because it does (arguably) give a servicable account of phenomena outside the event horizon, at least for an eternal static configuration. Historically the two most common conceptual models for general relativity have been the "geometric interpretation" (as originally conceived by Einstein) and the "field interpretation" (patterned after the quantum field theories of the other fundamental interactions). These two views are operationally equivalent outside event horizons, but they tend to lead to different conceptions of the limit of gravitational collapse. According to the field interpretation, a clock runs increasingly slowly as it approaches the event horizon (due to the strength of the field), and the natural "limit" of this process is that the clock asymptotically approaches "full stop" (i.e., running at a rate of zero). It continues to exist for the rest of time, but it's "frozen" due to the strength of the gravitational field. Within this conceptual framework there's nothing more to be said about the clock's existence..."



The author doesn't favour it, but it squares with what Einstein said about the speed of light varying with gravitational potential. The other interpretation doesn't. And note that Einstein didn't refer to the "coordinate" speed of light. He simply referred to the speed of light. So we can reasonably say that at the event horizon, the speed of light is zero, and that this is why a vertical light beam can't get out. The distant observer sees the infalling observer freeze at the event horizon. But the infalling observer doesn't see himself as frozen, because the speed of light at that location is zero. He sees nothing.



NB: SR time dilation is symmetrical, but GR time dilation isn't. If you and I passed each other in gravity-free space at some relativistic speed, we would each claim that the other's clock was slower. But when we're at different elevations, we both agree that the lower clock is going slower.

Saturday, 9 October 2010

Can a meteor shower happen simultaneously all around the planet?

Realistically, no, that cannot happen. Most meteor showers that we experience on Earth - the most famous being the Perseids and the Leonids - are a result of comets passing roughly through our orbit and leaving behind debris that was burned off as the comet passed by the Sun. We then come along, sweeping through this debris and from our perspective on Earth, we see a shower of meteors raining down on us.



That implies two important things. First, we primarily see meteors as we sweep through the debris from comets. They are not objects which have made a trajectory for Earth and hit us head on. Second, we only see meteors when we're on the side of the planet facing the orbital direction of the Earth.



That being said, for the sake of the story, one could envision an alien race of sufficiently advanced technology surrounding our planet with asteroid and comet debris and subsequently "dropping" these meteroids on us from everywhere all at once. I'm not sure if aliens are involved in the story you've linked, but aside from third party intervention, there's now way you're going to see meteors from every place on Earth all at the same time.

observational astronomy - How to determine the ellipticity of galaxies in SDSS

What is the best approach to determine the ellipticity of galaxies in the SDSS DR12. I have read this page. Are those really good methods?



Do flux-weighted second moments (as given in the stokes parameters) really give the correct value for ellipticity?



Also I am not sure what table field they actually refer too (in the Galaxy view) when talking about the method using the Adaptive Moments. For example where can I find the m_rr_cc value for the method described here?



Update



I have calculated the ellipticity of a spiral galaxiy with two different methods using data from SDSS DR7. First I used the method using the stokes parameters and second I used the values for isoA and isoB. Both calculations where done in the r-band.



The data can be obtained with the following CAS Query:



SELECT
g.objid as objId,
/* stoke parameters*/
q_r,u_r,
/* iso_a, iso_b */
isoA_r,isoB_r
FROM Galaxy AS g
WHERE
g.objid = 587722982832013381


yielding the result



objId: 587722982832013381   
q_r : -0.1206308
u_r: 0.002988584
isoA_r: 169.3949
isoB_r: 84.78848


The ellipticity by stoke parameters is given by



$$
e = 1 - frac{b}{a} = 1 - frac{1 - sqrt{Q^2 + U^2}}{1 + sqrt{Q^2 + U^2}}.
$$



The ellipticity from isoA and isoB is very given by



$$
e = 1 - frac{isoB}{isoA}.
$$



But I get completely different results when I use isoA and isoB or the stoke parameters. I get:



ellipticity from stoke parameters: 0.2153498356
ellipticity from 1-(isoA/isoB): 0.4994626166


The results can be reproduced with the following python code:



import math

def ellipticityStokes(q, u):
e = 1 - ((1-math.sqrt((q**2) + (u**2)))/(1+math.sqrt((q**2) + (u**2))))
return e

def ellipticityNormal(a, b):
e = 1 - (b/a)
return e

q = -0.1206308
u = 0.002988584

isoA = 169.3949
isoB = 84.78848

print(str(ellipticityStokes(q,u))) #0.2153498356239003
print(str(ellipticityNormal(isoA,isoB))) #0.4994626166431221


Looking at the image





(and compare with this website from Buta) I would say the value 0.4994626166 (from isoA and isoB) makes more sense. The value calculated from the stokes parameters is just wrong I think.



What is the mistake here? I would expect both calculations to yield similar results.



However for newer data releases isoA and isoB are not available any more because they seem not to trust those value? What should I do with newer data releases as DR12?

Thursday, 7 October 2010

biochemistry - Why does cyanide inhibit CuZnSOD, but not MnSOD or FeSOD?

I'm just taking a stab, because my inorganic chemistry is rusty.



cyanate (CN-) is well known to bind to iron (it inhibits hemoglobin and causes anoxia), and it also forms a pretty strong complex with Zinc:



http://en.wikipedia.org/wiki/Zinc_cyanide



I would guess that in the case of the SOD that the other liganding amino acids and the pocket of the enzyme are such that the affinity for cyanide is lowered to the point that oxygen can displace it.



In the case of hemoglobin, CN will coordinate to the heme iron in much the same fashion that O2 does, but it doesn't come off. In the case of SOD, the enzyme will be looking to coordinate to superoxide O2(2-) which would be at a roughly tetrahedral angle to the metal. This is not a favored geometry for CN coordination, which prefers to linearly coordinate to things.



In the case CuZnSOD it seems likely that this enzyme has enough room and its Zn and Cu binding envirionments don't forbid CN binding. it would be an evolutionary accident really as to why one is sensitive and another is not.



some structures at rcsb:



SODs:
http://www.rcsb.org/pdb/results/results.do?qrid=C0740111&tabtoshow=Current



thiocyanate bound. the pocket is pretty open between the two ions. no CN bound structure.
http://www.rcsb.org/pdb/explore/jmol.do?structureId=1SXS&bionumber=1

Moving from computerised to normal amateur telescope?

A Celestron Nexstar 130 SLT is not suitable for spectroscopy and the only photography that it will be mush use for would be afocal or DSLR imaging of the Moon (or Sun with a suitable filter).



For more serious work you will need a equatorial mounted scope with a better focuser.



Exactly which one to choose will depend of your budget and what you want to do.



When you say the computer does not work can you be more specific.

Sunday, 3 October 2010

the moon - Visibility of Apollo 11 module

I wasn't around when man landed on the moon in 1969. When I see the moon, I always wonder, were people able to see the rocket?



Yesterday, I looked at the moon in daylight and wondered again.



My question is: how far were people able to see the rocket after it launched?

Saturday, 2 October 2010

astrophysics - Dust mass-loss rate from a massive star given a set of parameters?

I've been looking for examples at how mass-loss rates are determined.



I'm studying a circumstellar dust shell ejected from a Wolf-Rayet star. I have some parameters like, expansion velocity of the shell (60km/s), the dust mass of the shell (0.1 M_sun), the radii of the shell (R_in=10000 AU and R_out=60000) and its age (T=26000 yrs).



I was wondering if there's a formula relating those parameters and if they are sufficient to determine the dust mass-loss rate or its dependent on other factors.



Could anyone help?