general relativity - Question about extreme space distortion and creation of a new dimension

Just because there's an extra dimension on a diagram doesn't mean that it's real. No such thing has been created. It's just an artifact of embedding a manifold with a non-Euclidean geometry into a Euclidean space.

One of the biggest stumbling blocks of intuition for people learning general relativity is that the physics only cares about the intrinsic geometry. For example, imagine an ordinary ball in three (Euclidean) dimensions, and take its surface. In jargon, the surface is the two-dimensional sphere $mathrm{S}^2$, and it has geometric properties that can be described by reference to it alone, e.g., lengths of curves drawn on it, angles between intersecting curves, that if one starts at some point and goes in a single direction, eventually one would be back to the starting point, etc.

One should consider the two-sphere as a valid geometry by itself, and the fact that it can be pictured as the surface of a three-dimensional Euclidean object as purely incidental--it $mathrm{S}^2$ can be embedded in the Euclidean space $mathrm{E}^3$ in that simple manner, yes, but that's a fairly arbitrary choice. It can be embedded in other spaces as well, or not embedded in anything at all.

For Riemannian manifolds (the purely spatial geometries), there are some nice mathematical results regarding embedding into Euclidean spaces, but nevertheless, they're usually impractical and not relevant to general relativity. Even worse, general Lorentzian manifolds (spacetimes) have no comparably 'nice' results regarding embedding spacetime geometries into flat pseudo-Euclidean spaces $mathrm{E}^{n,m}$, so it's doubly useless to worry about it, except perhaps in the cases where the spacetime is extraordinarily simple.

In the end, such extra dimension(s) are just artifacts of making a picture. They're not physically real, so they're not 'created' in any physical sense.

space time - Can a black hole rip spacetime

There is a useful model of spacetime as a rubber sheet that is bent by masses laying on it. But it should be remembered that this is an analogy (Obligatory xkcd) and most analogies fail if pushed too far. Spacetime isn't made of something that can rip.

A rotating black hole, a "Kerr black hole" is stranger than a static one, as it pulls spacetime around it and can accelerate objects passing close to it. Even so there are no rips, no singularities outside the event horizon.

However in another way, at the centre of every black hole there is a singularity, and a singularity is a single point "rip" in spacetime. General relativity can be used to predict what happens to spacetime around the black hole, and even inside the event horizon. But at the singularity the gravity becomes infinite, and at that one point, spacetime doesn't exist.

It is suggested that a "naked" singularity, not surrounded by an event horizon, cannot exist.

Relation between black hole mass and radius, and our universe's

According to the standard ΛCDM cosmological model, the observable universe has a density of about $rho = 2.5!times!10^{-27};mathrm{kg/m^3}$, with a cosmological consant of about $Lambda = 1.3!times!10^{-52};mathrm{m^{-2}}$, is very close to spatially flat, and has a current proper radius of about $r = 14.3,mathrm{Gpc}$.

From this, we can conclude that the total mass of the observable universe is about
$$M = frac{4}{3}pi r^3rho sim 9.1!times!10^{53},mathrm{kg}text{.}$$
Sine the universe at large is nonrotating and uncharged, it's natural to compare this to a Schwarzschild black hole. The Schwarzschild radius of such a black hole is
$$R_s = frac{2GM}{c^2}sim 44,mathrm{Gpc}.$$
Well! Larger that the observable universe.

But the Schwarzschild spacetime has zero cosmological constant, whereas ours is positive, so we should instead compare this to a Schwarzschild-de Sitter black hole. The SdS metric is related to the Schwarzchild one by
$$1-frac{R_s}{r}quadmapstoquad1 - frac{R_s}{r} - frac{1}{3}Lambda r^2,$$
and for our values we have $9Lambda(GM/c^2)^2 sim 520$. This quantity is important because the black hole event horizon and the cosmological horizon become close in $r$-coordinate when it is close to $1$, a condition that creates a maximum possible mass for an SdS black hole for a given positive cosmological constant. For our $Lambda$, that extremal limit gives $M_text{Nariai} sim 4!times!10^{52},mathrm{kg}$, smaller than the mass of the observable universe.

In conclusion, the mass of the observable universe cannot make a black hole.

---

Well, we don't fully comprehend black matter, do we? And it was just "yesterday" that we discovered the "black energy", wasn't it?

If GTR with cosmological constant is right, we don't need to "fully comprehend" it to know its gravitational effect, which is what the calculation is based on. If GTR is wrong, which is of course quite possible, then we could be living in some analogue of a black hole. But then it's rather unclear what theory of gravity you wish for us to use to try to answer the question. There's no remotely competitive theory that's even approaching general acceptance.

From the perspective of our huge ignorance, I think that 14.3Gpc and 44Gpc are not even one order of magnitude apart, which I consider a good approximation.

Actually, the point of that calculation was to show that it's at least prima facie plausible. The Schwarzschild radius calculation doesn't rule out the black hole--quite the opposite. However, it's also not appropriate for reasons I explained above. The more relevant one actually does have mass more than one order of magnitude apart, and shows inconsistency. So if GTR with Λ is correct, it's unlikely because the ΛCDM error bars aren't that bad.

However, even if we still treat it as "close enough", that does not by itself imply what you want. The question of what kind of black hole all the mass of the observable universe would make, if any, is quite different from whether or not we're living in one. The black hypothetical needs to be larger still.

The biggest point of uncertainly, though, is the cosmological constant, even if GTR is otherwise correct. If we're allowed to have very different conditions outside our hypothetical black hole, then we could still have one, but then we get into very speculative physics at best, and just complete guesswork at worst.

So treat the above answer as conditional on the mainstream physics; if that's not what you want, then there can be no general answer besides "we don't know". And that's always a possibility, although not a very interesting one.

black hole - Can we see the past image of Earth because of curving light by massive objects?

Light can be made to do this around black holes, sort of. At a certain distance from a black hole you have what's called a photon sphere. At this distance, determined by the mass of the black hole, photons travel in orbits because of the space-time curvature.

So to answer your question, maybe.

Firstly, a massive entity would need to be located at some point in space in order to influence the photons into an orbit - okay yes, possible. Secondly, this massive object would need to have a mechanism of instantaneously disappearing from space-time in order for the photos to tangentially come out of orbit and back towards earth.

I read on a previous question on Physics.SE that there is no such solution to the Einstein Field Equations which allows for a massive object to just disappear - the closest thing would be for the density distribution of the object to change but I can't say if that would yield the result you are asking about.

Rob Jeffries mentioned that images do exist which suggest that photons experience such a change in trajectory - this could happen on paths which are very close to a photon sphere-like one, but deviate enough from it to allow photos to escape.

the moon - Multiple aerobraking

I moved this question to more related : Space exploration

I spent a lot hours with Kerbal Space Program recently and I am courious about one thing.

I got into orbit of moon and then I was able to get back to orbit of "Kerbal" (the home planet in this game, similar to Earth) and land succesfully.

I did not have much fuel left, therefore I wanted to save as much as possible for anything that can happen during landing. When I was at my Apoapsis (which was as far away as moon) from Kerbal I used fuel to get my periapsis to only 50km, while Apoapsis remained the same. The trajectory was very "ellipse-like".

What happend? When I was aerobraking at 50km with more than 3000m/s speed, the apoapsis decreasing, while periapses remains almost the same. Then I was catapulted "back to the space", but with shorter trajectory around kerbin.

I did this multiple times and after some time, my speed at periapsis decreased to 2400m/s and after that, the apoapsis got as low that I stayed in atmosphere and landed.

The point is - I did not have to use as much fuel to get low orbit, I slow down a lot with repeating "slow a little with aerobreaking and then go again to space".

I am curious - why this is not used in reality? At least I did not hear about it. I am thinking that probably real materials do not take lightly "burn and freeze" multiple times...?

observation - Is there a way to tell what the surface of a planet is like?

Kepler-442b

I'm doing a project in which I need to find a planet within our galaxy that might be habitable. I found this planet that is within its stellar system's habitable zone, and due to research I have found that this planet is one of the closest in similarity to Earth, in terms of size, and temperature. But I don't really know anything about the surface or physical features.

I think there is an equation to find if there is life on another planet, but I cant find anything about water...

If you can help me, thank you so much.

I need to know the physical and chemical features on a planet's surface. I know the planet is within the habitable zone and is thought to contain water, but I need more detail, if there is any way to get the specific or at least highly probable features of the surface. Sorry I didn't make this clear enough.

telescope - Why aren't secondary mirrors offset to get rid of diffraction spikes due to the support vanes?

Yes that does seem like the best possible scope design. They've been built by hobbyist astronomers (even with truss tubes holding the secondary/eyepiece section and a "tube" made out of cloth to block stray light).

I'm not sure if they did over 4 times the primary mirror grinding just to use an off-axis circle-ellipse that's less than a quarter the area of the mirror. That sounds like a lot of work and waste to get rid of diffraction spikes. Imagine grinding a mirror about a yard wide just to make a 12-16" telescope. Although you could cut and sell the rest of the mirror or use it to build an imaging scope for each CCD or something and still make a small regular scope from the middle. Maybe if you're really careful it'd be less work to grind the side of a parabola without grinding the rest of it. Especially if they sell off-axis mirror blanks now so you don't have to buy the costlier huge mirror blank, cut it and then be burdened with a mirror blank that has a circle cut out of it. I don't know, I know almost nothing about telescope making.

Other hobbyist telescope makers have built 1-vane secondary mirror supports cause they prefer fewer and worse spikes; and curved secondary mirror vanes, which have every possible angle somewhere along it in equal amounts instead of just orthogonal or triangular one so they spread out the spikes into a less bothersome halo.

Why do black holes have jets and accretion disks?

When it comes to accretion disks, nothing is coming out of the black hole. That's just orbiting matter, though it is swirled around a bit by frame dragging. Even at high gravity, the ability to orbit around a massive body still exists. The gravitational force is already being "used up" to cause the orbiting (it accounts for the centripetal force), so there is no need for the gas to fall in.

As for jets, as far as I can tell there is no single explanation (I am not sure of this). One candidate explanation is the Blandford-Znajek process¹

The following image is from Black Holes and Time Warps: Einstein's Outrageous Legacy, by Kip S. Thorne:

Basically, most black holes rotate, and sometimes the intense rotation can cause forces that overcome gravity, even by a few orders of magnitude.

When a black hole spins, magnetic field lines anchored to it² spin along with it. Plasma (from the accretion disk) is then flung out along these lines, similar to what happens when you put a marble in a conical cup and rotate it. This is depicted in the first image.

In the second image, current passes through the field lines (I do not understand this one as well as the first, however this post has a reasonable explanation), accelerates plasma with a mechanism similar to an electromagnetic railgun. This is another way of creating jets.

Note that the energy here comes from the rotational energy of the BH, not the mass-energy of the "contents" of the BH (which is lost to the universe unless we consider Hawking radiation)

(I shall have a closer look at the paper when I have time and update the answer accordingly. Comments appreciated)

^{1. Blandford, R. D., & Znajek, R. L. (1977). Electromagnetic extraction of energy from Kerr black holes. Monthly Notices of the Royal Astronomical Society, 179, 433-456.}

^{2. While the no-hair theorem forbids a naked black hole from posessing magnetic field lines, one with an accretion disk may have them as the field lines cannot "escape" through the disk.}

imaging - Why are radar images of comets shaded only on one side?

NASA published a picture of a comet the other day. The image shows the comet being lit from above. See NASA's picture below.

However, since it is a radar image, I would have assumed to get a shading on all sides and grazing angles, like velvet or scanning electron microscopy. Or alternatively to have the sides facing the radar dish being shaded, and the edges being dim. For the velvety look, compare to this picture from Wikipedia:

So why is the comet's image shaded only from above?

telescope - Why do we use FITS format for scientific images especially in astronomy? How is it different from formats such as JPEG, PNG etc?

File formats tend to be industry/field-specific, with the format, tools, and expectations of the field coevolving to become more dependent on each other over time. JPEG co-evolved with amateur digital photography, PNG co-evolved with the web. Likewise, FITS co-evolved with astronomical data processing, and so is naturally more suited for that purpose than formats that had entirely different goals and communities involved in their development.

(Pedantic nit: JPEG is the name of the compression method and the group that designed it. The file format is technically "JFIF", but everybody colloquially calls it JPEG.)

As an image format, FITS has a number of desirable qualities that are lacking in JFIF and PNG, that are crucial for scientific data:

• Storage of more bits per pixel (CCDs can records 12, 14, or more bits), and also floating point values.


• Storage of arbitrary number of data channels (scientific data may have many, or other, frequency bands than the RGB that JPEG and PNG are limited to).


• No lossy compression as is typical (though not strictly required) for JPEG.


• Higher resolution (JPEG/JFIF, for example, is limited to 65,535 pixels in each direction), and FITS is also capable of storing 3D data volumes.


• Support for unlimited metadata in the header, for example the sky coordinates, information about the telescope, etc. JPEG and PNG don't have the fields and aren't set up to record arbitrary metadata.

Also, the use of FITS for astronomy predates the existence of both JPEG/JFIF and PNG -- FITS was standardized in 1981, JPEG in 1992, PNG in 1996. So even if those formats were suitable (which they are not), by the time they were invented there was already widespread use and existence of astronomical image processing tools that were geared to FITS files (and growing archives of astronomical data in FITS format), so it would have been a major effort to switch formats that would never be undertaken unless a new format not only met but exceeded all the prior requirements of the field.

cosmology - Difference in redshift between 2 interacting galaxies

I have a galaxy 'A', say at redshift 1. Let's suppose this galaxy has no peculiar velocity. What would be the redshift of another galaxy 'B', that has a radial velocity of 500 km/s, relative to A?

All this in the standard (Planck) model.

Any tools to do get this easily (an python code, astropy?)

Origin of the magnetic field of neutron stars

The strong magnetic fields in neutron stars are supposed to come from magnetic flux conservation. If we have:

$Phi_B = int B dS = const$

where $Phi_B$ is the magnetic field flux, $B$ is the magnetic field strength, and $dS$ is the elemental closed surface; then, this integral is constant through the surface.

If we consider the star surface over which take the integral, than

$S = 4pi R^2$

where $R$ is the star radius. This can be translated, altogether with the magnetic flux conservation law, as:

$B_f = B_i (frac{R_i}{R_f})^2$

where $i$ and $f$ are the indices for initial and final stages.
We know that the star implodes from a whatever star size to $sim10$ km. So the radii ratio is huge. You just need a starting magnetic field of $10-100$ G, to get a final magnetic field of the order of $10^{12}$ G, that is typical in neutron stars.

During night on the Moon is there Earth light and Earth phases?

Yes, there are Earth phases, viewing from the Moon. Full earths, half earths, quarter earths, waning and waxing earths. The easiest way to visualize this is, imagine the earth is still, one half of the Earth facing the sun, the other half away from the sun, so you have half the Earth is light, half is dark, now, imagine you're on the moon orbiting the Earth every 28 days. When you're over the sunny half of the Earth (night on your part of the Moon) the Earth is full. When you're over the dark side of the Earth (day on your part of the Moon), the Earth is new. As the moon takes 28 days to orbit the earth, like the moon in our sky, every 28 days would complete one cycle.

What's different is the Earth wouldn't move in the night sky. It would actually go back and forth a bit, but it would stay in the same general area, every day, every year, every century, because the Moon is tidally locked to the Earth, but apart from not moving, it would be similar to the Lunar cycles.

Another difference is that you could observe the Earth's rotation. Here's a pretty good video on what it would look like. 28 days squeezed into about 1 minute.

https://www.youtube.com/watch?v=-HgHEO0DUig

I would imagine the Earth looks quite bright from the point of view of the Moon, and I'd guess the pictures don't really do it justice, but that's just a guess. I've never seen it for myself. I'm also not sure it would be pitch black and not visible as a "New Earth" either. I remember reading that you can see stars from the moon even during the day, that's because there's no atmosphere to diffract the light so you could probably see the Earth even at new earth too. We can see the new moon from Earth sometimes, so I would think a "new earth" would be visible but dark.

Here's a discussion on being able to see the new moon. I would think, seeing a new earth from the moon would be even easier.

Short discussion: http://scienceline.ucsb.edu/getkey.php?key=26

Long discussion: http://physics.stackexchange.com/questions/1907/why-can-we-see-the-new-moon-at-night

the sun - Why there is no smoke around the Sun?

Fire is actually the rapid oxidation of a combustible material. Smoke is the airborne particulates and gases that result from the combustion, or from pyrolysis.

The sun is not undergoing an oxidation reaction, so it's not producing particulates that one might refer to as smoke.

The process the sun is undergoing is nuclear fusion, where hydrogen are combined and create helium. This reaction is very energetic and releases heat, visible radiation, and other radiation along a wide swath of the electromagnetic spectrum. The reaction is self-sustaining - as long as there is fuel, the emissions of fusion reactions cause nearby fuel to react as well. No oxygen or oxidizing agent is required.

Further, the resulting helium is comparatively heavy, and the sun being a huge mass keeps both the hydrogen that is the fuel and the helium that is the product nearby - they don't leave like hot smoke does from a fire.

So there is no smoke as we might consider it - just helium gas (plasma), and even if there were it would simply drop into the sun, there is no such thing as "rising" from it as one might consider smoke does on earth-borne fires.

You might find this music video provides further instruction on composition and reaction of the sun.

tidal forces - How can we tell that a short-period binary is tidally locked?

The phenomenon of tidal locking occurs primarily in short period binary systems with convective envelopes. The tides raise bulges in the stars and these interact with convective motions to dissipate energy and to synchronise the orbital and rotational periods of stars and circularise their orbits. The process is much less efficient in stars with radiative envelopes.

How can we tell observationally? Well, we have to show that the rotation period of a star is the same as its orbital period. The orbital period is easily found, either by measuring eclipses, in the case of binary systems with a high inclination, or by measuring the motions of the star(s) using the doppler shift in their spectra.

Measuring rotation periods is also not so difficult. Fast-rotating stars with convective envelopes have magnetic activity and starspots (analogous to sunspots, but covering a larger area). As the star rotates then the light from the star goes up and down depending on what starspots are on the visible surface. This modulation gives the rotation period. In eclipsing binaries it canbe difficult to separate out spots from eclipses, but here one can use the known orbital inclination and known radii of the stars to see whether their projected rotation velocities (measured fro the widths of spectral absorption lines) matches the predictions from a tidally locked orbit.

There are many known tidally locked bnaries. Any low-mass binary with an orbital period of less than a few days is almost certainly tidally locked, but the situation at orbital periods greater than about about 6-8 days becomes more complicated (e.g. Meibom et al. 2006). The plot below is taken from a review by Mazeh (2008), showing the rotation vs orbital periods from some small, but well-defined samples.

The absolutely classic, must-read, theoretical work on this is by Zahn (1989). The timescale for synchronisation depends on structural properties of the stars, their moments of inertia, the binary mass ratio and the ratio of their separation to their radii to the power of 6. It is this latter property that means tidal locking is nearly bimodal; binaries with periods shorter than some threshold (about 6 days) are usually tidally locked, but those with slightly longer periods almost never are.

gravity - Is a black hole a perfect sphere?

As I understand it, general relativity says that there's gravitation everywhere in the universe, and this gravitation creates dips in space, so to speak, often represented in 2D as a weight on a rubber sheet, like the picture below.

Source,

so, a black hole might generate a perfectly spherical event horizon, but it generates it on a not perfectly flat 3 dimensional surface, and I think the gravitation from other objects makes it not quite a perfect sphere. For example, a star or planet that orbits a black hole would drag around a ripple on the event horizon as it orbits the black hole.

Precisely what shape that ripple would be . . . I'm not sure. That said, if you had a black hole as the only object in a universe, then I think the event horizon might be a perfect sphere, or as perfect as possible, given quantum fluctuation, hawking radiation and the impossibility of precise observation and all that good stuff.

Non black holes tend to have lumpy/inconsistent gravity - see here. Even Neutron stars have some inconsistency, but Black holes probably avoid that because the matter is condensed to a point, so there's equal gravitational pull from all directions from the singularity.

Now a Kerr black hole, that's a whole different question. I'm not smart enough to try to answer that one. That might not be a sphere at all.

observational astronomy - About bias in sidereal time (used by astronomers)

I came up on this stanza in various websites when looking up on experimenter bias. It says

"If the signal being measured is actually smaller than the rounding error and the data are over-averaged, a positive result for the measurement can be found in the data where none exists (i.e. a more precise experimental apparatus would conclusively show no such signal). If an experiment is searching for a sidereal variation of some measurement, and if the measurement is rounded-off by a human who knows the sidereal time of the measurement, and if hundreds of measurements are averaged to extract a "signal" which is smaller than the apparatus' actual resolution, then it should be clear that this "signal" can come from the non-random round-off, and not from the apparatus itself. In such cases a single-blind experimental protocol is required; if the human observer does not know the sidereal time of the measurements, then even though the round-off is non-random it cannot introduce a spurious sidereal variation."

I understand that sidereal time is something related to the time measurement used by astronomers as a time-keeping system, but I don't understand by "someone who knows the sidereal time of measurement" can then unconsciously influence the results, because say if you saw the clock as 00.56 s and you rounded it to 00.6s, I don't think it will have any effect. But moreover, by rounding that, how can you tell it will eventually have effect on your results?

Please advise.

Sorry for any wrong tags. I'm new here, so still learning.

density - What is the most dense object in the universe?

Let us define this as the largest observable density of a stable object, in order to exclude black holes which may have a very large (infinite) density at their centers or objects collapsing towards a black hole status.

If we restrict the definition in this way, then the answer should be the core of the most massive neutron star that we know about.

At present there are a couple of neutron stars with mass of about $2M_{odot}$ (Demorest et al. 2010; Antoniadis et al. 2013. Depending on the exact composition and equation of state at their centres these should have densities of around $2 times 10^{18}$ kg/m$^{3}$ at their centers and average densities of $sim 10^{18}$ kg/m$^3$.

Note that these densities are around 3 times the density of a proton or neutron or 5-10 times the density of nuclei at zero pressure.

In principle, the density of a single electron is much higher.

Could human life thrive on a planet in a pulsar star system?

Only five pulsar planets have been confirmed or have garnered enough evidence to make a strong case for their existence. None of them are like the terrestrial planets in the Solar System insofar as the way they formed and their orbital movements.

• PSR B1620-26 b. This planet is in a circumbinary orbit around a pulsar and a white dwarf, with a semi-major axis of ~23 AU. It is theorized that it formed around a Sun-like star, which became a binary star system after encountering a neutron star (another partner must have been ejected from the system, as is the case with triple-star encounters). The Sun-like star promptly became a white dwarf, and the planet's orbit would have widened after the encounter with the neutron star, which is a pulsar.

  
  
  

  Basically, this planet did not form around the pulsar. The system was captured so to speak, and so neither the Sun-like star/white dwarf or planet would have been affected by the supernova. Additionally, the planet is fairly far away from the two stars in the system, meaning that effects from the pulsar are even less negligible.

  
  
  

  The point of this is that life on this planet is possible - especially given the planet's old age - but it would be hard for it to survive. If life formed before the encounter with the original binary system, it would have been disrupted a bit by the orbital changes, reducing the amount of starlight it received. Today, it would receive little light, as the white dwarf has a low luminosity. There wouldn't be enough energy to sustain life - as is the case for most pulsar planets.

  
  
  

  Also, as Rob Jeffries pointed out, this planet is a gas giant, not a terrestrial planet, meaning that human-like life could not develop on it and any arriving humans bent on colonization would not be able to live on it normally.




• The planets of PSR B1257+12. Three planets orbit this pulsar, at distances of about 0.19, 0.36, and 0.48 AU - fairly close to the pulsar and within range of a decent amount of harmful radiation, relative to PSR B1629-26 b. The major reason that these planets are not a good place for life is that they are thought to have formed after the supernova that led to the evolution of the star to a pulsar, from remnants of debris from the supernova. The star itself is quite young - ~3 billion years - and the supernova would have happened relatively soon thereafter. The planets themselves are less than 1 billion years old, meaning that life has not had time to develop.

  
  
  

  My issue with the age of the planets isn't based on the short time period for life to develop so much as on the idea that conditions might be hellish, comparable to those experienced in the Hadean eon. Geologically, these planets would have had different starts than Earth did, and they most likely would not have encountered the problems of the Late Heavy Bombardment, but life might not have a good chance at starting at all for another 500 million years.



• PSR J1719-1438 b. This planet is unusual because it is thought to be the remnant of the companion star - severely damaged by the supernova. It is also composed of exotic materials, for a planet - crystallized carbon. The conditions here are not conducive to life.

Looking at this list we see three distinct types of pulsar planets emerge:

• Captured planets


• Newly formed planets


• Former companion stars

The latter two are not good places for life, in part because they may not be composed or compounds life can use. The first group may be the best chance, as their formation could have been relatively normal. The problem lies in the fact that they might have been rogue planets, floating along without good sources of energy from stars.

All of this isn't even taking into account the radiation from the pulsar itself. The only escape is to have an orbit far from the pulsar, as with PSR B1620-26b, but then there's no outside source of energy. The best chance would be to have the planet in a circumbinary orbit around the pulsar and a companion star, as in this planet's case. At least then there would be some helpful light.

solar system - What is space temperature around Earth?

Assume you have a spherical blackbody.

The solar flux at the radius of the Earth is given to a good approximation by $L/4pi d^2$, where $d = 1$ au. This is $f=1367.5$ W/m$^2$ (though note the distance between the Earth and the Sun has an average of 1 au).

If it is a blackbody sphere it absorbs all radiation incident upon it. Assuming this is just the radiation from the Sun (starlight being negligible), then an easy bit of integration in spherical polar coordinates tells us that the body absorbs $pi r^2 f$ W, where $r$ is its radius.

If it is then able to reach thermal equilibrium and it entire surface is at the same temperature, then it will re-radiate all this absorbed power. Hence
$$pi r^2 f = 4 pi r^2 sigma T^4,$$
where $T$ is the "blackbody equilibrium temperature". Hence
$$ T = left( frac{f}{4sigma}right)^{1/4} = 278.6 K$$

telescope - Can angular information be known more precisely than the diffraction limit?

The diffraction limit deals with the ability to determine if two things are separate. I am interested in the ability to find the the centroid of a single object.

Imagine a star with no near neighbors would it be possible to determine the centroid of that star at a resolution higher than the diffraction limit?

The star is essentially a point source and I wonder if looking at the edges of the airy disk would allow a centroid to be determined if I had a magical camera with infinite pixels.

It seems to me that I should be able to see the disk and then calculate a centroid that is smaller than the airy disk.

Thanks for any help you can provide =)

galaxy - How can ionized emission line flux decrease as a function of increasing metallicity or abundance?

Metallicity and abundance

Metallicity

Without specifying a given metal, the term "metallicity" — abbreviated $Z$ — usually refers to the total metallicity of all elements, i.e. the mass fraction of all metals to the total mass of some ensemble of elements, e.g. a star, a cloud of gas, a galaxy, etc. (as usual, the term "metal" refers to all elements that are not hydrogen or helium). For instance, the mass of all metals in the Sun, divided by the Sun's mass, is 0.02:
$$
Z_odot equiv frac{M_mathrm{C} + M_mathrm{N} + M_mathrm{O} + ldots}{M_odot} = 0.02.
$$

Sometimes we talks about the metallicity of a given element, e.g. oxygen. The mass fraction of oxygen in the Sun is 0.005 (i.e. oxygen comprises 1/4 of all metals by mass), so we say $Z_mathrm{O} = 0.005$.

Unfortunately it is not uncommon to implicitly talk about the metallicity of an object, divided by Solar metallicity, such that a galaxy which has one-tenth of Sun's metallicity is said to have $Z=0.1$, rather than $Z=0.002$.

Abundance

The term "abundance" is only used for a single element. It basically expresses the same thing as metallicity, and is often used interchangeably, but is expressed in terms of the number $N$ of element nuclei, and as the ratio not to all nuclei but to hydrogen nuclei. For wacky historical reasons, we also take the logarithm and add a factor of 12. Taking again oxygen as an example, the mass fraction of 0.005 corresponds to a nuclei fraction of roughly $5times10^{-4}$, so we say that the abundance of oxygen is (e.g. Grevesse (2009))
$$
A(mathrm{O}) equiv log left( frac{N_mathrm{O}}{N_mathrm{H}} right) + 12 = 8.7.
$$

Metallicity of a given species vs. total metallicity

In general, the ratio of a given element to all metals is roughly constant. That is, various elements are produced by stars approximately by the same amount. But various processes may cause elements to exist in various forms. For instance, metals deplete to dust, but some elements tend not to form dust, e.g. Zn. For this reason, Zn is a better proxy of the total metallicity than, e.g. Mg, since half of the Mg may be locked up in dust.

Metals increases cooling

Elements also appear in various excitation states, which depend on various processes. The lines you mention, [O II] and [O III], arise from collisionally ionized oxygen, which subsequently recombines (in my first answer I wrote, wrongly, that it was excited), and thus depend on the temperature of the gas. As the metallicity of the gas in a galaxy increases, the ratio of the intensity of these lines to that of hydrogen lines (e.g. H$beta$) first increases, as expected. However, the increased metallicity also allows the gas to cool more efficiently. The reason is that metals have many levels through which electron can
"cascade" down.
If the electron recombines to the level where it was before, a photon of the same energy will be emitted, which itself may radiatively ionize another atom. But the many levels in metals makes de-excitation to intermediate level more probable, such that the electron cascades down, emitting several low-energy (infrared) photons, which are incapable of ionizing atoms and thus escape. The result is that energy leaves the system, i.e. the system is cooled.

This in turn means that, above a certain metallicity threshold — which is specific to a given species — the abundance of the collisionally excited lines begin to decrease. The following figure is taken from Stasińska (2002), and shows the turnover for the two oxygen lines:

This means that measuring the metallicity of a single species in general gives two solutions for the total metallicity. Luckily, as the turnover is different for different elements, measuring the metallicity for several species can constrain the total metallicity.

Gaia: What is the difference between CCDs used for astrometry, photometry, and spectroscopy?

CCDs are optimized for a certain wavelength range, and for a certain expected signal level. In astronomy, we tend to be short of light, so here we almost always want them to be as sensitive as possible (an exception may be observations of the Sun, which I don't know much about). But for instance, the Nordic Optical Telescope has a CCD which is optimized for blue wavelengths, but has quite a lot fringing in the near-infrared. And further out in the IR, CCDs aren't even used, instead using something which are just called "detectors".

However, whether the CCD is used for imaging (photometry and astrometry) or spectroscopy does not have anything to do with the CCD; it's just a matter of inserting a grism or not. I'm not really into the instruments of Gaia, but I assume that differences in the CCDs are due to different wavelength regions being probed. There may be a difference in how its sub-parts (it's actually an array of CCDs) are positioned (for instance, for spectroscopy in principle you don't need a large field of view, but can do with a long array rather than a more square one), but the design of the individual CCDs are the same.

gravity - Do asteroids have a gravitational field?

You asked two questions.

Do asteroids have a gravitational field.

Of course. Even a microscopic grain of dust has a gravitational field.

Do they gravitationally attract each other to form planets?

Not any more. During the formation of the solar system, asteroid-like and comet-like objects collided to build larger objects, which in turn collided to form even larger objects, and so on, eventually building the cores of giant planets and later, the terrestrial planets. But that stage ended long ago, shortly after the solar system formed.

Asteroids do of course gravitationally attract other objects, but this attraction is so weak due to the small masses of asteroids that it is easily overwhelmed by other perturbing forces. The vast majority of the asteroids lie between Mars and Jupiter, and Jupiter is the primary culprit in explaining why no planet exists in that gap.

When two astronomical bodies collide, one of the outcomes is a purely inelastic collision that makes two bodies form a single body. This only happens with a rather mild collision. A more energetic collision will result in some mass being expelled. An even more energetic collision will result in lots of mass being expelled; the colliding bodies become many smaller bodies. With a few exceptions, the latter is what is what is happening amongst the asteroids today, and for the last four-plus billion years or so.

Jupiter is such a huge perturbing body that collisions in the asteroid belt are generally very energetic. Instead of forming ever larger bodies, the asteroid belt is gradually being broken up into smaller and smaller bodies. Some of these collisional bodies are ejected from the solar system thanks to interactions with Jupiter. The smallest results of these collisions migrates sunward thanks to the Poynting-Robertson effect.

Measuring star distance by parallax using a small telescope

Short answer - not really, parallax for the closest stars is right on the limits of resolution for a good ground-based amateur equipment.

The nearest star would show a parallax angle of under an arcsecond. (The Parsec even takes its definition based on one arcsecond of parallax.)

Ground based amateur observations are probably limited to an arcsecond at most, so it's probably not possible to measure them.

A special (and totally fictional) case - if a nearby star like Alpha Centauri were exactly between us and a much more distant star, and it were possible to see them both without Centauri overpowering that distant star, then maybe it would be possible to observe them as a close double. That's not true in practice however.

Update:

Prompted by Rob's much better answer, the earliest parallax determination I've been able to find was using

the 6.2-inch (157.5 mm) aperture Fraunhofer heliometer at Königsberg

(Source: Heliometer article on Wikipedia.)

In purely aperture terms, that's an amateur-sized instrument (with some admittedly very precise measuring gear built in.) So my suggestion now is to ignore my initially doubtful response and try measurements, perhaps using a camera at fairly high magnification as Rob suggests...

rotation - Is axial tilt critical for life?

I agree with David Hammen. Hyperphysics is mostly a very good site but they dropped the ball on that page IMHO. Hope you don't mind a partially speculative answer, but here goes:

Why does it matter if there are some areas of a planet with extreme
temperatures, as long as there are other spots on the planet that are
not extreme?

It shouldn't matter if part of the planet is uninhabitable. There are deserts on Earth which are all but uninhabitable but that doesn't effect life elsewhere. Prior to 5.3 million years ago the Mediterranean sea evaporated and that entire region could have had a salty basin and been hugely hot, but I've not read of life on Earth having any problem with that. Source

Why are "moderate" seasons required for life to exist?

There's a boatload we don't know about the evolution of life on other planets but seems perhaps universally true that life adapts, so I find it difficult to believe that moderate seasons are necessary. Very extreme changes could be difficult, but change can force adaptation.

If humans can live at the equator on Earth where there is the least
amount of tilt, why would an exoplanet with less tilt or no tilt be
necessarily non-inhabitable?

The tilt is planet wide but the lowest variation happens near the equator, but animals that thrive near the poles adapt by hibernation or migration and smaller stuff can be frozen and then come back to life, so, I don't agree with the article on this point.

Even if humans could not live on a planet without axial tilt, are
there no other forms of known "advanced" life that can? We know that
extremophiles exist, such as tardigrades' ability to survive in the
vacuum of space. What is the most "advanced life" that could live on a
planet without axial tilt?

One of the interesting historical facts of life on Earth, at least to me, is how long it took what we might consider advanced life to develop. One celled life in various forms was around for over 3 billion years but the first fossils are about 650 million years old. It took life a very long time on earth to get from too small to see to large enough to leave a footprint . . . but, I digress.

I agree 100%, one celled life or Tardegrades could live on a planet with no tilt or 90 degree tilt. Easy. Ocean life in general should be fine cause oceans are more adaptive. Evaporation keeps ocean surfaces colder than land gets during peak heat and while a completely frozen over ocean isn't great for life, cold oceans hold more oxygen and CO2 which can be good for life. Oceans also circulate as an effective means of temperature moderation and fish don't really care how windy it is or how much or little it rains. The tilt question, I think, is really just about life on land.

Land life could be more vulnerable to high wind, extreme temperature shifts, droughts or floods, which could be driven by greater axial tilt, but I find it hard to believe that Axial Tilt is the be-all and end all. Day length and year length are key factors too.

One point I agree with the article on, is that a close to 90 degree tilt might not be ideal with one part of the planet always facing the sun and the other part never facing it but outside of extreme tilts, I don't see why it would be a big deal.

A thick cloud cover, for example, reduces seasonal changes. There's a number of factors.

galaxy - How probably is it that galaxies will extinguish?

Galaxies are gradually being extinguished. Most star formation activity occurs near the start of a galaxy's life, or in response to merger activity with other galaxies.

The star formation rate of the universe peaked at redshifts of around 3, corresponding to a look-back time of around 9 billion years. Since then the star formation rate has declined as the universe expands; mergers are less frequent, gas is driven out of galaxies by supernovae and active galactic nuclei.

However, most of the stars that have been formed are of lower mass (K- and M-dwarfs) than the Sun and will live on for tens or hundreds of billions of years. They are however much fainter than the Sun. So although high-mass, luminous O- and B stars live their short lives and are not replaced at the same rate, the low-mass stars continue to shine. This means that galaxies will get fainter on average as the remaining stellar populations increase in average age and decrease in average mass.

It is a slow process though. High mass stars are still being formed in our galaxy after 12 billion years, and most spiral galaxies have ongoing star formation. However, star formation has more-or-less ceased in gas-poor elliptical galaxies.

solar system - What are gravitational waves actually?

It's a "ripple" in spacetime. Imagine the traditional picture of planets as marbles sitting on a sheet--heavier bodies push the sheet down farther and deeper. If you place a bowling-ball on one end of the sheet, the entire sheet is affected, but not instantly. It takes an amount of time for the sheet to be moved--this movement can be seen as a wave, with a wavefront, a speed, etc.

In reality, massive gravitational bodies don't magically appear in space, (that we know of... maybe God is planning to drop a giant marble in our Solar System) but gravitational waves are still generated by any change in placement of mass or any acceleration (Einstein's theories that show that gravity and acceleration are indistinguishable).

So you get gravitational waves caused by movements of planets around the sun, etc. That's going to be relatively smooth though, so hard to detect. What we really need is something that creates sharp waves--two giant black holes which are orbiting each other at an incredibly fast RPM is perfect since they're massive and will create a lot of high/low waves as they rotate that we can then detect.

As we improve the detection technology, we may be able to use gravitational waves to detect all sorts of things besides black holes, it just depends on how much we can measure. The nice thing is that you don't need "line of sight" to measure with a gravity wave. So if God drops a giant marble on the other side of the Sun, we could detect it even though we couldn't see it.

How to tell a pulsar is rotation-powered or accretion-powered?

The spin behaviour of the two types of pulsar would usually be quite different. The $dot{P}$ for a rotationally powered pulsar is always positive, and higher order time derivatives of $P$ are quite small. This is because the rotational kinetic energy is powering the pulsar emission and the neutron star continually spins down as it loses rotational energy.

Accretion powered pulsars can have very variable spin down or spin up characteristics, because they are powered by mass transfer and accretion in binary systems, and are influenced by a variety of factors affecting the accretion flow and how it couples to the pulsar magnetic field. The rates of spin down can be much higher than can be plausibly be accounted for by magnetic spin down as in a rotationally powered pulsar. Spin up cannot be accounted for in a rotationally powered model and neither can sign reversals or dramatic variability in $dot{P}$.

Additionally, an accretion-powered pulsar would necessarily need to be in a short period binary system - therefore there would be a very obvious periodic modulation in the pulse period caused by the doppler shift as it travels in ts orbit.

habitable zone - Better than Earth habitability

Note: I am self-answering my own question in hope that someone post another answer that beats this one.

---

Earth is near the inner edge of the Sun's habitable zone. And since the Sun is expected to grow and increase it luminance, Earth might be unhihabitable for any life somewhere between 1 or 3 billion years in the future. So, a planet that have longer time to develop before its parent star moves it beyond the inner edge of the habitable zone is more favorable than Earth.

Since Earth itself always was inside the habitable zone since it formed 4.6 billion years ago and will still be for lets say more 1.4 billion years, with a large uncertanity factor, this gives roughly 6 billion years of time for complex life to develop. Given that it is unlikely to form early due to the time needed for evolution to take place and due to an elevated level of large bollides collisions, we could discount the first 2 billion years from any life bearing planet, including Earth, as unlikely to develop complex life. Further, it is unlikely that complex life would finally evolve out from simpler forms when the planet is already overheated and already crossing the inner edge of the habitability zone, so lets take out the finishing 10% of that period for any planet (probably something more than 10%, but lets keep this as a conservative estimative). So, for Earth, this gives a window of a size of 3.4 billions years to complex life evolve. Similar planets with larger windows have better probabilities.

Stars larger and more luminous than the Sun tends to be more unstable and live shorter. As a result, it is expected that planets around stars larger than the Sun has less time to develop complex life, and thus a shorter time-window. On the other hand, this means that stars smaller and less luminous than the Sun gives a larger time-window to the planets to develop life.

For stars smaller than the Sun (a G-type yellow star), we could consider the K-types (aka, orange dwarf) and the M-types (aka, red dwarf) as specially favorable.
An orange dwarf star may live for 10 to 30 billions years in the main sequence. A red dwarf star may live in the main sequence for trillions of years.

However, planets in the habitable zone of red dwarfs are likely to become tidally lock, and we don't know if this is really that bad or not for life biodiversity. Lets assume that this is really bad, so a planet orbiting an orange dwarf in the habitable zone is likely to have a better habitability than Earth.

Accordingly to this, a planet with two times the mass of the Earth, will have stronger gravity, and thus it is likely to be flatter. Further, it is likely to have a ticker atmosphere that would protect the surface from UV radiation better than Earth. It would be geologically active for a longer time, resulting in more carbon cycling. With the right quantity of water (not a desert nor a global very deep ocean), it might be an archipelago world, since its flatness would not allow the ocean to be very deep nor the continents to be very large. As a result, life would flourish in a number of rich biologically favourable environments significantly larger than Earth. Further it magnetic field is likely to be stronger than Earth's one, protecting the surface from cosmic rays.

As a result, a planet with two Earth masses orbiting an orange dwarf star in the habitable zone has a good chance to be more habitable to life than Earth itself.

Needless to say, near-circular orbits are more favourable than excentric ones, since excentric orbits may make the planet enter in periods of freezing or boiling. However, a reasonably excentricity that periodically changes the environment in a significant manner, but not as too much that it would extinguish non-extremophile life, might give to the planet life a selective pressure needed for developing rapid evolution to face the always changing climate.

supernova - Why does the Chandrasekhar limit affect white dwarfs differently?

Whether a white dwarf responds to the accretion of material by exploding or collapsing depends on the competition between energy being released in fusion reactions and energy being locked away by endothermic electron capture (neutronisation) reactions.

It is thought that most white dwarfs of moderate mass have a C/O composition. They will need to accrete a lot of mass to get to a density (at about $4times 10^{13}$ kg/m$^3$, reached at $1.38M_{odot}$ in a non-rotating WD) where neutronisation becomes energetically feasible. It is possible, that before this happens, that fusion reactions are ignited (due to high density, rather than temperature). The threshold density for ignition is lower for nuclei with lower atomic number (He < C < O) and the ignition threshold densities for He and C are probably lower than the neutronisation threshold for C.

In a C/O WD that has accreted a lot of matter, ignition could take place in C at the core, or it could be triggered in He (at even lower densities) at the base of a deep accreted shell of material. The electron degenerate matter has a pressure that is independent of temperature, leading to runaway fusion and the complete destruction of the star.

O/Ne/Mg WDs are made as the final stages of more massive stars ($8-10M_{odot}$) and are born as remnants with much higher mass $>1.2M_{odot}$ than typical C/O WDs. More massive WDs are smaller, with higher density. The neutronisation thresholds for O, Ne, Mg are only $1.9times10^{13}$, $6times 10^{12}$ and $3times 10^{12}$ kg/m$^3$ respectively (all lower than for C). This means that a O/Ne/Mg WD may have to accrete very little mass to reach this central density, begin neutronisation, which leads to collapse. In addition if such densities are insufficient to trigger C burning in a C/O WD, then they certainly won't be high enough to trigger burning in O/Ne/Mg because of stronger coulomb repulsion. Further, if little mass is accreted, then there won't be a deep envelope of accreted material in which to ignite burning off-centre.

For all these reasons, O/Ne/Mg WDs may be more likely to collapse than explode (the collapse would cause a type of core-collapse supernova though).

EDIT: Actually looking at the paper you reference (which is a bit dated), although some of the numbers have changed slightly, the semi-quantitative argument I give above is exactly how it is explained there. So I'm not sure whether my answer helps you.

general relativity - Would a Einstein–Rosen bridge change size and/or position in an expanding Universe?

The latter is closest to the truth, although I wouldn't use the phrasing "stretch". The "mouths" of the wormhole are (more or less) fixed in comoving coordinates (i.e. the coordinate system that expands with the Universe, and in which galaxies lie approximately still). But the bridge is sort of outside our three-dimensional space, and doesn't necessarily follow the same expansion.

If the mouths retained a fixed physical distance, they would accelerate in comoving coordinates beyond bound, eventually moving through space at superluminal velocities, which is forbidden. For example, consider the Milky Way and the galaxy GN-z11 which today lies at a distance of 32 billion lightyears from us. Roughly 13.3 billion years ago, the size of the Universe was 1/10 of today's value; that is $a=0.1$, and the distance to GN-z11 was only 3.2 billion lightyears. If at that time you created a wormhole$^dagger$ bridging MW and GN-z11, and if the distance between the mouths were fixed in physical coordinates, then today, at least one of the mouths would have moved so far outside its host galaxy that it must have traveled superluminally.

---

$^dagger$_{Creating a wormhole 13.3 billion years ago is left as an exercise for the reader.}

Telescope choice - Astronomy

Which one is better?
Features:

1)Magnification: 18x - 90x
Eyepiece Diameter: 5mm (18x), 10mm (60x)
Objective Diameter: 50mm
Focal Length: 36cm
Tripod Height: 38cm
Main tube Color: Silver
Net Weight: 1.78kg
Case Size: 43cm(L) x 10cm(W) x 26cm(H)

2)Optical Design: Refractor Mount Type: Altazimuth Ideal Usage: Astronomical and land observation Focal Length: 500 mm Aperture: 40 mm Focal Ratio: 13 Highest Useful Magnification: 94x Finderscope: 2 x 20 Lens Coating: Fully coated Light Gathering Power: 33x Limiting Stellar Magnitude: 10.5

3) 2x Barlow Lens and Enhanced Stability The Celestron Powerseeker 40AZ black comes with two extra eyepieces, one which is 20 mm and magnifies up to 25x and the other one is 8 mm which magnifies up to 63x. It also includes a 2x Barlow lens that almost doubles the magnifying capacity.

4)Meade NG60-SM Altazimuth Refractor Telescope Meade's value priced NG60-SM Altazimuth Refractor is an affordable entry level telescope that features an easy to use Altazimuth mount with slow motion controls for precise tracking. The complete package includes a sturdy metal tripod, a red dot viewfinder, two 1.25 inch eyepieces and a star diagonal, and a software DVD with instructional video. The NG60-SM Refractor Telescope comes disassembled in a compact box, but the instructional DVD video guides you through all the steps required for assembly. Go ahead and try it out in the daytime, that's the best time to align the red-dot finder scope while looking at a distant tree or telephone pole. The optics of Meade's NG60-SM produce an image that is right side up but the diagonal mirror reverses the image left-to-right. That's no problem most of the time, but an optional correct image diagonal is available. The low power 25mm eyepiece produces a magnification 28X which is just right for spotting the Moon or the planets, while the 9mm eyepiece (78X magnification) can be used to zoom in for more detail. The MH25 eyepiece at 28X shows a lovely view of the Lunar disk in a dark sky, while the MH9 eyepiece at 78X shows literally hundreds of craters on the Moon and begins to show the rings of Saturn and the cloud bands of Jupiter. For even better views it's easy to add better eyepieces. My best view of Saturn, for example, came with an optional 6.4mm Super Plossl eyepiece, about 110X magnification. The Altazimuth mount included with Meade's NG60-SM is lighter and easier to use than an Equatorial mount, yet the slow motion knobs make it easy to keep objects in view. The rotation of the Earth causes the Moon or planets to appear to drift out of the eyepiece, but the slow motion knobs make it easy to keep the telescope on target, even with a high power eyepiece. The telescope is sensitive to vibration, however, so a high power eyepiece can be difficult to focus.

So which 1 is best?????

I wanna see Saturn its rings,Jupiter and its moons,mars,venus, basically planets and their moons

planet - Years, Months, Day, and Weeks?

The synodic period of the moon is $29.53$ days, a little shorter than a calendar month, which is on average about $30.4$ days. This is slightly longer than its orbital period, but corresponds to the periodic visual appearance of the moon as viewed from Earth. I mention this to make it clear that we should be forgiving of a little imprecision.

Conventionally, the moon's appearance is divided into four phases: first quarter, full, last quarter, new. That means that on average, each phase lasts about $7.4$ days. Since calendars count days in integer amounts, a $7$-day period seems to be a natural choice.

The social importance of the seven-day period in Western cultural probably has much more to do with its religious significance in Abrahamic religions than astronomy per se (although certainly not unique to it). But its ultimate origin probably does lie in the natural division of the moon's appearance into four phases, which correspond to an apparent geocentric celestial latitude difference between the Moon and Sun of $0^circ$, $90^circ$, $180^circ$, and $270^circ$.

That, the explicit answer to your question is

• 1 week = 7 days = one lunar phase.

Trying to understand the way Saturn's ring look in this famous Cassini image

What's going on with the distortion of the rings on the upper half when they (presumably) cross in front of Saturn?

The brownish areas you see on Saturn are ring light, analogous to seeing the Earth by moonlight. Saturn's rings light up Saturn's night sky, particularly just after sunrise and just before sunset. The two dark bands across the face of Saturn are the A ring (upper dark band) and B ring (lower dark band). Of all of Saturn's rings, these two are the most opaque. The parts of Saturn underneath those two dark bands are brilliantly lit by ring light, but Cassini can't see those parts of Saturn because the opaque A and B rings block that light.

For a much better and more thorough explanation of all of the features in this incredible image, I suggestion you spend eleven minutes watching this youTube video by Emily Lakdawalla.

---

^{_{This is one of the very few times I will post a link to a youTube video. It is worth every eleven minutes.}}

Do the terrestrial planets form later than gas giants in our solar system?

The current ideas are that both terrestrial planets and giant planets start their formation in a similar manner. Dust settles towards the mid-plane of a predominantly gaseous disk, starts to stick together and eventually small (km-sized) planetesimals are formed. This process may be quicker in the outer parts of the solar systems where the gas is colder and the condensed material (ices) is probably "stickier". The planetesimals then interact with each other gravitationally and can grow by merger. The rate of growth is controlled by their spatial densities and their relative velocities, but is thought to occur quite quickly in both the inner and outer parts of the solar system ($sim$ 1-3 million years, e.g. Righter & O'Brien 2011).

Thereafter, the inner and outer parts of the solar system differ. The gas is cold enough in the outer parts of the solar system to accrete onto rocky cores and build giant planets on timescales of another few million years.
In the inner solar system the gas is too hot to be accreted and instread the next tens of millions of years are characterised by high velocity collisions between planetary embryos and planetesimals. This indeed may be sculpted and influenced by the early migration inwards of Jupiter to about 1.5 au, followed by migration outwards - the so-called "Grand Tack model" (Raymond & Morbidelli 2014).

Overall it probably takes the inner solar system and terrestrial planets of order 100 million years to settle down into its final configuration. The collision that formed the moon may have been several tens of millions of years after the formation of the Sun and certainly long after the giant planets formed.

black hole - What are the biggest problems about the numerical, finite-element GR models?

If you can provide examples of numerical methods in GR you've seen/heard of that would help focus the question.

From the article you linked to: "The technique keeps track of a vast number of quarks and gluons by describing the space and time inside a proton with a set of points that make up a 4D lattice". This almost gets to the main issue with Numerical Relativity. There is no natural computational grid on which to simulate space-time. The whole game with GR is that gravity is space-time so first you have to simulate the space-time and then you have to simulate the objects (neutron stars, black holes, gravitational waves) on top.

As the links below go into, its very difficult to create a consistent computational grid since the physical space-time your trying to simulate for a black hole has "funny" things in it like singularities, or an event horizon pas which we can't really know what's going on.

I think this article: http://astronomy.com/magazine/2016/02/putting-einstein-to-the-test?page=1

does a good job of summing up the field, and its quite accessible.

For something more rigorous please see: http://arxiv.org/pdf/1010.5260v2.pdf
That paper gets into some of the math behind the article linked to above.

co.combinatorics - Which lattices have more than one minimal periodic coloring?

The lattice $mathbb{Z}^n$ has an essentially unique (up to permutation) minimal periodic coloring for all $n$, namely the "checkerboard" 2-coloring. Here a coloring of a lattice $L$ is a coloring of the graph $G = (V,E)$ with $V = L$ and $(x,y) in E$ if $x$ and $y$ differ by a reduced basis element. (NB. I am not quite sure that this graph is the proper one to consider in general, so comments on this would also be nice.)

The root lattice $A_n$ has many minimal periodic colorings if $n+1$ is not prime (I have sketched this here, and some motivation is in the last post in that series); if $n+1$ is prime, then it has essentially one $n+1$-coloring. Two minimal periodic colorings for $A_3$ are shown below (for convenience, compare the tops of the figures):

The generic ("cyclic") coloring.

A nontrivial example.

The lattices $D_n$ are also trivially 2-colored.

So: are there other lattices that admit more than one minimal periodic coloring? I'd be especially interested to know if $E_8$ or the Leech lattice do.

(A related question: does every minimal periodic coloring of $A_n$ arise from a group of order $n+1$?)

lo.logic - Model of ZF + $neg$C in which Solovay's Theorem on stationary sets fails?

It is a theorem of Solovay that any stationary subset of a regular cardinal, $kappa$ can be decomposed into a disjoint union of $kappa$ many disjoint stationary sets. As far as I know, the proof requires the axiom of choice. But is there some way to get a model, for instance a canonical inner model, in which ZF + $neg $C holds and Solovay's Theorem fails?

I am interested in this problem because Solovay's theorem can be used to prove the Kunen inconsistency, that is, that there is no elementary embedding j:V -->V, where j is allowed to be any class, under GBC. The Kunen inconsistency may be viewed as an upper bound on the hierarchy of large cardinals. Without choice, no one has yet proven the Kunen inconsistency (although it can be proven without choice if we restrict ourselves to definable j). So if there is hope of proving Solovay's Theorem without choice, we could use this to prove the Kunen inconsistency without choice.

solar system - Ninth planet - what else could it be?

The introduction of the paper mentions some alternative interpretations put forward in earlier papers. It seems as if this "problem" has been noted earlier. This new solution to the problem beats earlier solutions. But maybe it is an evolving discovery process which again will invent new better explanations?

In addition to those mentioned in the paper, I suggest the following candidate alternatives:

-) Too few observations in order to hold up in a soon dramatically increased discovery rate. 4 or 5 or maybe even 8 TNOs which, as they claim, give a 0.007% significance level is certainly resting on some assumptions which will be challenged if thousands of similar objects are found in the next decade. So, as Andy has answered: possibly a coincidence due to small sample. (The paper's 0.007% probability is, well, we'll see)

-) Weird unthought-of science bias which, in math more than in lens, tends to select inadvertently among the (candidate) observations made and their characteristics. Since the authors are at the top in their fields, and Michael Brown has discovered loads of distant objects, including Sedna, because of that maybe one could suspect that some one thought has created a bias somehow.

fundamental astronomy - Calculating azimuth from equatorial coordinates

I try to get the azimuth of an object from its equatorial coordinates using this formula:

$a = arctan2(sin(θ - α), sin φ * cos(θ - α) - cos φ * tan δ)$

Where
φ = geographic latitude of the observer (here: 0°)
θ = sidereal time (here: 0°)
δ = declination
α = right ascension

I made two JavaScript functions to implement this calculation:

  obliq = deg2rad(23.44);  // obliquity of ecliptic
  lat2  = deg2rad(0);  // observer's latitude
  lmst  = deg2rad(0);  // siderial time

  function equatorial(lat, lon) {
    // returns equatorial from ecliptic coordinates
    dec = Math.asin( Math.cos(obliq) * Math.sin(lat) + Math.sin(obliq) *
          Math.cos(lat) * Math.sin(lon));
    ra  = Math.atan2(Math.cos(obliq) * Math.sin(lon) - Math.sin(obliq) * Math.tan(lat),
          Math.cos(lon));
    ra += 2 * Math.PI * (ra < 0);
    return [dec, ra];
  }

  function horizontal(lat, lon) {
    // returns horizontal from ecliptic coordinates
    coords = equatorial(lat, lon);
    dec = coords[0];  // δ
    ra  = coords[1];  // α
    alt = Math.asin(Math.sin(lat2) * Math.sin(dec) + Math.cos(lat2) *
          Math.cos(dec) * Math.cos(lmst - ra));
    azm = Math.atan2(Math.sin(lmst - ra), Math.sin(lat2) * Math.cos(lmst - ra) -
          Math.cos(lat2) * Math.tan(dec));
    azm += 2 * Math.PI * (azm < 0);
    return [alt, azm];
  }

I cannot see any error, but I get strange results for azimuth (a) as can be seen in this table (the other values seem correct):

   λ       δ         α         h         a
    0    0.0000    0.0000   90.0000    0.0000
   15    5.9094   13.8115   75.0000  246.5600
   30   11.4723   27.9104   60.0000  246.5600
   45   16.3366   42.5357   45.0000  246.5600
   60   20.1510   57.8186   30.0000  246.5600
   75   22.5962   73.7196   15.0000  246.5600
   90   23.4400   90.0000    0.0000  246.5600
  105   22.5962  106.2804  -15.0000  246.5600
  120   20.1510  122.1814  -30.0000  246.5600
  135   16.3366  137.4643  -45.0000  246.5600
  150   11.4723  152.0896  -60.0000  246.5600
  165    5.9094  166.1885  -75.0000  246.5600
  180    0.0000  180.0000  -90.0000  248.3079
  195   -5.9094  193.8115  -75.0000   66.5600
  210  -11.4723  207.9104  -60.0000   66.5600
  225  -16.3366  222.5357  -45.0000   66.5600
  240  -20.1510  237.8186  -30.0000   66.5600
  255  -22.5962  253.7196  -15.0000   66.5600
  270  -23.4400  270.0000   -0.0000   66.5600
  285  -22.5962  286.2804   15.0000   66.5600
  300  -20.1510  302.1814   30.0000   66.5600
  315  -16.3366  317.4643   45.0000   66.5600
  330  -11.4723  332.0896   60.0000   66.5600
  345   -5.9094  346.1885   75.0000   66.5600
  360   -0.0000  360.0000   90.0000   68.3079

Does anyone see the error? Thank you.

lo.logic - When can we prove constructively that a ring with unity has a maximal ideal?

I suspect that the most general reasonable answer is a ring endowed with a constructive replacement for what the axiom of choice would have given you.

How do you show in practice that a ring is Noetherian? Either explicitly or implicitly, you find an ordinal height for its ideals. Once you do that, an ideal of least height is a maximal ideal. This suffices to show fairly directly that any number field ring has a maximal ideal: The norms of elements serve as a Noetherian height.

The Nullstellensatz implies that any finitely generated ring over a field is constructively Noetherian in this sense.

Any Euclidean domain is also constructively Noetherian, I think. A Euclidean norm is an ordinal height, but not at first glance one with the property that $a|b$ implies that $h(a) le h(b)$ (with equality only when $a$ and $b$ are associates). However, you can make a new Euclidean height $h'(a)$ of $a$, defined as the minimum of $h(b)$ for all non-zero multiples $b$ of $a$. I think that this gives you a Noetherian height.

I'm not sure that a principal ideal domain is by itself a constructive structure, but again, usually there is an argument based on ordinals that it is a PID.

universe - Is time itself speeding up universally?

The rate at which a clock ticks is a local property. There is no universal rate of time which could speed up, so the answer is no.

A clock stationary in the same frame of reference, and the same gravitational field as me will tick at the same rate, (one second every second) It is only clocks that are moving, or in different gravitational fields that that tick at a different rate. If I am falling into a black hole, and I stop the check the time, I wouldn't see my watch slow down.

So there is no standard clock rate. Clocks in the early universe (or at least things that depend on time, such as nuclear decay) ran at the same rate, in their local gravitational field (at one second per second)

If we were to observe a clock from the early universe, it would be receding at great velocity (and so it would be redshifted and time dilated)

Now, the gravitational time dilation does not depend on the amount of mass directly, but on the intensity of the gravitational field. A galaxy has huge mass, but the only place in which the gravitational field is significant (from a GR perspective) is in the neighbourhood of Neutron stars and black holes.

To get gravitational time dialtion you need an intense gravitational field. Just having a lot of mass is not enough. After the period of inflation, the universe was homogeneous, and there were no significant "lumps" to give a net gravitational field. As the gas in the universe collapsed into galaxies and stars, and eventually black holes regions of intense gravity formed, in which clocks would run slowly.

However there is no general speeding up of clocks in the universe.

the sun - What is the shape of the Sun's orbit around the Earth taking into account elliptical orbits?

Consider the non-inertial frame that is at rest with respect to the Earth's revolution around the Sun. (Ignore the Earth's rotation around its own axis.) My question is, what is the shape of the Sun's orbit around the Earth in this reference frame?

Now if we assume the Earth moves in a circular orbit around the Sun, then the Sun's orbit around the Earth will also be circular. (Just use geometric the definition of a circle - the set of all points a fixed distance from a given point.). Specifically it's the great circle formed by the intersection of the ecliptic plane with the celestial sphere.

But the Earth does not move in a circular orbit - Kepler's first law states that the Earth moves in an ellipse around the Sun, with the Sun at one of the foci of the ellipse. So if we consider the Earth's elliptical orbit around the Sun, what is the shape of the Sun's orbit around the Earth. I doubt it's an ellipse, so would be it a more complicated-looking curve.

Note that I'm not interested in gravitational influences from the moon and other planets - this is pretty much a purely mathematical question: if we assume the Earth moves in an ellipse, what would be the shape produced?

milky way - Why is the Solar Helical (Vortex) model wrong?

It isn't correct, because a vortex is not a helix, and so while the planets do trace a helical path as they move through the galaxy, this is not evidence of a vortex.

Yes, the sun actually is moving through space, as it traces a path around the centre of the galaxy. The whole mass of the solar system moves with it, so the planets are not left behind as the sun moves.

Rhys Taylor and Phil Plait have comprehensive smackdowns debunking this vortex idea and other misunderstandings/delusions by the author.

stellar evolution - How massive does a main sequence star need to be to go type 1 supernova?

We know the mass a white dwarf needs to be. That's well defined by the Chandrasekhar limit, but before a main sequence star turns into a white dwarf it tends to lose a fair bit of its matter in a stellar nebula.

According to this site, the white dwarf that remains is about half the mass of the main sequence star, with larger stars losing a bit more.

So, the question: Is it correct to say that a star with a mass of about three solar masses will eventually go supernova, similar to a type 1 supernova, even when it's not part of a binary system? Has that kind of supernova ever been observed?

Or does something else happen like in the final stages of that star? Does it keeps going though collapse and expand cycles, losing enough mass that when it finally becomes a white dwarf it's below the Chandrasekhar limit in mass?

Mostly, what I've read on supernovae says that type 1 supernovae happen when a white dwarf accretes extra matter and reaches the limit and type 2 supernovae are much larger and require about 8-11 solar masses to generate the iron core which triggers the supernova. What happens with the death of the star between three solar masses and eight solar masses?

what is the percentage of stars with planetary systems?

For the purposes of the Drake equation you may as well assume that every star has a solar system.

At present, the exact fraction is unknown since the search techniques are limited to finding planets with certain characteristics. For example, transit searches tend to find close-in, giant planets; doppler shift surveys are also most sensitive to massive planets with short orbital periods, and so-on.

One can try and account for this incompleteness, but although some examples have been found, we are in the dark about the fraction of stars that have an Earth-sized planet at distances further than the Earth is from the Sun, or the fraction of stars that have any kind of planet orbiting beyond where Saturn is in our solar system.

For the Drake equation, you need to know the fraction of stars with planetary systems times the number of "habitable" planets per system is. There have been attempts to estimate this from Kepler data.

Possibly the best estimate at present is from Petigura et al. (2013), who estimate that 22% of sun-like stars have an earth-sized (1-2 Earth radii) planet that receives between 0.25-4 times the radiative flux of the Earth.

Obviously this is a lower limit because it doesn't include planets smaller than the Earth. It also doesn't include the moons of giant planets, which may also be habitable.

We still know very little about this fraction when it comes to M-dwarf stars, which are the most common type of star in the Galaxy...

history - Acquirable Raw Data in Amateur Astrophotography

First off, pairing a classic dob with a DSLR is a bit like a shotgun marriage. A dobsonian is fundamentally a visual telescope. Most manufacturers don't even consider the possibility that these instruments could be used for data collection via a sensor. There are 2 issues here:

1. The dobsonian is not tracking

The sky is moving, the dob stays still. You have to push the dob to keep up with the sky. Any long-exposure photo would be smeared. To remedy this, you'll need an equatorial platform, which will move the dob in sync with the sky.

Please note that only the best platforms allow reasonably long exposure times. Then the results can be fairly good.

2. There isn't enough back-focus

The best photos are taken when you remove the lens from the camera, plug it into the telescope directly, and allow the primary mirror to focus the image directly on the sensor. This is called prime focus photography. But most dobs can't reach the sensor within the camera, because their prime focus doesn't stick out far enough. There are several remedies for this, like using a barlow, moving the primary up, etc.

The bottom line is that it takes some effort to make a dob and a DSLR play nice together. Is it doable? Yes. Is it simple and immediate? No. So the literal answer to your question is that there isn't much you can do with just a dob and a DSLR.

You can take photos of the Moon and the Sun, because the short exposure there does not require tracking, but that's pretty much it. Here is an image of the Moon I took with a home-made 6" dob (with home-made optics) and a mirrorless camera (prime focus, about 1/320 sec exposure):

Makes a cute little desktop background, I guess, but it's definitely not research-grade.

Now add a tracking platform and things become more interesting, and the possibilities open up quite a lot.

---

In a more general sense:

There are telescopes that are specifically made for astrophotography. They have lots of back-focus, they are short and lightweight and therefore can easily be installed on tracking mounts. More importantly, there are tracking mounts made specifically for imaging - very precise, delicate mechanisms that follow the sky motion with great accuracy. In fact, the mount is more important than the scope.

A typical example would be a C8 telescope installed on a CGEM mount, or anything equivalent. Barring that, a dob with lots of back-focus sitting on a very smooth tracking platform (probably not as accurate as a GEM, but good enough for many purposes).

Make sure you don't exceed the load capacity of the mount. If the mount claims it can carry X amount of weight, it's best if the telescope weight doesn't exceed 1/2 of that amount. Close to the weight load limit, all mounts become imprecise.

Once you have: a tracking mount, a good camera, and a telescope (listed here from most important to least important), you can start imaging various portions of the sky for research. There are 2 main classes of objects that you could image:

1. Solar system objects

They're called "solar system objects" but the class includes anything that's pretty bright, not very big, and it's high resolution. Tracking is important but not that crucial.

You need a sensitive, high speed camera that can take thousands of images quickly (a movie, basically). These are called planetary cameras. As a cheap alternative in the beginning you could use a webcam, there are tutorials on the Internet about that. A DSLR in video mode in prime focus might work, but it's going to do a lot of pixel binning, so resolution would be greatly reduced unless you use a very powerful barlow (or a stack of barlows).

You'll load all those images in a software that will perform "stacking" to reduce them all to one single, much clearer image.

The scope needs to operate at a long focal length, f/20 being typical, so a barlow is usually required. The bigger the aperture, the better.

2. Deep space objects (DSO)

These are anything that's pretty faint and fuzzy, like galaxies, but some comets are also DSO-like in their appearance. You need to take extremely long exposures; usually a dozen or a few dozen images, each one between 30 sec and 20 min of exposure. Extremely precise tracking is paramount, so you need the best tracking mount you could buy. Autoguiding is also needed to correct tracking errors.

The scope needs to operate at short focal ratios, f/4 is pretty good, but as low as f/2 is also used; focal reducers (opposite of barlows) are used with some telescopes, like this or like this. Aperture doesn't mean much; small refractors are used with good results.

The camera needs to be very low noise; DSO cameras use active cooling that lowers their temperature 20 ... 40 C below ambient. DSLRs can also provide decent results, but their noise is typically higher than dedicated cameras, so you need to work harder for the same results.

Specific software is used for processing, stacking, noise reduction, etc.

---

So what can you do with such a setup?

Comet- or asteroid-hunting works pretty well. Terry Lovejoy has discovered several comets recently using equipment and techniques as described above. Here's Terry talking about his work.

Tracking variable stars is also open to amateurs. This could also be done visually, without any camera, just a dob, meticulous note-taking, and lots of patience.

With a bit of luck, you could also be the person who discovers a new supernova in a nearby galaxy. You don't need professional instruments, you just need to happen to point the scope in the right direction at the right time and be the first to report it. This also could be done purely visually, no camera, just a dob.

How big do comets get?

If my understanding is correct, Chury, the best explored comet so far is rather on the smallish side as comets go. Most comets observable with naked eye were so because they flew close to Earth, not because they were so big. I wonder, though, how big are the biggest ones.

Let's get it in two variants:

• How big is the biggest known comet? (mass, diameter)


• Is there an estimate, or a theoretical limit on how big a comet can be? (for whatever reason, e.g. becoming a planet, breaking up due to tidal forces or whatever)

cosmology - Strong force and metric expansion

Excepting a Big Rip scenario, there is no eventual 'clash'.

Consider a Friedmann–Lemaître–Robertson–Walker universe:
$$mathrm{d}s^2 = -mathrm{d}t^2 + a^2(t)left[frac{mathrm{d}r^2}{1-kr^2} + r^2left(mathrm{d}theta^2 + sin^2theta,mathrm{d}phi^2right)right]text{,}$$
where $a(t)$ is the scale factor and $kin{-1,0,+1}$ corresponds to a spatially open, flat, or closed cases, respectively. In a local orthonormal frame, the nonzero Riemann curvature components are, up to symmetries:
$$begin{eqnarray*}
R_{hat{t}hat{r}hat{t}hat{r}} = R_{hat{t}hat{theta}hat{t}hat{theta}} = R_{hat{t}hat{phi}hat{t}hat{phi}} =& -frac{ddot{a}}{a} &= frac{4pi G}{3}left(rho + 3pright)text{,} \
R_{hat{r}hat{theta}hat{r}hat{theta}} = R_{hat{r}hat{phi}hat{r}hat{phi}} = R_{hat{theta}hat{phi}hat{theta}hat{phi}} =& frac{k+dot{a}^2}{a^2} &= frac{8pi G}{3}rhotext{,}
end{eqnarray*}$$
where overdot denotes differentiation with respect to coordinate time $t$ and the Friedmann equations were used to rewrite them in terms of density $rho$ and pressure $p$. From the link, note in particular that $dot{rho} = -3frac{dot{a}}{a}left(rho+pright)$.

If dark energy is described by a cosmological constant $Lambda$, as it is in the standard ΛCDM model, then it contributes a constant density and pressure $rho = -p = Lambda/8pi G$, and so no amount of cosmic expansion would change things. Locally, things look just the same as they ever did: for a universe dominated by dark energy, the curvature stays the same, so the gravitational tidal forces do too. Through those tiny tidal forces, the dark energy provides some immeasurably tiny perturbation on the behavior of objects, including atomic nuclei, forcing a slightly different equilibrium size than would be otherwise. But it is so small that that it has no relevance to anything on that scale, nor do those equilibrium sizes change in time. The cosmological constant holds true to its name.

On the other hand, if dark energy has an equation of state $p = wrho$, then a flat expanding universe dominated by dark energy has
$$dot{rho} = -3frac{dot{a}}{a}left(rho+pright) = -3sqrt{frac{8pi G}{3}}left(1+wright)rho^{3/2}text{,}$$
and immediately one can see that there is something special about $w<-1$, leading to an accumulation of dark energy, while the cosmological constant's $w = -1$ leads to no change. This leads to a Big Rip more generally, as zibadawa timmy's answer explains.

---

? If the metric expands surely objects get further away from one another and that would include the stars inside galaxaies as well as the galaxies themselves?

Not at all. It wouldn't even make any sense: if you have an object like an atom or a star in gravitational freefall, by the equivalence principle only tidal forces across it are relevant. The tidal forces stretch the object until the internal forces balance them. But for a Λ-driven accelerated expansion, dark energy contribution to tidal forces is constant. Hence, an object already in equilibrium has no reason to further change its size, no matter how long the cosmic acceleration occurs. This also applies to galaxies, only that the internal forces are also gravitational and balance the dark energy contribution.

Looking at this in more detail, imagine a test particle with four-velocity $u$, and a nearby one with the same four-velocity, separated by some vector $n$ connecting their worldlines. If they're both in gravitational freefall, then their relative acceleration is given by the geodesic deviation equation, $frac{D^2n^a}{dtau^2} = -R^alpha{}_{mubetanu}u^mu u^nu n^beta$. The gravitoelectric part of the Riemann tensor, $mathbb{E}^alpha{}_beta = R^alpha{}_{mubetanu}u^mu u^nu$, represents the static tidal forces in a local inertial frame comoving with $u$, which will drive those particles apart (or together, depending). Hence, keeping those particles at the same distance would require a force between them, but it's not necessary for this force to change unless the tidal forces also change.

Galaxies don't change size through cosmological expansion. Stars don't either, nor do atoms. Not for Λ-driven expansion, at least. It would take an increase in tidal forces, such as those provided by a Big Rip, for them to do so.

A related way of looking at the issue is this: according to the Einstein field equation, the initial acceleration of a small ball of initially comoving test particles with some four-velocity $u$ (in the comoving inertial frame), which is given by the Ricci tensor, turns out to be:
$$lim_{Vto 0}left.frac{ddot{V}}{V}right|_{t=0}!!!!= -underbrace{R_{munu}u^mu u^nu}_{mathbb{E}^alpha{}_alpha} = -4pi G(rho + 3p)text{,}$$
where $rho$ is density and $p$ is the average of the principal stresses in the local inertial frame. For a positive cosmological constant, $rho = -p>0$, and correspondingly a ball of test particles will expand. That's cosmic expansion on a local scale; because of uniformity of the FLRW universe, the same proportionality works on the large scale as well, as we can think of distant galaxies as themselves test particles if their interactions are too minute to make a difference.

Again, we are led to the same conclusion: if the ball has internal forces that prevent expansion, then those forces don't need to change in time unless the dark energy density also changes, which doesn't happen for cosmological constant. If they're not 'clashing' now, then they won't need to in the future either.