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.

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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.