Neutron Stars are real!

"alone"? What other spectrum is there? :D. In any case, I'm speaking solely of the visual spectrum, which is only a small slice of all that incident radiation. The image above showing the ships & large prominences looks more like what I would expect, bright but not so intense as to completely swamp any detail.
It's not taking into account the emitted particles.

Note that I edited above. It seems I don't know the formula for the area of a sphere. :rolleyes: It makes rather a large difference.

Fundamentally, all you need is some dark glasses and that rather goes without saying; we can't even see details like that on our own sun without significant filtering. We would need to know how the energy was emitted in order to know what the visual brightness was.

It's also interesting to note that our ships can easily withstand terawatts per sq metre of starlight, but a few megawatts from a laser immediately causes serious damage.
 
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It's not taking into account the emitted particles.

Note that I edited above. It seems I don't know the formula for the area of a sphere. :rolleyes: It makes rather a large difference.

Fundamentally, all you need is some dark glasses and that rather goes without saying; we can't even see details like that on our own sun without significant filtering. We would need to know how the energy was emitted in order to know what the visual brightness was.

It's also interesting to note that our ships can easily withstand terawatts per sq metre of starlight, but a few megawatts from a laser immediately causes serious damage.
The "emitted particles" are principally photons, i.e., EM radiation of various wavelengths. Other particles are not registered visually though they can generate additional EM via processes like synchrotron radiation. Our sun's peak emissions are centered in our visual perception range, so it appears extremely bright visually; a neutron star emitting mostly in X-rays (black body peak @ 3 angstroms) could appear much less bright visually, pretty much identical in color to O & A stars because they emit much more in blue and violet light than in lower wavelengths, but dimmer because of size and overall emission strength in those colors (O & A stars peak in blue and violet). Since ED's portrayals of most stars are inherently less bright than they should be, it follows that how NSs are shown is a break in that paradigm, as they seem much brighter than they should be.
 
The "emitted particles" are principally photons, i.e., EM radiation of various wavelengths.
Wikipedia said:
The solar luminosity, L☉, is a unit of radiant flux (power emitted in the form of photons) conventionally used by astronomers to measure the luminosity of stars, galaxies and other celestial objects in terms of the output of the Sun. One nominal solar luminosity is defined by the International Astronomical Union to be 3.828×10^26 W. This does not include the solar neutrino luminosity, which would add 0.023 L☉.

I have always assumed our canopies contain an automatic filtering mechanism that would work in much the same way as audio compression. All stars appear to be much the same apparent brightness because they are "optically compressed" differently. Therefore a NS would appear about as bright as any other.

Bending the fourth wall a little, you can see this in operation if you take a high res screenshot of a bright star. The exposure appears to be calculated for each tile separately and you end up with square artifacts. You can see them in the first post in this thread.
 
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The idea of canopy filtration has crossed my mind as well, but Camera Suite images aren't from inside the cockpit and show the same graphics, so that puts paid to that idea. The visual presentation of all stars is more poetic than super-scientific and that's fine, it makes for a much more interesting game, but neutron stars and white dwarfs seem to break out of the usual treatment in a way that doesn't jibe well with physics. That in itself is not unusual for ED, but it does present some imaging issues, as the OP's work demonstrates.
 
The camera has the same mechanism.
Riiiiight. Funny how this wonderful filtration system doesn't seem to affect the brightness of other objects, unless it's extremely selective, and can't seem to cope with amplifying incident starlight on planetary nightsides. It's almost as if it's really just some kind of computer simulation made to make things look cool... oh, wait... ;)
 
I'm pretty sure, that you can "see" the actual NS in this image, which I once made by stacking a few hundred different amateur photos of Messier 1. They were all captured in the visible wavelengths. It's the star roughly at the center of the image, below/right of the bright yellow star:

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Of course, what you see is not the star, but emission and probably reflection from the surrounding gas.

Edit: Ok, with an apparent magnitude of 16.5 it's not the brightest star out there :cool:
 
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Stacking such images is a form of amplification, and says little about the actual visual brightness of a NS, just that it can emit in visible wavelengths, which is a given. The apparent size of the stacked image is much more likely to be due to atmospheric distortion than emissions at the source, unless amatuers have been able to build their own adaptive mirror optics. Being only 10 to 50 km in diameter the star itself is of course not resolvable, and any emission or reflection from surrounding material would likely be much more extended, as is the supernova remnant itself. Even in large 'scopes NSs require extended time exposures to capture their visual output, and can be somewhat ambiguous even then -- the exact identity of the Crab Pulsar was in doubt and contention for quite a while. They are inherently faint from Earth in visible light, and likely not as bright as normal stars even up close, because their greatest emissions are not in the visible spectrum.
 
The stacked image I made of M1, was from a series of images I made, using a method I developed, called "Crowd Imaging" by one of my fellow amateur astronomers. I took all the creative commons jpg images I could find of a chosen object, and stacked them using PixInsight. The idea was that they all contained information and noise, so stacking them carefully you would be able to extract "the sum of the information", while reducing the noise. With many popular objects like M1, it was possible to gather several hundred images. With Andromeda I had 1500+. Stacked with different weights depending on noise, FWHM etc. and using deconvolution, the method ended up making some pretty remarkable results, even though each individual image was shot using relatively small 3"-16" telescopes with no adaptive imaging. Any astroimaging is like peeling a grape with a razor blade anyhow. The colors were very precise and rather accurate, because one astrophotographers image being too green was compensated by another being too red, etc. I started out making my own statistics algorithms, but realized that I was just reinventing the wheel.

With one of the images I made, I tried to see how deep I could go, apparent magnitude wise. I gathered 180 good images of the Leo Triplet, and used Simbad to look up the faintest stars I could visually identify in the image. The faintest one I found was mag 25.2 (green), 24.2 (red). The Keck Telescope (10 meter) is able to resolve mag 24-26 stars using a one hour exposure. The reason why "Crowd Imaging" can resolve equally faint stars is the sum of the exposure time for all the images being extreme, up to several thousand hours. The signal to noise ratio only increase with the square root of the exposure time, but the square root of say 5000 hours is still a decent number.

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I also tried to look for distant quasars. The one furthest away was QSO B1118+1352 with a redshift of z=2.43:

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I haven't looked through an eyepiece for a long time now, but I think it should be theoretically possible to see the Crap Pulsar using a very large Dobsonian telescope from a very dark place. A project for a star party perhaps, if it hasn't already been done. Those Dobsonian people are crazy ;)

 
The stacked image I made of M1...
Thank you for that explanation, I wasn't aware that kind of analytical astro-ware was publicly available, but it doesn't surprise me now I think on it. Still, even with noise reduction ad absurdum, what's being processed is a smeared image due to atmospheric turbulence (if not a Hubble or other space-based image, that is, or one made with adaptive optics), though the resolving of very low apparent magnitude objects would improve as you demonstrate.

Re: intrinsic NS brightness -- For Geminga (mentioned above), if m = 26 @ 160 parsec, it works out to about Mv =~20, still very dim by comparison to what we consider "bright" stars, and far below naked eye visibility. Formula used is Mv = m - 2.5 log[ (d/10)^2 ], found on the web.
 
They are inherently faint from Earth in visible light, and likely not as bright as normal stars even up close, because their greatest emissions are not in the visible spectrum.

This response isn't aimed directly at you, but rather for those reading along:

Just to clarify here, in case of confusion... The visible part of the black body spectrum doesn't actually get more dim for an overall brighter object. If you take a given object and gradually heat it up, well beyond 5000 Kelvin, it gets brighter at all wavelengths, but the peak output (and thus the bulk of the energy emitted) continues to move to higher frequencies (shorter wavelengths). The portion that is within the visible part of the spectrum gets more "tilted" toward the blue, and still gets brighter, but not proportionately to how much energy it is emitting overall, since the bulk of it is at higher and higher frequencies, outside the visible part of the spectrum.

Neutron stars are more dim due to being small, however they're extremely hot due to left over heat from being the core of a star, plus all of the gravitational compression they experienced.


CNX_UPhysics_39_01_BBradcurve-1.jpg


figure05-10.jpg
 
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I'm pretty sure, that you can "see" the actual NS in this image, which I once made by stacking a few hundred different amateur photos of Messier 1. They were all captured in the visible wavelengths. It's the star roughly at the center of the image, below/right of the bright yellow star:

View attachment 147318

Of course, what you see is not the star, but emission and probably reflection from the surrounding gas.

Edit: Ok, with an apparent magnitude of 16.5 it's not the brightest star out there :cool:
If that star has Apparent Magnitude 16.5, and it is 6500 ly (1993 parsecs) away, and the web sources I found are accurate, and if I understood them, then its Absolute Magnitude is 5. Which just happens to be the same as Sol. So the Neutron star is putting out the same amount of light as Sol, but from a far smaller surface area.

So just to be clear, if our sun was replaced by that neutron star, then we would have the same light levels on Earth as we do currently, from an object about 10km across, 150 million kilometres away.

It is insanely bright, in the visible spectrum.

In the X-Ray spectrum, I'm less sure of my figures or what they mean, but I think if it follows a black-body curve, that gives it a peak at 3 angstroms (0.3nm) of about ten billion times more intensity. Is intensity a power thing or an amplitude thing?

It's fairly obvious that anything 270km from it would be fried by X-rays, but it's visible light we're interested in.

In my calculation above, the luminosity of the white dwarf was 0.0295
Sol has a luminosity of 1, centered in the visible spectrum.
So the M1 neutron star has an equivalent visible luminosity of 1, and it emits approximately 3.828E+26 watts in the visible spectrum.
And at a distance of 270km, that gives us 4E+14 watts per square metre at (waves hands) sort-of sun equivalent wavelengths.

In comparison, the incident power on Earth from the sun is 1360 watts per square metre.

That's such an outrageous answer that I checked the spreadsheet by putting in Earth's figures -- luminosity 1, distance 150Gm -- and got 1353 W/m2, which is close enough.

At 270km, neutron star M1 would be 300 billion times brighter than the sun is from Earth. That power would increase through the ultra-violet and into the x-rays until it reached a peak of a further ten billion times more intensity.
 
It's fairly obvious that anything 270km from it would be fried by X-rays, but it's visible light we're interested in.
I have engineered the shield generator ;)

A lot of the radiation could be concentrated in the jets. Those are instantly deadly at 270 km. But I agree that the UV, X-ray and gamma radiation that close to the star would rip anything apart faster than the blink of an eye.

I think the next object I will try to get up close and personal with is a white dwarf, but I'm pretty sure that one will be dangerous. I have brought no AFMU.
 
How does the NS create all the radiation? There is no fusion going on inside it. Is it just the rest heat, and if that is the case, will it lose it's glow far into the future?
 
Re: intrinsic NS brightness -- For Geminga (mentioned above), if m = 26 @ 160 parsec, it works out to about Mv =~20, still very dim by comparison to what we consider "bright" stars, and far below naked eye visibility. Formula used is Mv = m - 2.5 log[ (d/10)^2 ], found on the web.

The visible magnitude drops by a factor of 100 per 5 magnitudes, so Geminga is a million times dimmer than the sun.
At 270km, it would only be 300,000 times brighter than the sun is from the Earth.

How does the NS create all the radiation? There is no fusion going on inside it. Is it just the rest heat, and if that is the case, will it lose it's glow far into the future?

There's your answer: two neutron stars that differ by a million times in their power output.
M1 was formed only a thousand years ago, Geminga was formed about 300,000 years ago.
 
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This response isn't aimed directly at you, but rather for those reading along:

Just to clarify here, in case of confusion... The visible part of the black body spectrum doesn't actually get more dim for an overall brighter object. If you take a given object and gradually heat it up, well beyond 5000 Kelvin, it gets brighter at all wavelengths, but the peak output (and thus the bulk of the energy emitted) continues to move to higher frequencies (shorter wavelengths). The portion that is within the visible part of the spectrum gets more "tilted" toward the blue, and still gets brighter, but not proportionately to how much energy it is emitting overall, since the bulk of it is at higher and higher frequencies, outside the visible part of the spectrum.

Neutron stars are more dim due to being small, however they're extremely hot due to left over heat from being the core of a star, plus all of the gravitational compression they experienced.


CNX_UPhysics_39_01_BBradcurve-1.jpg


figure05-10.jpg
Despite the overall increase in visual "intensity" per your curves, the overall luminosity of NSs is hugely reduced by their incredibly small diameters. By comparison, look at typical white dwarf stars on the H-R diagram:
Though they have very high surface temperatures and peak emissions mostly higher than visual, they are of low luminosity because they are very small. NSs are much, much smaller, and thus would have even lower intrinsic luminosity. Perceived brightness and luminosity are not equivalent but are related, objects of lower luminosity appearing dimmer than those of higher when viewed at a standard distance. This is the basis for Mv -- absolute visual magnitude -- which uses 10 parsecs as the standard distance.
"The idea of a neutron star was developed in 1939 when calculations were made of a star that was composed solely of degenerate neutrons. If the mass of a normal star were squeezed into a small enough volume, the protons and electrons would be forced to combine to form neutrons. For example, a star of 0.7 solar masses would produce a neutron star that was only 10 km in radius. Even if this object had a surface temperature of 50,000 K, it has such as small radius that its total luminosity would be a million times fainter than the Sun. '' -- http://abyss.uoregon.edu/~js/ast122/lectures/lec19.html
When entering a NS system in ED, the star is incredibly, nearly blindingly bright, even though the drop-in distance is not greatly different than that for Sol. Yet Sol is not blinding when entering its system, though it has a higher luminosity. This indicates to me that the apparent brightness (dependent on luminosity and distance) of the NS is greatly exaggerated, which is dramatic, but I think physically inaccurate. Its apparent brightness would increase as you approach the NS, but would not exceed the visual brightness of other 50,000 K stars such as type O, because they share a similar black body radiation curve.
 
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[When entering a NS system in ED, the star is incredibly, nearly blindingly bright, even though the drop-in distance is not greatly different than that for Sol. Yet Sol is not blinding when entering its system, thougjh it has a higher luminosity. This indicates to me that the apparent brightness (dependent on luminosity and distance) of the NS is greatly exaggerated, which is dramatic, but I think physically inaccurate. Its apparent brightness would increase as you approach the NS, but would not exceed the visual brightness of other 50,000 K stars such as type O, because they share a similar black body radiation curve.

Your argument only works for point sources.

When you drop into a sol-type system, the brightness is spread over a disk as large as your canopy but when you drop onto a neutron star it is still a point source. The entire brightness from that massive plane is concentrated into a single speck.

The maths of spheres gets complicated but imagine that you are at the centre (inside) a sphere that is emitting light toward you. You receive x watts per square metre in total. If the sphere has a surface area of y square metres, then every square metre is responsible for delivering x/y watts per square metre to you. You will perceive the brightness of the sphere according to how bright each part of it is. It's apparent brightness will be proportianal to x/y. In the case of a neutron star it is just one tiny dot on the sphere that delivers all x watts per square metre. Its apparent brightness will be proportional to x. The bigger y is, the dimmer it appears relative to the neutron star, but it still delivers the same total brightness to you.
 
Your argument only works for point sources.
Absolutely not so. Luminosity is intrinsic, proportional to the surface area and temperature of an emitting body, and is independent of distance. It therefore applies to all kinds of luminous extended objects (not point sources), from neutron stars to blue supergiants. Apparent visual brightness is luminosity as modified by the viewer's distance and any diminishing of brightness due to intervening matter (negligible in the cases we are discussing), and is also affected to some degree by the object's black-body spectrum, as it may emit more brightly in non-visual wavelengths. In the case of two objects with the same or very similar surface temperatures (50,000 K, for example), that which is vastly larger than the other, say a type O star compared to a neutron star, would be vastly brighter when both are viewed from the same distance, because the type O has a vastly greater luminosity.

I have yet to encounter a type O star but the codex entry for it looks much like a very bright (but not blindingly bright) type A (those I have seen), with discernible surface features. The neutron star I encountered appeared a great deal brighter, and the OP's images show that NSs show no surface detail when seen closer in. It is portrayed as vastly brighter in the visual wavelengths, even when seen at distances comparable to those at which Type O stars are easily seen in detail. This contradicts the entire idea of stellar luminosity (and thus apparent brightness) being dependent on object size, even when allowing for ED's "poetic" treatment of star appearance.
 
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