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Mosfet
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17 Dec 2016 05:04

heheh, they add intentionally those compounds to make it smelly so it's easy to know if there's a leak.
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21 Dec 2016 10:40

This may be more of a gameplay question, or maybe not.  

What are the canal or riverbed looking geological phenomena that I often see over ice worlds?  Are they actually ancient riverbeds, or are they old tectonic fissures?  I'm curious because I see them all the time but I don't know what they are.  Part of my joy of playing space engine is learning some astronomy.  
 
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21 Dec 2016 11:39

These features are prominent on Europa in our own solar system and they're called lineae. We don't know for sure how they are formed, but the leading hypothesis is volcanic activity and tidal flexing.
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21 Dec 2016 14:03

Thanks, Hornblower.  And thanks to Harbinger Dawn who answered the same question in the other forum two years ago.  

Next question, definitely more gameplay related: Are the lineae randomly generated or is there a method behind it? 
 
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22 Dec 2016 10:43

Would you all agree that if you were to look at the event horizon of a black hole, it would be absolutely black?
I may have found that you might be able to see light coming out of the event horizon. Tell me if I'm right, but if I'm wrong, explain to me the flaw of my reasoning. Basically, at the event horizon, some photons from objects that fell inside would be traveling in the other direction of the singularity at the speed of light, so it would not be falling. Rather, it would be in equilibrium. But isn't there a chance that photon could escape as the radius of the event horizon shrinks due to decay? If this were true, you would be able to detect photons trapped at the event horizon as the black hole decays...

Also, wouldn't material in an accretion disk appear to be more are more red shifted the closer it was to the event horizon?
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22 Dec 2016 19:18

Hornblower wrote:
Would you all agree that if you were to look at the event horizon of a black hole, it would be absolutely black?

It would be the blackest black that was ever black.  

Your idea reminds of a more common question: "if material falling into the hole is frozen in time at the event horizon, then would it be possible for it to escape or become visible again as the hole evaporates?"  The answer to that is unequivocally "no".  But here you're asking about the light, and that works a bit differently and your intuition on it is pretty good.  Unfortunately, the answer is still "no", but let's examine why:

Material falling into a black hole crosses the horizon in a finite amount of time according to its own frame of reference.  It emits light along the way, and the light which was emitted at the horizon and moving upwards/outwards is frozen on the horizon.  The explanation given on PBS Space-Time is apt: Think of the photon as trying to go up a downward-moving escalator.  At the horizon, this escalator goes down at the speed of light, so relative to someone far away, the photon seems to stay still.  Or it's like trying to swim against a current which moves at the same speed you can swim.

Problem:  To actually trap a photon on the horizon is an improbable and unstable situation!  The time it takes for the object to cross through the horizon is extremely short.  Therefore the amount of light which is emitted at the horizon is extremely small.  Furthermore, for the photon to be frozen at the horizon, it needs to be emitted precisely at the horizon and precisely in the radially outward direction.  Any bit of deviation in terms of the radius where it was emitted, or the direction it was emitted in, will result in it either falling into the singularity or escaping to infinity, and it will do so very quickly -- much more quickly than the time it takes for the horizon to shrink an appreciable distance due to Hawking radiation.  
And as if that was not enough, even if the photon somehow does manage to find itself just outside the horizon and escape, then it still has to climb out of the gravitational well to reach you.  If you're at a safe distance, then it will reach you almost an infinite time later, with nearly zero energy and infinite wavelength.  So you won't actually see it.

Also, wouldn't material in an accretion disk appear to be more are more red shifted the closer it was to the event horizon?


Yes, but with some caveats.  The accretion disk can only get so close to the event horizon before material stops orbiting and simply plunges straight in (this is given by the "innermost stable circular orbit", or "ISCO").  If the black hole is not spinning, then this limit lies at three times the event horizon radius, and at that distance the gravitational redshift relative to an observer very far away is not that strong.

For a spinning black hole, the ISCO lies closer to the horizon, and in the limit that the black hole is spinning as quickly as possible, the ISCO reaches to the event horizon.  So we would see more gravitational redshift there.  But the temperatures are also higher, and since the radiation emitted by this material is essentially a blackbody spectrum (hotter materials emit more light and at shorter wavelengths), that more or less cancels out the effect of the gravitational redshift.

The rendition of accretion disks you see in Space Engine is quite accurate in showing this.  You might think for example that the side of the disk rotating towards you should be bluer than the side rotating away from you, due to the Doppler shift.  Instead it mostly just looks a lot brighter, as if the material itself were at a higher temperature.
 
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22 Dec 2016 19:38

Watsisname, interesting. So in order for a photon to escape from inside the event horizon, it would have had to be omitted in a very narrow window of time, and even then, the red-shift alone would make it almost impossible to detect. However keep in mind, the photon being omitted doesn't have to come from an object falling in, it could have been omitted anywhere over the event horizon as the collapsing star's surface passed the horizon. This would logically greatly increase the chances of this, right (even though those photons would escape only shortly after the black hole had formed)?

Thanks for addressing my questions  :D

P.S. When you say the event horizon is the blackest black that was ever black, would it be the exact hex value of #000000 or would it be even more black? In other words, is it possible for the luminosity is negative?
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23 Dec 2016 01:36

Hornblower wrote:
Watsisname, interesting. So in order for a photon to escape from inside the event horizon, it would have had to be omitted in a very narrow window of time, and even then, the red-shift alone would make it almost impossible to detect. 

Precisely.  If emitted very close but just above the horizon, then it is detected almost an eternity later, with such feeble energy and long wavelength as to be a ghostly whisper of radio.  If emitted very close but just below the horizon, then it pulled into the singularity, never to be known to anyone in the external universe. And if it was emitted very close and just above the horizon, but in a direction not close enough to directly outward, then it still is pulled in.

So what about the light from the star itself, as it collapsed to form the black hole in the first place?  It's a very bright surface, to be sure. Being trapped in time for the distant observer, would it help provide light from the event horizon later?  Even in the sense of the occasional random photon?

No.

The thing is, the same effect which is causing the star surface to be frozen in time as it collapses is also redshifting it to invisibility.  And it happens fast.  It's really quite astounding how fast it is, so let's put numbers to it.  

Suppose it's our own Sun collapsing to become a black hole (it won't, but let's just say it did.)  How long would it take for the luminosity, as seen by the distant observer, to drop from the Sun's present value (3.85x1026 watts) to just one watt?  

Solution:  About 1.6 milliseconds!  The fade to black is faster than the blink of an eye!

Why?  When we run through the mathematics of the collapse using general relativity, it turns out that the gravitational redshift causes the observed luminosity of the collapsing star to drop exponentially.  Exponential functions are very fast, and in this case, when the collapsing star is close to its Schwarzschild radius (where the event horizon forms), the luminosity drops by a factor of e (2.71828...) for each 26 microseconds for each solar mass of material in the collapse.  To go from one solar luminosity to one watt is 61.2 reductions of e (the natural log of 3.85x1026 is 61.2), and 61.2 times 26 microseconds is 1.6 milliseconds.

This calculation makes the assumption that light is continuous and we can divide it up in to arbitrarily many little bits.  But it's not -- light is actually quantized.  That makes it even worse.  There are only so many photons that can escape.  After about 1 millisecond, the last photon that will ever escape the star's surface is seen by the distant observers.  After that, it is truly black.

Some might also ask of the light of Hawking radiation.  But for any black hole massive enough to be meaningful (not a "micro black hole" with less than the mass of a planet)... it's ridiculously dim.  The wavelength of the photons is comparable to the diameter of the hole, and the power emitted is like that of an object near absolute zero.  For a solar mass black hole, it radiates with 9x10-29 watts, which is pretty much zero.  The intensity of the cosmic microwave background radiation is blindingly bright compared to it.

So yes, even while making a quip, I'm quite serious when I say it is the blackest black that was ever black. :)  The blackness of #000000 on your monitor is not even close -- the screen still emits thermal radiation, and even if you discount that, it still reflects a small amount of ambient light.  There is no material in nature as dark as a black hole's event horizon.  Not the darkest coal, not a starless night sky, not even that weird Vantablack.

Thanks for addressing my questions  :D

It's absolutely my pleasure.  Black holes are amazing things. :)
 
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04 Jan 2017 06:05

Is there a relationship between brightness and distance ?
For the gravitation, it is :
[math]
[math]
g in m/s2, G the constant of the gravitation in N⋅m2⋅kg-2, m the mass of the object in kilogrammes (kg) and d the distance between the center of the object and the position in meters (m).
I don't know but I thought, taking as a starting point from this formula of gravitation, a formula resembling that :
[math]
l the apparent luminosity or received luminosity in the position actual in W or Sol's luminosity, L the luminosity of the object in W or Sol's luminosity, R the radius of the object in m and d the distance actual in m, n a power like 2 (d2). I am not a scientist, I have no proof for the formula to be good so I think it may be completely false.
And I don't know how much does the Earth receive from the brightness of the Sun or what is the apparent luminosity of the Sun seen from Earth.
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04 Jan 2017 21:43

Yeah, there's a formula for it, and there is indeed insight in comparing with the law of gravity. :)  Your formula is close, but not quite right.  

First let's go over the terminology.  The "apparent luminosity" actually isn't a meaningful term.  Luminosity is an intrinsic property of an object, and means the total amount of light it emits per unit time.  Light is energy, and energy per time is power, measured in watts.  What you actually want to find is something which describes the intensity of the light that reaches us. This has a different name which we'll see shortly.  It does obey the inverse square law like your formula does, but it has a slightly different form overall.  Let's see how to figure it out.

Start by assuming the object emits light uniformly in all directions.  As the distance increases, the light is spread out over the surface of larger and larger spheres, so the amount of light per unit area is decreasing with distance.  The name for this is "flux", meaning how much stuff passes through a surface of unit area per unit time.  If the "stuff" in question is light, its energy per unit time is again power, and so the flux has units of watts per square meter.

Since the surface area of a sphere is proportional to the square of its radius, if the radius is tripled, then the same amount of light is now spread out over nine times as much surface area, and the light intensity that we receive at that radius is reduced a factor of nine.  There's the inverse square law, which your formula correctly shows as [math].  

But how about the L*R?  That tells us the amount of energy we receive grows in proportion to the radius of the object and it's luminosity.  Let's think about that.  If I have two light bulbs, both of them rated at 100 watts, and one is the size of a baseball and the other is the size of a marble... which one looks brighter at a distance of, say, 10 meters?  

They are the same!  Their apparent sizes will be different, and so will their surface brightnesses (how much light is emitted by each unit of area on their surfaces), but the amount of light we receive from them is the same.  They will illuminate a room equally well.

So what do we need to do to fix the formula?  Since flux is the power per area, and we said the area is increasing as spheres, let's divide the luminosity of the object by the surface area of a sphere with radius [math].  The surface area of a sphere is [math].  That gives us:

[math]

Let's apply it with the Sun and Earth.  The Sun has a luminosity of 3.85x1026 watts.  The Earth is 1AU away, or 1.5x1011 meters.  Then the formula tells us that the intensity of sunlight at Earth is about 1370 watts per square meter.  This is a useful measure -- for example, we can use it to find out how much power we could obtain with a solar panel of a given size and efficiency.


Now I said there is a useful connection between the ideas used here and with the gravitational law.  Both obey the inverse square law.  But why?  In both cases, we can think of a "source object" emitting something in all directions.  A star emits light energy.  We can also say a mass emits... "something"... which causes other masses to accelerate.  What is it?  Truthfully, nobody knows. :)  Unlike light, it's not something we can see.  What we have are models that describe it, and in Newton's model, we can imagine a "force field" which emanates from the source mass, and assigns to every point in space a gravitational force which would be exerted on another mass if it was placed there.  You can visualize the force field as being like the lines of magnetic field that you see when iron filings are placed around a magnet.  Except in this case, instead of curling, they spread straight out in all directions.  The magnitude of the gravitational field is proportional to how many "field lines" pass through a surface, so the "gravitational flux" decreases with the square of the distance.  Just like with light.
 
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05 Jan 2017 01:56

If I understand :
[math] W
[math] m
[math] W/m2 is the flux of luminosity that the Earth receives.

If I make with the Sun :
[math] W
[math] m
[math] W/m2
But if I want to find the distance, just do it [math], so if [math] W and [math] W/m2, so [math] m.
Last edited by Quarior on 10 Jun 2017 03:12, edited 1 time in total.
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05 Jan 2017 08:27

Your calculations are all correct.  But what is the interpretation for what you calculated in the last part?

By setting [math], you're saying let the amount of light per square meter at a distance [math] from the object be equal to all of the light that the object emits in all directions. The Sun's surface area is a lot bigger than one square meter, so [math] would have to be a lot smaller than the Sun itself.

In fact, the result of that calculation will be the same for anything.  It is simply telling us how big is the radius of a sphere whose surface area is one square meter.  That's ~0.28 meters.

We can also calculate what is the amount of light that leaves the Sun for each square meter of its surface -- that is, what is the flux at the sun's surface?  Divide the luminosity by the surface area of a sphere the size of the Sun, and we get 6.33x107 W/m2.
 
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08 Jan 2017 16:34

What if Venus formed in the place of Mars and Mars formed in the place of Mars, along with Uranus and Neptune switching places and Pluto and Ceres switching places at the start of the solar system?
 
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08 Jan 2017 17:28

Image
 
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08 Jan 2017 18:10

Starlight Glimmer,
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