Ultimate space simulation software

 
User avatar
Watsisname
Science Officer
Science Officer
Posts: 2105
Joined: 06 Sep 2016
Location: Bellingham, WA

Astrophotography

20 Aug 2020 22:02

A-L-E-X wrote:
Source of the post Why is it called the "forbidden transition" Wat?    It doesn't seem to be all that forbidden

Yes, it's a bit of misnomer, although it almost means what it implies. "Forbidden" transitions are not impossible, but they are profoundly less probable than "allowed" transitions, and perhaps counter-intuitively it's because they are so much less probable that they show up in a lot of important places and applications.

To explain, you probably know that it is possible for an electron in an excited hydrogen atom to spontaneously drop to a lower energy level, emitting a photon with that difference in energy in the process. But it can't drop from just any excited state to any lower energy state. The transition must also obey certain "selection rules", like that the angular momentum quantum number must change by 1. So for example an electron can't drop from the 2S to 1S state, since both states have the quantum number l=0. It has to instead go from 2P to 1S. That means a hydrogen atom in the 2S state is "stuck" like that, unable to lose that energy by spontaneous emission.

Now a "forbidden transition" is one which occurs even though it violates the selection rules. The catch is that it is many orders of magnitude less probable! Because it is so improbable, it statistically takes a much longer time before it will happen on average. Usually allowed transitions happen in less than nanoseconds, while in the case of the transition that makes the green aurora or green meteor afterglow, it takes about a second, which is why the afterglow is visible for about that long. It's also why those glows aren't visible below about 90km altitude. At lower altitudes, collisions between atoms are frequent enough to de-excite them long before the forbidden transitions could occur. 

Similar forbidden transitions are also responsible for many glow in the dark materials. It's because those transitions are so improbable that the glow lasts for so long, which makes it so useful. :)

‡Actually the 2S --> 1S transition is not possible even by forbidden transitions, but it is possible for the 3S --> 1S transition to occur, which violates the aforementioned ∆l = +/-1 selection rule and is thus "forbidden".
 
User avatar
midtskogen
Star Engineer
Star Engineer
Posts: 1237
Joined: 11 Dec 2016
Location: Oslo, Norway
Contact:

Astrophotography

21 Aug 2020 00:59

A-L-E-X wrote:
Source of the post maybe having started out at around 80 km when I first saw it. 

Slow meteors (< 20 km/s) become visible at 70-90 km.  Fast meteors, such as the Perseids travelling at 59 km/s, become visible around 120 km above ground, and will often have evaporated completely by they reach 70-80 km.
NIL DIFFICILE VOLENTI
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 2764
Joined: 06 Mar 2017

Astrophotography

24 Aug 2020 07:46

Watsisname wrote:
A-L-E-X wrote:
Source of the post Why is it called the "forbidden transition" Wat?    It doesn't seem to be all that forbidden

Yes, it's a bit of misnomer, although it almost means what it implies. "Forbidden" transitions are not impossible, but they are profoundly less probable than "allowed" transitions, and perhaps counter-intuitively it's because they are so much less probable that they show up in a lot of important places and applications.

To explain, you probably know that it is possible for an electron in an excited hydrogen atom to spontaneously drop to a lower energy level, emitting a photon with that difference in energy in the process. But it can't drop from just any excited state to any lower energy state. The transition must also obey certain "selection rules", like that the angular momentum quantum number must change by 1. So for example an electron can't drop from the 2S to 1S state, since both states have the quantum number l=0. It has to instead go from 2P to 1S. That means a hydrogen atom in the 2S state is "stuck" like that, unable to lose that energy by spontaneous emission.

Now a "forbidden transition" is one which occurs even though it violates the selection rules. The catch is that it is many orders of magnitude less probable! Because it is so improbable, it statistically takes a much longer time before it will happen on average. Usually allowed transitions happen in less than nanoseconds, while in the case of the transition that makes the green aurora or green meteor afterglow, it takes about a second, which is why the afterglow is visible for about that long. It's also why those glows aren't visible below about 90km altitude. At lower altitudes, collisions between atoms are frequent enough to de-excite them long before the forbidden transitions could occur. 

Similar forbidden transitions are also responsible for many glow in the dark materials. It's because those transitions are so improbable that the glow lasts for so long, which makes it so useful. :)

‡Actually the 2S --> 1S transition is not possible even by forbidden transitions, but it is possible for the 3S --> 1S transition to occur, which violates the aforementioned ∆l = +/-1 selection rule and is thus "forbidden".

Thanks, Wat! It sounds like these forbidden reactions last so long because, for them to happen, requires a lot of energy!  It also proves that, "forbidden" just means "occurs much less often but will eventually happen anyway because of the sheer number of total particles" (although "forbidden sounds much more catchy! :P)  It sounds a lot like "inert gases"  later renamed "noble gases" because although they react with other elements very rarely, it does happen on occasion (particularly with Fluorine, since Fluorine seems like a particularly aggressive reactor!)  Although the two lightest of them, Helium and Neon, still haven't been shown to react with any other element, because they hold onto their electrons the "tightest"?  And is 2S--->1S truly forbidden since it is not possible at all (in nature?)  Also, regarding forbidden reactions, isn't there a forbidden "triple carbon" reaction that happens in some stars under extreme pressure?  I remember reading an Asimov novella that was based around this.
Last edited by A-L-E-X on 24 Aug 2020 08:20, edited 2 times in total.
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 2764
Joined: 06 Mar 2017

Astrophotography

24 Aug 2020 07:50

midtskogen wrote:
A-L-E-X wrote:
Source of the post maybe having started out at around 80 km when I first saw it. 

Slow meteors (< 20 km/s) become visible at 70-90 km.  Fast meteors, such as the Perseids travelling at 59 km/s, become visible around 120 km above ground, and will often have evaporated completely by they reach 70-80 km.

Ah, perhaps that's why they and the Leonids are the most colorful?  Putting yours and Wat's posts together, the ones that evaporate higher up are more prone to show "forbidden" colors.  The Geminids on the other hand, dont show much color at all.
 
User avatar
midtskogen
Star Engineer
Star Engineer
Posts: 1237
Joined: 11 Dec 2016
Location: Oslo, Norway
Contact:

Astrophotography

24 Aug 2020 11:58

Geminids do indeed go much deeper.  Sometimes down to 35 km.
NIL DIFFICILE VOLENTI
 
User avatar
Watsisname
Science Officer
Science Officer
Posts: 2105
Joined: 06 Sep 2016
Location: Bellingham, WA

Astrophotography

24 Aug 2020 14:20

A-L-E-X wrote:
Source of the post Thanks, Wat! It sounds like these forbidden reactions last so long because, for them to happen, requires a lot of energy!

It is not really a function of energy. Think about it: if a particular forbidden transition produces a photon of visible light, that's basically the same energy as any allowed transitions that produce visible light. Allowed transitions may even be much higher energy, such as hydrogen's Lyman series which are all in the ultraviolet.

What is instead happening are very small magnitude effects that normally we brush under the rug when we do calculations. For example, the finite size of the atom when encountering or emitting an electromagnetic wave. If we make the approximation that the atom is much smaller than the wave, then the electric field it "sees" is just a sinusoidal oscillation in time. This approximation makes the math easier and gives almost the right answers for the kinds of electronic transitions that may occur. But in truth, the finite size of the atom means it also sees a very small spatial variation in the electric field (one side of the atom will at times be in a slightly higher or lower potential than the other), and that variation means there is a small but nonzero probability to excite additional transitions -- those which we would call "forbidden" because the probability they occur if there was no spatial variation would indeed be zero.
 
John Done
Observer
Observer
Posts: 11
Joined: 31 Aug 2020

Science and Astronomy Questions

01 Sep 2020 04:40

Has anyone heard about the latest news about a black hole fire in the heart of Phoenix Galaxy Cluster? Does every galaxy have black hole jets in it? In how many years our galaxy will be swallowed up by a black hole or will it be destroyed by a collision with another planet? 
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 2764
Joined: 06 Mar 2017

Science and Astronomy Questions

04 Sep 2020 22:01

Watsisname wrote:
A-L-E-X wrote:
Source of the post Thanks, Wat! It sounds like these forbidden reactions last so long because, for them to happen, requires a lot of energy!

It is not really a function of energy. Think about it: if a particular forbidden transition produces a photon of visible light, that's basically the same energy as any allowed transitions that produce visible light. Allowed transitions may even be much higher energy, such as hydrogen's Lyman series which are all in the ultraviolet.

What is instead happening are very small magnitude effects that normally we brush under the rug when we do calculations. For example, the finite size of the atom when encountering or emitting an electromagnetic wave. If we make the approximation that the atom is much smaller than the wave, then the electric field it "sees" is just a sinusoidal oscillation in time. This approximation makes the math easier and gives almost the right answers for the kinds of electronic transitions that may occur. But in truth, the finite size of the atom means it also sees a very small spatial variation in the electric field (one side of the atom will at times be in a slightly higher or lower potential than the other), and that variation means there is a small but nonzero probability to excite additional transitions -- those which we would call "forbidden" because the probability they occur if there was no spatial variation would indeed be zero.

hmm that actually reminds me of certain models we use to simplify reactions.  If the reactions were indeed as simple as the models indicate, then certain outcomes would be "forbidden" or "not possible."  But because our model is just an approximation, there are more possibilities than in the simplified version.
It actually reminds me a bit of the Law of Conservation of Mass-Energy, virtual energy can be created or destroyed (quantum fluctuations, Casimir Effect and Uncertainty Principle) but not in the simplified model.
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 2764
Joined: 06 Mar 2017

Science and Astronomy News

04 Sep 2020 22:21

How close can an asteroid get without actually hitting the earth or burning up?  Can one make it into (let's say) at the same altitude as our cirrus clouds and not get burned up or hit the planet? Could a sufficiently large one cause tidal impacts or even influence our weather?
 
User avatar
Watsisname
Science Officer
Science Officer
Posts: 2105
Joined: 06 Sep 2016
Location: Bellingham, WA

Science and Astronomy Questions

05 Sep 2020 04:41

A-L-E-X wrote:
Source of the post Can one make it into (let's say) at the same altitude as our cirrus clouds and not get burned up or hit the planet?

Maybe, if it is large (100s of meters across or more) and dense enough, but even then it might not survive the journey for other reasons (the ram pressure may fragment it or even result in an airburst.) Otherwise more typical smaller asteroids penetrating that low into the atmosphere at such a shallow angle lose their cosmic velocity.

A useful technique to apply here is Newton's Impact Depth approximation. Basically, a high velocity projectile will be stopped after it has swept out its own mass worth of target material in its path. For high velocities the impact depth turns out to be roughly independent of the velocity, and the most important factors are the projectile's size and density relative to the target. If the target's density is uniform then this is simple to calculate: the impactor will have swept out its own mass worth of target once it has penetrated to a depth equal to its own length times the ratio of its density to the target density.

For an asteroid passing through the atmosphere, however, we must account for how the density of air decreases exponentially with altitude. The density of air at the Earth's surface is ρ0 = 1.225kg/m3, and this decays with a scale height H of about 8km (decreases by a factor of e for each 8km increase in height), according to the formula

Image

Now with a bit of calculus we can find out the depth a particular impactor should reach before losing its cosmic velocity if it comes in vertically.

Let the impactor have a mass mi, approximate length L, and density ρi. The mass of a column of atmosphere it sweeps out will be its cross sectional area L2 times the density ρ of the air at an altitude z times a sliver of altitude dz, integrated over the range of altitudes (from the impactor's final altitude to infinity where the air density trends to zero). Setting the mass of the column of atmosphere equal to the mass of the impactor, we have:

Image

Now solving for the final altitude zf

Image

Before we plug in any numbers, let's think about what this is saying. The scale factor is a multiplier in front, with a minus sign that might seem weird. But rest assured that for small impactors that are not stupidly dense, the answer will be positive, because the impactor's length times density will be much smaller than the scale height times the surface air density, so the natural logarithm will be negative which cancels the negative in front. That also leads to one nice piece of insight right away: the conditions needed for the impactor to be able to reach the ground before being slowed down fully. Set zf=0, and that means the argument of the natural log must be 1, and therefore Lρi must equal Hρ0. That is, the impactor length times density must equal the atmosphere's scale height times the air density at the surface!  If the impactor is even larger or more dense, then it will be able to hit the surface with some of its original cosmic velocity. 

For example, this result tells us that an iron asteroid (density ~8000kg/m3) must be at least about 1.3 meters across in order to reach the surface with some of its cosmic velocity. Anything smaller or less dense than that will instead fall to the ground at a much slower terminal speed (if it reaches the ground intact at all -- it could also fully ablate or airburst.) Or consider an icy meteoroid with a density of, say, 1000kg/m3. That would need to be at least about 10 meters across to reach the ground at a high speed.

What about Earth-grazers coming in at a shallow angle, so that they might neither be destroyed by the atmosphere nor hit the surface? Once again we can use the impact depth approximation, but integrated over the whole path. The key idea is that with such a shallow angle, the path through the atmosphere will be much longer, so for a given asteroid size and composition the minimum altitude it could reach before getting seriously slowed down will be greatly increased. Unless it was something 100s to 1000s of meters across, then the only way it could survive the passage and go back out into space is if its minimum altitude was many 10s of kilometers above the surface -- much higher than cirrus clouds.


A-L-E-X wrote:
Source of the post Could a sufficiently large one cause tidal impacts or even influence our weather?


Weather and climate could certainly be modified if the passage deposits a lot of soot into the atmosphere. But impacts are far more effective for that, since they eject a far greater amount of vaporized target material.  


As for tidal effects, no, since for the tidal effect to be significant you'd need something like a minor moon (100s of km across), but such an object is not well held together except by its own gravity, which means by being close enough to graze Earth's atmosphere it would first be torn apart near the Roche limit.
 
User avatar
midtskogen
Star Engineer
Star Engineer
Posts: 1237
Joined: 11 Dec 2016
Location: Oslo, Norway
Contact:

Science and Astronomy Questions

05 Sep 2020 09:39

Sounds reasonable.  My cameras have recorded a few Earth grazers, meteors that enter the atmosphere and "bounces" out again (it doesn't really bounce, but travel in a straight line and gain altitude again because of Earth's curvature).  In one instance it dipped down to about 60 km altitude.  Much deeper than this, the atmosphere gets too thick and meteor is probably doomed.  These were rocks less than 100 kg.  I think the mass requirement will increase rapidly below 50 km, and then other factors also come into play.

An asteroid whose trajectory would take it to just 10 km altitude and leaving Earth again, would likely have to be so large that in reality I think it would disintegrate completely even if it came in at minimum speed (~11 km/s), since asteroids that large would not be a solid block of iron.  It would be an incredible catastrophy, the heat and shockwave over a large area causing much more damage than the fragments actually hitting the ground.
NIL DIFFICILE VOLENTI
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 2764
Joined: 06 Mar 2017

Science and Astronomy Questions

08 Sep 2020 18:34

I would like you all to read this paper 

https://www.vox.com/platform/amp/scienc ... e-equation

This person encapsulates my thoughts very well about why we may not find any technological civilizations out in space......she addresses humankind's unsustainable and warlike practices and presents a mathematical equation that seems to indicate that there is a reasonable chance that we may be the only such civilization in our observable universe.  In the paper she even addresses the question of other intelligent animals on our planet.  
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 2764
Joined: 06 Mar 2017

Science and Astronomy Questions

08 Sep 2020 18:57

Watsisname wrote:
A-L-E-X wrote:
Source of the post Can one make it into (let's say) at the same altitude as our cirrus clouds and not get burned up or hit the planet?

Maybe, if it is large (100s of meters across or more) and dense enough, but even then it might not survive the journey for other reasons (the ram pressure may fragment it or even result in an airburst.) Otherwise more typical smaller asteroids penetrating that low into the atmosphere at such a shallow angle lose their cosmic velocity.

A useful technique to apply here is Newton's Impact Depth approximation. Basically, a high velocity projectile will be stopped after it has swept out its own mass worth of target material in its path. For high velocities the impact depth turns out to be roughly independent of the velocity, and the most important factors are the projectile's size and density relative to the target. If the target's density is uniform then this is simple to calculate: the impactor will have swept out its own mass worth of target once it has penetrated to a depth equal to its own length times the ratio of its density to the target density.

For an asteroid passing through the atmosphere, however, we must account for how the density of air decreases exponentially with altitude. The density of air at the Earth's surface is ρ0 = 1.225kg/m3, and this decays with a scale height H of about 8km (decreases by a factor of e for each 8km increase in height), according to the formula

Image

Now with a bit of calculus we can find out the depth a particular impactor should reach before losing its cosmic velocity if it comes in vertically.

Let the impactor have a mass mi, approximate length L, and density ρi. The mass of a column of atmosphere it sweeps out will be its cross sectional area L2 times the density ρ of the air at an altitude z times a sliver of altitude dz, integrated over the range of altitudes (from the impactor's final altitude to infinity where the air density trends to zero). Setting the mass of the column of atmosphere equal to the mass of the impactor, we have:

Image

Now solving for the final altitude zf

Image

Before we plug in any numbers, let's think about what this is saying. The scale factor is a multiplier in front, with a minus sign that might seem weird. But rest assured that for small impactors that are not stupidly dense, the answer will be positive, because the impactor's length times density will be much smaller than the scale height times the surface air density, so the natural logarithm will be negative which cancels the negative in front. That also leads to one nice piece of insight right away: the conditions needed for the impactor to be able to reach the ground before being slowed down fully. Set zf=0, and that means the argument of the natural log must be 1, and therefore Lρi must equal Hρ0. That is, the impactor length times density must equal the atmosphere's scale height times the air density at the surface!  If the impactor is even larger or more dense, then it will be able to hit the surface with some of its original cosmic velocity. 

For example, this result tells us that an iron asteroid (density ~8000kg/m3) must be at least about 1.3 meters across in order to reach the surface with some of its cosmic velocity. Anything smaller or less dense than that will instead fall to the ground at a much slower terminal speed (if it reaches the ground intact at all -- it could also fully ablate or airburst.) Or consider an icy meteoroid with a density of, say, 1000kg/m3. That would need to be at least about 10 meters across to reach the ground at a high speed.

What about Earth-grazers coming in at a shallow angle, so that they might neither be destroyed by the atmosphere nor hit the surface? Once again we can use the impact depth approximation, but integrated over the whole path. The key idea is that with such a shallow angle, the path through the atmosphere will be much longer, so for a given asteroid size and composition the minimum altitude it could reach before getting seriously slowed down will be greatly increased. Unless it was something 100s to 1000s of meters across, then the only way it could survive the passage and go back out into space is if its minimum altitude was many 10s of kilometers above the surface -- much higher than cirrus clouds.


A-L-E-X wrote:
Source of the post Could a sufficiently large one cause tidal impacts or even influence our weather?


Weather and climate could certainly be modified if the passage deposits a lot of soot into the atmosphere. But impacts are far more effective for that, since they eject a far greater amount of vaporized target material.  


As for tidal effects, no, since for the tidal effect to be significant you'd need something like a minor moon (100s of km across), but such an object is not well held together except by its own gravity, which means by being close enough to graze Earth's atmosphere it would first be torn apart near the Roche limit.

Thanks for this elegant explanation, Wat!  Cirrus clouds are around 40,000 ft up....so let's say we reframed the question to include the limits of the atmosphere....the exosphere extends upward to about 1,000 miles I think?  So if we reframed the question to that level, how likely do you think such a "graze" would be?  Also, you mentioned that a very high density changes the equation- and that brought to my mind the idea of micro black holes.....I remember for awhile some believed the Tunguska event was actually a micro black hole collision event.  We now know better.  Assuming that micro black holes do exist, how close would one have to get to our planet's surface for its effects to be noticeable?
 
User avatar
Watsisname
Science Officer
Science Officer
Posts: 2105
Joined: 06 Sep 2016
Location: Bellingham, WA

Science and Astronomy Questions

09 Sep 2020 00:35

A-L-E-X wrote:
Source of the post let's say we reframed the question to include the limits of the atmosphere....the exosphere extends upward to about 1,000 miles I think?  So if we reframed the question to that level, how likely do you think such a "graze" would be?

As midtskogen mentioned, they happen pretty often and his meteor observing network has recorded a few. If you observe a meteor shower when the radiant is near the horizon, you might spot the occasional meteor skim the upper atmosphere and go back out into space, again not because they 'bounce' but because they did not get slowed down enough, and their path continues on in close to a straight line. Midtskogen can probably give a good description or imagery of what such a grazer will look like and how to know if you see one.

Grazing meteors, and most meteors in general, are usually visible between around 80 to 120km altitude. Sometimes they can be seen a little lower if they are bigger, or little higher if they are faster. Above around 200km the air density is too low for aerodynamic effects to be visible or important for the meteor.


Assuming that micro black holes do exist, how close would one have to get to our planet's surface for its effects to be noticeable?



That depends very significantly on the mass. On the smaller end of the scale the Hawking radiation would be noticeable, and on the larger end the gravitational effects would be noticeable. And if it passes through the atmosphere or the Earth's surface, then luminosity due to accretion would be very important. I refer to my earlier post on microscopic black hole effects on the Earth for the details. :)
 
User avatar
JackDole
Star Engineer
Star Engineer
Posts: 1874
Joined: 02 Nov 2016
Location: Terra

Science and Astronomy Questions

14 Sep 2020 18:38

How do I calculate the gravitation at the Schwarzschild radius of a black hole?
I tried to create a 'Birch Planet'. I have read that the black hole must have about 1.5 trillion solar masses so that the gravitation at the Schwarzschild radius is the same as on earth. https://space-engine.fandom.com/wiki/Birch_Planet
But how do I calculate that exactly?
scr01221.jpg
JackDole's Universe 0.990: http://forum.spaceengine.org/viewtopic.php?f=3&t=546
JackDole's Archive: http://forum.spaceengine.org/viewtopic.php?f=3&t=419
JackDole: Mega structures ... http://old.spaceengine.org/forum/17-3252-1 (Old forum)

Who is online

Users browsing this forum: No registered users and 1 guest