Starship Asterisk*

Solar Flare in the Gamma ray Sky

Explanation: What shines in the gamma-ray sky? The answer is usually the most exotic and energetic of astrophysical environments, like active galaxies powered by supermassive black holes, or incredibly dense pulsars, the spinning remnants of exploded stars. But on March 7, a powerful solar flare, one of a series of recent solar eruptions, dominated the gamma-ray sky at energies up to 1 billion times the energy of visible light photons. These two panels illustrate the intensity of that solar flare in all-sky images recorded by the orbiting Fermi Gamma-ray Space Telescope. On March 6, as on most days, the Sun was almost invisible to Fermi's imaging detectors. But during the energetic X-class flare, it became nearly 100 times brighter than even the Vela Pulsar at gamma-ray energies. Now faded in Fermi's view, the Sun will likely shine again in the gamma-ray sky as the solar activity cycle approaches its maximum.

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This might be of interest to those who have the means to overlay this

http://apod.nasa.gov/apod/image/0907/Fe ... s_2048.jpg

I am a "newbie" to APOD, but enjoy the images. My question is "How was today's image taken"? What is THAT far out in space that makes our solar system look THAT small?

SoonieKoo wrote:I am a "newbie" to APOD, but enjoy the images. My question is "How was today's image taken"? What is THAT far out in space that makes our solar system look THAT small?

Good question, and I'm not the right person to answer it. But the Fermi telescope is a space telescope, which is either orbiting the Earth or possibly stationary in one of the Lagrange points between the Earth and the Moon.

The thing is that a space telescope doesn't have to be very far away from Earth to take impressive all-sky pictures. Does it look as if the Sun is light-years away from the telescope that took the picture? Maybe the answer is more mundane, namely that the Sun isn't overwhelimingly bright in gamma rays even during a powerful X-ray outburst. If an optical telescope is going to take pictures of astronomical objects in opitical light, then that telescope must always "look away" from the Sun, because the Sun is blindingly bright in optical light and will wash out the optical light signals from all other objects. Or that is what it looks like from our point of view.

The Sun is never blindingly bright in gamma rays, so the Fermi telescope doesn't have to "look away" from the Sun. It can look at the sky and the Sun at the same time without any problems at all.

To put things in perspective, I have to wonder if a similar X-ray outburst that produced a similar flash of gamma rays on the nearest star from the Sun, Alpha Centauri, would even have been detectable in today's APOD?

Ann

you can find out more about the fermi space scope here http://fermi.gsfc.nasa.gov/

in the second picture, the lower left light is the sun, not the entire solar system;

the reason it looks so small, is simply, because the oval picture you see is in fact what is called a mollweide equal area map projection of a sphere, the entire sky in this case (but could also be the surface of a planet if you like); so obviously, if you are going to take a picture that doesn't include just what is in front of you, but also what is above you, below, behind, to the right and left, and put it all in a single picture, then what you see in front of you appears smaller, because it is relative to the entire view all around you; it is just a matter of perspective;

as for the mollweide projection itself, the reason NASA likes to use it is because it has the special property that a square centimeter of map anywhere on the map is proportional to the actual sphere - so if a square cm of map is a square km of planet or sky, then it is the same all over the map; comparatively, a rectangular map like your ordinary wall map, which also shows the entire planet (or sky) also maps a sphere onto a picture, rectangular instead of an oval mollweide, however, it is distorted - in a rectangular map, a square centimeter near the pole covers far more square km than near the equator, so it is stretched and does not represent the surface correctly - but it is easy to map onto a sphere by a computer; if you just want to see the entire sky or planet in one picture, mollweide is better;

Ann wrote:

SoonieKoo wrote:I am a "newbie" to APOD, but enjoy the images. My question is "How was today's image taken"? What is THAT far out in space that makes our solar system look THAT small?

Good question, and I'm not the right person to answer it. But the Fermi telescope is a space telescope, which is either orbiting the Earth or possibly stationary in one of the Lagrange points between the Earth and the Moon.
Ann

So now we're being bombarded by gamma rays! That must be par; but how much does that affect our evolution as man?

Are you saying that during the CME the sun emitted 1 billion times more energy in gamma rays than in visible light? Or are you saying that an individual gamma ray photon is a billion times more energetic than a visible light photon?

Because I don't think the latter statement is correct. I believe gamma rays are more in the order of 10^15 more energetic than visible light.

And if the former is correct--that the sun emitted a billion times more energy in gamma rays than in visible light--it seems to me that the planet would be fried to a cinder!

Did I misunderstand?

I had no idea that solar flares could emit gamma rays. That totally surprises me, actually.

I know that there are lots of gamma photons generated in the sun's core due to fusion going on there, but that makes sense because nuclei are so tightly bound and atomic level interactions are so energetic.

The upper layers of the sun though are very cool in comparison. What kind of mechanism is generating gamma photons in solar flares? Is that understood at all?

-s

K1NS wrote:
Are you saying that during the CME the sun emitted 1 billion times more energy in gamma rays than in visible light? Or are you saying that an individual gamma ray photon is a billion times more energetic than a visible light photon?

Because I don't think the latter statement is correct. I believe gamma rays are more in the order of 10^15 more energetic than visible light.

And if the former is correct--that the sun emitted a billion times more energy in gamma rays than in visible light--it seems to me that the planet would be fried to a cinder!

Did I misunderstand?

I believe gamma rays are more in the order of 10¹⁵ more energetic than radio waves:

K1NS wrote:Are you saying that during the CME the sun emitted 1 billion times more energy in gamma rays than in visible light? Or are you saying that an individual gamma ray photon is a billion times more energetic than a visible light photon?

The energy of a visible light photon is around 2 eV. The energy of gamma ray photons ranges from about 1e6 eV to 1e10 eV. So if this event produced hard gamma rays, it is reasonable to assume the photons were about a billion times more energetic than the visible light photons produced by the Sun.

And if the former is correct--that the sun emitted a billion times more energy in gamma rays than in visible light--it seems to me that the planet would be fried to a cinder!

Yes, it would! The total energy of the gamma rays was only a fraction of the total energy output of the Sun. "Brightness" is determined by the photon flux, not the photon energy, and the Sun didn't emit very many gamma ray photons compared with the much less energetic ones is emits all the time.

What is the object in the center of the image above the galactic plane? It appears brighter on the 7th than on the 6th.

Chris Peterson wrote:

And if the former is correct--that the sun emitted a billion times more energy in gamma rays than in visible light--it seems to me that the planet would be fried to a cinder!

Yes, it would! The total energy of the gamma rays was only a fraction of the total energy output of the Sun. "Brightness" is determined by the photon flux, not the photon energy, and the Sun didn't emit very many gamma ray photons compared with the much less energetic ones is emits all the time.

Yes, as a whole and on average, the sun emits very little energy in the gamma range. Gamma photons are MUCH more energetic than visible photons, but the sun rarely spits them out. This is because the gamma photons generated in the core end up getting absorbed and reduced to multiple "smaller" photons repeatedly before they ever leak out to the surface layers. The surface of the sun is relatively cool and generally does not emit gamma photons.

You can go to google images and search for "solar energy spectrum" to see how much energy the sun emits in the various frequency ranges. Solar energy peaks in the visible. The vast majority of solar light is emitted in the visible and infrared range. Energy emission falls off sharply in the UV range and that well below the super energetic gamma photon range.

Here's a link to a suitable image that I scraped from a google images search:



It's remarkable, and very surprising to me that solar flares are so bright in the gamma range. Is that typically the case or are gamma rich flares relatively special?

-s

Psnarf wrote:What is the object in the center of the image above the galactic plane? It appears brighter on the 7th than on the 6th.

I think it's the quasar PKS 1510-089

http://www.nasa.gov/mission_pages/GLAST ... scope.html

bystander wrote:

Psnarf wrote:
What is the object in the center of the image above the galactic plane? It appears brighter on the 7th than on the 6th.

I think it's the quasar PKS 1510-089

http://www.nasa.gov/mission_pages/GLAST ... scope.html

http://iopscience.iop.org/1538-3881/135/6/2212/fulltext/aj_135_6_2212.text.html wrote:
<<The GeV γ-ray-loud blazar PKS 1510-089 is one of the best-monitored active galactic nuclei (AGNs). The most interesting feature of the PKS 1510-089 optical light curve is a 336 ± 14 d (0.92 yr) quasi-periodic flux minimum (Xie et al. 2002). A hypothesis has been proposed, suggesting that there is a supermassive binary black hole (SMBBH) system in the center of the source and the deep flux minimum is caused by the eclipse of the system (Xie et al. 2002). This minimum re-occurred almost exactly as we predicted according to this binary star theory, resulting in a quasi-periodic deep flux minimum, i.e. the main minimum; the primary black hole could also eclipse a secondary black hole, leading to one quasi-periodic deep flux minimum, i.e., the secondary minimum. In a word, the quasi-periodic deep flux minimum suggests that the orbital periodicity of the SMBBH system is ~672 ± 28 days.>>

Even though my local atmosphere is full of water vapor, and thus prevents me from seeing the Venus - Jupiter conjunction, I'm still grateful to live on a planet with a magnetic field and dense atmosphere, especially when there are solar x-ray flares and coronal mass ejections! Here's a question for the mathematically inclined: How much more of an effect would these gamma rays have on the surface of Mars, compared to the effect here on Earth's surface?

Anthony Barreiro wrote:Even though my local atmosphere is full of water vapor, and thus prevents me from seeing the Venus - Jupiter conjunction, I'm still grateful to live on a planet with a magnetic field and dense atmosphere, especially when there are solar x-ray flares and coronal mass ejections! Here's a question for the mathematically inclined: How much more of an effect would these gamma rays have on the surface of Mars, compared to the effect here on Earth's surface?

Mars is about 1.5 times farther from the Sun than the Earth, so the intensity of the radiation will be reduced to about 40% of that here. But the energy of each photon is unchanged, and with hard gamma rays even one photon can do serious biological damage. Since most of the radiation will make it to the ground on Mars (as opposed to virtually none on Earth), this is a very real concern for any future Mars missions. Some type of shielding will be essential to protect people on the Martian surface.

In follow up then to Chris Peterson's response about gamma rays hitting Mars' surface, what about the astronauts in the Space Station? The thin metal of their vessel would absorb much of the gamma rays and re-radiate them as lower energy x-rays, but more of them (conservation of energy). They would be absorbed too and re-radiate with even more photons of yet lower energy, etc. It all depends upon the thickness of the metal shield and what the shield is made of. Decades ago I read a small book that studied the practical problems of Solar System exploration and this issue was one of those presented.

Thanks Chris. I read that the Curiosity rover is monitoring radiation in flight on its way to Mars and will continue to monitor radiation on the surface, to help the engineers plan how to shield astronauts, should we ever have a human mission to Mars.

Chris Peterson wrote:

K1NS wrote:
[If] the sun emitted a billion times more energy in gamma rays than in visible light-
-it seems to me that the planet would be fried to a cinder!

Yes, it would! The total energy of the gamma rays was only a fraction of the total energy output of the Sun.

"Brightness" is determined by the photon flux, not the photon energy, and the Sun didn't emit very many gamma ray photons compared with the much less energetic ones is emits all the time.

Brightness = lumens per steradian = energy flux/steradian as observed visually.

The most common use of brightness involving "invisible light" radiation
refers to energy flux/steradian and NOT to photon flux/steradian.

http://en.wikipedia.org/wiki/Candela wrote:
<<Prior to 1948, various standards for luminous intensity were in use in a number of countries. These were typically based on the brightness of the flame from a "standard candle" of defined composition, or the brightness of an incandescent filament of specific design. The term candlepower was originally defined in England by the Metropolitan Gas Act 1860 as the light produced by a pure spermaceti candle weighing one sixth of a pound and burning at a rate of 120 grains per hour. Spermaceti is found in the head of sperm whales, and once was used to make high quality candles. Germany, Austria and Scandinavia used the Hefnerkerze, a unit based on the output of a Hefner lamp.

The candela (symbol: cd) is the SI base unit of luminous intensity; that is, power emitted by a light source in a particular direction, weighted by the luminosity function (a standardized model of the sensitivity of the human eye to different wavelengths, also known as the luminous efficiency function). A common candle emits light with a luminous intensity of roughly one candela:

The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540×10¹² hertz and that has a radiant intensity in that direction of 1⁄683 watt per steradian.

A 25 W compact fluorescent light bulb puts out around 1700 lumens; if that light is radiated equally in all directions, it will have an intensity of around 135 cd. The luminous intensity of light-emitting diodes is measured in millicandela (mcd), or thousandths of a candela. "Ultra-bright" LEDs can reach 15 000 mcd, or higher.>>

--------------------------------------------------------
1 KING HENRY VI Act 1, Scene 1

Messenger: A Talbot! a Talbot! cried out amain
And rush'd into the bowels of the battle.
--------------------------------------------------------
1 lumen = 1 talbot per second = 1⁄683 watt
1 candela = 1 lumen per steradian
1 nit = 1 candela per square meter

Code: Select all

Quantity              Symbol 	SI unit 	Symbol 	Dimension 	Notes
-------------------------------------------------------------------------------------------------------
Luminous energy	      Qv	 lumen second 	lm⋅s 	T⋅J [nb 3] 	units are sometimes called talbots
Luminous flux 	       Φv	 lumen (= cd⋅sr) 	lm 	J 	also called luminous power
Luminous intensity 	  Iv 	candela (= lm/sr) 	cd 	J 	an SI base unit, luminous flux per unit solid angle
Luminance 	           Lv 	candela per m^2 	cd/m^2 	J/L^2 	units are sometimes called nits
Illuminance 	         Ev 	lux (= lm/m^2) 	lx 	J/L^2 	used for light incident on a surface
Luminous emittance 	  Mv 	lux (= lm/m^2) 	lx 	J/L^2 	used for light emitted from a surface
Luminous exposure 	   Hv 	lux second 	lx⋅s 	T⋅J/L^2
Luminous energy density ωv 	lumen second per metre^3 	lm⋅s/m^3 	: T⋅J/L^3

Opening narration The Incredible Hulk (tv series) wrote:
Dr. David Banner: physician; scientist. Searching for a way to tap into the hidden strengths that all humans have. Then an accidental overdose of

gamma radiation

alters his body chemistry. And now when David Banner grows angry or outraged, a startling metamorphosis occurs. The creature is driven by rage and pursued by an investigative reporter. [Banner:] "Mr. McGee, don't make me angry. You wouldn't like me when I'm angry." The creature is wanted for a murder he didn't commit. David Banner is believed to be dead, and he must let the world think that he is dead, until he can find a way to control the raging spirit that dwells within him.

neufer wrote:1 nit = 1 candela per square meter

Hmm... So if you're a 'nitpicker', you should end up being a very bright person after a while, providing you don't lose your nits, in which case you could end up like this

Beyond wrote:

neufer wrote:
1 nit = 1 candela per square meter

Hmm... So if you're a 'nitpicker', you should end up being a very bright person after a while,
providing you don't lose your nits, in which case you could end up like this

Hmm... You're definitely one of the brightest nit wits here on the Starship Asterisk*

• Love's Labour's Lost Act 4, Scene 1

COSTARD: By my soul, a swain! a most simple clown!

• Lord, Lord, how the ladies and I have put him down!
  O' my troth, most sweet jests! most incony vulgar wit!
  When it comes so smoothly off, so obscenely, as it were, so fit.
  Armado o' th' one side,--O, a most dainty man!
  To see him walk before a lady and to bear her fan!
  To see him kiss his hand! and how most sweetly a' will swear!
  And his page o' t' other side, that handful of wit!
  Ah, heavens, it is a most pathetical nit!
  Sola, sola!

I think nit-picking is good. Shows you give a candela about things. Very much like a Physics University professor and specially those who have been chosen to be their TA.

neufer wrote:The most common use of brightness involving "invisible light" radiation
refers to energy flux/steradian and NOT to photon flux/steradian.

http://en.wikipedia.org/wiki/Candela wrote:

I disagree. In practice, "brightness" is usually casually synonymous with "intensity", are refers to the particle (typically photon) flux.

Chris Peterson wrote:

neufer wrote:
The most common use of brightness involving "invisible light" radiation
refers to energy flux/steradian and NOT to photon flux/steradian.

http://en.wikipedia.org/wiki/Candela wrote:

I disagree. In practice, "brightness" is usually casually synonymous with "intensity", are refers to the particle (typically photon) flux.

You're thinking about particle accelerators; I'm talking about light.