kamaz

2012-Mar-15, 09:55 PM

I can see Spica shining through my window right now... how to calculate how many photons per second from that star hit my retina?

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kamaz

2012-Mar-15, 09:55 PM

I can see Spica shining through my window right now... how to calculate how many photons per second from that star hit my retina?

slang

2012-Mar-15, 10:04 PM

I found a rather interesting site that has multiple discussions on the topic... see if these (http://www.bautforum.com/archive/index.php/t-41457.html) two (http://www.bautforum.com/archive/index.php/t-159.html) help?

kamaz

2012-Mar-15, 11:22 PM

Wow, cool site you got there... I am tempted to join :D

After reading the threads, it looks like a seriously complicated way of calculating that... How about this: convert apparent magnitude to W/m^2 and divide by photon energy...

After reading the threads, it looks like a seriously complicated way of calculating that... How about this: convert apparent magnitude to W/m^2 and divide by photon energy...

StupendousMan

2012-Mar-16, 02:36 AM

Quick and simple number: a star of magnitude zero will produce roughly one million photons per second per square centimeter in the V-band. Ignoring factors of three, just multiply by the area of your pupil (pi * 0.25cm*0.25cm ~= 0.2 sq.cm.) and you get a few hundred thousand photons per second.

George

2012-Mar-21, 03:26 PM

Wow, cool site you got there... I am tempted to join :D

After reading the threads, it looks like a seriously complicated way of calculating that... How about this: convert apparent magnitude to W/m^2 and divide by photon energy...

Yes, but Spica is extremely hot, so spectral photon distribution is significantly different than say the Sun.

Assuming we see about 400 w/m^2 of visible energy from the Sun at sea level -- I think the visible poriton from space is about 45% of the total -- and we chop this by the brightness difference of 27.74 magnitudes between the Sun and Spica (or 8x10^-12), then we would have about 3.2x10^-9 w/m^2, or 3.2x10-12 w/cm^2.

Since Spica is very, very hot (18,500K) [added: primary is 22,000K], the peak wavelength of energy from Spica is around 150nm [130 nm]. The ratio of "blue" photons compared to the red ones is close to 5 to 1, unlike the Sun, which is almost a flat photon distribution across the visible spectrum. Since the "blue" photons have almost twice the energy of the red ones, then things are little more complicated to get to a more accurate photon flux density.

Assuming a Planck distribution, playing with an average photon energy value within Spica's spectrum of ~ 3.7 x10^-15 Joules/photon, which when divided into the enegy flux above, yields about 850 photons per second per sq. cm. So for a 6 mm pupil, my flux is about 250 photons per second into an eye. [Added: This seems too low to be right, so where am I wrong?]

After reading the threads, it looks like a seriously complicated way of calculating that... How about this: convert apparent magnitude to W/m^2 and divide by photon energy...

Yes, but Spica is extremely hot, so spectral photon distribution is significantly different than say the Sun.

Assuming we see about 400 w/m^2 of visible energy from the Sun at sea level -- I think the visible poriton from space is about 45% of the total -- and we chop this by the brightness difference of 27.74 magnitudes between the Sun and Spica (or 8x10^-12), then we would have about 3.2x10^-9 w/m^2, or 3.2x10-12 w/cm^2.

Since Spica is very, very hot (18,500K) [added: primary is 22,000K], the peak wavelength of energy from Spica is around 150nm [130 nm]. The ratio of "blue" photons compared to the red ones is close to 5 to 1, unlike the Sun, which is almost a flat photon distribution across the visible spectrum. Since the "blue" photons have almost twice the energy of the red ones, then things are little more complicated to get to a more accurate photon flux density.

Assuming a Planck distribution, playing with an average photon energy value within Spica's spectrum of ~ 3.7 x10^-15 Joules/photon, which when divided into the enegy flux above, yields about 850 photons per second per sq. cm. So for a 6 mm pupil, my flux is about 250 photons per second into an eye. [Added: This seems too low to be right, so where am I wrong?]

Hornblower

2012-Mar-21, 05:22 PM

Yes, but Spica is extremely hot, so spectral photon distribution is significantly different than say the Sun.

Assuming we see about 400 w/m^2 of visible energy from the Sun at sea level -- I think the visible poriton from space is about 45% of the total -- and we chop this by the brightness difference of 27.74 magnitudes between the Sun and Spica (or 8x10^-12), then we would have about 3.2x10^-9 w/m^2, or 3.2x10-12 w/cm^2.

Since Spica is very, very hot (18,500K) [added: primary is 22,000K], the peak wavelength of energy from Spica is around 150nm [130 nm]. The ratio of "blue" photons compared to the red ones is close to 5 to 1, unlike the Sun, which is almost a flat photon distribution across the visible spectrum. Since the "blue" photons have almost twice the energy of the red ones, then things are little more complicated to get to a more accurate photon flux density.

Assuming a Planck distribution, playing with an average photon energy value within Spica's spectrum of ~ 3.7 x10^-15 Joules/photon, which when divided into the enegy flux above, yields about 850 photons per second per sq. cm. So for a 6 mm pupil, my flux is about 250 photons per second into an eye. [Added: This seems too low to be right, so where am I wrong?]

This looks like a units conversion foulup. In the blue end of the visible spectrum the energy per photon is about 3 electron volts which equals about 5x10^-19 joule.

I estimated about 2,000 photons per second through a dilated pupil from a 6th magnitude star. I got that starting with the Sun delivering about 1,300 joules per square meter with about half of that in the form of visible light, at roughly 2 electron volts per photon. That would be about 200,000 photons per second for a first magnitude G star, or about 2/3 of that for a B star where the average energy per photon would be closer to 3 electron volts.

Assuming we see about 400 w/m^2 of visible energy from the Sun at sea level -- I think the visible poriton from space is about 45% of the total -- and we chop this by the brightness difference of 27.74 magnitudes between the Sun and Spica (or 8x10^-12), then we would have about 3.2x10^-9 w/m^2, or 3.2x10-12 w/cm^2.

Since Spica is very, very hot (18,500K) [added: primary is 22,000K], the peak wavelength of energy from Spica is around 150nm [130 nm]. The ratio of "blue" photons compared to the red ones is close to 5 to 1, unlike the Sun, which is almost a flat photon distribution across the visible spectrum. Since the "blue" photons have almost twice the energy of the red ones, then things are little more complicated to get to a more accurate photon flux density.

Assuming a Planck distribution, playing with an average photon energy value within Spica's spectrum of ~ 3.7 x10^-15 Joules/photon, which when divided into the enegy flux above, yields about 850 photons per second per sq. cm. So for a 6 mm pupil, my flux is about 250 photons per second into an eye. [Added: This seems too low to be right, so where am I wrong?]

This looks like a units conversion foulup. In the blue end of the visible spectrum the energy per photon is about 3 electron volts which equals about 5x10^-19 joule.

I estimated about 2,000 photons per second through a dilated pupil from a 6th magnitude star. I got that starting with the Sun delivering about 1,300 joules per square meter with about half of that in the form of visible light, at roughly 2 electron volts per photon. That would be about 200,000 photons per second for a first magnitude G star, or about 2/3 of that for a B star where the average energy per photon would be closer to 3 electron volts.

StupendousMan

2012-Mar-21, 07:30 PM

Take note: I wrote

a star of magnitude zero will produce roughly one million photons per second per square centimeter in the V-band. Ignoring factors of three, just multiply by the area of your pupil (pi * 0.25cm*0.25cm ~= 0.2 sq.cm.) and you get a few hundred thousand photons per second.

and Hornblower wrote

I estimated about 2,000 photons per second through a dilated pupil from a 6th magnitude star.

Since a sixth-magnitude star is about 250 times fainter than zeroth magnitude star, these two rough values agree. This is a good sign that these two people are doing a good job.

a star of magnitude zero will produce roughly one million photons per second per square centimeter in the V-band. Ignoring factors of three, just multiply by the area of your pupil (pi * 0.25cm*0.25cm ~= 0.2 sq.cm.) and you get a few hundred thousand photons per second.

and Hornblower wrote

I estimated about 2,000 photons per second through a dilated pupil from a 6th magnitude star.

Since a sixth-magnitude star is about 250 times fainter than zeroth magnitude star, these two rough values agree. This is a good sign that these two people are doing a good job.

George

2012-Mar-22, 04:23 AM

This looks like a units conversion foulup. In the blue end of the visible spectrum the energy per photon is about 3 electron volts which equals about 5x10^-19 joule. Thanks. Yes, I think I muffed the conversion in E= h nu, but I haven't access to my prior work to see what rushing error I created. [I get 4.32E-19 Joules for 460nm.]

Not having time to pull actual sp. irr. data, I'll use my available Planck distribution values, which will average about the same, though a little less for visible light, I think.

Per sq. meter, the Sun produces about 400 watts of visible light at the surface (1000 watts total surface energy). The sp. irr. numbers average around 1700 watts, so the actual values will be about 23% of sp. irr. values. Using the photon energy values for each wavelength, the Sun produces about 400/1700 of ~ 4.5E21 photons per second per sq. meter. To reduce this to a 1.04 mag. level, we must reduce this by a factor of 1.25E11.

Reducing these units per sq. meter to that of a 6mm eye (~28mm^2) gives me about 250,000 photons per seconds, at least if it was a G2 star. The 22,000 K temp. changes things some because a much larger portion of the flux is in the blue portion, but the blue color cones are only about 70% as effective. The net effect may not alter the photon flux that much, but it might. [Must rush off before answering this last part.]

Not having time to pull actual sp. irr. data, I'll use my available Planck distribution values, which will average about the same, though a little less for visible light, I think.

Per sq. meter, the Sun produces about 400 watts of visible light at the surface (1000 watts total surface energy). The sp. irr. numbers average around 1700 watts, so the actual values will be about 23% of sp. irr. values. Using the photon energy values for each wavelength, the Sun produces about 400/1700 of ~ 4.5E21 photons per second per sq. meter. To reduce this to a 1.04 mag. level, we must reduce this by a factor of 1.25E11.

Reducing these units per sq. meter to that of a 6mm eye (~28mm^2) gives me about 250,000 photons per seconds, at least if it was a G2 star. The 22,000 K temp. changes things some because a much larger portion of the flux is in the blue portion, but the blue color cones are only about 70% as effective. The net effect may not alter the photon flux that much, but it might. [Must rush off before answering this last part.]

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