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Philippe Lemay
2010-Sep-23, 09:18 PM
I'm trying to use Hubble as a baseline sensor platform for modern technology in detecting space objects, and I'd like to know what the dimmest object it can detect is. The only unit I know to work with right now is apparent magnitude, but if someone could also give me a way to convert apparent magnitude into watts per square meter, that would help.

Philippe Lemay
2010-Sep-23, 09:24 PM
Hmmm, Wikipedia gave me this.

The Hubble Space Telescope has located stars with magnitudes of 30 at visible wavelengths and the Keck telescopes have located similarly faint stars in the infrared.Which helps. But still, I would still like to find a way to convert the apparent magnitude method into a watts per square meter, if it's possible.

Philippe Lemay
2010-Sep-23, 10:07 PM
Ok I ran across a rather involved process, and I'd like to run it by an expert.

The Sun's apparent magnitude is -26.73. This means the total difference between the sun's radiance and the dimmest object Hubble can detect is 56.73 apparent magnitude. The difference from one level of apparent magnitude to another is... 2.5119, which gives us about 4.92x10^22 more brightness in the sun than in our mag 30 object.

The Sun, supposedly, has 1366 watts per square meters at the Earth's orbit. So... if I divide 1366 by (4.92x10^22), I should get the amount of watts per square meter coming from the dimmest detectable object by Hubble. Or about 2.776x10^-20 watts per square meter. Anyone with better math skills who is will to check my calculations, please by all means do.

ngc3314
2010-Sep-23, 10:54 PM
Or, a bit more directly and depending on the filter system used for a particular magnitude, there are values for the mean flux per unit wavelength within that band corresponding to magnitude 0. ("Mean" defined in a particular way, because the flux is integrated through a filter with a defined response curve and the detection is in generally photons rather than energy units). If this reference flux is F0, the flux corresponding to magnitude m will be F0 X 10(-0.4m). There is a table of flux zero points for various HST filter bands here (http://www.stsci.edu/documents/dhb/web/c32_wfpc2dataanal.fm1.html).

Philippe Lemay
2010-Sep-23, 11:18 PM
The energy value of a photon changes depending on it's wavelength, right?

For the purposes of the discussion, I think I should focus on the average wavelength of infrared emissions.

AndreasJ
2010-Sep-24, 06:21 AM
The energy value of a photon changes depending on it's wavelength, right?
Indeedy. Specifically:

http://www.codecogs.com/eq.latex?E=\frac{hc}{\lambda}

where h is Planck's constant, c the speed of light, and λ the wavelength.

Hornblower
2010-Sep-24, 12:33 PM
My rough estimate for a magnitude 30 object is less than a photon per second into Hubble's aperture. That means long integration times of many hours to build up a useful signal in a state of the art detector.

Philippe Lemay
2010-Sep-24, 07:31 PM
Wow, 1 photon a second. I should probably go for something a little more reservist for my baseline of modern technology. Especially if I want to build a whole network of these hypothetical sensors (the Hubble's pretty big).

This one (also found on wikipedia) seems pretty consistent:
Apparent Magnitude 22. Approximate limiting magnitude of a 24" Ritchey-Chrétien telescope with 30 minutes of stacked images (6 subframes at 300s each) using a ccd detector[35]

Re-doing the numbers, I get our dimmest detectable object at (3.1e19) times dimmer than the sun (being 1366 watts per m˛), so this new object would throw... (4.4e-17) watts per square meter. Now I just need to calculate how many infrared photons that is...

Also, there's a relation to the kind of wavelength being emitted and an object's temperature, right? Does anyone know that equation?
Also... the wavelengths in the equation are expressed in meters, or nanometers?

George
2010-Sep-24, 08:26 PM
My rough estimate for a magnitude 30 object is less than a photon per second into Hubble's aperture. That means long integration times of many hours to build up a useful signal in a state of the art detector. That is an interesting result. Crunch...crunch...yep. For a solar twin, it is 1.14 photons per sec.

George
2010-Sep-24, 08:32 PM
Also, there's a relation to the kind of wavelength being emitted and an object's temperature, right? Does anyone know that equation? Use the one AndreasJ gave. E represents the engery of each photon, which can be easily expressed as a Joule and a Joule per sec is a watt, so you can convert any wattage to photon flux, just be very careful with your units.

neilzero
2010-Sep-24, 08:41 PM
You likely need to specify a portion of the Infrared spectrum as the whole band is quite wide, and significantly dependent on the star's color temperature of it's photosphere. Neil

Jeff Root
2010-Sep-25, 02:11 AM
Hornblower,

Just to be sure: That is "less than a photon per second" through
Hubble's aperture, rather than striking a detector or actually
detected by the detector?

George,

Knock yourself on the side of the head. Some gear or something
must have got stuck. Philippe wants the Stefan-Boltzmann law,
Planck's law of blackbody radiation (also called the Wien-Planck
law), Wien's displacement law, and the Wien distribution law
(also called Wien's law). The latter is useful for approximate
calculations in certain cases.

None of which I am competent to apply accurately.

An individual photon does not have a temperature, since
temperature is a characteristic of the motions of large numbers
of particles giving off a distribution of different wavelengths
(or frequencies) of light as described by the above-named laws.

-- Jeff, in Minneapolis

Hornblower
2010-Sep-25, 02:42 AM
Hornblower,

Just to be sure: That is "less than a photon per second" through
Hubble's aperture, rather than striking a detector or actually
detected by the detector?

-- Jeff, in Minneapolis

It is through the aperture and striking a detector, as a result of being reflected by the main mirror and focused onto the detector.

From the aforementioned calculations I found about 2.5 x 10^-20 joule per second per square meter, or about 10^-19 joule per second striking Hubble's main mirror. That is less than one electron volt per second, with visible light being 2 or 3 ev per photon and only part of the total flux. I don't know how efficient the detectors are.

WayneFrancis
2010-Sep-25, 05:50 AM
I'm trying to use Hubble as a baseline sensor platform for modern technology in detecting space objects, and I'd like to know what the dimmest object it can detect is. The only unit I know to work with right now is apparent magnitude, but if someone could also give me a way to convert apparent magnitude into watts per square meter, that would help.

The problem with "apparent magnitude", as I understand it, is that it is a term used by optical astronomers. As an objects are further and further away we receive less photons from them due to the inverse square law. Now with actual objects we don't just have that inverse square law happening. We also have photons red shifting out of the optical range reducing their apparent brightness in most cases. and in the cases where the star peak luminosity was in the ultra violet and was red shifted into optical light then I think they'd be very dim just from the total distance.

Ack I just reread your question...its pretty straight forward as I'd be surprised if this isn't in the Wiki entry for the HST

WayneFrancis
2010-Sep-25, 05:51 AM
Hmmm, Wikipedia gave me this.
Which helps. But still, I would still like to find a way to convert the apparent magnitude method into a watts per square meter, if it's possible.

Ah this is what I get for reading and responding to posts in a thread as I read them :)

Wasn't there just a astronomy cast episode discussing this?

WayneFrancis
2010-Sep-25, 05:56 AM
Ok I ran across a rather involved process, and I'd like to run it by an expert.

The Sun's apparent magnitude is -26.73. This means the total difference between the sun's radiance and the dimmest object Hubble can detect is 56.73 apparent magnitude. The difference from one level of apparent magnitude to another is... 2.5119, which gives us about 4.92x10^22 more brightness in the sun than in our mag 30 object.

The Sun, supposedly, has 1366 watts per square meters at the Earth's orbit. So... if I divide 1366 by (4.92x10^22), I should get the amount of watts per square meter coming from the dimmest detectable object by Hubble. Or about 2.776x10^-20 watts per square meter. Anyone with better math skills who is will to check my calculations, please by all means do.

That sounds like a good generalisation but the problem is higher frequency photons have more energy even if we don't perceive them as brighter. Hence another problem with apparent brightness in today's astronomy. Again the astronomy cast episode that recently discussed this is pretty good.

WayneFrancis
2010-Sep-25, 05:58 AM
The energy value of a photon changes depending on it's wavelength, right?

For the purposes of the discussion, I think I should focus on the average wavelength of infrared emissions.

Man I keep answering questions you've already answered for yourself :P

WayneFrancis
2010-Sep-25, 06:01 AM
My rough estimate for a magnitude 30 object is less than a photon per second into Hubble's aperture. That means long integration times of many hours to build up a useful signal in a state of the art detector.

Any idea of how efficient the CCD is that captures those photons? I think that factors into it too. IE we get a dimmer image just from the loss of some of the photons hitting the CCD but between actual sensors on the CCD. Much like a screen door dims the light coming through it.

WayneFrancis
2010-Sep-25, 06:06 AM
Use the one AndreasJ gave. E represents the engery of each photon, which can be easily expressed as a Joule and a Joule per sec is a watt, so you can convert any wattage to photon flux, just be very careful with your units.

Someone should ask Google to put that in its converter. :)

George
2010-Sep-25, 03:27 PM
Knock yourself on the side of the head. Some gear or something
must have got stuck. Philippe wants the Stefan-Boltzmann law,
Planck's law of blackbody radiation (also called the Wien-Planck
law), Wien's displacement law, and the Wien distribution law
(also called Wien's law). The latter is useful for approximate
calculations in certain cases.
There was a shift in this thread to photon flux, so I went with the flow. :) But I did use the Planck distribution approach to get to my value, which seems to differ only slightly with Hornblower's result. For the Sun, at 1000nm, the SED value (per unit wavelength) is about 757 w - m-2, which is 757 J-s-1-m-2. Then the apparent magnitude of the Sun must be lowered to 30, which is a big drop. The resulting brightness reduction multiplied by the photon flux (using AndreasJ's equation) will yield about 1 photon per the 2.4 meter aperture Hubble mirror.

ngc3314
2010-Sep-25, 05:36 PM
An alternate estimate uses one of those old rough rules in the biz - a V=0 star has 1000 photons per (cm2 s A) arriving. (More precisely, it's about 3% higher, but at the level of accuracy it depends on details of the stellar spectrum). So at V=30, we would have 10-9 per (cm2 s A), with a typical filter bandwidth of 1500 A and mirror area 40700 cm2 or so. So that gives about 0.06 photon per second into the aperture. The CCDs are pretty efficient these days, peaking above 90%, so with a broadband filter most of that would be captured. (Part of the difference with the above estimates must be the filter width; the photon rate goes up by maybe a factor 4 if you use something like the STIS "white-light" mode which passes from 400-1000 nm at once, as they used in the HDF-S).

Those numbers make sense with typical HST data - looking at a single-orbit WFC3 image in the I band, the detection limit is for stars around 0.2 detected photon/second. Long stacks such as the Deep Fields could, in principle, go nearly 10 times deeper.

George
2010-Sep-26, 07:08 PM
The following values I obtained came from using the Planck distribution. Since the visiual maginitude is set for 30, then the change in brightness is required, and, being visual, I used the 430nm to 720nm range -- is there an astronomical standard here? -- and found the percent (visual portion/total) that allowed me to convert from a 1AU level that I use for the known visual mag. of the Sun (-26.7 @ 1 AU), then calculated the delta B for a visual 30 mag. result. I then determined the percent of energy in the IR band ngc3314 mentions (150nm band width around 1 micron wavelength). Since the average IR photon energy in this band is 3.18 e21 Joules I found the total photon flux density and divided it by the brightness change. Here's what I found...

In the IR band (920nm to 1070nm) photon flux of various stars at a visual magnitude 30 scooped-up by Hubble's 2.4m mirror:


Stelar Temperature:............................. 2,000 2,500 2,900 5,777 20,000
Total Wattage Est. (fictious):..................... 1,690 4,400 8,150 135,400 19,633,000
Stefan Boltzman Wattage (1AU):....................... 19.6 47.8 86.6 1,363.0 195,803.8
Visible Total (430 to 720nm)......................... 19.76 192.73 701 49517 1,796,590.0
Visibile Wattage %:................................. 1.17% 4.38% 8.60% 36.57% 9.15%
Visual Wattage:...................................... 0.229 2.094 7.449 498.461 17,917.746
Average Energy of visual photon:................. 5.60E-20 5.60E-20 5.60E-20 5.60E-20 5.60E-20
Photons (visual) per sec per m^2:.................. 4.09E+18 3.74E+19 1.33E+20 8.90E+21 3.20E+23

Relative brightness to Sun in visual (delta B)... 0.00046 0.00420 0.01494 1.00000 35.94611
Change in Solar magnitude at 1AU:............... 8.34 5.94 4.56 0.00 -3.89
Mag. at 1AU:........................................... -18.36 -20.76 -22.14 -26.70 -30.59
Delta Change in M to 30M, visual:.................. 48.36 50.76 52.14 56.70 60.59
Delt B change:........................................ 2.20E+19 2.01E+20 7.15E+20 4.79E+22 1.72E+24
Visual Photon flux per m^2:......................... 0.19 0.19 0.19 0.19 0.19

Total IR wattage (920 to 1070nm)................ 96 407.5 909 12326 127465
Percent of IR band in total wattage:............. 5.68% 9.26% 11.15% 9.10% 0.65%
Actual IR wattage est.:............................. 1.113 4.427 9.659 124.079 1271.234
Average Energy of IR photon:...................... 3.18E-20 3.18E-20 3.18E-20 3.18E-20 3.18E-20
Photon flux at 1AU:................................. 3.50E+19 1.39E+20 3.03E+20 3.90E+21 3.99E+22
Change in Brightness (see above):................ 2.20E+19 2.01E+20 7.15E+20 4.79E+22 1.72E+24
Flux at 30M:............................................ 1.59 0.69 0.42 0.08 0.02
Hubble 2.4m mirror area:............................. 4.52 4.52 4.52 4.52 4.52
IR Photon flux at M= 30:.............................. 7.19 3.13 1.92 0.37 0.10

[I deleted the former values since they were goofed-up. The above should be correct. I used 5777K for the Sun, but used 5850K for the Planck distribution percentages.]