I just meant “terrible” in the sense of terribly hard to detect because of attenuation of signal strength Chris. Thanks for your answer.Chris Peterson wrote:That's not terrible at all. That's the divergence you get with a perfectly collimated Gaussian beam. You can't do better- this is just the result of the way the beam interferes with itself. A geometrically collimated (parallel) beam can't exist in nature.BDanielMayfield wrote:A beam spreading to 88 AU over just 4 LY is terrible
Whether radio or optical communications are better over a few light years depends on details of the technology. Certainly, we should be able to use either, depending on engineering tradeoffs.
APOD: A Laser Strike at the Galactic Center (2013 Dec 01)
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
Just as zero is not equal to infinity, everything coming from nothing is illogical.
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
I guess there’s just no way to get around the inverse square law then, is there? The apparent intensity of any radiating source will always be four times dimmer with every doubling of distance, won’t it? Even if the source is a laser pointer.
It’s the laws:
Thou shall not go faster than light.
Thou shall not talk faster than light.
Thou must pay for long distance calls. (With more energy and larger receivers.)
It’s the laws:
Thou shall not go faster than light.
Thou shall not talk faster than light.
Thou must pay for long distance calls. (With more energy and larger receivers.)
Last edited by BDanielMayfield on Sun Dec 01, 2013 11:46 pm, edited 1 time in total.
Just as zero is not equal to infinity, everything coming from nothing is illogical.
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
Well, the divergence is much greater at longer wavelengths. A biggish radio antenna, say around 10 m, has a divergence of around a degree for microwaves, about 50 times greater than the green laser. So the radio energy is spread out over a much greater area when it gets to the next star.BDanielMayfield wrote:I just meant “terrible” in the sense of terribly hard to detect because of attenuation of signal strength Chris. Thanks for your answer.
Chris
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
There is a region near the focusing optics where you don't have an inverse-square relationship. In that region, the beam has a curved profile. But far from the optics, it is just a conical profile, so you have the inverse-square law.BDanielMayfield wrote:I guess there’s just no way to get around the inverse square law then, is there? The intensity of any radiating source will always be four times dimmer with every doubling of distance, won’t it? Even if the source is a laser pointer.
Chris
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
At a distance of 10,000 ly, the beam diameter from a 70 m radar dish (at half the 21 cm wavelength) will be about 37 ly.Chris Peterson wrote:
At a distance of 10,000 ly, the beam diameter will be at least 4 ly.
For a , 100W source, that means that any given square meter aperture will receive one photon every 39 days.
For a 126kW source, that means that any given square meter aperture will receive about 400 microwave photons per second.
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
May all beings be happy, peaceful, and free.
Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
But in a couple of weeks it might make an excellent long-exposure photo of Earth, from Sinus Iridum, by Chang'e 3.
http://asterisk.apod.com/viewtopic.php? ... 3&start=42
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
My calculations aren't getting anywhere near either of yours. Chris, your divergence is much larger than I calculate, and Art, your photon collection rate per m2 is much higher than I calculate. The discrepancy is so large I'd like to find out why. Maybe I'm missing something. Regarding the photon count rate per m2 I'm assuming one calculates it by taking the irradiance at distance (W/m2) ÷ Photon Energy which gives 1/[m2·sec]neufer wrote:At a distance of 10,000 ly, the beam diameter from a 70 m radar dish (at half the 21 cm wavelength) will be about 37 ly.Chris Peterson wrote:
At a distance of 10,000 ly, the beam diameter will be at least 4 ly.
For a , 100W source, that means that any given square meter aperture will receive one photon every 39 days.
For a 126kW source, that means that any given square meter aperture will receive about 400 microwave photons per second.
Chris - What is the beam diameter your starting with to arrive at 4 ly diameter? Are there other assumptions you've made to arrive at that (e.g. interstellar extinction)?
Art - Are you using 10.5cm wavelength / 2.86GHz? I assume you're calculating antenna diffraction ≈ λ/Diameter (the Half-Power Beam Width)?
Thanks.
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
I assumed 1mm, which would be about right for the cavity beam in a 100W gas laser.alter-ego wrote:My calculations aren't getting anywhere near either of yours. Chris, your divergence is much larger than I calculate, and Art, your photon collection rate per m2 is much higher than I calculate. The discrepancy is so large I'd like to find out why. Maybe I'm missing something. Regarding the photon count rate per m2 I'm assuming one calculates it by taking the irradiance at distance (W/m2) ÷ Photon Energy which gives 1/[m2·sec]
Chris - What is the beam diameter your starting with to arrive at 4 ly diameter? Are there other assumptions you've made to arrive at that (e.g. interstellar extinction)?
I ran all the calculations off very quickly and didn't check them. It's entirely possible I made an error. Lots of very big and very small numbers; plenty of opportunities for exponent errors.
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
Yes to all that.alter-ego wrote:
Art - Are you using 10.5cm wavelength / 2.86GHz?
I assume you're calculating antenna diffraction ≈ λ/Diameter (the Half-Power Beam Width)?
I get a beam an order of magnitude wider than Chris's laser beam
but with a power 3 orders of magnitude stronger so the beam intensity is ~10 times stronger.
I extrapolated from Chris's photon count based upon photons that are 200,000,000 times weaker
when they should have been 200,000 times weaker
![Exclamation :!:](./images/smilies/icon_exclaim.gif)
so I really should have gotten 0.4 microwave photons per second. What do you get?
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
First off, I haven't ignored your question. Between the holidays and unfortunate timing of forum access problems, I haven't responded.neufer wrote:Yes to all that.alter-ego wrote:
Art - Are you using 10.5cm wavelength / 2.86GHz?
I assume you're calculating antenna diffraction ≈ λ/Diameter (the Half-Power Beam Width)?
I get a beam an order of magnitude wider than Chris's laser beam
but with a power 3 orders of magnitude stronger so the beam intensity is ~10 times stronger.
I extrapolated from Chris's photon count based upon photons that are 200,000,000 times weaker
when they should have been 200,000 times weaker![]()
so I really should have gotten 0.4 microwave photons per second. What do you get?
Well, care is needed to keep track of the details which mostly amounts to orders of magnitude here. If I did my calculations correctly, your and Chris' final photon collection rates per m2 are off by orders of magnitude. This was significant enough that I wanted to review my calculations first. So to try to keep things clearer, I'm posting calculation details below that you and Chris can review if you wish.
OK, cutting to the chase, I get the radio telescope photon collection time per m2 ~66 hours at a distance of 10,000 ly, and a laser photon collection time ~382,000 yrs. To validate this, I first wanted to first calculate the photon collection ratio using the ratio approach as you did above. I believe the cyan-highlighted equation below (incorporating the constants) correctly predicts the laser-to radio photon collection time to be ~50,000,000. This ratio approach above yields the same answer as the ratio of specific cases calculated below.
The case-specific photon collection time calculations are:
These are simplified calculation that don't attempt to account for the next issue of interstellar extinction which attenuates the photon counts. I've also looked at this but no need to go there here.
Last edited by alter-ego on Sun Dec 08, 2013 9:22 am, edited 1 time in total.
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Re: APOD: A Laser Strike at the Galactic Center (2013 Dec 01
Your divergence is in the right ballpark. I calculate a 7-ly diameter for a 532nm beam. Originally, I was thinking divergence from a 14m diameter beam which is a whole bunch less. Also, in my reply to Art, my example uses 532nm for the wavelength. If you were using a real guide star wavelength, I missed that. In this case, the wavelength difference is really of no significance to what I calculate the photon collection rate per m2 is.Chris Peterson wrote:I assumed 1mm, which would be about right for the cavity beam in a 100W gas laser.alter-ego wrote:My calculations aren't getting anywhere near either of yours. Chris, your divergence is much larger than I calculate, and Art, your photon collection rate per m2 is much higher than I calculate. The discrepancy is so large I'd like to find out why. Maybe I'm missing something. Regarding the photon count rate per m2 I'm assuming one calculates it by taking the irradiance at distance (W/m2) ÷ Photon Energy which gives 1/[m2·sec]
Chris - What is the beam diameter your starting with to arrive at 4 ly diameter? Are there other assumptions you've made to arrive at that (e.g. interstellar extinction)?
I ran all the calculations off very quickly and didn't check them. It's entirely possible I made an error. Lots of very big and very small numbers; plenty of opportunities for exponent errors.
As I posted above, for the 100-W laser with a 1mm beam diameter (532nm), the photon density 10,000 ly appears greatly lower than maybe you calculated. I calculate the time between photons per m2 is an amazing 382,000 years
![Exclamation :!:](./images/smilies/icon_exclaim.gif)
The calculation details are in my reply to Art.
A pessimist is nothing more than an experienced optimist