And since the main problem with surface telescopes( athmosphere's refraction) has been overcome using active optics, i don't think it's necessary to put another space telescope.
You've obviously never worked with an AO system, but think you can make wild claims about it anyway. So I've never worked with one on a night-time telescope, but I do have one about one floor down from me right now on a solar telescope.
So let me tell you: AO isn't perfect. If the atmosphere is disturbing the image too much, the AO won't even lock. It'll make good images better, but it can't fix terrible images. Even if you have good seeing conditions, the AO won't get the image to the diffraction limit of the telescope. I doubt any ground-based telescope can get anywhere near Hubble quality regularly. Second, you need to correct the atmosphere in real time, like 500 times per second. You can't integrate a couple of hours like you do when observing faint objects. So you'll need a bright object sufficiently close to the faint object you're observing. Third, the AO can only correct a part of the image, because the atmosphere isn't distorting the image the same way everywhere. If the object you're observing is too large, the AO can't correct the whole image. Fourth, the atmosphere distorts different colors differently. Since typically you want to observe at different wavelengths (i.e., colors), you'll only get one sharp one. Finally, there's no clouds in space.
In short: AO helps when it's not cloudy, the seeing is reasonable to good, you have a bright object close to your target, your target is not too large, and you only want data at one wavelength. There are plenty of valid scientific reasons to want to have space telescopes. Observing at wavelengths that don't get through the atmosphere is just one.
There's also an interview with Dr. Hayley Bignall, [...], where he discusses [...].
He? I know astrophysicist is a male-dominated profession. But a name like 'Hayley' should at the very least make one wonder. See this page to accurately determine Dr. Bignall's gender.
Furthermore, this is nothing really new; see this/. story, amongst others.
Still, it's a very creative way of increasing resolution! Not to mention difficult and time-consuming. I wonder how believable the results are. I use a similar technique (called Speckle Masking) to eliminate earth-atmosphere scintillation from Solar observations, with astounding results. These, however, can be checked against single 'lucky shot' images of extrodinary quality or observations from space...
Like the subject says. Optical astronomy, and building good telescopes in particular, is extremely difficult. Let's go though it step by step.
The primary mirror. The biggest night-time telescope (which is not yet in operation) has a 10.4-meter diameter primary, which is formed from 36 hexagonal segments. Search Google for "Gran Telescopio Canarias". Your amateur telescope has about a 30-cm aperture. You'll need 1200 of those alone to just capture the same amount of light. Here's the second point.
Primary mirrors are big so they collect a lot of light. This is good, because it means you don't have to expose for 10 years to actually see some faint object. The second reason for making a primary large, is because the theoretical resolution you can get with your telescope goes down as the size of the primary goes up. Bigger mirror equals better theoretical resolution.
You may have heard of interferometry as a means of getting a high resolution by using lots of small telescopes. It's used extensively in radio astronomy. It's been tried in optical astronomy. ESO's VLT at Paranal Observatory is, as far as I know, the most advanced with it's VLTI instrument. I'm not sure, but I think they've got it working at infrared wavelengths now.
Interferometry becomes more difficult as the wavelength gets smaller. Infrared sits around 1000 nanometers. Optical is around half that. For radio, it's about a centimeter. You'll need to know the distances between your telescopes to a fraction of the wavelength accurately. Also, you can't combine the signals later. Radio astronomy can, because they can easily record both amplitude and phase information of their signal. Then, using custom hardware (DSPs), the signals of a number of telescopes are combined. A CCD only records amplitude information. You'll have to combine the light in real-time. Is that hard? You'd better believe it.
Last point I'll make: seeing. Seeing is the reason why the Hubble makes such nice pictures. It's above the Earth's atmosphere, thus it's view is not disturbed by the same cruft you can see above a road on a hot day. You, on the other hand, are just about as low as you can get. Telescopes aren't just built anyplace. Extensive testing is done to select those sites that have the best seeing. Typical sites are high (think above 2.5km above sea level), near a large, quiet body of water (since it stabalizes the air temperature) and in areas that don't have a stratospheric jet stream. Did I mention clouds? Oh, and in remote areas with little light polution from cities. You'll be observing in less than ideal conditions, giving you a much reduced resolution.
So what about adaptive optics? A bright object, say star close by, with no resolvable extent (e.g., not a galaxy, supernova remnent, you name it), allows the use of AO. The AO needs this bright object, because it needs to adjust the mirror in real time (seconds, at the very most), and it can't wait an hour or so to actually see something. Usually, if you're observing a faint object near a bright one, you'll lock the AO on the bright one. But what if you don't have a bright object near your faint object of interest? The solution it to shoot a rather high power laser up in to the sky. It'll form a bright dot somewhere high in the atmosphere. Aim it close to your faint object, and presto: you've got yourself an artificial star. Search Google for "telescope laser" and you'll find a few nice images. AO is child's play compared to interferometry. That doesn't mean just anyone can do it, though.
Call me a pessimist, but I don't see how any group of amateurs can hope to achieve the quality of the images recorded by professional observatories.
I just skimmed through the abstract of the article to be published, and I think the post on the front page is a bit disorienting. They're not using a bubble of gas the way one uses a lens (or mirror) in a telescope. Fat chance of getting a blob of gas aligned in between the object and you eye, and if that does happen purely by chance, then that blob is likely to be shaped unregularly, making a very, very poor lens.
The big idea is that you can deduce extra information from what you see when a blob of gas passes in front of the object you're observing. Basically, the gas fudges the image in much the same way as the Earth's atmosphere does (called seeing) but on a longer timescale. The lack of atmosphere, as you all know, is why the Hubble is such a good telescope. If you know how the object you're observing was creamed, then possibly you can reconstruct the original from what you've observed. Extra information has to come from somewhere, and that means you're going to be observing for a long time to get some statistics together.
I know it works for solar observations, since I've written code that does it myself. I can't find a good before and after example right now, but it's pretty impressive. I guess this will work. Neat.
You have to include the mass of its kinetic energy as part of its mass of course, but why should kinetic mass be treated any differently than any other kind of mass?
Because `kinetic mass' does not exsist. Saying `the mass of an object increases when it goes faster' is wrong. Mass has absolutely nothing to do with speed. The mass of an object is the same in any system, inertial or not. That's the way it's defined. You can't change that.
From now on I will refer to mass as the physical defenition of mass.
So photons do not have mass. They do not have kinetic energy either. Their energy is actually in an electro-magnetic field. So their energy is dependant on their frequency.
There is no problem in stating energy and momentum conservation laws in Special Relativity, or General Relativity. If there were, the theory would have been abandoned by Einstein himself. In General Relativity the conservation law of energy and momentum is... P^{\mu\nu}P_{\mu\nu}=0 (tensor math). Done. In Quantum Mechanics, yes, there is a possibility of `borrowing' energy from nowhere really for such a time that it cannot be detected in any way. Every (sane) physisist you will talk to will tell you that energy is globally conserved.
The sail temperature issue is non-trivial.
Let's look at this sail temperature problem, shall we? It's not all that difficult really. The total power emitted by an object of temperature T and surface area A is A \sigma T^4 where \sigma equals the Stefan-Boltzman constant. In the article, they say approximately 10 kilowatts was beamed to the sail. In equilibrium these figures must be the same, which yields T=1152 Kelvin for a sail of 1 square meter total, that's front and back. Yeah, that's probably hotter than you'd want it to be.
So no, the energy is not 0.5mv^2
No, it's not. With your defenition of mass, it would be (m-m0)c^2, m0 the rest mass. Your own strange defenition of mass is killing you here. Use E=Gmc^2, G=1/sqrt(1-v^2/c^2) instead. If you don't believe this results in Ekin=0.5mv^2 for v small compared to the speed of light (Newtonian), make a Taylor series to the linear term, and substract mc^2.
The relation of energy to momentum is E=c*sqrt(p^2+m^2 c^2). Assuming p to be small in comparison to mc and v again small compared to the speed of light (Newtonian), yes this results in Ekin=0.5pv.
Incidentally the wavelength is irrelevant to how efficient light sails are.
True. The efficiency is only dependant on the percentage of photons missing the target, the percentage of photons being absorbed in the interstellar matter, and the reflection coefficient of the sail.
Newton's statement of his three laws is actually completely correct in SR.
I can't come up with a reason why Einstein would come up with his Theory of Relativity if this statement is correct. Can you?
I've just read your follow up on a reply to this post, and here's what I have to say about it.
First off, I read the "random decent reference", and found it to be in accordance with my remarks, and not yours. Read it again, and remember that when it talks about mass, it is talking about what you would call rest mass. Then note that for v=c the gamma factor goes to infinity. Since the rest mass is zero, gamma*m is undefined. Your background in math should tell you that much. By the way, a mathematician should be able to apperciate an invariant quantity. I know, because I am not only a physics student, I am also a mathematics student. And a astrophysics student.
And from a purist's point of view it is not better to look at mass as the relativistic mass. The energy in your body as heat is most certainly not measured when you step on a scale. What you need to do is forget mass, and go for energy! Mass is not conserved. If it were, nuclear fusion or fission wouldn't work. Stars wouldn't exsist. Energy is conserved, and a far better quantity to work with when relativity is needed.
The interesting thing about/. is how little basic science a lot of the people who think they are qualified to answer your question have.
True.
Momentum is mass*velocity.
I have to disagree on this one. Since photons (light "particles") move with the speed of light c in vacuum (duh), they CANNOT have mass. This is a direct consequence of the Special Theory of Relativity. All particles that move with the speed of light must be massless. Yet photons do have momentum, and therefore your definition of momentum cannot be correct.
Momentum is defined as p=dE/dv, with E=Gmc^2, G=1/(1-(v/c)^2)^0.5 for matter. This results in p=Gmv which is allmost equal to mv for non-relativistic particles.
For light, the definition gives p=d(hc/l)/dc, with h the Planck constant and l the wavelenght (which, btw, relates to the color) of the light. So we have p=h/l.
Good, now we see that we want to use short wavelenghts, since those photons have a lot of momentum.
The mass is dependent on the frequency.
Photons have no mass.
So light coming from a gravity well, or from something receeding from us loses energy, and the photon is "red-shifted".
No. Photons coming from a gravity well (curved spacetime) are not shifted because of their enery loss, because they can't lose energy, enery is conserved -- allways. Gravity itself causes the redshift.
Also, redshift can be caused by objects moving away from the observer, or towards (then, it's blueshift, actually). This is again a result of the General Theory of Relativity.
Photons just travelling throught the cosmos experience redshift, allthought spacetime is rather flat. This is a result of the universe expanding at a steady rate. Ever heard of the cosmic 3 Kelvin background radiation? It was created in a time when no planets or stars were around. The ambient temperature was in the order of a million Kelvin. The universe much much smaller, and much much hotter. Now, 20 or so Billion years later, the universe is bigger, and all the energy in those photons created long ago is spread out over an enourmous volume.
They would want to avoid absorbing photons since that would burn up the sail very fast.
This is not the reason they want to avoid absorbing phonons, IMHO. Yes, the sail would heat up, since it is pretty thin, it would pretty fast at that too. But then again, it's pretty big as well, so it will also irradiate thermal photons (so called black body radiation). So I can't say if it will melt. It certainly won't burn, you need oxygen for that.
But there is a better reason not to absorb photons: less momentum is transferred in comparison to reflection. If the photons moves back to where it came from, the momentum transfer is twice the momentum of the photon, once for `stopping' it, once for getting it going again in the opposite direction.
Energy scales as velocity squared, and light moves pretty fast, so it delivers a *lot* of heat for the momentum.
Again, you're trying to use Newtonian mechanics on a photon. The energy of a photon is simply hc/l. Nothing E=0.5mv^2 about it; m is zero for a photon.
Slashdot Mirror
And since the main problem with surface telescopes( athmosphere's refraction) has been overcome using active optics, i don't think it's necessary to put another space telescope.
You've obviously never worked with an AO system, but think you can make wild claims about it anyway. So I've never worked with one on a night-time telescope, but I do have one about one floor down from me right now on a solar telescope.
So let me tell you: AO isn't perfect. If the atmosphere is disturbing the image too much, the AO won't even lock. It'll make good images better, but it can't fix terrible images. Even if you have good seeing conditions, the AO won't get the image to the diffraction limit of the telescope. I doubt any ground-based telescope can get anywhere near Hubble quality regularly. Second, you need to correct the atmosphere in real time, like 500 times per second. You can't integrate a couple of hours like you do when observing faint objects. So you'll need a bright object sufficiently close to the faint object you're observing. Third, the AO can only correct a part of the image, because the atmosphere isn't distorting the image the same way everywhere. If the object you're observing is too large, the AO can't correct the whole image. Fourth, the atmosphere distorts different colors differently. Since typically you want to observe at different wavelengths (i.e., colors), you'll only get one sharp one. Finally, there's no clouds in space.
In short: AO helps when it's not cloudy, the seeing is reasonable to good, you have a bright object close to your target, your target is not too large, and you only want data at one wavelength. There are plenty of valid scientific reasons to want to have space telescopes. Observing at wavelengths that don't get through the atmosphere is just one.
Cheers, Alfred
There's also an interview with Dr. Hayley Bignall, [...], where he discusses [...].
/. story, amongst others.
He? I know astrophysicist is a male-dominated profession. But a name like 'Hayley' should at the very least make one wonder. See this page to accurately determine Dr. Bignall's gender.
Furthermore, this is nothing really new; see this
Still, it's a very creative way of increasing resolution! Not to mention difficult and time-consuming. I wonder how believable the results are. I use a similar technique (called Speckle Masking) to eliminate earth-atmosphere scintillation from Solar observations, with astounding results. These, however, can be checked against single 'lucky shot' images of extrodinary quality or observations from space...
Cheers,
Alfred
Like the subject says. Optical astronomy, and building good telescopes in particular, is extremely difficult. Let's go though it step by step.
The primary mirror. The biggest night-time telescope (which is not yet in operation) has a 10.4-meter diameter primary, which is formed from 36 hexagonal segments. Search Google for "Gran Telescopio Canarias". Your amateur telescope has about a 30-cm aperture. You'll need 1200 of those alone to just capture the same amount of light. Here's the second point.
Primary mirrors are big so they collect a lot of light. This is good, because it means you don't have to expose for 10 years to actually see some faint object. The second reason for making a primary large, is because the theoretical resolution you can get with your telescope goes down as the size of the primary goes up. Bigger mirror equals better theoretical resolution.
You may have heard of interferometry as a means of getting a high resolution by using lots of small telescopes. It's used extensively in radio astronomy. It's been tried in optical astronomy. ESO's VLT at Paranal Observatory is, as far as I know, the most advanced with it's VLTI instrument. I'm not sure, but I think they've got it working at infrared wavelengths now.
Interferometry becomes more difficult as the wavelength gets smaller. Infrared sits around 1000 nanometers. Optical is around half that. For radio, it's about a centimeter. You'll need to know the distances between your telescopes to a fraction of the wavelength accurately. Also, you can't combine the signals later. Radio astronomy can, because they can easily record both amplitude and phase information of their signal. Then, using custom hardware (DSPs), the signals of a number of telescopes are combined. A CCD only records amplitude information. You'll have to combine the light in real-time. Is that hard? You'd better believe it.
Last point I'll make: seeing. Seeing is the reason why the Hubble makes such nice pictures. It's above the Earth's atmosphere, thus it's view is not disturbed by the same cruft you can see above a road on a hot day. You, on the other hand, are just about as low as you can get. Telescopes aren't just built anyplace. Extensive testing is done to select those sites that have the best seeing. Typical sites are high (think above 2.5km above sea level), near a large, quiet body of water (since it stabalizes the air temperature) and in areas that don't have a stratospheric jet stream. Did I mention clouds? Oh, and in remote areas with little light polution from cities. You'll be observing in less than ideal conditions, giving you a much reduced resolution.
So what about adaptive optics? A bright object, say star close by, with no resolvable extent (e.g., not a galaxy, supernova remnent, you name it), allows the use of AO. The AO needs this bright object, because it needs to adjust the mirror in real time (seconds, at the very most), and it can't wait an hour or so to actually see something. Usually, if you're observing a faint object near a bright one, you'll lock the AO on the bright one. But what if you don't have a bright object near your faint object of interest? The solution it to shoot a rather high power laser up in to the sky. It'll form a bright dot somewhere high in the atmosphere. Aim it close to your faint object, and presto: you've got yourself an artificial star. Search Google for "telescope laser" and you'll find a few nice images. AO is child's play compared to interferometry. That doesn't mean just anyone can do it, though.
Call me a pessimist, but I don't see how any group of amateurs can hope to achieve the quality of the images recorded by professional observatories.
Alfred
I just skimmed through the abstract of the article to be published, and I think the post on the front page is a bit disorienting. They're not using a bubble of gas the way one uses a lens (or mirror) in a telescope. Fat chance of getting a blob of gas aligned in between the object and you eye, and if that does happen purely by chance, then that blob is likely to be shaped unregularly, making a very, very poor lens.
The big idea is that you can deduce extra information from what you see when a blob of gas passes in front of the object you're observing. Basically, the gas fudges the image in much the same way as the Earth's atmosphere does (called seeing) but on a longer timescale. The lack of atmosphere, as you all know, is why the Hubble is such a good telescope. If you know how the object you're observing was creamed, then possibly you can reconstruct the original from what you've observed. Extra information has to come from somewhere, and that means you're going to be observing for a long time to get some statistics together.
I know it works for solar observations, since I've written code that does it myself. I can't find a good before and after example right now, but it's pretty impressive. I guess this will work. Neat.
Alfred
Because `kinetic mass' does not exsist. Saying `the mass of an object increases when it goes faster' is wrong. Mass has absolutely nothing to do with speed. The mass of an object is the same in any system, inertial or not. That's the way it's defined. You can't change that.
From now on I will refer to mass as the physical defenition of mass.
So photons do not have mass. They do not have kinetic energy either. Their energy is actually in an electro-magnetic field. So their energy is dependant on their frequency.
There is no problem in stating energy and momentum conservation laws in Special Relativity, or General Relativity. If there were, the theory would have been abandoned by Einstein himself. In General Relativity the conservation law of energy and momentum is... P^{\mu\nu}P_{\mu\nu}=0 (tensor math). Done. In Quantum Mechanics, yes, there is a possibility of `borrowing' energy from nowhere really for such a time that it cannot be detected in any way. Every (sane) physisist you will talk to will tell you that energy is globally conserved.
The sail temperature issue is non-trivial.
Let's look at this sail temperature problem, shall we? It's not all that difficult really. The total power emitted by an object of temperature T and surface area A is A \sigma T^4 where \sigma equals the Stefan-Boltzman constant. In the article, they say approximately 10 kilowatts was beamed to the sail. In equilibrium these figures must be the same, which yields T=1152 Kelvin for a sail of 1 square meter total, that's front and back. Yeah, that's probably hotter than you'd want it to be.
So no, the energy is not 0.5mv^2
No, it's not. With your defenition of mass, it would be (m-m0)c^2, m0 the rest mass. Your own strange defenition of mass is killing you here. Use E=Gmc^2, G=1/sqrt(1-v^2/c^2) instead. If you don't believe this results in Ekin=0.5mv^2 for v small compared to the speed of light (Newtonian), make a Taylor series to the linear term, and substract mc^2.
The relation of energy to momentum is E=c*sqrt(p^2+m^2 c^2). Assuming p to be small in comparison to mc and v again small compared to the speed of light (Newtonian), yes this results in Ekin=0.5pv.
Incidentally the wavelength is irrelevant to how efficient light sails are.
True. The efficiency is only dependant on the percentage of photons missing the target, the percentage of photons being absorbed in the interstellar matter, and the reflection coefficient of the sail.
Newton's statement of his three laws is actually completely correct in SR.
I can't come up with a reason why Einstein would come up with his Theory of Relativity if this statement is correct. Can you?
I've just read your follow up on a reply to this post, and here's what I have to say about it.
First off, I read the "random decent reference", and found it to be in accordance with my remarks, and not yours. Read it again, and remember that when it talks about mass, it is talking about what you would call rest mass. Then note that for v=c the gamma factor goes to infinity. Since the rest mass is zero, gamma*m is undefined. Your background in math should tell you that much. By the way, a mathematician should be able to apperciate an invariant quantity. I know, because I am not only a physics student, I am also a mathematics student. And a astrophysics student.
And from a purist's point of view it is not better to look at mass as the relativistic mass. The energy in your body as heat is most certainly not measured when you step on a scale. What you need to do is forget mass, and go for energy! Mass is not conserved. If it were, nuclear fusion or fission wouldn't work. Stars wouldn't exsist. Energy is conserved, and a far better quantity to work with when relativity is needed.
Cheers, Alfred.
True.
Momentum is mass*velocity.
I have to disagree on this one. Since photons (light "particles") move with the speed of light c in vacuum (duh), they CANNOT have mass. This is a direct consequence of the Special Theory of Relativity. All particles that move with the speed of light must be massless. Yet photons do have momentum, and therefore your definition of momentum cannot be correct.
Momentum is defined as p=dE/dv, with E=Gmc^2, G=1/(1-(v/c)^2)^0.5 for matter. This results in p=Gmv which is allmost equal to mv for non-relativistic particles.
For light, the definition gives p=d(hc/l)/dc, with h the Planck constant and l the wavelenght (which, btw, relates to the color) of the light. So we have p=h/l.
Good, now we see that we want to use short wavelenghts, since those photons have a lot of momentum.
The mass is dependent on the frequency.
Photons have no mass.
So light coming from a gravity well, or from something receeding from us loses energy, and the photon is "red-shifted".
No. Photons coming from a gravity well (curved spacetime) are not shifted because of their enery loss, because they can't lose energy, enery is conserved -- allways. Gravity itself causes the redshift.
Also, redshift can be caused by objects moving away from the observer, or towards (then, it's blueshift, actually). This is again a result of the General Theory of Relativity.
Photons just travelling throught the cosmos experience redshift, allthought spacetime is rather flat. This is a result of the universe expanding at a steady rate. Ever heard of the cosmic 3 Kelvin background radiation? It was created in a time when no planets or stars were around. The ambient temperature was in the order of a million Kelvin. The universe much much smaller, and much much hotter. Now, 20 or so Billion years later, the universe is bigger, and all the energy in those photons created long ago is spread out over an enourmous volume.
They would want to avoid absorbing photons since that would burn up the sail very fast.
This is not the reason they want to avoid absorbing phonons, IMHO. Yes, the sail would heat up, since it is pretty thin, it would pretty fast at that too. But then again, it's pretty big as well, so it will also irradiate thermal photons (so called black body radiation). So I can't say if it will melt. It certainly won't burn, you need oxygen for that.
But there is a better reason not to absorb photons: less momentum is transferred in comparison to reflection. If the photons moves back to where it came from, the momentum transfer is twice the momentum of the photon, once for `stopping' it, once for getting it going again in the opposite direction.
Energy scales as velocity squared, and light moves pretty fast, so it delivers a *lot* of heat for the momentum.
Again, you're trying to use Newtonian mechanics on a photon. The energy of a photon is simply hc/l. Nothing E=0.5mv^2 about it; m is zero for a photon.
cheers, Alfred.