scalbers Posted July 13, 2008 Posted July 13, 2008 (edited) True, the pattern builds up from individual photons. Each photon though has to assess the probability function based on the shape of the entire mirror, then implement this when it decides to show up in one of the various Airy disk rings. So perhaps it is the E/M wave that reflects off of the mirror and the photon is just manifest when it hits (or the wave function collapses at) the detector at the focal point? On the broader questions is there really any evidence that a photon oscillates? Or does it just have a certain amount of energy and momentum, derived from the E/M waves that in turn oscillate at the characteristic frequency? This is kind of like the Figure 3 in the "How Long Is A Photon" paper. Also, as a review, what is the evidence that photons travel as particles any appreciable distance? Can the waves do the traveling and the photons just be a sort of an interacting virtual particle? Does this get into the question of real vs virtual photons? Edited July 13, 2008 by scalbers
Norman Albers Posted July 13, 2008 Author Posted July 13, 2008 Excellent discussion, scalbers. My understanding fails where we try to compare physics of one photon to that of the ensemble falling at different reflection points on a parabolic mirror. You are saying that one photon "knows all" as I think John Wheeler espoused. I say, one photon lands with some uncertainty of momentum, and I am asking if many of them distributed around a mirror of many square meters, then establish the diffraction clarity or lack thereof?
scalbers Posted July 13, 2008 Posted July 13, 2008 (edited) Hard to say for sure how to answer your question. I can say that either one photon knows all (perhaps by being as large as the mirror), OR alternatively it's the wave that does the reflecting. I wouldn't think that you have various individual photons reflecting off of different parts of the mirror and somehow interacting with each other. Recall the aspects of post #24, 2nd paragraph. I can also recall that diffraction patterns will still develop with a very low photon flux, say from a 30th magnitude star that would be something on the order of one photon per square meter per second. Each photon somehow has the information on the probability function communicated to it - how? Edited July 13, 2008 by scalbers
Norman Albers Posted July 14, 2008 Author Posted July 14, 2008 (edited) If the mirror surface is perturbed, like you described, say, built of little flats at angles to the overall curve, is there no loss of some photons, a loss of luminosity or light collected? . . . . Clearly the individual photon detections do not interact. They all individually do whatever it is they do, and the interference patterns build up in discrete steps. The quantum wave probability function describes the overall effect. It seems you point out that if photons were only local EM wave packets at a small (centimeter, say) region, then the diffraction would be great. This is something that I would think takes many events to determine, however. Is there no way we can add the larger interference patterns from all parts of the mirror and come out with the finer result? Edited July 14, 2008 by Norman Albers
Norman Albers Posted July 17, 2008 Author Posted July 17, 2008 (edited) I think we can agree that the photon with unit angular momentum is what is detected by identifiable detectors which respond to quantum quantities. What, then, do we know of the 'detectors' and what do we know of the vacuum radiation field??? This question drives my studies and pushes me to further dimensions of understanding, I hope. I realize that when quantum electrodynamics and field theory make you feel queasy, you are starting to get it. It seems that quantum wave functions demand a spacelike reality outside the lightcone, in the language of relativity. Is this what others see? I figure the larger theory deals with such waves on a stochastic background. The fact is, there was a detectable coherent photon event. Edited July 18, 2008 by Norman Albers
scalbers Posted July 19, 2008 Posted July 19, 2008 If the mirror surface is perturbed, like you described, say, built of little flats at angles to the overall curve, is there no loss of some photons, a loss of luminosity or light collected? . . . . The total number of photons should be conserved that are "reflected" off of the mirror. If the perturbations are big enough more of the photons will show up in the outer rings of the diffraction pattern instead of the central disk. The interesting rule of thumb I recall in optics is that (seemingly regardless of the slope of the perturbations), if the total amplitude of the perturbations is less than 1/4 wavelength you will end up with a high quality diffraction pattern. I had often been puzzled by this assertion when making zonal measurements of slope (and integrating them over the mirrors's radius) in doing foucault knife-edge testing of my telescope mirrors.
Norman Albers Posted July 19, 2008 Author Posted July 19, 2008 Scalbers, thank you so much for your practical experience and observations. This is helping me move forward in challenging my thinking about the nature of the radiation field. Especially, since I have focused on possible electrodynamics of photon localization, I have underestimated the wave nature of the larger field. It makes sense that it takes the whole mirror to focus one photon. Can you get me into realworld magnitudes? I ran calcs on a star like our sun in power output, generalizing from 1,300 watts/square meter at Earth orbit radius. If such a star were one billion lightyears distant, I calculate that a 9-meter mirror would see only one photon per hour. Do we only observe galaxies here?
scalbers Posted July 19, 2008 Posted July 19, 2008 (edited) We might see if this sounds consistent with what I mentioned in post #28 2nd paragraph, since a 30th magnitude star (or galaxy) is very roughly the limit of what the Hubble telescope can see with a 2.4m mirror. The sun for reference is -26.7 magnitude, and every 5 magnitudes is a factor of 100 in brightness. I had also found a web page somewhere that equated zero magnitude to something like 10^5 Janskys, where 1 Jansky is 10^7 photons per square meter per second. That's how I came up with the info in post #28. We can recheck all of these relationships. In terms of seeing stars vs galaxies, the brightest individual stars are nearly 15 magnitudes brighter than the sun. I'm trying to recall what the absolute magnitude of a galaxy is, perhaps near -30? The absolute magnitude of the sun is +4.8. Distant galaxies are also seen with the aid of gravitational lensing and perhaps some sort of magnification effect of space? This starts to get off topic perhaps. Edited July 19, 2008 by scalbers
Norman Albers Posted July 19, 2008 Author Posted July 19, 2008 (edited) Janskys, I love it. Now tell me the answer! . . . . . OK, helpful conversions: energy of one EV (photon) is 1.6 E-19 Joule. Edited July 19, 2008 by Norman Albers
scalbers Posted July 19, 2008 Posted July 19, 2008 (edited) I suppose I earlier addressed a question in post #28 that is closely related to yours though perhaps not identical. Literally speaking I did indeed address your question of getting into "real world magnitudes" Your second question was about seeing stars vs. galaxies If we use the info in #28 as a starting point, namely 30th magnitude is one photon per sq m per sec, I assume you now are asking about the magnitude of the sun at 1 billion light years. The sun is +4.8 (absolute) magnitude at 32 light years (ten parsecs). The sun at 1 billion ly would thus be 43rd magnitude. This is thus 1000 seconds to wait for photon to hit a 9 meter mirror (quick back of the envelope calculation, without the envelope). Good - that's very close to what you get to the nearest power of ten. I still like my 30th magnitude number in post #28 as a more "real world" telescopic example. Edited July 19, 2008 by scalbers
Norman Albers Posted July 19, 2008 Author Posted July 19, 2008 Cool, it seems we agree that a single moderate star at a billion LY is 'pretty weak'?
scalbers Posted July 19, 2008 Posted July 19, 2008 (edited) Yes, it would also seem that the brightest stars might be near the limit of what can be seen in a big telescope at 1 billion ly. The sun would be quite invisible. I think the Hubble Ultra Deep Field magnitude limit (seeing galaxies) is somewhere between 30-32. The Keck has 10m mirrors, though looking with shorter integration times and the atmosphere to look through might be roughly the same limiting magnitude. Edited July 19, 2008 by scalbers
Norman Albers Posted July 19, 2008 Author Posted July 19, 2008 In the inimitable words of my brother, when faced with the stupendous, that's pretty cool !!!
scalbers Posted July 19, 2008 Posted July 19, 2008 Janskys, I love it. Now tell me the answer! . . . . . OK, helpful conversions: energy of one EV (photon) is 1.6 E-19 Joule. Yes, we might also note that a Jansky is flux per hertz of frequency, so this has to be integrated over the range of visible light to help get a photon count. http://en.wikipedia.org/wiki/Jansky
Norman Albers Posted July 21, 2008 Author Posted July 21, 2008 Clarifying my statements in post #13, when I shorten the "fuzziness" or falloff envelope of my assumed wave packet, the first order terms yield a picture of two opposed dipoles. I have the first four orders of integration by parts and the other set of terms is at right angles and out a bit in the 'X' dimension, namely maxima in an expression of: [math] Xe^{-a^2 X^2} [/math]. So each is served with a twist. This is more like Fig. 3 in the paper offered by Jacques. It is a poor choice of words to say "opposed dipoles". This is so in an axial sense, but to look at it there are similarly oriented dipoles above and below the axis.
scalbers Posted July 21, 2008 Posted July 21, 2008 Cool, it seems we agree that a single moderate star at a billion LY is 'pretty weak'? Clarifying this a bit more, the sun would be +43 magnitude at 1 billion light years. It would be at a +30 magnitude HST visibility limit if it were located 2 million light years away.
Norman Albers Posted July 21, 2008 Author Posted July 21, 2008 Do we observe only galaxies, then, at such distances? Thinking high luminosity, maybe quasars? How far out do we measure the population of novae?
scalbers Posted July 21, 2008 Posted July 21, 2008 (edited) The relative brightness of objects (and by implication their respective visibility at various distances) is easiest for me to express in terms of absolute magnitude. The brightest stars as mentioned before are around absolute magnitude -10 (visible around 3 billion light years away with a +30 apparent magnitude limit). The Milky Way Galaxy is -20 and the brightest galaxies are actually -22 (instead of what I said earlier). In this context a supernova might be around -15. A quasar I'm unsure of though it appears some can outshine entire galaxies. It seems like galaxies can POTENTIALLY be easily seen at distances well beyond 10 billion light years, however they APPEAR dim due to the red shift. The red shift might even play a role in the visibility of the more distant supernovae. Another issue might simply be that stars wouldn't be isolated very far away, so one in practice might only resolve the larger galaxy, even in cases where one could in theory see something the brightness of a star. So the exact situation might depend on what particular distance and object you want to talk about and whether the object is by itself or immersed (as is typical) in a galaxy. Edited July 21, 2008 by scalbers
Norman Albers Posted July 21, 2008 Author Posted July 21, 2008 Aren't Type I supernovae our standard candle for figuring density?
scalbers Posted July 21, 2008 Posted July 21, 2008 (edited) The high red-shift supernovae are perhaps most notably the key to the accelerating universe, and they've been seen to a red-shift of 1.0 or so. Edited July 21, 2008 by scalbers
Norman Albers Posted July 21, 2008 Author Posted July 21, 2008 I calculate about 2 billion light-years; yah, no?
scalbers Posted July 21, 2008 Posted July 21, 2008 I would suggest trying out Ned Wright's cosmology calculator to see what you get: http://www.astro.ucla.edu/~wright/CosmoCalc.html
Norman Albers Posted July 21, 2008 Author Posted July 21, 2008 Very cool and , yea, idiot-proof! Thanks, scalbers. I get about three times what I first said, or about half the way through evolution. "You can make a calculation for a man, or you can get him an abacus to do all his own calcs!"
scalbers Posted July 21, 2008 Posted July 21, 2008 (edited) Yes that would be roughly 3 times. Here's what I get, note the various measures of the distance (light travel, angular size, luminosity) are all diverging as we are really getting cosmological. This underscores the fact that there's more than mere "distance" that determines how visible distant supernovae and galaxies are. Glad you like this abacus! For Ho = 71, OmegaM = 0.270, Omegavac = 0.730, z = 1.000 * It is now 13.666 Gyr since the Big Bang. * The age at redshift z was 5.935 Gyr. * The light travel time was 7.731 Gyr. * The comoving radial distance, which goes into Hubble's law, is 3317.2 Mpc or 10.819 Gly. * The comoving volume within redshift z is 152.895 Gpc3. * The angular size distance DA is 1658.6 Mpc or 5.4096 Gly. * This gives a scale of 8.041 kpc/". * The luminosity distance DL is 6634.3 Mpc or 21.638 Gly. 1 Gly = 1,000,000,000 light years or 9.461*1026 cm. 1 Gyr = 1,000,000,000 years. 1 Mpc = 1,000,000 parsecs = 3.08568*1024 cm, or 3,261,566 light years. Edited July 21, 2008 by scalbers
Norman Albers Posted July 21, 2008 Author Posted July 21, 2008 In my GR textbook I puzzled for a while before understanding the integration back in "time" as you go further, is into states of higher Hubble "constant". That's where they expand the first two orders of terms, and the second is the R" deceleration parameter.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now