What does this even mean?
Which is why I said “fewer than”. 
(Don’t lecture me on cluster observations, missy! Five years of analyzing Chandra data are the reason that, well, I don’t analyze Chandra data any more!)
No kidding. I started to try to reply, tried to figure out what a parabolic lens (vs. any other kind) had to do with it, and gave up.
If you have a wave and a lens you’ll get only an instance of a wave. The one vector from the light source straight to your location. But you can see the same light in different locations because of the wave nature - everywhere. That is the reason why parabolic lenses even produce one image (read instance) and the rest of it remains unseen. With photons you should count the number of them to have them on film. And now you have to think of a) wavelenght b) speed of light c) quality of lens compared to wavelength d) not to mention shutter speed
But Chandra data are fun!! I’ve spent counts 8 years analyzing Chandra data of galaxy groups, clusters and radio galaxies. And lemme tell you, it beats the interferometry! But you get prettier pictures with interferometry. Sssh. Don’t tell my boss I said that!
Of course, I now realise that its incredibly likely that I know you from a conference or have worked with you at some point (unless you left the field before I got into it)
Me too. My brain still hurts from trying to parse it.
The phrase as clear as mud comes to mind. We see an image because we have the optics to see it with? Cause we can detect it? Right, OK. But say if Doc Alien on the other side of the Galaxy points his/her telescope at the exact same spot, Doc Alien will see the same object.
The inverse square law applies regardless of whether the emissions are waves (or sums of waves) or bunches of particles. In either case, you have some “stuff” spread over the surface of a sphere, which was emitted during some interval of time. As distance from the source increases, the stuff is spread over a larger and larger area — that area increasing as the square of the distance.
Only when the emitter is omnidirectional — emitting light in all directions. Lasers, for example, emit along a line (with some small angular spread). There are other sources too that are constrained along one direction, or a set of preferred directions. In these cases, you can’t see the emitted light unless you’re in the right place for it.
Like others, I’m having trouble making sense of your words.
You might be interested in this image of Jupiter, taken in the X-ray band. Now, Jupiter isn’t much of an X-ray source. It emits those photons only sporadically. But, if you aim your instrument at it and keep the “shutter” open long enough, you can accumulate enough hits to make an image.
This is exactly what’s happening with visible light, except that there’s so many of these photons arriving every microsecond that you might as well regard the emitted light as smooth and continuous.
Oh! And one more!
Katunari, some of the people posting to this thread have described experiments where an image is formed by single photons. A detector senses a single photon, a few seconds go by, and then the detector senses another. If you add up all the individual detection events over many hours, an image gradually emerges. If light is just a wave, how do you explain this?
Effectiveness of a wave is in relation to viewpoint and an emitter. In my mind the same “photon” could be seen elsewhere, too. The original emitter is not steady, if the electrons are not in the same place and stage before returning to their lower orbits.
I have been talking about waves, but now I understand that you may be thinking of a different waveform. My waves are spheres - 3D circles growing from the emitter through space. You can see only light with respect to the center of the emitter/reflector.
Your mind is wrong, and a “3D circular waveform” would not negate Hamster King’s question.
Please draw an emission point and a circle and radius - you will get as many “photons” as you wish. Actually this circle is a sphere and the radius the normal from your viewpoint. If you use film you’ll get random results.
If you mean the earlier question HK is talking about objects with several emission/reflection points.
Huh? If I use film? What sort of film? Photographic film? And you’re saying that that’s what we perceive as photons?
Three questions.
The formula for the area of the surface of a sphere is 4 pi r[sup]2[/sup]. Therefore your spreading spheres are subject to the inverse square law. Do you disagree? If you agree what are the consequences of this?
Since your spreading spheres strike all surfaces within reach equally, how do you explain darkness?
Then why isn’t the “spherical wave” seen by adjacent cells in the detector? The detector is a large grid of CCDs. If a spherical wavefront “washed” across it, you’d expect to see to cells detect it in an expanding ring. But that’s not what you see. One isolated cell registers the impact of one isolated photon. Then a long time passes. Then a different isolated cell registers the impact of a different isolated photon. And on and on. Eventually if you sum up all these individual events, a picture of a distant object appears.
I somehow missed the answer to an earlier part of this thread. Please give it to me straight…
Did you ever agree whether energy added to a state of matter causes matter to release energy in a different state?
A wave of quantum spacetime energy exhibits two dimensions. What waves are you talking about?
ItS
peace
rwjefferson
Oh good there I was worried this thread hadn’t gone on long enough.
Quantized spacetime (QST) is typically thought to exist not necessarily in 11 M dimensions but rather in a eigendimenional state of 4. Strangely enough this actually allows for the collapse of multidimensional wave states into recognizable events. In short, the way we always thought we understood nature actually is how nature works. The artificial scaffolding thrown up by a complexity loving 20th century can be seen for what it was … a deception aim squarely at placing the elitist scientist above the rest of us.
So the photoelectric effect is actually the result of packetized quantum spacetime dimensionally crossing a barrier and having its energy refracted. Very simple when you think about it. It also allows for the sun to “burn” electrically since the contained qst packets are crossing into space from the sun and refracting again!
I feel a little guilty now.
“What if Einstein got it all wrong?”
Not many wanted to answer my question, but have a huge need to defend curved space-time. That’s OK, but I didn’t ask for it. Now it seems that I have to use all my time to answer questions.
I agree - I had the same formula in mind. If radius doubles the intensity of light will be less (because you don’t want to create energy). Think the sphere as a soap bubble - the walls of it get weaker in relation to it’s size. Inverse-square-law. You see light only from your viewpoint and darkness everywhere else. I know that it’s difficult to get big spheres trough small holes, but shit happens. Of course there is not light everywhere, because of physical obstacles.
I would be more than happy, if people just answered the original question.