Attention to detail lacking
Jim, I think you often don't word things carefully enough. The resulting discussions get pointless in a big hurry.
The optics used for focusing are NOT mirrors, they are (hopefully) transparent at the frequency under use. A mirror on the other hand is required to be OPAQUE with respect to transmission, we want full, 100%, reflectivity. That means that every photon that hits that mirror interacts, loses some energy, and gets re-emitted. ^^^^^^^^^^^^^^^^^
Are you implying that the wavelength for incident photons changes upon interaction with the mirror?
The energy loss at the mirror is lost photons not altered wavelengths. The lost photons have varying fates.
You stated that every photon interacts, loses energy and is re-emitted.
I think the reflected beam has the same wavelength as the incident beam. Your blurb about absorption and cascades is only true for some fraction of the lost photons that constitute the inefficiency of the mirror. Others have a different fate. Maybe that's what you meant but you did say "every photon." And here's an exchange with Tim :
At 6:30 PM -0500 7/24/01, Jim Choate wrote:
And these are reasonably low power lasers...
http://www.ee.surrey.ac.uk/SSC/IJSSE/issue1/unwin/unwin.html
The simple fact is that the thermodynamic impact of a laser beam that is several feet across and emitting more photons than the surface of the sun will not be easy to reflect unless immense cooling is taken. Cost/weight factors alone argue it in the negative.
"More photons than the surface of the sun" for HOW LONG?
A minute? A second? A millisecond? A microsecond?
You confuse fluence with flux, a classic mistake.
(A pulse "brighter than the sun" but lasting only milliseconds will have far less heating effect than other flux level pulses lasting longer. Calculations matter. And, yes, I used to do these calculations when I was refuting Kosta Tsipis' calculations of the late 70s. Fluence matters.)
--Tim May
The sun produces shitloads ( check your CRC Handbook for conversions between the shitload and more familiar units ) of power : http://seds.lpl.arizona.edu/nineplanets/nineplanets/sol.html says 386 billion billion megawatts If we know the spectral characteristics of the sun ( the black body spectrum perhaps? ) we could come up with a photon count. I'm not sure whether you mean to talk about photon counts and adjust the power and wavelength variables or you really mean to discuss something that operates somewhere between IR and UV. Let's assume the latter. It is after all a LASER. You did say "surface of the sun". To me that means integrate over 4 pi. 3.86E26 W regardless of the radius. I doubt if anyone has made a laser that operates at that power level even for one fs. Let's try the other approach... The power output from the sun is about 1350 W/m^2 as measured here. Maybe that was what you meant as a reference power level. Let's see, 1350 W/m^2 -> 1.35E-3 W/mm^2 so a 1 mW laser with a beam area of .74mm^2 is "as bright as the sun" at least in terms of gross energy density. That disregards spectral effects. Not too tough to be "brighter than the sun". I don't think you could even light a bucket of gasoline 1 m away with it no less knock down a rocket. It's also pretty easy to handle with a basic mirror. I'd say that's a pretty wussy power level for something that needs to melt a rocket in flight. Focussed to a spot that is 1/1000 the area of the parent beam it starts to get interesting but let's see you hold that spot steady from a 747 in turbulence long enough to burn a hole in a nice shiny casing going 8000kph 200km away. So if we're going to discuss physics let's do it with a bit of care. Maybe it will be more interesting. I'm no expert but I'm willing to try. Yawn, Mike
At 8:24 PM -0700 7/24/01, mmotyka@lsil.com wrote:
You stated that every photon interacts, loses energy and is re-emitted.
I think the reflected beam has the same wavelength as the incident beam. Your blurb about absorption and cascades is only true for some fraction of the lost photons that constitute the inefficiency of the mirror. Others have a different fate.
Maybe that's what you meant but you did say "every photon."
And here's an exchange with Tim :
....
So if we're going to discuss physics let's do it with a bit of care. Maybe it will be more interesting. I'm no expert but I'm willing to try.
Yawn, Mike
Yawn, indeed. Arguing physics with Choate is even worse than arguing math or history or law with him. He has peculiar notions of what energy, mass, reflectance, and a hundred other physics concepts are. Consult the archives for dozens of examples. Photons hitting a surface most definitely do not "lose some energy" and get "re-emitted." There are some very particular configurations that can act as wavelength doublers, but this is a particular, and hard to set up, configuration. Photons hitting a mirror either are re-emitted with the same energy as before or interact via the photoelectric effect and are thermalized (converted to phonons). That colors are preserved in mirrors, absent tints (special absorbers), is a Physics 1 clue that mirrors do not downshift photon energies!. I had a technician who once worked for me who had the same idiosyncratic grasp on physics concepts. Now I don't expect techs to have physics degrees, but I _do_ expect them not to develop their own personal notions of what a "semiconductor" is, or what a "Fermi level" is, and then reason with great confidence from these flawed concepts. My tech used to do just this. He even "corrected" some of the equations in the copy of the book he'd asked to borrow fro me, Andy Grove's "Physics and Technology of Semiconductor Devices." I told Frank, my tech, to knock it off, and to NEVER deface one of my books with his weird notions of how physics ought to work. I think Choate is much like this tech of mine: lacking a solid grounding and overly reliant on his own private notions of what "mass" and "energy" and "group velocity" and so on are. All the best cranks view the world this way. I don't know Choate's educational background, but I would not be at all surprised if he is self-taught and moved into computers out of some technician training school. (Not that college physics is needed. When I was in high school I knew enough about physics and math not to have made some of the boners Choate has come out with.) --Tim May -- Timothy C. May tcmay@got.net Corralitos, California Political: Co-founder Cypherpunks/crypto anarchy/Cyphernomicon Technical: physics/soft errors/Smalltalk/Squeak/agents/games/Go Personal: b.1951/UCSB/Intel '74-'86/retired/investor/motorcycles/guns
At 8:35 PM -0700 7/24/01, Tim May wrote:
At 8:24 PM -0700 7/24/01, mmotyka@lsil.com wrote:
I think Choate is much like this tech of mine:
Have you ever seen the two of them together?
(Not that college physics is needed.
I should hope not, I've got a Fine Art degree with a smattering of philosophy and English. Which is why I work with computers for a living.
When I was in high school I knew enough about physics and math not to have made some of the boners Choate has come out with.)
I don't know enough math, but I know that I don't, so where I get confused I ask.
Tim May Wrote:
I think Choate is much like this tech of mine: lacking a solid grounding and overly reliant on his own private notions of what "mass" and "energy" and "group velocity" and so on are. All the best cranks view the world this way.
maybe Choate is the long lost son of oedipa maas. phillip
At 10:29 AM -0400 7/25/01, Phillip H. Zakas wrote:
Tim May Wrote:
I think Choate is much like this tech of mine: lacking a solid grounding and overly reliant on his own private notions of what "mass" and "energy" and "group velocity" and so on are. All the best cranks view the world this way.
maybe Choate is the long lost son of oedipa maas.
What a w.a.s.t.e. --Tim May -- Timothy C. May tcmay@got.net Corralitos, California Political: Co-founder Cypherpunks/crypto anarchy/Cyphernomicon Technical: physics/soft errors/Smalltalk/Squeak/agents/games/Go Personal: b.1951/UCSB/Intel '74-'86/retired/investor/motorcycles/guns
On Tue, 24 Jul 2001 mmotyka@lsil.com wrote:
You stated that every photon interacts, loses energy and is re-emitted.
Sure, it has it's momentum changed. Think about it. The photon comes in from one direction and is absorbed/interacts with the atoms. As a result they get re-emitted (reflected) in the exact opposite direction. The point is the photons that get re-emitted ARE NOT THE SAME PHOTONS THAT WERE ABSORBED. You can't do that without losing something. photons only have one thing, energy as represented in their wavelength. The beam that gets re-emitted is less energetic than the beam that came in. Even if it does have the same phase and time coherence as the incident one. 2nd law of thermodynamics. You're confusing the intermediate vector boson as the carrier of information with the information itself. -- ____________________________________________________________________ Nature and Nature's laws lay hid in night: God said, "Let Tesla be", and all was light. B.A. Behrend The Armadillo Group ,::////;::-. James Choate Austin, Tx /:'///// ``::>/|/ ravage@ssz.com www.ssz.com .', |||| `/( e\ 512-451-7087 -====~~mm-'`-```-mm --'- --------------------------------------------------------------------
At 10:45 PM -0500 7/24/01, Jim Choate wrote:
On Tue, 24 Jul 2001 mmotyka@lsil.com wrote:
You stated that every photon interacts, loses energy and is re-emitted.
Sure, it has it's momentum changed. Think about it. The photon comes in from one direction and is absorbed/interacts with the atoms. As a result they get re-emitted (reflected) in the exact opposite direction. The point is the photons that get re-emitted ARE NOT THE SAME PHOTONS THAT WERE ABSORBED.
You can't do that without losing something. photons only have one thing, energy as represented in their wavelength. The beam that gets re-emitted is less energetic than the beam that came in. Even if it does have the same phase and time coherence as the incident one. 2nd law of thermodynamics.
You're confusing the intermediate vector boson as the carrier of information with the information itself.
You're gibbering about things you have no clue about. Babbling about "the intermediate vector boson" when you clearly don't even understand high school physics is especially bizarre. Photons are _quanta_, as in quantum theory. Their energy is given by the usual E = hv (v is nu, frequency). They aren't "less energetic" when they scatter (i.e., are reflected). A photon fired at a surface will scatter/reflect with precisely the energy it had when it hit the surface, unless it is absorbed (in which case it knocks electrons out of atoms...the photoelectric effect in a vacuum, thermalized in ordinary solids). This is the essence of Planck's and Einstein's work in the first decade of the 20th century. Photons don't lose a "little" bit of their energy. They either get completely absorbed, aka the photoelectric effect, or they hold _all_ of their energy. Ironically, the key experiment was done with photons of varying energies striking (scattering) off of a metal plate. The photoelectric effect established that a photon only gives up its energy when it is energetic enough to knock an electric out of a orbital. Photons do not "give up some energy" except in this all or nothing way. The only way photons change their (apparent) energy is through Doppler shift, which is really a frame of reference situation. Red shift, for example. (Ditto the Mossbauer Effect, where gamma photons alter energies slightly.) Whether a photon moving through a medium is the "same" photon or a series of absorbed/emitted photons is an interesting topic to discuss. But the one thing we _know_ is that such photons do not "lose energy" in the way you describe. If they did, then blue light would turn into red light. It doesn't. A blue photon is a blue photon is a blue photon. It's not your ignorance of high school physics (or high school math, or high school history, etc.) that's annoying, it's your oracular pronouncements of flawed theory. This is why people call you a crank. --Tim May -- Timothy C. May tcmay@got.net Corralitos, California Political: Co-founder Cypherpunks/crypto anarchy/Cyphernomicon Technical: physics/soft errors/Smalltalk/Squeak/agents/games/Go Personal: b.1951/UCSB/Intel '74-'86/retired/investor/motorcycles/guns
On Tue, 24 Jul 2001, Tim May wrote:
You're gibbering about things you have no clue about. Babbling about "the intermediate vector boson" when you clearly don't even understand high school physics is especially bizarre.
Photons are _quanta_, as in quantum theory. Their energy is given by the usual E = hv (v is nu, frequency). They aren't "less energetic" when they scatter (i.e., are reflected). A photon fired at a surface will scatter/reflect with precisely the energy it had when it hit the surface, unless it is absorbed (in which case it knocks electrons out of atoms...the photoelectric effect in a vacuum, thermalized in ordinary solids).
<groan> Here is what actually happens. It's called "The Radiated Electric Field". Some 1st year engineering physics books will have it listed in the index under 'mirror'. The incident photons strike the mirror. A current is induced. That current is electrons moving in a resistor. Making heat, losing energy. Note, we are NOT talking about photons here but J/C. That current re-emits photons that retain both frequency and temporal/time related coherence (see Maxwell's Equations for more detail). However, the total number of photons MUST be reduced from the incident beam. This also means the incident photons can not be the same as the emitted photons. The photons (as opposed to 'a photon') lose energy. Blipverts strike again! </groan> Physics J. Orear ISBN 0-02-389460-1 In numbers, z_o is the area of the incident beam j_o is the surface current per unit distance I_o is the induced current I_o = (z_o)(j_o) j_o = (c * E_o)/(2 * 3.14159... * k_o) Go look those up yourself. -- ____________________________________________________________________ Nature and Nature's laws lay hid in night: God said, "Let Tesla be", and all was light. B.A. Behrend The Armadillo Group ,::////;::-. James Choate Austin, Tx /:'///// ``::>/|/ ravage@ssz.com www.ssz.com .', |||| `/( e\ 512-451-7087 -====~~mm-'`-```-mm --'- --------------------------------------------------------------------
At 12:39 AM 7/25/2001 -0500, Jim Choate wrote:
On Tue, 24 Jul 2001, Tim May wrote:
You're gibbering about things you have no clue about. Babbling about "the intermediate vector boson" when you clearly don't even understand high school physics is especially bizarre.
Photons are _quanta_, as in quantum theory. Their energy is given by the usual E = hv (v is nu, frequency). They aren't "less energetic" when they scatter (i.e., are reflected). A photon fired at a surface will scatter/reflect with precisely the energy it had when it hit the surface, unless it is absorbed (in which case it knocks electrons out of atoms...the photoelectric effect in a vacuum, thermalized in ordinary solids).
<groan>
Here is what actually happens. It's called "The Radiated Electric Field".
Some 1st year engineering physics books will have it listed in the index under 'mirror'.
The incident photons strike the mirror.
A current is induced.
That current is electrons moving in a resistor. Making heat, losing energy. Note, we are NOT talking about photons here but J/C.
That current re-emits photons that retain both frequency and temporal/time related coherence (see Maxwell's Equations for more detail). However, the total number of photons MUST be reduced from the incident beam. This also means the incident photons can not be the same as the emitted photons.
The photons (as opposed to 'a photon') lose energy.
The photons don't lose energy: the beam or flux is diminished in intensity. Your improper choice of terms is what's getting creating the misunderstandings. steve
On Tue, 24 Jul 2001, Steve Schear wrote:
The photons don't lose energy: the beam or flux is diminished in intensity. Your improper choice of terms is what's getting creating the misunderstandings.
Really? I'd say that any distinction between the 'photons energy' and the 'current energy' is a false distinction. The 'current' is realy a 'virtual' current since it exists 'instantaniously' (ie there is no discernable delay between the incident photons and the emitted one) otherwise your 'mirror' wouldn't be a 'mirror' (ie the reflected image would not retain both phase and time coherence) as the reflected waves would be in-coherent. For a similar 'virtual' effect look at the current flow on the input of a op-amp (ie virtual ground). In either case, your distinction is irrelevant to the issue at hand. Will a mirror, even the most effective mirror on the planet, work as a defence against a high-level optical laser for a rocket (ie ICBM)? Answer, no. Even the small losses of such a perfect mirror at these beam intensities would be too great. There are three components to this failure. The first is the simple losses in the mirror itself, the second is the massive cooling it would require to compensate for these loses would completely destroy the ability of the rocket to deliver its warhead, and then some since it would be roughly equivalent to another rocket being strapped onto this one. The impact on Isp is catastrophic in this scenario. Then we can look at operational issues, will it even fit in a submarine now effectively? Will it fit on a mobile launcher? I doubt it considering the size changes that would need to take place. And thirdly, even if we allow that the mirroring worked, it is a moot point. The rocket is now too large and slow to deliver its warheads effectively because the increase in mass has so reduced its range and payload. By defeating the laser, you've created a missile that you can't launch. The goal of the laser operator is obtained, in either case. -- ____________________________________________________________________ Nature and Nature's laws lay hid in night: God said, "Let Tesla be", and all was light. B.A. Behrend The Armadillo Group ,::////;::-. James Choate Austin, Tx /:'///// ``::>/|/ ravage@ssz.com www.ssz.com .', |||| `/( e\ 512-451-7087 -====~~mm-'`-```-mm --'- --------------------------------------------------------------------
On Wed, 25 Jul 2001, Jim Choate wrote:
The incident photons strike the mirror.
A current is induced.
That current is electrons moving in a resistor. Making heat, losing energy. Note, we are NOT talking about photons here but J/C.
That current re-emits photons that retain both frequency and temporal/time related coherence (see Maxwell's Equations for more detail). However, the total number of photons MUST be reduced from the incident beam. This also means the incident photons can not be the same as the emitted photons.
No. The electromagnetic field, as described by Maxwell's equations, is a statistical abstraction arising out of quantum electrodynamics, with strict limits on its applicability. When you try to deal with individual photons, you're going outside these limits, just as surely as you would be going outside the limits of classical thermodynamics if you e.g. tried to argue from the 2nd law in a simple enough quantum system like an isolated electron. The classical ED only gets you statistical results, and as such does not allow you to reason about the behavior of individual quanta. Reflection of light in a mirror happens at a scale beyond the reach of classical electrodynamics. The only thing that happens is that some photons are converted to heat, while others scatter as-is. (Besides, the above stuff is nonsense at its face, as one can clearly see from the fact that insulators can be reflective, and that an incident magnetic field does not visibly affect the reflectance of a conductive mirror.) Sampo Syreeni, aka decoy, mailto:decoy@iki.fi, gsm: +358-50-5756111 student/math+cs/helsinki university, http://www.iki.fi/~decoy/front
Jim Choate
That current re-emits photons that retain both frequency and temporal/time related coherence (see Maxwell's Equations for more detail). However, the total number of photons MUST be reduced from the incident beam. This also means the incident photons can not be the same as the emitted photons.
The photons (as opposed to 'a photon') lose energy.
That still means your original post was wrong.
Jim Choate
The optics used for focusing are NOT mirrors, they are (hopefully) transparent at the frequency under use. A mirror on the other hand is required to be OPAQUE with respect to transmission, we want full, 100%, reflectivity. That means that every photon that hits that mirror interacts, loses some energy, and gets re-emitted.
See? "Every photon that hits the mirror," etc. You were under the impression that each photon lost energy. You were wrong. It's not hard to admit it. C'mon. If you don't want to type it, you can just cut and paste the following into a message: ---Begin cut area--- I, James Choate, was wrong. My statements concerning the interactions of photons with mirrors showed a clear misunderstanding of the underlying physics. ---End cut area--- -- Riad Wahby rsw@mit.edu MIT VI-2/A 2002 5105
On Wed, Jul 25, 2001 at 09:10:00AM -0400, Riad S. Wahby wrote:
---Begin cut area---
I, James Choate, was wrong. My statements concerning the interactions of photons with mirrors showed a clear misunderstanding of the underlying physics.
---End cut area---
My working theory is that Choate is using the assembled, um, wisdom of the cypherpunks as an alternative to classical forms of education (note I'm not saying he's actually learning anything). Or, alternatively, Choate is simply a Markov chain that uses the cypherpunks archives, the U.S. Constitution, and a high school physics book as its base text. Perhaps eventually the Book of Choate will become as well known as the Book of Beak: http://www.talltree.net/~aad/beak.html -Declan
participants (9)
-
Declan McCullagh -
Jim Choate -
mmotyka＠lsil.com -
Petro -
Phillip H. Zakas -
Riad S. Wahby -
Sampo Syreeni -
Steve Schear -
Tim May