Taking a Fresh Look at the Physics of Radiometers.
in [Physics]
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From: Sam Wormley on 8 Jun 2010 13:23 On 6/8/10 5:49 AM, NoEinstein wrote: > Dear Sam: You are this credential-less armchair, blow-hard who tries > to elevate your standing by attacking the person (me) who has made a > greater contribution to the understanding of science than all other > physicists combined. You may be a nice guy, but your posting record does not support the notion that you have much understanding of physics. Can you name one person who has benefited from you "contributions" to science?
From: Sam Wormley on 8 Jun 2010 13:27 On 6/8/10 6:05 AM, NoEinstein wrote: > ...pressure is NEVER from the photons. Radiation pressure http://en.wikipedia.org/wiki/Radiation_pressure "Radiation pressure is the pressure exerted upon any surface exposed to electromagnetic radiation. If absorbed, the pressure is the power flux density divided by the speed of light. If the radiation is totally reflected, the radiation pressure is doubled. For example, the radiation of the Sun at the Earth has a power flux density of 1,370 W/m2, so the radiation pressure is 4.6 �Pa."
From: Sam Wormley on 8 Jun 2010 13:55 On 6/8/10 3:58 AM, kado(a)nventure.com wrote: > On Jun 7, 1:04 pm, Sam Wormley<sworml...(a)gmail.com> wrote: > >> > >> > From the quantum mechanical perspective, >> > >> > 1. photons are emitted (by charged particles) >> > 2. photons propagate at c >> > 3. photons are absorbed (by charged particles) >> > >> > Photon momentum >> > p = hν/c = h/λ >> > >> > Photon Energy >> > E = hν > I would like to point out an inconsistency in your post > about the photon momentum > > Momentum is a mathematically calculated dynamic property > of an uniformly moving mass, i.e., momentum is mass times > velocity by definition. > > The photon is commonly given as a 'massless' quality, i.e. > a photon has no mass. > > Thus any idea of photon momentum is an oxymoron. Physical Properties http://en.wikipedia.org/wiki/Photon#Physical_properties "Einstein showed that, if Planck's law of black-body radiation is accepted, the energy quanta must also carry momentum p=h/λ, making them full-fledged particles. This photon momentum was observed experimentally[40] by Arthur Compton, for which he received the Nobel Prize in 1927".
From: Sam Wormley on 8 Jun 2010 14:04 On 6/8/10 5:40 AM, NoEinstein wrote: > Photons are emitted by all masses whether charged or not. Strike 1 2. Photons propogate at 'c' plus or minus the velocity of the source. Strike 2 3. Photons are absorbed by all masses whether charged or not Strike 3 Yer Out!
From: Timo Nieminen on 8 Jun 2010 15:00
On Jun 8, 9:34 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote: > On Jun 8, 2:30 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Mon, 7 Jun 2010, Tim BandTech.com wrote: > > > On Jun 7, 3:55 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > [a very quick point, will return later] > > > > > Of course the energy hasn't disappeared! Momentum isn't energy, energy > > > > isn't momentum! They're not the same thing! > > > > When the momentum is absorbed is not the energy likewise absorbed? > > > No! (I think "absorbed" is the wrong word here.) > > Is this your new careful use of language Timo? This 'no' is to mean > that you wish to discuss the reflected case. Well, I am discussing the > absorbed case. So please do not change the argument as a means of > falsifying my argument. So, be clear! We have been discussing the case of reflection as well, not just absorption. If you're restricting yourself here to absorption only, say so. In the reflecting case, there can be a transfer of momentum with no transfer of energy. You can have transfer of momentum without absorbing the photon. If you absorb the photon, yes, you transfer all of the momentum and energy. Most of the energy (all of the energy, for a stationary absorber) _doesn't_ become work done on the absorber. This is the next step (2a,b) in the other line of this thread, let us deal with it there. > This 'No' above can be miscontrued. I will > take this wording as support of my argument, for there is no actual > falsification content provided here. That's very sloppy. A lack of falsification is not support (for example, a complete lack of comment is also a complete lack of support). > > Bounce a ball off a wall. KE_in = KE_out. No loss of KE. Change in > > momentum = 2 * momentum_in. |