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You have an "interesting" definition of a perfect mirror; it absorbs energy from the light it reflects.Would you like to try again with a better definition?
In fact, the more energy is being reflected, the stronger is the pressure of a reflected EM wave - energy is still being transferred into the reflective surface, despite the complete lack of energy absorption.
Quote from: CrazyScientist on 07/06/2021 13:28:02In fact, the more energy is being reflected, the stronger is the pressure of a reflected EM wave - energy is still being transferred into the reflective surface, despite the complete lack of energy absorption.Not quite.The force which the photons exert does not necessarily mean there is transfer of energy. Energy is transferred when the force moves through a distanceIn order to be perfectly reflective the walls have to be infinitely massive (this causes other problems).
However, there's another way to do it.You can imagine54 a nearly massless mirror.When a photon hits it, it will move and take some energy from the photon. But that means that, when another photon hits it on the other side, it will add energy to that photon.Overall, the sum of the energies will be conserved The wavelengths of the photons will be "scrambled" and will settle down to a black-body distribution.The energy (on average) imparted to the light, rigid mirror will be Boltzmann's constant times 3 times the temperature. (That's the same energy as would be carried by an electron or proton at that temperature.)As more photons are added, the temperature will rise. If you do that slowly, you can easily calculate the energy of the mirror.
However, in the end, one of two things will still happen; the sphere will break or it will collapse into a BH
But as long as the pressure is being applied to the reflective surface, there is some transfer of energy between photons and the mirror.
I'm sure, that if it would be the case, someone already use this process, to create some kind of weapon of mass destruction, or just to make a cheap source of heat.
ideal optical diode to let light in but keep it inside
My question is, if EM waves reflected from a perfect mirror keep their original wavelenght?
Quote from: CrazyScientist on 07/06/2021 15:11:43But as long as the pressure is being applied to the reflective surface, there is some transfer of energy between photons and the mirror.Only if it moves.And, if it moves then, in due course, it transfers energy back to other photons.Quote from: CrazyScientist on 07/06/2021 15:11:43I'm sure, that if it would be the case, someone already use this process, to create some kind of weapon of mass destruction, or just to make a cheap source of heat.Do you understand that the requirements- both the perfect mirror and the Quote from: CrazyScientist on 07/06/2021 12:37:40 ideal optical diode to let light in but keep it inside do not, and can not exist?
Quote from: CrazyScientist on 07/06/2021 15:11:43My question is, if EM waves reflected from a perfect mirror keep their original wavelenght?They must, by definition.Because a change in wavelength would mean a change in energy and that is inconsistent with the definition of a "perfect" mirror.
And you are peffectly aware, that this makes each single photon an almost ideal source of a constant and infinite kinetic energy?
All you have to do, is to keep each photon "alive" as long as it is possible.
And you are peffectly aware, that this makes each single photon an almost ideal source of a constant and infinite kinetic energy?Since real life mirrors can use up to 99% of that energy to produce a kinetic energy. All you have to do, is to keep each photon "alive" as long as it is possible.
Your problem is, that you think that light emmision inside a spherical shell, is like inflating an empty balloon - and this is totally wrong.
I guess this settles the issue of photon momentum transfer during reflection.
Think about this scenario as about some source of a buzzing sound
Quote from: CrazyScientist on 07/06/2021 17:41:58Think about this scenario as about some source of a buzzing soundWhy think about a really bad analogy when we can think about the actual science?
So you don't consider quantum physics as an actual science?
what should happen if we apply here your way of thinking
EM waves which propagate in vacuum don't interact with each other -
Quote from: CrazyScientist link=topic=82373[quote author=CrazyScientist link=topic=82373.msg642439#msg642439 date=1623083186I guess this settles the issue of photon momentum transfer during reflection.It was never in dispute; you have shown that I am right.