If we observed a body which absorbed all light as well as emitted it, it would be an observation of two exactly equal oppositional flows, which means a cancellation of energy? Like how a black hole cancels out all light at the horizon. A body which absorbs and emits all light would appear like a black hole. If emission equals the flow of energy from the source(s), we get a potential of -1 from flow in opposite direction.

Annonser

Waves traveling in outdoor directions don’t cancel each other out. While at a particular point the peak of one might meet the valley of the other and thus form a zero point, each wave, or photon, would continue after crossing paths unaffected. This would happen until the collide with something that can absorb them. Take for example sound waves. Noise canceling head phones work by projecting a sound that is in the opposite phase in the same direction as the incoming noise is traveling, ie into your ear drum. No amount of broadcasting sound waves toward each other will permanently cancel each other out. If you set up to sound sources projecting at each other, for the sake of simplicity, the exact same pitch you would observe along the line connecting these 2 points alternating points of cancellation and peaks of double the broadcast intensity. As the waves pass each other there will be points at which the pressure valley meets the pressure peak and would create a dead point. Likewise, there are also points where valley would meet valley (or peak meets peak) and create they same pitch but at double the intensity. Off the top of my head I’m going to speculate that the distance between these points is going to be 1/4 of the wavelength.

Also, if you were to do the experiment by and release a single packet of sound, a very short tick, you would find that you could stand at different positions and sometimes hear neither, or hear both with a varying degree of loudness. This would be based on the wavelength of the pitch and your distance from the sources. No matter where you stood on the line though, unless it was at the single dead point, you would hear sound from both sources. This is no different than light.

The comments about what a black body would appear is inaccurate as well. A black body emits based on its temperature. If it’s hot enough it would look like a star, which is the closest thing we have in the known universe to black body’s. While it’s accurate that a black body absorbs all wavelengths equally, this isn’t the same as a black hole which pulls all light into it. Black holes don’t emit, therefore they appear black. Black bodies emit. Although I’m not site that it is part ood the definition of a black body, or is generally assumed in discussion that the black body doesn’t have enough mass to inhibit light leaving it’s gravitational field.

GillaGilla

I apprecciate your comment, since I put this idea on here because I want to discuss it. Through discussion we refine ideas, and find the flaws. Kind of the opposite approach than what climate science has. So, thank you!

I´ll answer what caught my eye first, and read through the rest more carefully.

Any body emits according to it´s temperature. But a blackbody emits all absorbed heat, which a ”not-blackbody” doesn´t. A ”grey” body doesn´t absorb all heat, but it emits according to it´s temperature. The draper point is proof of that, that all solids emits equally at the same temperature. This means that the heat flow is independent of the mass it flows through, because all masses emit equally at this temperature. Emission depends on the internal state only, and the internal state is what we measure as temperature of the emitter. Prevost stated this a long time ago, and nobody has questioned it. Of course, because who dares to question the old witch of thermodynamics, she would punish you hard for it.

GillaGilla

All bodies emit radiation according to temperature, including black bodies and gray bodies, though they are only theoretical. A black body will emit according to the Planck curve based on the black bodies temperature. Higher temp moves the peak toward shorter wave lengths. Stars are near black body in their emittance.

A gray body is different in that it’s absorptivity, and consequently, emissivity, is less than 1. A theoretical gray body will absorb/reflect all wavelength in equal proportion. If it has an a of .7, this means it will absorb 70% of all incoming radiation, regardless of wavelength. The other 30% is either reflected or transmitted. Krischoff’s law says that absorptivity must equal emissivity. So the body will emit with a peak in the same place as a black body of the same temp as it, but at only 70% of the black bodies intensity.

I’m not sure of the draper point proof you speak of and couldn’t find it through a quick google search. But based on your statement that all solids emit equally at the same temperature, I’m going to have to say I’m either misunderstanding the comment, or it’s false. Metal that is heated to 65C is going to feel much different, both through touch and radiated heating, than foam heated to 65C. So as I read that statement it is not correct, but I may be misunderstanding. Please link me to the proof you speak of.

GillaGilla

I skimmed through your post now, and I think I get your point. If you imagine a wave that travels in one dimension, along a line, what difference does that make? If the wave can`t expand or move in three dimensions?

That is my view of a photon, since it is unaffected by time and space. It has the same energy when it arrives at earth, as when it left the sun. So, in space, it doesn´t expand, or change direction. To me, that means that it can only be a point, travelling along a line. And a perfect emission of exactly equal energy would travel along the same line in opposite direction as a point. Knowing that it is a wave that can´t move in space, or expand, then my conclusion is cancellation.

I have begun to realize that this concept would be like holography, with the black body/hole as the projector.

GillaGilla

Your right that it doesn’t matter if the wave expands in other directions. For simplification and visualization I wanted to not worry about everywhere else, just the line connecting the 2 sources. In reality, ER, sound, and many other waves propagate in 2 or 3 dimensions.

Light is both considered to be particles and waves. Focusing on just 1 aspect or property will lead people astray from the reality of what light is.

GillaGilla

Light is considered both wave and particle because we don´t know what it is. My point of view is that the wave exist outside matter and the particle is the finite, measured response we get in measurement. The truth is, that nobody has ever observed a photon. What we observe is the response in matter, the response in an eye, in a device measuring, or in some kind of matter. We can not be sure that a photon exists outside matter. And my belief is that it doesn´t, because there is no proof of it.

So, if a photon is an impulse of energy that only visit matter in a black hole/body before being reversed back, what would the effect be? Is cancellation out of the question?

”In reality” you write, and I´m not sure a free photon exists in relaity as we know it.

GillaGilla

I began to type up my thoughts on photon collision. After doing a bit of research my thoughts were kin accurate. Turns out when discussing photons it is said that they don’t interact with one another.

https://physics.stackexchange.com/questions/54323/what-would-the-collision-of-two-photons-look-like

GillaGilla

Since I have your attention. This post is just speculation, what I really want to be seen and reviewed, is this: https://lifeisthermal.wordpress.com/2017/04/13/just-numbers-no-blankets-2/

Wherever I post the results in there, everyone goes quiet. Please have a look, and give me any *relevant* opinions.

GillaGilla

Will check this out tomorrow.

GillaGilla

What you say about two waves travelling in opposite directions isn’t correct, apart from at specific points. See ”Two sine waves travelling in opposite directions create a standing wave” on this site.

GillaGilla

A standing wave if there is constructive interference, yes. But destructive interference would cancel it. Destructive interference would be what happens if the wave turns to opposite direction at a fixed end-node.

I`m not saying this is what a black hole does, just that the concept of a blackbody has some properties that I didn´t think of before. Mainly that perfect emission from perfect absorption means that no work is done by the heat, which is what differs it from other radiating bodies. And, if no work is done by the heat flow, the emission must be exactly equal in the exact opposite direction, for every absorbed and emitted photon. The behaviour at a node of a wave that change to opposite direction, would be destructive interference. Perfect emission from perfect absorption would have to be exactly equal in effect, to a wave turning at a node and switching the antinode from +1 to -1. Like the second example in this link: http://www.phys.uconn.edu/~gibson/Notes/Section5_2/Sec5_2.htm

GillaGilla

Those are travelling in the same direction.

GillaGilla

Yes, but the same rules apply for opposite direction. The link you provided shows it.

GillaGilla

It shows that they exactly cancel only at certain points, not that it cancels everywhere (you get a standing wave).

GillaGilla

And now you add the fact that a photon is a wave packet, not an endless wave. Since the energy is finite, there is no energy to keep going at the point where it cancels to zero. It cannot create constructive interference, because you need to add two separate quantities of energy for that. The emitted wave packet must be exactly equal in quantity of energy as the absorbed packet. Nanometer for nanometer. Constructive interference would mean double the energy. A single photon absorbed without any work done, would have to produce destructive interference in emission. Constructive interference would mean that it amplified itself, violating energy conservation. It would be a strange situation half way through the absorption of the wave, with instantaneous emission. When half of the wavepacket is absorbed and emitted instantaneously, the outgoing half would amplify the rest of the incoming wavepacket.

GillaGilla

You’ve got me sufficiently confused by this, that I’m no longer sure, but I don’t think what you’ve said is correct. What we observe from an object is what it emits, or reflects, not what it absorbs – they don’t cancel as you seem to be suggesting. So, something that absorbed all the incoming light and then emitted an equal amount of light, would not look like a black hole because a black hole emits nothing.

GillaGilla

Think about it, complete absorption of every photon from every direction with emission of exactly the same energy levels of the photons, require emission in the exact same angle as incoming energy, with the exact same power.

It would have to mean cancellation.

A black hole would then act like a ball bearing instead of purely attractive.

I’m fumbling on this: wouldn’t that solve the problem of dark matter?

A pressing force of emission.

GillaGilla

No, if something emits a photon, then we can observe it. This would be true even if it happened to have absorbed a photon of exactly the same energy coming from exactly the same direction (but travelling the other way). A black hole is very simply something for which the escape velocity exceeds the speed of light.

GillaGilla

It was just an idea that came to me. But I am still not convinced that you are right.

If two photons travels in opposite direction along a line, what we are talking about is: two waves, exactly equal in energy, that collides. When two mirrored waves collide, they cancel. Just go and sit in your bathtub and make equally large waves going in opposite direction. They will not pass through each other if they are equal in power and travels in exactly opposite direction. I can´t see why this wouldn´t apply to photons, which in space is best described as waves.

The situation is a bit hard to grasp though, because we have never observed perfect absorbers which also emits all light/heat without any conversion of energy into work. This is central to this idea, a perfect emission of light/heat from perfect absorption, must be a process with no work involved. If no work is involved, emission needs to happen along the exact same line which the photon arrived from. Emission in another angle of an absorbed photon, requires that the energy moves inside the body to change location or angle. Which means work. That is what happens in a heat engine, energy moves through the system and displaces mass, and that is what we call work. So, for perfect emission from perfect absorption, the same energy that came in as a point of energy from an exact direction, has to go out as equal energy in the exact opposite direction. I see no way to avoid them cancelling each other out.

GillaGilla