# Naked Science Forum

## General Science => General Science => Topic started by: MikeL on 26/10/2017 06:08:20

Title: How do we get higher than ignition temperature?
Post by: MikeL on 26/10/2017 06:08:20
I'm once again revealing my ignorance here, but I would appreciate your help. After redirection to the photoelectric effect with my discussion on quanta I learnt that the two main concepts of frequency and amplitude (intensity) of the wavelength (of a photon) affect the electron discharge from a metal plate.

With increasing amplitude (same frequency) we are able to clear more of the electrons at that energy level off the plate. But we cannot increase the energy of the electrons. Thus if we are hosing the board with 'n' energy level photons, they will not summate to cause 2n energy value electrons to shoot off. Instead the excess energy not being used is just reflected back out again.

Contrary to this, if we hose the board with a certain frequency n, 2n, 3n, photon, then only electrons of that energy level or below will be discharged from the board. Energy in equals energy out.

Perhaps I missed something in the translation.

My question becomes, how can we ever get above ignition temperature when creating a fire? If I strike my match at a 2n energy level and it sets all the 2n energy particles (and below) in the newspaper to shoot off as heat, then how do they ever reach 3n? The heat from a log is much hotter than that from the kindling that ignited it.

It can't be that 2 2n's summate as that would be the same as increasing amplitude, which only hoses off more 2n. It can't create 4n.

This means it can't be that the Work Function (the initial ground state of the electron) is a spectrum of threshold n values (3n, 4n, 5n), and that adding 2n boots them all up 2 values, causing energy emission. If it did somehow mean that, then why isn't the temperature increase instantaneous. Why can't we light a block of wood with a match?

Or is the gradual temperature increase of a fire because 2n energy fields of the kindling and wood enclose higher energy fields - that once the 2n restraint dissolves, the 4n energy is free to radiate? When we think about a fire that burns itself into charcoal, we seem to have a runaway radiation effect decreasing mass.

What's going on?
Title: Re: How do we get higher than ignition temperature?
Post by: jeffreyH on 26/10/2017 12:25:49
You need to stop investigating quanta and start investigating heat.
Title: Re: How do we get higher than ignition temperature?
Post by: MikeL on 26/10/2017 12:36:04
Thanks for your reply. I am after a little more than fuel, air and spark though. Heat at the quantum level is the exchange or release of quanta. I am inquiring how that can escalate through temperature without a frequency change. If there is a frequency change, how does it occur?
Title: Re: How do we get higher than ignition temperature?
Post by: chiralSPO on 26/10/2017 14:51:52
This is an excellent question!

I only have time to give a brief answer, but a discussion is better than a lecture anyway, so please do follow up on any points of confusion, or things that I left out!  :)

There are a few different, but related, topics to touch on:

1) Changing energy levels of electrons is not the only way to receive or transfer energy.
Another very important factor (especially when thinking about things that are "hot") is motion of the atoms themselves. In gasses, the atoms or molecules (gas particles) are free to move about through space, and (as far as we can tell) can have any kinetic energy (a continuous range of possibilities from essentially zero up to almost the speed of light). These gas particles transfer energy between each other like tiny billiard balls, exchanging kinetic energy and momentum through collisions. The kinetic energy of the gas particles is where most of the energy of a flame goes (and comes from), especially when considering reactions between two gasses (say oxygen and hydrogen). In a log, things are a bit more complicated: the atoms are no longer free to move about, but they do have enough freedom to wiggle and jiggle (vibrate). These vibrations ARE quantizied, but the spacing of the energy levels is much, much tighter than typical energy level gaps, which actually bring us to our next point.

2) Electronic and vibrational energies can be coupled together (we call these "vibronic" energy levels)
In the picture below (from wikipedia: https://en.wikipedia.org/wiki/Vibronic_spectroscopy ), we see two electronic energy levels, represented by the two curves (these are labelled E0 and E1), and then each of the curves is populated with multiple vibrational energy levels (v0, v1, v2, etc.) The vertical axis represents energy, so you can see that there is a large energy gap between electronic states, and relatively minor gaps between the vibrational states. So there are many (still discrete) energies that can get from E0 to E1: you can excite from E0v0 to E1v2 (as shown in the diagram), or you can go from E0v5 to E1v0, with a lower energy photon (effectively converting some of the vibrational energy to electronic energy). Note that for single atoms, no vibration is possible, so this topic doesn't come up when thinking about the spectroscopy of atomic gasses.

250px-Franck-Condon-diagram.png (44.52 kB . 250x309 - viewed 2219 times)

The color of sulfur gives very extreme example of vibronic coupling. The electronic transition in an S8 molecule is well out of the visible region (in the ultraviolet), and when sulfur is very cold (like liquid nitrogen temps) it is essentially white. However, as the temperature increases, the molecule begins vibrating, and some of the more vibrationally excited molecules can absorb light at the very bluest (violet) end of the visible range, thus appearing yellow (this is the form we typically see it in). But it doesn't stop there. As the temperature of sulfur increases further, it is able to absorb visible photons of lesser energy, and just around its melting point, sulfur turns a dark orange/brick red color (absorbing UV, blue, and green). Upon further heating the sulfure become blood red, and eventually black, as it begins absorbing all of the visible spectrum!

3) And finally, both relativity and the Uncertainty Principle allow for a very tiny amount of wiggle room
Let's imagine that there is an atom with energy transition of exactly 1.98934839510399003 eV. There's no vibronic coupling involved because it is a single atom. Will it absorb a photon with an energy of 1.98935149305 eV (0.0000030979 eV higher in energy)? Possibly. Because the energy that a photon has is actually ever so slightly uncertain--ie it can be treated as a range of energies. It's typically very narrow, but there are ways to make the range of uncertainty so large that single photons can act as if they were white light. (https://en.wikipedia.org/wiki/Ultrashort_pulse ) Also, because atoms are moving, any photon they interact with will be Doppler shifted one way or another. Typically atoms move at very small fractions of the speed of light, so the effect is usually minimal. But in, for instance, a hydrogen flame, H2 molecules can be traveling well over 7500 m/s which is just enough to change the apparent frequency of a photon by a few ppm, if I calculated that right. Not very impressive, but at more extreme temperatures, like in a high energy plasma, this can become important.

(so much for brevity)
Title: Re: How do we get higher than ignition temperature?
Post by: chiralSPO on 26/10/2017 20:57:24
Why can't we light a block of wood with a match?

Or is the gradual temperature increase of a fire because 2n energy fields of the kindling and wood enclose higher energy fields - that once the 2n restraint dissolves, the 4n energy is free to radiate? When we think about a fire that burns itself into charcoal, we seem to have a runaway radiation effect decreasing mass.

What's going on?

This bit has to do with how the energy is "stored." Although you can think of wood as having electrons in high (ish) energies, there are no available empty orbitals for them to drop into. Wood is not an excited electronic state. Instead, it must wait for an oxygen molecule to hit it hard enough to knock an electron loose into the empty orbital on oxygen (this is a massive oversimplification, but I provide it just to make sure this discussion goes down the right track).

The apparent decrease in mass has essentially nothing to do with radiation, and is observed due to the fact that the products of wood combustion are mostly gases. Flames can go the other way too: silane is a colorless and highly flammable gas, but when it burns, it produces silica (basically glass)--but even though we end up with a bunch of solid residue that wasn't there before the fire, there is not actually any gain in mass (it's just harder to appreciate the mass of the gaseous reactants)
Title: Re: How do we get higher than ignition temperature?
Post by: evan_au on 26/10/2017 21:02:52
Quote from: MikeL
The heat from a log
A log is very definitely a macroscopic object.
The heat from a log (or even kindling) consists of an immense number of quanta.
10 Watts from the kindling produces around 10000000000000000000 photons of 1 eV each, every second.
10 kiloWatts from a log emits even more photons.

It is very difficult to detect the quantum effects of 1 photon when it is lost in a flood of photons.

Quote
My question becomes, how can we ever get above ignition temperature when creating a fire?
The point about ignition is that it applies to exothermic reactions. Take a mixture of oxygen and hydrogen (a fairly dangerous mixture...).
- Assuming the mixture is at room temperature, you only need to break the bonds of a small* number of oxygen and hydrogen molecules (eg with a tiny electric spark).
- The free atoms recombine as steam, producing a lot of heat.
- This heat raises the temperature of nearby molecules enough them to react, producing even more heat.
- This produces a runaway reaction which we call an explosion.
- This is how internal combustion engines work via a spark plug.
- *"Small" here means almost invisible to a human, but that is still a large number of molecules!

Your match, the newspaper kindling and the log all use the same logic, but at a much slower rate.
Title: Re: How do we get higher than ignition temperature?
Post by: MikeL on 27/10/2017 10:35:29
a discussion is better than a lecture

I am still processing a lot of what this all entails. It's incredibly interesting and no doubt I have it ass-end backwards still at this stage. Just to clarify some of the concepts you have shared with me, would it be fair to say that in di-atomic and poly-atomic molecules, vibrational (including rotational) energy is acting as the range that seems to be so missing from (my introductory understanding of) Quantum Mechanics? And yet this range is itself a series of quanta?

Photons themselves provide some range value due to the blurry nature of the acceptable values of a photon. Is this how we get different value work functions of electrons initially?

The energy graph x-axis shows that nuclear coordinates increase with increasing energy. This is not because of Heisenberg uncertainty (which I would imagine has an inverse relationship with increasing energy) but due to vibrational energy itself?

The wavelengths of the vibration are able to interact with the wavelengths of the electrons because they are a (rough) multiple of the wavelength even though their sizes are so different? That doesn't seem right.

So, thinking aloud as I type, based on the apparent range value of vibration, could you argue that quanta itself is simply a boundary value that allows energy transmission? It seems this thinking brings me back to the photoelectric effect I was initially directed to where range values are not allowed.

I'll keep turning it over. Thanks for your discussion. I would  love hear any other comments or answers you have to offer.
Title: Re: How do we get higher than ignition temperature?
Post by: MikeL on 27/10/2017 11:00:50
A log is very definitely a macroscopic object.
The heat from a log (or even kindling) consists of an immense number of quanta.
10 Watts from the kindling produces around 10000000000000000000 photons of 1 eV each, every second.
10 kiloWatts from a log emits even more photons.

It is very difficult to detect the quantum effects of 1 photon when it is lost in a flood of photons.

Hi Evan_au, but surely this is just amplitude - not frequency? The photoelectric effect I was directed to reveals that increasing the amplitude only serves to remove more of that value energy, not higher value states. Which is why I asked how it was impossible to achieve higher than ignition temperature. To remove energy locked in higher value states (coal burns hotter than wood) we need to change the frequency instead, right.

The point about ignition is that it applies to exothermic reactions. Take a mixture of oxygen and hydrogen (a fairly dangerous mixture...).
- Assuming the mixture is at room temperature, you only need to break the bonds of a small* number of oxygen and hydrogen molecules (eg with a tiny electric spark).
- The free atoms recombine as steam, producing a lot of heat.
- This heat raises the temperature of nearby molecules enough them to react, producing even more heat.
- This produces a runaway reaction which we call an explosion.
- This is how internal combustion engines work via a spark plug.
- *"Small" here means almost invisible to a human, but that is still a large number of molecules!

I wonder if two ideas might be being conflated here - amplitude and frequency?
To produce a lot of heat, is an amplitude function isn't it?

The exothermic consideration is a good one. Do you think the exothermic reaction would qualify as an example of a 2n energy having been acting as a restraint for a 4n radiation process? That when the 2n energy was released the binding energy was dissolved allowing the 4n transmission and thus that frequency transmission?

Title: Re: How do we get higher than ignition temperature?
Post by: evan_au on 27/10/2017 11:59:37
Quote from: MikeL
I wonder if two ideas might be being conflated here - amplitude and frequency?
When you are talking about a burning log, you are not talking about individual quanta, you are talking about a spectrum of radiation. Most of it is at infra-red frequencies that you can't see, but you can still feel it when you close your eyes.

Physicists try to produce general rules for things that convey the essential idea of something.
The general idea of the heat from a blackened, burning log is described in physics as "Black Body Radiation".

It shows that for macroscopic objects, there is a relationship between amplitude and frequency.

Depending on the temperature of the log, it can be glowing a bright orange, a dull red, or not glowing at all (but still putting out a lot of heat). This shows that the spectrum (the relationship between amplitude and frequency) changes its shape as the temperature changes.

If you read the following section, you will see that the only way they could explain this relationship accurately was by assuming some quantisation of light. Assuming that light carried continuous energy simply gave the wrong result. This was another input that led to quantum theory.
Title: Re: How do we get higher than ignition temperature?
Post by: MikeL on 27/10/2017 12:55:50
When you are talking about a burning log, you are not talking about individual quanta, you are talking about a spectrum of radiation. Most of it is at infra-red frequencies that you can't see, but you can still feel it when you close your eyes.

I totally take your point, Evan_au. I think we just had different starting positions for our discussion. My primary question was about how we moved frequency - from a burning match head to a log, based on what I read in my book about black body radiation and thereafter the photoelectric effect I was redirected to.

It is very interesting to hear about a relationship at the macroscopic level between amplitude and frequency. No doubt this is part of the reason there is a lack of unification between GR and QM. I will give the black body link a good read through. Thanks for your direction.
Title: Re: How do we get higher than ignition temperature?
Post by: chiralSPO on 27/10/2017 14:57:44
a discussion is better than a lecture

I am still processing a lot of what this all entails. It's incredibly interesting and no doubt I have it ass-end backwards still at this stage. Just to clarify some of the concepts you have shared with me, would it be fair to say that in di-atomic and poly-atomic molecules, vibrational (including rotational) energy is acting as the range that seems to be so missing from (my introductory understanding of) Quantum Mechanics? And yet this range is itself a series of quanta?

Yes. That is a huge part of it. The diagram I showed earlier was for a diatomic molecule (so there is only one degree of freedom), which already increases the number of allowed energies of photons by a significant amount. Even just going up to a simple molecule like methane (CH4) the number of vibrational modes increases the vibrational modes to 4 (there would be more, but the symmetry of the molecule makes many of them equivalent see here: http://www2.ess.ucla.edu/~schauble/MoleculeHTML/CH4_html/CH4_page.html )

Once you get to polymeric materials like wood there are millions of possible vibrations (note that these largely serve to broaden the absorption they don't really shift it that much because adding a few meV of vibrational energy to 0.5–2 eV doesn't make a huge difference.

Photons themselves provide some range value due to the blurry nature of the acceptable values of a photon. Is this how we get different value work functions of electrons initially?

I think that's a "yes" on the blurriness, but I'm not sure what you mean by "different value work functions of electrons"...

The energy graph x-axis shows that nuclear coordinates increase with increasing energy. This is not because of Heisenberg uncertainty (which I would imagine has an inverse relationship with increasing energy) but due to vibrational energy itself?

Yes. Because this is a diatomic molecule, the nuclear coordinates indicate the length of the bond between the atoms (the distance between nuclei). Because they have accounted for an unsymmetrical potential energy well, the average distance between the atoms increases slightly (anharmonic vibration). As shown in the figure the excited state has a longer bond, but this isn't always the case: bonds can also shrink as electrons are promoted out of antibonding molecular orbitals or in to bonding molecular orbitals.

The wavelengths of the vibration are able to interact with the wavelengths of the electrons because they are a (rough) multiple of the wavelength even though their sizes are so different? That doesn't seem right.

No. Try thinking of it like this: the electron energy levels are caused by the interaction between the electron and the nucleus, but we only think about the electron moving because even the least massive nucleus is nearly 2000 times as massive as an electron. Also in an atom, we typically choose a coordinate system that places the nucleus at the origin, so we don't think of it s moving. But in a molecule it become apparent that the nuclei move too (albeit more slowly). An electromagnetic wave will interact with all of the charged particles in the molecule (electrons and nuclei), adding energy to the whole molecule, which we choose to think of as a specific electron being excited and a specific vibrational mode being stimulated. It is better to think of it as a change in vibronic state.

So, thinking aloud as I type, based on the apparent range value of vibration, could you argue that quanta itself is simply a boundary value that allows energy transmission? It seems this thinking brings me back to the photoelectric effect I was initially directed to where range values are not allowed.

I'll keep turning it over. Thanks for your discussion. I would  love hear any other comments or answers you have to offer.

I don't think that the photoelectric effect (observed when high energy photons hit a metal with a low work function) is a good way of thinking about combustion (a process that involves the interaction of different molecules, each with their own energy levels and kinetic energy).
Title: Re: How do we get higher than ignition temperature?
Post by: chiralSPO on 27/10/2017 15:04:59
To get an idea of how broad absorptions can be, here is the UV-Vis spectrum of Cr(III) ions dispersed in crystalline alumina (this is commonly referred to as ruby)
[ Invalid Attachment ]

(note, rubies look red because they absorb both blue and green light very strongly. Rubies look especially red because they also fluoresce when illuminated with blue or green light, emitting red photons (694 nm).
Title: Re: How do we get higher than ignition temperature?
Post by: MikeL on 28/10/2017 03:01:57
An electromagnetic wave will interact with all of the charged particles in the molecule (electrons and nuclei), adding energy to the whole molecule, which we choose to think of as a specific electron being excited and a specific vibrational mode being stimulated.

Fascinating. I wish I could give you a more cerebral response in turn.

The vibrational energy could be considered shorthand for an irregular pulse relationship set up between consecutive electrons in a molecule with the addition of EM energy. This pulse relationship between the electrons in turn creates a pulse relationship with each nucleus and its electrons. This pulse relationship might be thought of as a compressive energy relationship (compression = repulsion between the electron and nucleus increases due to a distance encroachment through momentum set up by the inter-atomic electron pulses). When the pulsing asymmetry/a-harmonic within an atom exceeds that allowed to maintain stability in the atom, a quanta of energy is released to settle it back down. Is that about right or did I take a wrong turn?
Title: Re: How do we get higher than ignition temperature?
Post by: chiralSPO on 30/10/2017 18:25:03
An electromagnetic wave will interact with all of the charged particles in the molecule (electrons and nuclei), adding energy to the whole molecule, which we choose to think of as a specific electron being excited and a specific vibrational mode being stimulated.

Fascinating. I wish I could give you a more cerebral response in turn.

The vibrational energy could be considered shorthand for an irregular pulse relationship set up between consecutive electrons in a molecule with the addition of EM energy. This pulse relationship between the electrons in turn creates a pulse relationship with each nucleus and its electrons. This pulse relationship might be thought of as a compressive energy relationship (compression = repulsion between the electron and nucleus increases due to a distance encroachment through momentum set up by the inter-atomic electron pulses). When the pulsing asymmetry/a-harmonic within an atom exceeds that allowed to maintain stability in the atom, a quanta of energy is released to settle it back down. Is that about right or did I take a wrong turn?

I'm not really sure what you mean by "pulse"...

It is very rare to have an excited electronic state that absorbs another photon and becomes more excited. (2 photon physics can happen in cases involving lasers, but otherwise its essentially unheard of)

My point about the vibronic states was really to offer a way that atoms can absorb light with energies slightly greater or slightly less than the energy difference between two idealized electronic states. This also offers a way for the energy from an absorbed photon to be added to the system (adding internal kinetic energy to the molecules by increasing the vibrations) without just re-emitting the same photon later (energy *can* build up in this way).
Title: Re: How do we get higher than ignition temperature?
Post by: jeffreyH on 01/11/2017 20:31:26
Heisenberg used an oscillator to do his first calculations in quantum mechanics. This is how he realised that oscillator energy is quantised. However the quantisation is too small to measure individually. Hence why I said concentrate on heat.