What is fire and why does it burn

Recently, I kindled a fire on the beach and realized that I did not know anything about the fire and how it works. For example - what determines its color? Therefore, I studied this question, and this is what I learned.

the fire

Fire is a stable chain reaction involving combustion , which is an exothermic reaction in which an oxidizer, usually oxygen, oxidizes fuel, usually carbon, resulting in combustion products such as carbon dioxide, water, heat and light. A typical example is the burning of methane:
')
CH ₄ + 2 O ₂ → CO ₂ + 2 H ₂ O

The heat generated during combustion can be used to power the combustion itself, and in the case when this is enough and no additional energy is required to sustain the combustion, a fire arises. To stop the fire, you can remove the fuel (turn off the burner on the stove), the oxidizer (cover the fire with a special material), heat (spray the fire with water) or the reaction itself.

Combustion, in a sense, is the opposite of photosynthesis , an endothermic reaction in which light, water and carbon dioxide enter, resulting in carbon.

It is tempting to assume that when burning wood, carbon found in cellulose is used . However, apparently, something more complicated is happening. If wood is exposed to heat, it undergoes pyrolysis (unlike burning, which does not require oxygen), which converts it into more combustible substances, such as gases, and these substances light up during fires.

If the tree burns long enough, the flame will disappear, but the corruption will continue, and in particular the tree will continue to glow. Smoldering is an incomplete burning , as a result of which, unlike complete burning, carbon monoxide occurs.

Flame

The flame is the visible part of the fire. With combustion, soot occurs (part of which is a product of incomplete combustion, and part - of pyrolysis), which heats up and produces thermal radiation . This is one of the mechanisms that give the fire color. Also, with the help of this mechanism, the fire warms up its surroundings.

Thermal radiation is produced due to the movement of charged particles: all matter of positive temperature consists of moving charged particles, so it radiates heat. A more common, but less accurate term is black body radiation . This description refers to an object that absorbs all incoming radiation. Thermal radiation is often approximated by the radiation of a blackbody, possibly multiplied by a constant, since it has a useful property — it depends only on temperature. The emission of AChT occurs at all frequencies, and as the temperature rises, the radiation rises at high frequencies. The peak frequency is proportional to the temperature of the Wien displacement law .

Everyday objects constantly emit heat, most of which is in the infrared . Its wavelength is longer than that of visible light, so you cannot see it without special cameras . The fire is bright enough to give out visible light, although it has enough infrared radiation.

Another mechanism for the occurrence of color in a fire is the emission spectrum of a burning object. In contrast to the radiation of the blackbody, the emission spectrum has discrete frequencies. This is due to the fact that electrons generate photons at certain frequencies, moving from a high-energy to a low-energy state. These frequencies can be used to determine the elements present in the sample. A similar idea (using the absorption spectrum ) is used to determine the composition of stars. The emission spectrum is also responsible for the color of fireworks and colored lights .

The shape of the flame on Earth depends on gravity. When a fire warms the surrounding air, convection occurs: hot air containing, among other things, hot ash, rises, and cold (containing oxygen), lowers, supporting the fire and giving the flame its shape. With low gravity, for example, on a space station, this does not happen. The fire feeds on the diffusion of oxygen, so it burns more slowly and in the form of a sphere (since burning takes place only where the fire touches oxygen-containing air. There is no oxygen inside the sphere).

Black body radiation

The ACT radiation is described by Planck’s formula , which relates to quantum mechanics. Historically, it was one of the first applications of quantum mechanics. It can be derived from quantum statistical mechanics as follows.

We calculate the frequency distribution in the photon gas at temperature T. That it coincides with the frequency distribution of the photons emitted by an absolutely black body of the same temperature follows from the Kirchhoff radiation law . The idea is that the blackbody can be brought into temperature equilibrium with the photon gas (since they have the same temperature). A photon gas is absorbed by THT, also emitting photons, so for equilibrium it is necessary that for each frequency at which THT emits radiation, it would absorb it at the same speed, which is determined by the frequency distribution in the gas.

In statistical mechanics, the probability of a system being in the microstate s, if it is in thermal equilibrium at temperature T, is proportional to

e ^{- β E _s}

where E _s is the energy of the state s, and β = 1 / k _B T, or thermodynamic beta (T is temperature, k _B is the Boltzmann constant ). This is the Boltzmann distribution . One explanation for this is given in Terence Tao's blog post . This means that the probability is

p _s = (1 / Z (β)) * e ^{- β E _s}

where Z (β) is the normalizing constant

Z (β) = ∑ _s e ^{- β E _s}

called the partition function . Note that the probabilities do not change if E _{s is} changed by ± constant (which as a result multiplies the partition sum by a constant). Only the energies of different states differ.

Standard observation indicates that the statistical sum, up to a constant factor, contains the same information as the Boltzmann distribution, so everything that can be calculated based on the Boltzmann distribution can also be calculated from the statistical sum. For example, the moments of a random variable for energy are described

<E ^k > = (1 / Z) * ​​∑ _s E ^k _s * e ^{- β E _s} = ((-1) ^k / Z) * ​​∂ ^k / ∂ β ^k * Z

and, up to the solution of the problem of moments , it describes the Boltzmann distribution. In particular, the average energy will be equal to

<E> = - ∂ / ∂β log Z

Boltzmann distribution can be used as a definition of temperature. It says that in a sense, β is a more fundamental value, since it can be zero (which means equal probability of all microstates; this corresponds to an “infinite temperature”) or negative (in this case, microstates with high energies are more likely; this corresponds to " negative absolute temperature ").

To describe the state of a photon gas, you need to know something about the quantum behavior of photons. With standard quantization of the electromagnetic field, the field can be considered as a set of quantum harmonic oscillations , each of which oscillates with different angular frequencies ω. The energies of the eigenstates of a harmonic oscillator are denoted by a non-negative integer n ∈ _{≥ 0} , which can be interpreted as the number of photons of frequency ω. The energies of their own states (up to a constant):

E _n = n ℏ ω

where ℏ is the reduced Planck constant . The fact that we only need to track the number of photons follows from the fact that photons belong to bosons . Accordingly, for a constant ω the normalizing constant will be

Z _ω (β) = ∑ [n = 0; ∞] e ^-nβℏω = 1 / (1 - e ^-βℏω )

Retreat: Wrong Classic Answer

The assumption that n, or, equivalently, the energy E _n = n ℏ ω, must be integer, is known as the Planck conjecture , and historically this may have been the first quantization (as applied to quantum mechanics) in physics. Without this assumption, using classical harmonic oscillators, the sum above turns into an integral (where n is proportional to the square of the amplitude), and we get the “classical” normalizing constant:

Z ^CL _ω (β) = ∫ [0; ∞] e ^{- n β ℏ ω} dn = 1 / βℏω

These two normalizing constants give very different predictions, although the quantum one approaches the classical one, when βℏω → 0. In particular, the average energy of all photons of the frequency ω, calculated through the quantum normalizing constant, is obtained

<E> _ω = - d / dβ * log 1 / (1 - e ^-βℏω ) = ℏω / (e ^βℏω - 1)

And the average energy calculated through the classical normalizing constant will be

<E> ^CL _ω = - d / dβ * log (1 / βℏω) = 1 / β = k _B T

The quantum response approaches the classical one as ℏω → 0 (at low frequencies), and the classical answer corresponds to the equidistribution theorem in classical statistical mechanics, but it is completely at variance with the experiments. It predicts that the average energy of the whip radiation at a frequency ω will be constant, independent of ω, and since radiation can occur at frequencies of any height, it turns out that the frequency response emits an infinite amount of energy at any frequency, which of course is not. This is the so-called. " ultraviolet disaster ".

In turn, the quantum normalizing constant predicts that at low frequencies (relative to temperature) the classical answer is approximately true, but at high frequencies the average energy falls exponentially, and the fall is large at lower temperatures. This is because at high frequencies and low temperatures the quantum harmonic oscillator spends most of its time in the ground state, and does not go so easily to the next level that the probability of which is exponentially lower. Physicists say that most of this degree of freedom (the freedom of an oscillator to oscillate at a certain frequency) is “frozen”.

Density of states and the Planck formula

Now, knowing what is happening at a certain frequency ω, it is necessary to sum over all possible frequencies. This part of the calculation is classical and no quantum corrections are needed.

We use the standard simplification that the photon gas is enclosed in a volume with a side of length L with periodic boundary conditions (that is, it will actually be a flat torus T = ℝ ³ / L ³ ). Possible frequencies are classified according to the solutions of the equation of electromagnetic waves for standing waves in a volume with the specified boundary conditions, which, in turn, correspond, to an accuracy of a factor, to the eigenvalues ​​of the Laplacian Δ. More precisely, if Δ υ = λ υ, where υ (x) is a smooth function T →, then the corresponding solution of the equation of an electromagnetic wave for a standing wave will be

υ (t, x) = e ^{c √λ t} υ (x)

and therefore, given that λ is usually negative, and therefore, √ λ is usually imaginary, the corresponding frequency will be equal to

ω = c √ (-λ)

Such a frequency occurs dim V _λ times, where V _λ is the λ-eigenvalue of the Laplacian.

We simplify the conditions with the help of periodic boundary conditions because in this case it is very easy to record all the eigenfunctions of the Laplacian. If for simplicity we use complex numbers, they are defined as

υ _k (x) = e ^ikx

where k = (k ₁ , k ₂ , k ₃ ) ∈ 2 π / L * ℤ ³ , the wave vector . The corresponding eigenvalue of the laplacian will be

λ _k = - | k | ² = - k ² ₁ - k ² ₂ - k ² ₃

The corresponding frequency will be

ω _k = c | k |

and the corresponding energy (one photon of this frequency)

E _k = ℏ ω _k = ℏ c | k |

Here we approximate the probability distribution over possible frequencies ω _k , which, strictly speaking, are discrete, with a continuous probability distribution, and calculate the corresponding density of states g (ω). The idea is that g (ω) dω should correspond to the number of available states with frequencies in the range from ω to ω + dω. Then we integrate the density of states and obtain the final normalizing constant.

Why is this approximation reasonable? The complete normalizing constant can be described as follows. For each wave number k ∈ 2 π / L * ℤ ^3, there is a number n _k ∈ _≥0 describing the number of photons with such a wave number. The total number of photons n = ∑ n _{k of} course. Each photon adds to the energy ℏ ω _k = c | k |, which means that

Z (β) = ∏ _k Z _{ω k} (β) = ∏ _k 1 / (1 - e ^{-βℏc | k |} )

over all wave numbers k, therefore, its logarithm is written as the sum

log Z (β) = ∑ _k log 1 / (1 - e ^{-βℏc | k |} )

and we want to approximate this sum by an integral. It turns out that, for reasonable temperatures and large volumes, the integrand changes very slowly with a change in k, so this approximation will be very close. It stops working only at extremely low temperatures, where Bose-Einstein condensate occurs.

The density of states is calculated as follows. Wave vectors can be represented as uniform lattice points living in “phase space”, that is, the number of wave vectors in a certain region of phase space is proportional to its volume, at least for regions large compared to the 2π / L lattice spacing. In fact, the number of wave vectors in the phase space region is V / 8π ³ , where V = L ³ is our limited volume.

It remains to calculate the volume of the region of the phase space for all wave vectors k with frequencies ω _k = c | k | in the range from ω to ω + dω. This is a spherical shell with a thickness of dω / c and a radius of ω / c, so its volume

2πω ² / c ³ dω

Therefore, the density of states for the photon

g (ω) dω = V ω 2/2 π ² c ³ dω

In fact, this formula is two times understated: we forgot to take into account the polarization of photons (or, equivalently, the spin of the photon), which doubles the number of states for a given wave number. Proper density:

g (ω) dω = V ω ² / π ² c ³ dω

The fact that the density of states is linear in the volume V works not only in the flat torus. This is a property of the eigenvalues ​​of the Laplacian according to the law of Weil . This means that the logarithm of the normalizing constant

log Z = V / π ² c ³ ∫ [0; ∞] ω ² log 1 / (1 - e ^{- βℏω} ) dω

The derivative of β gives the average energy of the photon gas

<E> = - ∂ / ∂β log Z = V / π ² c ³ [0; ∞] ℏω ³ / (e ^βℏω - 1) dω

But for us it is important the integrand, which gives "energy density"

E (ω) dω = Vℏ / π ² c ³ * ω ³ / (e ^βℏω - 1) dω

describing the amount of photon gas energy, originating from photons with frequencies in the range from ω to ω + dω. As a result, we got the form of Planck’s formula, although you need to play a bit with it in order to turn it into a formula that relates to ACHT and not to photon gases (you need to divide by V to get density in a unit of volume, and do something else to get measure of radiation).

Planck’s formula has two limitations. In the case when βℏω → 0, the denominator tends to βℏω, and we get

E (ω) dω ≈ V / π ² c ³ * ω ² / β dω = V k _B T ω ² / π ² c ³ dω

This is a variant of the Rayleigh-Jeans law , a classical prediction on the emission of the blackbody. It is approximately performed at low frequencies, but at high frequencies it diverges from reality.

Second, as β ℏ ω → ∞, the denominator tends to e ^{β ℏ ω} , and we get

E (ω) dω ≈ V ℏ / π ² c ³ * ω ³ / e ^βℏω dω

This is a variant of the approximation of wine . It is approximately performed at high frequencies.

Both of these limitations historically arose before Planck’s formula itself.

The law of displacement of wine

This kind of Planck formula is enough to find out at what frequency the energy E (ω) is maximum at temperature T (and, therefore, what color will be approximately the blackbody at temperature T). We take the derivative with respect to ω and find that it is necessary to solve the following:

d / dω ω ³ / (e ^βℏω - 1) = 0

or, which is the same (taking the logarithmic derivative)

3 / ω = βℏe ^βℏω / (e ^βℏω - 1)

Let ζ = βℏω, then rewrite the equation

3 = ζ e ^ζ / (e ^ζ - 1)

Or

3 - ζ = 3e ^-ζ

With this form of the equation, it is easy to show the existence of a unique positive solution ζ = 2,821 ..., therefore, given that ζ = βℏω and the maximum frequency

ω _max = ζ / βℏ = ζ k _B / ℏ * T

This is the law of Wien's bias for frequencies. Rewrite using wavelengths l = 2πc / ω _max

2πc / ω _max = 2πcℏ / ζ k _B T = b / T

Where b = 2πcℏ / ζ k _B ≈ 5,100 * 10 ^-3 mK (meter-Kelvin). This calculation is usually done slightly differently, first expressing the energy density E (ω) dω in terms of wavelengths, and then obtaining the maximum of the resulting density. Since dω is proportional to dl / l ² , ω ^{3 is} changed to ω ⁵ , and ζ is replaced with a unique solution '

5 - ζ '= 5e - ζ ^'

which is approximately equal to 4.965. This gives us the maximum wavelength.

l _max = 2πcℏ / ζ 'k _B T = b' / T

Where

b '= 2πcℏ / ζ' k _B ≈ 2,898 * 10 ^-3 mK

This is Wien's law of displacement for wavelengths.

The temperature of a burning tree is about 1000 K, and if we substitute this value, we get the wavelength

2πc / ω _max = 5.100 * 10 ^-3 mK / 1000 K = 5.100 * 10 ^-6 m = 5100 nm

AND

l _max = 2,898 * 10 ^-3 mK / 1000 K = 2,898 * 10 ^-6 m = 2898 nm

For comparison, the wavelengths of visible light are in the range from 750 nm for red and 380 nm for violet. Both calculations suggest that most of the radiation from the tree occurs in the infrared, this radiation heats, but does not shine.

But the surface temperature of the sun is about 5800 K, and substituting it into the equations, we get

2πc / ω _max = 879 nm

AND

l _max = 500 nm

which means that the sun emits a lot of light in the entire visible range (and therefore it seems white). In a sense, this argument works backwards: perhaps the visible spectrum during evolution has become such, because at certain frequencies the Sun radiates the most light.

And now a more serious calculation. The temperature of a nuclear explosion reaches 10 ⁷ K, which is comparable with the temperature inside the Sun. We substitute this data and get

2πc / ω _max = 0.51 µm

AND

l _max = 0.29 µm

This is the X-ray wavelength . Planck's formula does not stop at a maximum, so nuclear explosions emit radiation with shorter wavelengths — namely, gamma rays . A nuclear explosion produces this radiation only because of its temperature — because of its nuclear nature, an explosion produces, for example, neutron radiation .