"Is Lighting a Butane Lighter A Nuclear Reaction?"
by Ted Huntington
01/19/2006, finalized 5/12/06
Surprisingly, the answer may in fact be yes, lighting a common butane lighter is initiating a nuclear reaction, because the light and heat coming from the burning gas might be the result of atoms being unwound, and that includes the neutrons and protons, the nuclear components of the Hydrogen, Carbon, and Oxygen, whose mass is completely disintegrated into photons.One other alternative is that the photons are made from separated electrons, which would make lighting a butane lighter non-nuclear.
The view established by Antoine Lavoisier back in 1777, is that Hydrogen + Oxygen + a spark = Water + energy.The view was and still is that all of the Hydrogen and Oxygen gas is converted to water. The light and heat that exit this equation were viewed simply as a loss of "energy".However, given the new theory that all matter is made of particles of light (photons), one interpretation of this simple combustion phenomenon is that the photons exiting with a frequency of infrared and visible, originated in the actual atoms themselves as electrons, protons, or neutrons. In other words, those photons are from electrons, protons, or neutrons that are separated into their source particles, which are photons.So in this view, (and even in the current popular view) the photons exiting any combustion reaction are equal to the amount of mass lost in the end products of the combustion reaction.
The traditional view is that some portion of the mass of the fuel and the mass of the gas the fuel is being burned in (most often oxygen), are being converted to energy in the form of light and heat, which everybody can relate to as simply being the hot flame we see and feel the warmth from.So, where does this mass, that was converted to light and heat come from?I think we should keep an open mind with every theory, but I think it is clear that much of that energy, in the form of light and heat comes from the conversion of the fuel and oxygen atoms being separated into their source photons.So, if that is true, we should add the number of those converted atoms to the chemical equations of combustion (or perhaps keep a separate equation for that specific reaction).
Why has such a stark simple truth, never been explained publicly before?Simple combustion is all around us almost everyday.Many people use gases like methane (the heating and gas stoves in many people's houses), butane (in many gas lighters), and propane (for example a gas grill on your patio).
In the current view of this simple process ofcombustion:
Burning Hydrogen in Oxygen is:
H2+O2 → H2O + light + heat
There is a general formula for hydrocarbon combustion in oxygen (methane, butane, propane, etc):
CxHy + (x + y/4)O2 → xCO2 + (y/2)H2O1
Here is the equation for propane:
C3H8 + 5O2 → 3CO2 + 4H2O1
We should not forget that some source of photons, or spark is needed on the left side of all combustion equations.In contrast to a popular Bruce Springsteen song, we can in fact start a fire without a spark, for example, by using a lens or mirror. Some people include the amount of energy in kilojoules, or ergs on the right hand side, but where does that energy balance on the left side?The same is true for equations that list a +/-∆ heat, a change in heat/temperature, I think that it is complex (in particular for heat loss), but any heat gained or lost needs to be explained on the left side in the form of actual matter.
They all basically state that all of the gas is completely converted to different molecules.But that is impossible, because some (and perhaps only a tiny fraction) of that gas is being converted to light and heat.I do not doubt that the above equations are partially true.Some of the atoms of gas are rearranged into CO2 and H2O, I doubt chemists like Antoine Lavoisier and many others would lie about such a finding, but clearly the above equations are not complete.There needs to be at least one more equation added or integrated and that is:
C3H8+O2+initial photons-> photons (light and heat)
So we need to change the equation:
2CxHy + 2(x + y/4)O2 → xCO2 + (y/2)H2O1+ 2xy mole of heat
or add another equation to describe combustion more completely:
CxHy = Xheat+light
Clearly, this basic reaction of H2 + O2 + initial photons => H2O + light + heat, and CH4 + O2 + initial photons => H2O + CO2 +light + heat, are definitely happening, as are numerous other reactions (for example there are thought to be many in between reactions, in addition to "incomplete combustion" where there is not enough oxygen and CO, carbon monoxide is formed instead of CO2), but there must be something else happening too.Those equations cannot explain everything that is happening, because they do not account for the heat and light on the left side of the equation.So I think we can say, in this theory that there are some possible sources of those photons we see and feel and light and heat which we must add to the left side of every and any combustion equation:
1) The photons come from the source electrons losing mass, perhaps in changing orbit electrons lose mass in the form of photons.This might happen as the atoms are adapting to a new molecular structure, for example from an H2 structure to the H2O structure. This requires that electrons have varying mass (electrons that come if a variety of sizes).In addition, this loss of mass cannot affect their "charge" (how many photons can an electron or proton lose before losing the "charge" effect?).This is a non-nuclear reaction since all original particles are accounted for, and in theory electrons are not in the nucleus (although I want to add that neutron decay into a proton, antineutrino and electron hints that there are electrons in the nucleus, bound up in the neutron).
2) The photons come from neutrons or protons which lose mass, but still maintain their particle nature.For example, if a proton loses some mass, it must still maintain a full positive 1 charge.This is perhaps a nuclear reaction, since it does involve neutrons and or protons, but the nucleus basically remains intact, so no nucleus is destroyed in this hypothetical explanation.
3) The photons come from the full conversion of electrons, protons, or neutrons.In other words some or all of these particles are completely separated into photons.It seems unlikely that only part of some atom would be separated into photons, because that would leave an atom unbalanced in terms of charge and structure.The more likely scenario is that a complete atom would be separated into photons.However, if a partial atom was ever separated we would expect to measure charged particles (electrons or protons) from a simple combustion, or neutrons, and to my knowledge, we do not detect such particles from simple combustion indicating that the entire atom conversion to photons is perhaps the more likely.If the entire atom is separated into photons of light and heat then the equation should account for those atoms (although the number of atoms may in fact be very small). This would be a nuclear process since an entire atom is being converted to photons.So that is kind of an exciting aspect of this hypothesis, that combustion of gas might be very similar to nuclear fission.In combustion, atoms might be being annihilated by a chain reaction initiated by photons.In nuclear fission, atoms are being divided (and potentially some are being annihilated) by a chain reaction initiated by neutrons (can neutrons initiate a simple combustion reaction?).
I think we should definitely do more experiments to publicly prove that no other particles (protons, neutrons, electrons, muons, pions, neutrinos, etc.) exit from simple combustion, in addition to very careful measuring of quantities of mass of before and after atoms and photons in various combustions.One very interesting finding is that Oxygen is not the only gas that can result in combustion, Fluorine and Chlorine gases function the same way as oxygen does, most people are not aware of that, and it is very interesting.Other gases like Hydrogen (presumably Hydrogen without Oxygen will not combust, even when sparked or heated with photons focused by a lens or mirror), Helium, Nitrogen and the inert gases can not do combustion.
In conclusion, I think that we need to go back to those 1800s combustion experiments, and think more about the source of the photons that exit such reactions in light and heat frequencies.We need to determine if those photons are from complete atomic separation, electrons losing mass, or from some other source.We should closely measure how much mass is lost to heat and light, understand how many photons are being released from atoms of both the fuel and combustible gas, and include those atoms in all of the combustion chemical equations.I am somewhat shocked that this simple truth has not been realized before now by the many people involved in science, and that the chemistry textbooks and videos have not yet been corrected.