Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Im confused about fire. The way I see it :

  1. Heat creates (kinetic) energy in mass and this creates stronger vibrations of atoms.
  2. When those vibrations are strong enough the electrons interact stronger due to electromagnetic forces.
  3. This causes the electrons to fly away. This is the creation of a plasma and the direction of most electrons is the direction of the flame.
  4. the shape and heat of the flame depends mainly on the amount of electrons and their velocity ( apart from the source of heat and the considered gas and solid material)
  5. the electrons move freely ( but not random ) in the flame.
  6. there is a chemical reaction going on that usually produces smoke or such.
  7. the entropy increases in all the above 6 steps and the process is repeated.
  8. the electromagnetic energy is converted into kinetic energy of the electron and conductivity in the plasma (flame).
  9. the moving electrons create an electromagnetic field.

That is more or less how I see it. What steps are ok ?

And the main question , why is there light involved ? I mean why cant we just have kinetic energy of the electrons and a changing magnetic field ? Why are there photons ? If the electrons are in a high energy state why cant they just move faster instead of emitting a photon ???

Why is the light usually visible ? is that because of our structure of air ?

If the light is an electromagnetic wave why does the flame go up ? does that imply all electromagnetic waves go up ?? But electromagnetic waves are not influanced by gravity or buoyancy are they ??

What is the final destiny of the electrons and photons if the heat is gone ??

I am puzzled by this.

share|cite|improve this question
There is lots wrong with your description and inferences; read these articles:,,, – DilithiumMatrix Jan 20 '13 at 23:00
up vote 7 down vote accepted

You should take some time to read the links zhermes suggests. As a starting point:

Fire, or more accurately a flame, is a gas phase reaction. When you look at wood burning you are actually seeing wood heated by the flame giving off gas, and the flame is this gas reacting with oxygen.

When a molecule of combustible gas reacts with a molecule of oxygen the reaction products fly away with more energy than the original molecules had. This extra energy comes from the energy of reaction. The increased velocity of the reaction products corresponds to an increase in temperature according to the Maxwell-Boltzmann distribution.

The light comes from two sources. Firstly, the typical red/orange /yellow glow comes from particles of unburnt carbon in the flame. These are heated by collisions with the rapidly moving reaction products and glow by black body radiation. Secondly the most energetic reaction products collide hard enough to excite electronic transitions or even ionise gas molecules. As the excited molecules relax they may give off light that is characteristic of the transition. That's why you get specific colours e.g. green from copper, yellow from sodium etc.

The flame is not a plasma. Only a tiny tiny fraction of molecules and/or atoms in the flame are ionised.

share|cite|improve this answer
Why do "particles of unburnt carbon in the flame" produce light at all? How are the photons being generated? Is it through the same electron-transition mechanism as 2)? – Kirk Woll Apr 13 '14 at 0:45
@KirkWoll: everything that is hot emits light, whether it's carbon particles in a flame or metal heated in a forge. This is black body radiation. See the link in my answer, or have a read through the answers to What are the various physical mechanisms for energy transfer to the photon during blackbody emission?. – John Rennie Apr 13 '14 at 5:50
perfect! I struggled to find the correct words to search for, but that's exactly what I was looking for. Thanks. – Kirk Woll Apr 13 '14 at 14:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.