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Can I burn a piece of wood by emitting single photons on it? (for example by emitting only one photon per second or per milisecond etc to the wood). How much should be the rate of emitting single photons? How long does it take for wood to catch fire in this way? Are the energy of photons important in this process?

mathLover
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    Interesting question, but how much energy is required to make wood burning? Normally there is a critical temperature given, so I guess power is more important than energy because you have to apply the energy very fast, otherwise it will be conducted away by re--emission, molecular collisions etc... – Charles Tucker 3 May 15 '21 at 10:38
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    Hmmm... You can set fire to a bit of wood using the sun and a magnifying glass. But it doesn't quite catch without the magnifying glass, which says that the sun does not send enough photons for the wood to burn ! – Vincent Fourmond May 15 '21 at 21:20
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    Given that there is a continual flux of gamma rays at sea level, and that there are forests at sea level, perhaps some other considerations are needed. – Andrew Morton May 15 '21 at 21:35
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    As long as the wood is not at absolute zero temperature it will emit (thermal) photons anyway (and a lot of them). – lalala May 17 '21 at 17:01

7 Answers7

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No. To set fire to a piece of wood you must deliver heat energy to it at a faster rate than it will dissipate. With visible light, the energy of a single photon arriving per second is negligible.

There is an interesting article on Wikipedia https://en.wikipedia.org/wiki/Ultra-high-energy_gamma_ray which illustrates the energy of extremely high frequency gamma ray photons. In theory it is possible for an individual gamma ray to have more energy than a burning match, say, but there are two practical difficulties that would prevent you from igniting wood that way- firstly there is no way to create such high energy photons at present, and secondly they would tend to pass straight through your stick without interacting with it.

Marco Ocram
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    thanks for your help. You said "they would tend to pass straight through your stick without interacting with it." Why? Very high energy photons cannot interact with the wood? – mathLover May 16 '21 at 03:00
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    @mathLover, a more familiar example of very high energy photons might be x-ray photons. Yes, they can pass through matter without interacting with it. Their likelihood of doing so depends on their energy and on the material. These are exactly the characteristics on which medical x-ray imaging relies. – John Bollinger May 16 '21 at 13:17
  • @JohnBollinger. Excuse me, a high energy photon can also produce a free electron of large energy in the wood? (I saw this in another answer, Charles Tucker 3). – mathLover May 16 '21 at 21:02
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    Yes, @mathLover. There is a probability that any given photon will interact with the material instead of passing through. That probability is very small for a single photon with energy comparable to that needed to heat a macroscopic amount of wood to its auto-ignition temperature, but non-zero. If such an interaction did happen then it would likely take the form of ejecting a free electron. – John Bollinger May 17 '21 at 00:24
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    And note that your photon isn't going to do it no matter how energetic. Your epic photon scores a bullseye and dumps it's energy into a nucleus comes apart and dumps it's energy into the surrounding atoms, say raising them to 10kK. You clearly get ignition--but you don't get fire. The problem is that fire has a minimum size and you aren't anywhere near that. The surrounding wood soaks up the heat too fast, the fire goes out before you even see it. You must light an area on fire, not a single point. – Loren Pechtel May 17 '21 at 00:50
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    @LorenPechtel doesn't the loss of heat to the surrounding wood depend on the wood geometry? Through a sufficiently narrow filament the heat generated by combustion should overtake the dissipation of heat from the initial ignition. We're already talking about very carefully engineered conditions, so I don't think a microscopic fuse leading into a wool-like structure made of wood, perhaps ventilated with warmed pure O2 should be off the table. – Will May 17 '21 at 09:24
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    OP does not ask whether a single photon can light up a fire, but if sending photons one-by-one (whatever that means) could do it... – AnoE May 17 '21 at 11:16
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    @mathLover A somewhat similar situation you can look at is the 'Personal Chernobyl of Anatoli Burgoski', who received the full force of a particle accelerator's 76 GeV beam through his head and survived ( https://en.wikipedia.org/wiki/Anatoli_Bugorski ). So he received a 200K+ Roentgen dose (where 1K is already 100% lethal), but vast majority passed right through [ https://www.youtube.com/watch?v=mD4J5VUwiAs ]. But the Bragg peak suggests you could make your wood very thin and long and shoot the photon along that axis? Bugorski would be dead if along the leg or torso, anything near 25cm depth. – user3445853 May 17 '21 at 15:27
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    @user3445853 The SI unit of absorbed dose is the Gray. 300,000 roentgen is 2.6 kGy – D Duck May 17 '21 at 16:40
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    @mathLover roughly speaking, high-energy Gamma photons do not interact with electron shells at all, the electrons simply cannot absorb enough energy. High gamma only interacts directly with nucleus. And at the sort of energy level needed to ignite wood from a single photon, that gamma will go right through a carbon atom like a cannonball through a ten-pin bowling alley. It won't scatter and heat, it will demolish. Repeatedly. – PcMan May 17 '21 at 18:16
  • @user3445853 10 Gray is a guaranteed kill for a whole-body dose. He received a horrendous dose but in a very small part of his body. IIRC that 10 Gray kills through messing with the digestive tract--and that was unaffected. – Loren Pechtel May 17 '21 at 22:27
  • @DDuck indeed bad/rough formulation, "received dose" should be something like "exposed to" to stress that it's not absorbed (as made explicit in latter part anyway)? Anyway gamma rays are not the topic, was meant as analogy; question is could you analogously to radiation align the photon path with the long axis of a very long (and thick and wide) piece of wood to absorb sufficient energy --- or would that just be too much surface and dissipate? – user3445853 May 18 '21 at 19:27
  • @LorenPechtel a less-horrendous-but-horrendous dose (read: with Bragg peak at say 12cm) in exactly the same location would have cooked his brain and instantly killed. The same incident sideways through his GI tract would have killed instantly, front-to-back probably not for same reason as he survived. [But anyhow maybe better example would have been neutrinos flowing through us and being painstakingly detected in swimming pools at the bottom of abandoned mines; it's the same high-velocity-stuff-missing-targets.] – user3445853 May 18 '21 at 19:34
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Let's try to calculate that:

The heat capacity of wood is (Wikipedia): $C=1700 \frac{J}{kg K}$ and the temperature at which wood starts to burn is approx $300^\circ C$. Assuming that the wood has approx $0^\circ C$ (it's winter...) we need to heat it by $\Delta T = 300 K$.

An assumption is needed for the mass of the wood we want to burn, let's assume a very small piece of wood of only $m=0.001$ kg.

The energy $Q$ needed to heat this piece of wood is: $Q = C \cdot m \cdot \Delta T = 510$ J $= 3.2 \cdot 10^{21}$ eV. That is massive, a Photon in the visible spectral range has only about 3 eV.

Anyways, the corresponding wavelength of a photon of that energy is $\lambda = \frac{h \cdot c}{Q} = 3.88 \cdot 10^{-28}$ m $= 3.88 \cdot 10^{-19}$ nm.

Thats really ultra short, much much shorter than X-rays or gamma rays, but still larger than the Planck length, so could exist in theory.

However, that's the mathematical answer. In the real world the energy of the photon would not 1:1 go into heat because in the first place the photon would produce a free electron of large energy that would in a cascade produce other electrons and ions and most of them would just escape from the piece of wood taking their energy with them.

Charles Tucker 3
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    @mathLover We can also turn this answer upside down to answer your second question. Assuming 1 s is short enough time (given number of zeroes below, it doesn't really matter) that the wood doesn't lose the heat, you could theoretically do it with individual 3 eV visual photons emitted one every 10^21 seconds, or every zeptosecond. But while this is (much) longer than Plack time, it is short enough that the photons' wavefunctions would be overlapping and it would become meaningless to ask if they are individual photons any more. So there is no rate fast enough for individual visible photons. – tylisirn May 16 '21 at 09:24
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    @tylisirn, But when using visual photons, the majority of it will be reflected from the wood (otherwise we couldn't see wood). No idea what the reflection coefficient of wood is, but I would guess that the amount of photons needs at least to be doubled... – Charles Tucker 3 May 16 '21 at 11:41
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    Thats [...] still larger than the Planck length, so could exist in theory. As far as we know there is no lower limit to the wavelength of photons (including not the Planck length), so that sentence makes no sense as far as I can tell. Also note that OP does specifically not ask whether a single photon can light up wood, but if many photons delivered one-by-one would do it... – AnoE May 17 '21 at 11:21
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    @AnoE: That's interesting and points to the question whether the wavelength of a photon is a real length. I asked a similar question here: https://physics.stackexchange.com/questions/636573/why-is-photon-energy-no-function-of-transition-time. May I ask you to have a look at it? – Charles Tucker 3 May 17 '21 at 12:06
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    @CharlesTucker3: I'm familiar with these topics, but by far not wise enough to add anything to the "real" physicists answering here. But the StackExchange algorithm gave me this in the sidebar: https://physics.stackexchange.com/questions/159922/are-there-any-theoretical-limits-on-the-energy-of-a-photon?rq=1 might be interesting as well to you. – AnoE May 17 '21 at 12:16
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    @AnoE: Thanks! I especially like the idea that a photon becomes a black hole if it's too energetic... My problem with the wavelength not being a real length is: How would you explain interference? Meaning, two photons below the Planck length - would they interfere constructively or destructively...? – Charles Tucker 3 May 17 '21 at 13:04
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    @CharlesTucker3: I don't know, but in my interpretation, the Planck units are primarily a construct of physicist who want to make their formulas easier. Some of them are incredibly small, others are incredibly large, one (momentum) is human-scale. Nothing in their construction supports the guess that they are a hard physical limit to anything. It just so happens that the Planck length and time are so very small that they are in the Q regime (because that's where they need them) so everything is very weird, and the zero's explode very fast ;). – AnoE May 17 '21 at 13:26
  • @AnoE Googled it, it's like I remember from lectures. Planck length is the fundamental length needed to decide which length is longer (if you have two different lengths). It's cause of uncertainty in quantum mechanics. Nice article here https://www.fnal.gov/pub/today/archive/archive_2013/today13-11-01_NutshellReadMore.html. So 2 photons which wavelength difference of less than one Planck length and you can't say which one is longer. Interesting but unclear to me: photon of less than one Planck length and you can't say if it has actually a wavelength larger zero or not...? – Charles Tucker 3 May 17 '21 at 18:10
  • @CharlesTucker3: yeah, it's a problem with measuring... but does it mean that for example a photon with a wavelength of 0.5 Planchk length cannot exist? I don't think we know that yet. – AnoE May 18 '21 at 06:43
  • Let us continue this discussion in chat. – AnoE May 18 '21 at 06:44
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The short answer is no. The energy of a photon is $$E=hf$$ where $h$ is Plancks constant and $f$ is frequency.

For visible light $E= 10^{-34}10^{15}$ about $10^{-19}J$.

This doesn't seem high enough to set fire to wood, as the heat energy generated would easily be conducted away before it could build up to a value high enough to cause a fire.

John Hunter
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    So, the plank's constant, then? – Richard May 15 '21 at 20:11
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    This isn't enough - because we know light DOES start fires; and the question isn't about a single photon, but a single photon at a time. You need to multiply by another 10^15 to get the power that you can apply - which gets us to about 0.1mW. The power is low enough that that it would be radiated away before it could get to the temperature required for ignition. – UKMonkey May 16 '21 at 18:29
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Define "burn". If you mean "set on fire", the answer in general is an obvious "no". Remember that even in space, far away from any stars, there is a constant barrage of ubiquitous background radiation. In order to provide an environment where only "single photons" (your "one per second") hit the wood, you need to shield it from the background radiation, radio, cosmic rays etc. with a thick-walled container that itself is cooled down close to 0K in order to avoid black body radiation. I can assure you that wood in near absolute darkness at ~0K would not spontaneously ignite.

If you instead mean "oxidize" the answer is probably "yes". Above a certain frequency threshold photons will create oxygen radicals which are very reactive and will oxidize the carbon and hydrogen molecules making up most of the wood. A gram of wood contains something like $10^{22}$ or so atoms; if you oxidize one of them every second you will need about $10^{14}$ years to oxidize the wood. If you speed up the process by a factor thousand you only need $10^{11}$ years, which may or may not constitute progress, depending on the observer. If that is any consolation: Versus the end of the experiment you'll not need the lead chamber any longer because you'll be alone with a very dark sky.

Peter - Reinstate Monica
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If you could make sure your system, which includes some oxygen, is perfectly heat insulated from anything else (which includes your emitter or you are somehow able to emit the photon while maintaining the heat insulation), then over a loooooong time your system would heat up enough.

George Menoutis
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  • You would have to have a highly reflective outer surface for the system in order to not loose the heat through radiation. But it would still have to be absorptive at the wavelength of your stream of photons. So basically, you would have to make use of the greenhouse effect. – Emil May 17 '21 at 15:49
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Yes. As others have correctly explained, you need to deliver enough energy per photon to ignite the wood. @charles-tucker-3 calculated that an ultra-high energy gamma ray with about 500 Joule would do the trick, which is incredibly high for a single photon energy.

Others argued that the photon will not interact with the wood, this is wrong. While wood is fairly transparent to X-rays, every material is opaque for ultra-high energy gamma rays. Even "empty" space itself is opaque to such gamma rays which react with photons from the cosmic microwave background.

Even if the probability of interacting was small, the probability is not zero. The question was whether it is possible to ignite wood with a flux of 1 photon/second at all. Taken literally, it means we have infinitely many trials (there is no time-limit on the flux). We can just wait until the interaction happens.

Others argued that the photon will do inverse Compton scattering and that the electron will escape the wood without delivering energy. That is unlikely. For such high energy photons, the most likely interaction is e+e- pair production. The electron-positron pair has a chance to escape the wood, but it also has a chance to interact again inside the material. The energy loss per unit length of such high-energetic particles is extremely high, so for sure energy will be dumped into the wood. Even if it is just a small fraction of the energy of the original photon, we can simply increase the energy of the photon a bit more in our gedankenexperiment to compensate for the losses.

In some cases a local bubble of high-energy plasma would be formed by the interaction. The plasma gets absorbed by the surrounding matter which is heated in turn. If this bubble is close enough to the surface so that the heated material has a chance to combust in contact with air, then yes you could ignite the piece of wood with a photon flux of 1 Hz.

olq_plo
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    This is perhaps the best answer. What is the average time such a high-energy photon exists before it changes into a matter-antimatter pair? Can you find out what the chance is that it interacts passing through, say, 10 cm of wood? – Peter - Reinstate Monica May 17 '21 at 17:58
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Sure, emit them while traveling toward the piece of wood at high velocity and they'll bunch up enough to make fire in any size of timber

amara
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