no i didnt

im just asking would it be kinda like a change of basis? like in otherwords, are the subspaces spanned by the two eigenfunctions of each operator equivalent?

just when A is also degenerate, then you have that two different linear combinations are eigenvectors of A with the same eigenvalue

@Loong so is it like x and y take on certain values, and the lowest value limits the overall rate of this reaction? if so, which is lower here? fire or oxygens exponent?

@Loong i was trying to parse through what happens in a fire and if its simply combustion or something more. i think it must be more complicated

@Relativisticcucumber the values of x and y are irrelevant for the pure oxygen scenario - increasing the concentration of either $[A]$ or $[B]$ speeds up the reaction regardless of what $x$ and $y$ are, and the 100% oxygen sends $[B]$ to $5[B]$.

@Allie well, the general statement is that there exists a common eigenbasis, i.e. a basis of vectors that are both eigenvector of A and of B

so this is not true?

then, of course, if you start with some eigenbasis of A that's not an eigenbasis of B, you can view switching to the common eigenbasis as a change of basis

ahhhhhh i see

@ACuriousMind darn that's true. the weird thing is when i looked this up, i struggled to find a good resource but the consensus seemed to be that oxygen is usually not the limiting reactant :P

so then i was puzzled about this statement

so maybe i misunderstand how limiting reactants work

@Loong so do you mean the limiting reactant just controls when the reaction stops?

if somethings on fire theres already a fuel source

@Allie hm?

@Relativisticcucumber yes, and how much can react in total before it stops, and what will be left after that

@Loong ah

also do not be confused between rate-limiting step and limiting reactant

limiting reactant is how much reactant you have available for a reaction

rate-limiting step is the "bottleneck" for the reaction rate

@Relativisticcucumber Just to give you an idea what can happen: do not grease the valves of your oxygen cylinder!

@ACuriousMind do you know the answer to your own comment here: physics.stackexchange.com/questions/337728/…

@SillyGoose nope

also that comment is 7,5 years old :P

oh so i think i see. the limiting reactant is indeed the fuel, but the rate of reaction (and therefore amount of fire) is determined by the rate limiting step, which is the consumption of oxygen, so the vastness of the fire is determined by the amount of oxygen and fuel, but how long the fire can burn is determined by the amount of fuel? this makes sense to me, but just making sure i got it right? @Loong

@ACuriousMind the question is timeless :)

@SillyGoose note that I left an equally frustrated comment on the accepted answer :P

@ACuriousMind or even more than just a factor of five. Ambient air is 20 % oxygen at 1 bar. A bursting oxygen cylinder has 100 % oxygen at 200 bar (quickly dropping, though).

@Relativisticcucumber in most cases, yes

@ACuriousMind oh i meant the comment on the accepted answer

pro tip: You can link comments directly by copying the link under the timestamp behind the comment:

@Loong ermmmm this leads me to an interesting q xD so why can i have a tank of gas touch the air and it doesnt combust

You mean "gas" as in gasoline, not a real gas?

not enough energy to start the reaction

@Loong like methane or propane or something like this

theres an activation barrier to combustion

@Allie why

because all chemical reactions have an activation barrier lol

...

??

i mean thats not an answer

the energy released from the combustion serves as the energy that allows other combustion reactions to occur

its a reiteration

okay

i dont know what about my statement is confusing but its oky

if someone says "why is there an activation barrier" and you say "all reactions have an activation barrier" im not sure how that's an answer

@Relativisticcucumber Because in order to make some new chemical bonds, first you have to break some old ones.

because in order for a reaciton to occur you need to go into a transiton state

which is higher in energy than the reactants or products

this is gen chem

@Loong this is a good explanation

@TobiasFünke Here is not a textbook, but a paper along the lines of discussion earlier: science.org/doi/10.1126/science.aaf5037. They seem to have done a Wigner-esque classification of fermions in a crystal via the method of induced representations (little group, stabilizer subgroup, etc. method) of crystal space groups.

meow

also the first author is (incidentally) a prof at UIUC :)

@Loong hm so shouldnt heat be a necessary requirement for a combustion reaction? i checked this but when i looked it up heat is only a byproduct

hence my confusion

that would explain why things dont just burn when exposed to air as my previous q was asking

yes it is

one of the 3 triangles

sorry, 3 parts of the fire tirangle

@SillyGoose thanks for the link. yes, that makes sense I think

lol

yo @Slereah tell Pythagoras a new triangle just dropped

this triangle is fire

@ACuriousMind Slereah will now link some weird ass geometry in which the fire triangle is the equivalent of a right triangle

why is it not in the reactions i dont understand

@HerrFeinmann I see you have understood exactly why I pinged him :P

what?

personally I'm hoping for some water, air and earth triangles

because heat isnt usually incldued in a reaction equation

2025 comes with a new level of comprehension

sometimes it is

but you always know that a reaction has an activation barrier

reactions dont just happen spontaneously

sorry

@Allie but systems should tend toward lower energy states, no? i mean sure sometimes maybe there's a barrier, but there must be cases where there isn't and things happen spontaneously, no?

yes, there is

@Relativisticcucumber but then they will, on average already have happened when you encounter them

by and large it's a safe assumption the stuff around us has settled into some kind of local equilibrium

@ACuriousMind true

i dont know enough thermodynamics/statistical mechanics to give you the exact picture but my understanding is that you CAN have a slight outlier of molecules that can overcome that energy barrier

@Allie fire by tunneling

but the vast majority of molecules before being ignited do not have enough energy to overcome that activation barrier almost all the time

not even tunneling, just they happen to have enough energy for it to occur

@Relativisticcucumber sometimes the light at the end of the tunnel is a flamethrower

LOL

but in a fire you have so many molecules that have enough energy to combust and that energy released from those combustions is enough to cause other molecules to overcome the activation barrier and so on

but if you dont have enough heat, most of the molecules just wont have the energy to get overthe barrier

(paraphrased quote stolen from Mort by Pratchett)

and just because one isolated combustion reaction can occur because that specific molecule happened to have enough energy to react, the energy released from that one single reaction is not heat enough to cause other molecules to start overcomign the barrier

ok i just have one last unresolved fire thing

meow

so i know that heat is a byproduct of combustion, but what is the relation to this and the flame? because i read that combustion reactions can be flameless or have flame

ok idk the ansswer to that one

darn

@User1865345 @TobiasFünke Got these 3 books!!!!!

@Relativisticcucumber the light from the frame is black-body radiation (from the heated combustion products) and spectral lines (because the high heat can knock atoms and molecules into excited states and then they emit the spectral line when the go back to their ground state)

Wiki is pretty good on the subject, I think

i still have a qualm with this discussion upon reflection. so say we have things A, B, and C. A is bound to B but it would be more stable if A was bound to C. Thus, we can say we have some "energy black box" where i say ok if i give some activation energy, A unbinds from B and then binds to C, but what i want to know is what's actually happening in this "energy black box". [...]

[...] i mean this is similar to if we have a bound state and it gets promoted to a free state but im confused about how this energy transfer actually works

@ACuriousMind yes but why would there ever be no flame is what i dont get

lemme peruse the wiki to see if it answers the no flame q

@Relativisticcucumber well, if you have a reaction without soot particles you don't get the blackbody radiation and if the combustion products don't have any strong spectral lines in the visible spectrum then you get a colorless (="no") flame

presumably it would be pretty bright in some other part of the spectrum

ah

also now it makes sense why people buy propane

ive been living in the dark for so long

@ACuriousMind by soot do you mean the heated combustion products?

@Allie hehe nice. have fun!

im trying to understand what you mean by "a reaction without soot particles"

@Relativisticcucumber by soot I mean something that's a solid (like small chunks of carbon) so it can exhibit blackbody radiation

@Relativisticcucumber It's just breaking the bonds between A and B. As you raise the temperature, more and more of the states that aren't the state "A strongly bound to B" become occupied. But still, as long as they sit inside that local equilibrium well of "A bound to B", they just fall back to that. But at some point a significant fraction of the particles occupies states so high in energy that they sit above the potential barrier and they roll down "the other side of the hill", i.e. "A bound to C"

@ACuriousMind yes i see this but where should i consider that energy going within the physical setup? i am having trouble phrasing this q :P

if you now think "isn't this all statistical and doesn't that mean the reaction constantly produces both "A-B" and "A-C" bindings?" then yes, all chemical reactions are stochastic and have chemical equilibrium

@Relativisticcucumber Which energy?

The initial heat goes into breaking some A-B bond. Then this forms an A-C bond with lower energy and that energy difference goes back to heat, breaking more bonds, etc.

wait also about the flame -- the wiki says "As the combustion temperature of a flame increases (if the flame contains small particles of unburnt carbon or other material), so does the average energy of the electromagnetic radiation given off by the flame (see Black body)." why is the "if the flame contains small particles of unburnt carbon or other material" necessary here?

@Relativisticcucumber because the reaction products might not include pure carbon or other solids

that's exactly my "reaction without soot particles" condition from above

if there's no solids in there there's no blackbodies that could radiate

oh so you mean the carbon is the blackbody here?

@ACuriousMind does that include heated fuel, like a candle wick or charcoal?

oh whoops chat happened while I was typing

I would not usually consider the wick part of the flame :P

but sure, the wick might blackbody radiate, too

@Relativisticcucumber yes

and about the energy, if u ask me "what will happen if A-B exists but C is around and A-C is more stable and i provide energy?" sure i can say "A-B will become A-C" but that is just me saying something. i dont actually understand how this happens i only know that it will

so is there a how :P

@Relativisticcucumber Didn't we just discuss the how? You need to break the bonds between and A and B and lift the whole thing over the potential barrier that separates A-B from A-C

I'm not sure what other kind of "how" you're looking for here

@ACuriousMind hm how do i know what physically can be a blackbody? i only know the idea of what a blackbody does

a solid :P

@ACuriousMind oh you meant particles/gases in the flame sorry I was not quite following along lol

@ACuriousMind is there a spectrum of something being a good or bad blackbody?

sure

so how do i know what can be a good blackbody

but I think in this context all that matters is that gases are very bad blackbodies so if there's only gases coming out of your combustion you don't get blackbody radiation

(this should be evident because gases already fail at the "being a body" part :P)

@ACuriousMind ok back to a comment you made earlier. can i say that if i have things bound to eachother, then i add energy, perhaps this channels into self energy of the constituents, then they start to spazz stochastically and so a given amount break the bonds with their self energy induced motions and they they float around and some of them encounter their new bondmate and those settle into a bond and this is all just a stochastic process triggered by the [...]

[...] increased self energy of the entities that were originally bonded?

this actually makes sense so i hope its right

@ACuriousMind lol

@ACuriousMind right i see.

@Relativisticcucumber sounds good to me

but I'm just faking any chemical expertise here :P

yippee

this blackbody discussion has reinvigorated an interest that i have in the difference btwn diamond and graphite -- it always comes back to CM

@Relativisticcucumber in the case of combustion im not exactl sure what kind of reaction mechanism would ttake place

but organic chemistry is entirely focused on the way these reactions occur

for example one of the reactions you first learn is SN2

you can see here the bond is broken and the new bond is formed at the same time, adn you have this in between "transition state". this is where the activation barrier comes from, that transition state is unstable

sn1 is another possible reaction mechanism, and this one happens more in the way you describe it: the bond first breaks, then the new bond comes in at a later time. of course it is clear here why the intermediate would be higher in energy, and why there would be an activation barrier for this reaction

the reason a reaction might prefer to occur via sn1 is if that intermediate state where the bond has broken is unusually stable, stabilized by some sort of chemical group. so that its not super unfavorable for the bond to just break. while sn2 would be the case where breaking the bond would take so much energy you need to just form the other bond at the same time

Anyone can suggest me where can I find a derivation of the solution of the equation $\partial_\mu \partial^\mu \phi (\vec x,t)=0$ for the massless scalar field?

The book I am currently reading just gives the solution as a sum over different $\vec p$ values, but it doesn't show a derivation