"The above definition though is not unique, but is only one of 480 possible definitions for octonion multiplication with e0 = 1."

Only 480

@Jim yes to "a or b"

@Slereah I'm not sure. I'm also not sure whether my 1,2,3,4 mod 8 was correct for Majorana fermions. There's a weird subtlety to Majorana spinors that they exist or don't exist depending on which signature convention of the metric you choose

let's check

what the hell is a Majorana spinor anyway

@Slereah Probability waves have wavelength, which means they have frequency. Frequency of what they have? What is osculating?

It's a real spinor

@Slereah It's done here rather thoroughly

I think it's made from the spacetime algebra rather than the Dirac algebra

@AnubhavGoel exactly

@ACuriousMind omg thank you

is that a Rigorous reference on spinors that does not assume I have a PhD in algebra

Only Majorana ones

dammit ACM

@Jim "Exactly" Please please complete sentence

Majorana spinors are best spinors, though

@Slereah what is "topological manifold" in El French

Variété topologique

The Tits group and the Wiener measure do not sound hilarious in French, but Boole does

what's a variety then

@Slereah You call manifolds varieties? What's a variety then?

(it sounds like balls)

also variété

Not sure of the etymology

let's check

@ACuriousMind I asked my advisor if I should learn algebraic geometry and he asked me why I would want to lean algebraic geometry and I had no answer

Nobody wants to learn it

seriously, why learn it?

Can an electron move in straight line?

Define "electron"

And "straight line"

@Slereah lowest mass, negatively charged lepton

@Slereah geodesic

@Slereah You know it.

Are we defining it as a point particle

@Slereah Yes, you may, as you wish.

And what are we talking about, the average measurement value or the real measurement

average

Then yes the electron can move in a straight line

real

Then no

@AnubhavGoel let's say that we can approximate the electron (field) with a very localized classical object that moves along a straight line, up to a certain extent (i.e. error)

The probability of a particle taking a smooth path is 0

And can't do smoother than a line

not with that attitude

@Slereah I'm not sure that's true. The smooth paths have zero measure, but the Wiener measure is not the measure that tells you how probable certain classical paths are, is it?

Well, it is ALMOST zero

If you want to use the linguo

But you know what I mean :p

all specific paths are almost zero

@Jim and all feynman heuristics is just....... heuristics

@Slereah No, I don't - I don't know how to derive from QM the probability that a certain classical path is taken, because that doesn't actually make sense - what we can derive are results of single measurements, not "paths"

@yuggib I love technicalities

Well the probability of two measurements being exactly along the same line are 0

Is that better

(almost zero)

@Slereah 100% since any two points make a line

@Slereah Two points always lie in a straight line.

Well THREE measurements

Gee

Don't go busting my chops

I doubt it's identically zero

It will be of measure 0

@ACuriousMind let's say that if a free quantum particle is localized, up to a certain extent, in a point of the phase space (i.e. if its quantum state converges, in the limit $\hbar\to 0$, to the measure concentrated in a point)

Well, actually @Slereah, you're right - because the probability to detect a particle at a single point is always zero

Where does the radial force to deviate from its path come from?

that would mean there's some fundamental thing preventing a particle from occupying any position along a line from its previous two measured positions

Yes

Which would not happen if it was classical

otherwise, if it isn't forbidden, it must occur

Classial particles are all nice dirac deltas

then its quantum evolution at time $t$ will yield a state that is localized (in the same sense above) in the phase space point obtained evolving the original one in a "straight line" for a time $t$ (i.e. along the free classical motion)

@Jim There is a difference between events that are forbidden and those that occur almost never (that is, with probability zero).

@AnubhavGoel That's not a meaningful question - quantum objects neither follow "paths" nor have they "forces" acting upon them.

@ACuriousMind let's split hairs here. There is a fundamental difference between almost never and probability zero

@yuggib Yes, that does seem to make sense

yes

@Jim No, there isn't :P

Although if you do infinitely many quantum measurements

There is still a pretty good chance that you won't find three points aligned

@Slereah What the hell does that mean?

Measure the position of the electron, repeat ad infinitum :p

Also, I think talking about actually performing the measurement is bad, since that changes the state anyway

but that destroys all the properties of quantum evolution...

^^

Not if the state is an eigenstate of the measurement

@yuggib's definition seems better to capture the notion of the movement of a quantum particle

@ACuriousMind I stand corrected, I wasn't aware that infinitesimal probability and probability zero were synonyms

@Slereah ...we are still talking about the position operator, no? :P

Well they are not quite the same

I mean, if you have a uniform distribution between 0 and 1

The probability of being 0.5 is almost 0

The probability of being 2 is 0

@Jim almost never means in a set of probability zero in maths, not in a set with very small probability

but 2 will never occur whereas 0.5 will almost never occur

@ACuriousMind "Quantum objects don't have forces acting upon them" Please share a link so I can read it a little before any further discussion.

Any value you get from the distribution when measured had a probability of almost 0 of occuring

But it still happened

the link provided explains probability zero as events that can happen but whose mathematical probability is 0 because the set is on an infinite domain (terminology might just have been butchered by me)

@AnubhavGoel What do you mean "share a link"? This is contained in every introduction to quantum mechanics - the Schrödinger equation knows only the Hamiltonian and has no notion of "force"

Well not quite true

The force is just $\hat F = i [\hat H, \hat p]$ :p

it is pretty rarely used, though

that's basically the same as saying the probability is infinitesimal. It happens, which means the theoretical probability of it happening is not identically zero, but it is a finite number divided by infinity which is zero, or an infinitesimal

@Jim it depends on the probability distribution you are considering...if you are on a space of values that are the real numbers, and consider a distribution of probability that is continuous with respect to the Lebesgue measure, then every isolated point has measure (probability) zero

@Jim Infinitesimal is usually either "non-zero object that squares to zero" or "object smaller than every positive real number but greater than zero" - but the probability of an almost never event really is identically zero, probabilities don't take infinitesimal values (you might want to argue that's a defect in the theory, but that's how it is)

of course there are probability distributions that are singular with respect to the lebesgue measure, and such that e.g. the probability of $1$ is $1$

Well that's because probability theory is usually defined on $R$

Not $R^*$ or whatever else

Not sure what happens if you define the theory on hyperreals

I think it would still be 0 measure

it all depends what you want to do...returning to the "probability distribution" of paths, even when it is well-defined it surely is neither continuous nor singular wrt the lebesgue measure (since there is no lebesgue measure for paths); so the question is essentially moot

@Slereah @ACuriousMind does the string COM obey the geodesic eq

COM?

@ACuriousMind You're right, if the theoretical probability of an event is identically zero, it should absolutely never be observed. If it is observed, then the theory was mistaken. So I would want to fight that. But if there's really nothing I can do about it, then fighting is a futile waste of effort

Strings obey the Polyakov equation

@Jim Things that are absolutely never observed just don't belong to the space of events, so they don't even get a probability assigned

Which I think for open strings might just be the geodesic equation of every point

Closed strings can oscillate, tho, I think?

Not big on strings tho

So not sure

@0celo7 The target space is flat space, the geodesic equation is just "moves in a straight line", which is conservation of c.o.m. momentum.

@ACuriousMind I find that statement dubious. But I haven't the wherewithal to argue against it at this time

@ACuriousMind You mean an electron does not experience any electrical and magnetic forces, it never deviates due to these forces?

@ACuriousMind well what if you do it on curved space

@AnubhavGoel I mean that to speak of "electric and magnetic forces" and of "deviating" is highly misleading as it paints a classical picture of the electron.

string theory should be on a fully dynamical background, no?

No

String theory is on a fixed background

Well you can do it on a dynamic background

@Slereah I know that

but why

Like if you do cosmic strings dynamics

Electric and magnetic fields do have influence on the behaviour of the electron. To say that this is by means of "forces" or that the electron "deviates" doesn't actually add anything meaningful.

Because string theory is basically a fancy version of covariant quantum gravity

Hey, can anyone tell me if this statement is correct?

It is correct, yes

"Protons and electrons will not always combine; electron capture only happens under high enough energies. That's why neutron stars form when the mass (and energy) of a white dwarf goes beyond a certain point."

I'm not quite sure what this means

@0celo7 That's a rather difficult question to answer.

The Polyakov action is on a fixed background, usually taken to be just flat $\mathbb{R}^{1,9}$

Er, is ACM here?

The $g_{\mu\nu}$ is not a dynamical variable of the theory, however, it plays the role of "coupling constants" for the worldsheet theory

You can have Polyakov on a dynamic background, tho

Conformality of the theory means that the beta functions are zero, i.e. the coupling constants don't really change

Cosmic strings and whatnot

although of course actual cosmic strings do not obey the Polyakov action

But it's a good approximation

What happens is that as you go to the low energy effective 10d SUGRA theory, the string states may manifest as non-trivial metrics on the target space of the effective 10d SUGRA

And that's where all sorts of crazy and not really coherent stuff starts to happen

@ACuriousMind Sorry to bug ya, but can you tell me if that statement is correct?

btw why do you always tell me that string theory isn't an action theory if we do not know yet what the actual form of M theory is :p

For all you know it could be an action theory!

People can't seem to agree how much of this theory is "classical" and how much is "quantum", and I'm not actually sure what really happens there

@Slereah Yes, it could be, but the part of string theory we know isn't - only the low-energy effective theories are

I wonder what happens if you try to do it that way

I'm sure people tried

I wonder what problem makes it unusable as a theory

@ACuriousMind I have no idea what that means.

@SirCumference I'm neither sure what exactly that statement is trying to say nor do I know whether it is correct

@ACuriousMind "Protons and electrons will not always combine; electron capture only happens under high enough energies. That's why neutron stars form when the mass (and energy) of a white dwarf goes beyond a certain point."

I've never understood what "low energy effective theory" means.

That is the same statement

Ya know about electron capture, right?

@0celo7 : Means it kinda works at low energy

@Slereah No shit.

Do you mean electron capture by atoms or by protons

@0celo7 It's the field theory that reproduces, when interpreted as an ordinary QFT, the tree-level string amplitudes

@ACuriousMind Well yes, I understand that.

You have a tough time understanding what I mean when I said I don't know what something means.

@Slereah What's the difference?

@0celo7 Then you understand as much about what "low-energy effective theory" means as I do

Reciting a definition is not understanding.

@ACuriousMind I have no idea how I would extract one.

@0celo7 You compute the tree-level massless string amplitudes and stare long and hard at them until you manage to come up with a QFT that fits them. You are guided in this process by supersymmetry - there aren't that many consistent supersymmetric theories, and their particle spectra all look rather distinctive, and by sheer "luck" it happens that the SUGRA theories with the correct spectrum indeed match the string amplitudes

hm, sugar theory

This "accident", I believe, is one of the older reasons to believe string theory is on to something - Nothing in the construction of string theory guarantees you get such a QFT at all, let alone that it is consistent.

I wish I had time to relearn QFT and ST.

Or learn QFT.

I never really learned it.

Nobody does

QFT is all bogus

"Went through the motions" is more like it.

This "accidental" business gets even more mysterious when one realizes that the only anomaly-free 10d SUGRA theories are actually such low-energy limits of string theory

@Slereah A compact manifold with harmonic curvature and nonnegative sectional curvature has parallel Ricci curvature.

Does it

I don't know what that means

Is harmonic curvature like

$\Box R = 0$

What is $\Box$

D'Alambertian

What I don't know is if "harmonic curvature" means harmonic Riem or harmonic R

Does harmonic Riem even make sense

@Slereah What do waves have to do with this?

Well $\Delta$, if we are doing a Riemannian manifold

also what is sectional curvature

and a parallel curvature

@Slereah Jesus...

Only the central concept in the study of curvature?

Well yes but use proper Physicist linguo, plz

$K(x,y)=R(x,y,x,y)/|x\wedge y|$

None of that math talk

Don't talk to me about fundamental forms or bilinear maps

It's the "induced" curvature in the plane spanned by two vectors

See was that so hard

It's pretty useless in Lorentzian geometry.

yeah I don't think I ever saw it

Usually it's more n-1 D hypersurface induced stuff

Yes because that's Riemannian.

But there are some interesting things about Lorentzian sectional curvature

Like what

I have to get out a PhD level geometry book, hold up

I don't trust "PhD level" from you

Are you gonna bust out a baby book teaching about shapes

BEE is PhD level

Ok, so if the sectional curvature of all nondegenerate planes is bounded then the curvature is constant

So I think I've asked this before

Why is the area spanned by a plane formed by the vectors $x,y$ equal to $(x\cdot x)(y\cdot y)-(x\cdot y)^2$

That looks awfully like a determinant to me

Do you want a rigorous answer or a dumb answer

dumb

That formula is $x^2 y^2 - x^2y^2 \cos^2(\theta) = x^2 y^2 (1- \cos^2 \theta) = (xy\sin\theta)^2$

That's the (squared) area of a parallelogram with sides $x$ and $y$

I think that is also $\| x \wedge y \|$

Which can be defined as a determinant

@Slereah how is that dumb

makes sense to me

Well then ur welcome

people.math.osu.edu/derdzinski.1/preprints/lnm81.pdf

good lord this notation

good old typewritten papers

I wonder what typewriter has greek characters on it

Do they put the paper upside down to write $\nabla$

@Slereah probably a Greek one???

Oh wait it has italics

Probably not a typewriter

> Raman spectroscopy has been applied to the study of basic Cu sulfates including antlerite, brochiantite, posnjakite, langite, and wroewolfeite and selected complex Cu sulfate minerals.

Those names are made up

Well there are a lot of minerals

I guess after a while you stop trying to make clever names

It's like mesons

Mesons used to have catchy names

Now they have codes

"It is conceivable of course that a particle will meet a traversible wormhole leading to Deneb or an appropriate distortion of space shortening its way, but one cannot hope to meet such a convenient wormhole each time one wants to travel"

@ACuriousMind You mean in quantum world no forces like electrical and magnetic and nuclear forces exist. Only fields exist. And there are no forces in these fields?

@AnubhavGoel I mean that it is not clear what, quantum mechanically, it even means to say that there "exists a force".

To say that there are forces is neither right nor wrong, it just doesn't actually mean anything precise

"Neglect the contribution of the coin to the metric of the world."

Krasnikov what are you talking about

@AnubhavGoel I would guess you're still thinking about scattering calculations like the electron-proton question you referred to earlier. Is that correct?

"The effect produced by the traveller on spacetime need not be weak. For example, by a (relatively) small expenditure of energy the spaceship can break the equilibrium in some close binary system on its way, thus provoking the collapse"

What kind of madman is Krasnikov

He wants to destroy whole stellar systems

I like how that type of paper always refers to an "advanced civilization"

@Slereah should I do that Geroch theorem exercise

Depends

I've read the proof of his theorem once

Did not understand it.

@JohnRennie Yes, I am thinking about why does wavelength matter if electron and proton were to strike. I need to look from each direction.

what is semi-continuous

take something continuous

cut it in half

@AnubhavGoel OK. If the energies are well below relativistic energies then it's usual to do the calculation assuming the electrical etc forces are given by the usual classical equations.

ACuriousMind's comments apply to relativitic calculations where we need to use quantum field theory. In that case there aren't extrenal forces because all forces are calculated using exchange of virtual particles.

@JohnRennie Virtual particles?

Aren't those mere analogies, woo?

But even in the simpler non-relativistic case the particles are described by their wavefunctions not as classical point objects. Typically we assume they are momentum eigenstates i.e. plane waves.

Everything is mere analogies, @0celo7

Physics is about model building, not truth

not according to @SirCumference

Go to philosophy SE if u want truth

@0celo7 He said "calculated using exchange of virtual particles", which is true - that's what they are, calculational tools.

So we don't conside classical point particles that have a precise position and trajectory. That is the point of all the previous comments made by our QM experts above.

@JohnRennie sorry! If I won't be able to keep my pace here on chat. It takes me a lot of time to understand every statement presented here.

@AnubhavGoel What do you mean "why does wavelength matter"? Are you talking about the deBroglie wavelength? In what way do you think it "matters" in collision?

Isn't Rayleigh formula dependent on wavelength?

@AnubhavGoel You said that "electron and proton were to strike". What has the Rayleigh formula to do with protons and do you mean the formula describing the Rayleigh scattering of light or the Rayleigh-Jeans law describing blackbody radiation?

@AnubhavGoel in scattering calculations the wavelength is important where you are scattering off extended objects. For example in molecular beam scattering where the extended objects are molecules the wavelength does affect the scattering.

@JohnRennie I suspect we're talking about the wrong thing - for instance, the frequency dependence of Rayleigh scattering arises from classical dipole behaviour, there is nothing quantum about it.

However for non-relaticvistic scattering of electrons and protons both can be treated as point masses. In that case the wavelength isn't as important.

@ACuriousMind Agreed.

I'm guessing that Anubhav is still thinking of scattering calculations in a "billiard-ball" fashion.

There's a great little animation of soliton scattering in Sine Gordon, btw

Look at it

You can decompose the Backlund superposition of two solitons into three parts

Two regular solitons and a little transfer soliton

It's great

You can just see the momentum transfer

Wavelength does matter for photons. Okay?

@AnubhavGoel You'll have to be more precise - matter for what?

@Slereah ::looks at it::

The decomposition isn't unique but I think it's swell

You can just picture the little blue soliton as some kind of virtual particle once quantized

@Slereah You still haven't told us how you consider any of these things as quantum states when quantized

Well I'd need to read that whole bloody soliton book, first

@AnubhavGoel If you pass light through a diffraction grating then the scattering is strongly dependent on the light wavelength, or more precise on the ratio of the light wavelength to the diffraction grating spacing.

From what I've skimmed it seems to be some path integral kind of deal

And if you scatter light of an extended object that is comparable in size to the ight wavelength then you see similar effects for similar reasons.

If a photon will scatter when it pass through a gas or goes undeviated depends on wavelength of photon. Okay? Then , why it doesn't for electrons?

However if you scatter light off e.g. an electron, that is effectively a point particle, then the wavelength matters far less.

@AnubhavGoel it's the ratio of the light wavelength to the size of the object doing the scattering that matters. Since electrons are point particles the light wavelength is always infinitely greater than the electrin size so the ratio is, well, infinity - sort of.

@AnubhavGoel Because that kind of "scattering" on gas molecules is not a quantum effect. It's not "a photon" that gets scattered, it's the classical electromagnetic wave that is light. In this classical regime, the electron doesn't have a wavelength, so the question makes little sense.

So, for electron and proton size ratio is 0. Does, it mean they would always strike or I have to take proton to be point sized too

"The notation $A\star B$ for points $A,B$ will mean that there exists a sequence $\{a_b\}$ : $a_{n(i)} \to A, a_{n(j)} \to B$

What?

@AnubhavGoel The electron and proton aren't point particles. When you get down to distance of the order of a Bohr radius both the electron and proton are behaving as waves. The waves interact with each other, but there is no collision as there would be between classical objects.

The proton isn't even a point if you assume classical things

@Slereah it is in the non-relativistic regime

well, effectively.

I've just realised I said The electron and proton aren't point particles. I meant the electron and proton aren't classical particles.

"Electron isn't a point particle" That's confusing me now. So long , you have been saying electrons are point particle. Now you are rejecting it

@AnubhavGoel: to take a step back: are you wondering how we go from the idea of the two particles heading towards each other to two particles bound in a hydrogen atom?

@AnubhavGoel I've just realised my mistake and corrected my statement.

Oh ! I see

We have to clearly state what situation we are considering here before saying more confusing things, I think: I still don't know if we are talking about electron-proton scattering, about electron-photon scattering or about Rayleigh scattering, and what exactly the question is about either of those

It all begun from here

@AnubhavGoel If we have an electron and a proton then we can calculate their total energy.

@MAFIA36790 Do you wish me to cry?😭 Decide one thing , they are or not?

The total energy is just the kinetic energy due to their motion plus the (negative) potential energy due to the attraction between them.

If the total energy is greater than zero then they will not join to form a hydrogen atom.

That's because their energy is great enough to immediately ionise the atom again.

@AnubhavGoel Well, that's a rubbish question. The electron and the proton don't "stick" because when they get very close we can't treat them as classical particles any more and in quantum mechanics saying that two things "stick" to each other doesn't exactly make sense. The "sticky" version of an electron and a proton is called a neutron or a hydrogen atom, depending on what you mean by "sticking".

So in the situation described in that question the electron and proton would just scatter off each other and head off to infinity.

If the total energy is less than zero then in effect we are starting with an excited state of a hydrogen - something like a Rydberg state.

The accepted answer is also rubbish, because it never really talks about the crucial flaw of the question: Saying that an electron and a proton "stick" doesn't actually mean anything.

@ACuriousMind It was question like why isn't a hydrogen a neutron.

@MAFIA36790 I had found enough duplicate to that question, including one of mine. It's what every physics students ponders over. physics.stackexchange.com/questions/261342/…

@AnubhavGoel Hydrogen is a bound state of a proton and an electron. The neutron can be the result of a proton-electron collision event, but it is not a bound state that you could decompose into proton and electron, it's a bound state of three quarks.

Which of these you get when an electron and a proton interact depends on a multitude of factors. I'm still not sure what exactly the problem here is.

@AnubhavGoel there is a process called electron capture, which is effectely the opposite of beta decay.

Yeah , I studied it in my school , electron capture, on chapter on nuclear physics

But the process e + p -> n + v is endothermic i.e. energetically unfavourable.

It only happens in some nuclei where the rearrangement of the nucleus can provide enough energy to make the interaction happen.

An isolated electron and proton won't interact that way because it would raise their total energy.

Yes, I know , binding energy stuff. But I don't know can this collision, an electron and proton form neutron , if required energy is given externally using kinetic energy of electron. It may form a neutron for a very small interval before it decays.

Yes, if there is sufficient energy, a neutron and a neutrino are a possible outcome of the collision.

And I think it's not a "very small interval", neutrons live for minutes

In principle yes, but the weak force is, well, weak. So the probability of it happening is very low.

For them at rest I had shown some calculations, that can provide this energy if photons are not emitted . physics.stackexchange.com/questions/261342/…

@ACuriousMind What's need of neutrino? It makes calculations difficult.

@AnubhavGoel I'm not sure how you computes the "max energy" there, but: If the distance between them is large enough to make up for the mass difference between proton+electron and neutron, then it is possible that they form a neutron. Otherwise, they will form a hydrogen atom (or just scatter off each other and remain proton and electron, until they interact again)

I used $kq^2/r$ to provide electric potential energy

@AnubhavGoel There is a conserved quantity called "lepton number". The electron has lepton number 1 while the proton and neutron have lepton number 0, so the r.h.s. of the reaction needs something else carrying lepton number 1, but it also needs to be uncharged because of charge conservation. The only such thing is a neutrino.

@Bass What is your email address?

Can we know Total Energy of a particle and it's position at same time