The h Bar

DanielSank

00:09

Hi, everybody.

@ACuriousMind This is even kind of true for infinite dimensional linear transformations.

ACuriousMind

07:59

@DanielSank if your operator is nice enough that its trace exists, it works there too, yes

B. Brekke

09:26

Spontanous symmetry breaking and explicit symmetry breaking are in principle very different. However, are they different in practice?

Nihar Karve

09:52

what does "in practice" mean?

B. Brekke

Hmm, if I model the spontanous symmetry breaking by introducing a small term which break the symmetry explicitly, would all calculated physical quantities remain the same?

Slereah

10:08

You could expand it as a function of the scale of the symmetry breaking, I guess

B. Brekke

@Slereah What does that mean?

Slereah

I would guess that you could probably expand the solution as a Taylor expansion in that parameter?

bolbteppa

10:27

Obviously explicitly breaking a symmetry is very different from ssb for which the symmetry is still there

B. Brekke

@bolbteppa In the sense that all physical properties still exhibit the symmetry?

ACuriousMind

@B.Brekke Explicitly breaking the symmetry by a small term produces pseudo-Goldstone bosons, which are light but not massless

That is, you can tell whether a spontaneously broken symmetry is also explicitly broken by looking at whether or not the corresponding goldston is massless or not

Slereah

10:46

IIRC pions are an example

Since the $SU(2)$ flavor symmetry isn't an actual symmetry

bolbteppa

Quartic interaction

In quantum field theory, a quartic interaction is a type of self-interaction in a scalar field. Other types of quartic interactions may be found under the topic of four-fermion interactions. A classical free scalar field φ {\displaystyle \varphi } satisfies the Klein–Gordon equation. If a scalar field is denoted φ {\displaystyle \varphi } , a quartic interaction is represented by adding a potential term ( λ / 4...

At the end they show the symmetry is still there (obviously discrete case, but still)

ACuriousMind

@Slereah yes, pions are more or less the canonical example

Slereah

Massive EM is probably a nice example, too

Nihar Karve

12:41

@ACuriousMind do you like rum

ACuriousMind

nah, I'm a whisk(e)y guy

my flatmate loves rum, though, why?

Nihar Karve

dang it

"Myself, ACuriousMind" is an anagram of "Disco Elysium/Rum Fan"

2

I thought all anagrams were true

ACuriousMind

lol

I mean, I don't hate rum

it's at least not tequila :P

LSS

12:58

@ACuriousMind

ACuriousMind

@LSS ?

LSS

Do you know that? I am a little confused in how to prove, using this, that ap dagger creates particle with momentum p

Actually, i was reading Peskin qft, and after show this equation he claims that the operator ap dagger creates a particle with momentum p, but i trying to see why. I think i am missing something

Good morning.

ACuriousMind

You have to show that $\mathbf{P}a_\mathbf{q}^\dagger\lvert 0\rangle = \mathbf{q}a_\mathbf{q}^\dagger\lvert 0\rangle$ - it's a straightforward use of the commutation relations of the $a_\mathbf{q}^\dagger$ with the $a_\mathbf{p}$ inside the integral.

LSS

Oh i see, it is in fact easy to prove it. Very good, thank you :)

Avantgarde

14:37

hey guys

long time

no c++

satan 29

15:07

Hi everyone

When we solve the SE in 3 dimensions using sep of variables in ploar coordinates

We conclude that the quantum number l can take on integer values 0 to n-1, and it relates to the orbital angular momentum :

$$L_{orb}=\sqrt{l(l+1)} \hbar$$

however, When we make use of Commutation relations, (I dont understand the details fully), we end up with the fact that l can be half integral as well. l= 0,1/2,1...etc

How is this resolved

I know this has something to do with the quantity spin but what exactly?

Charlie

The values of $l$ are definitely integers in the standard treatment, perhaps you mean the values of $m_s$ can be half integers?

RewCie

the value of $i$ are integers and $m$ are naturals

satan 29

Griffiths QM, 4th edition, page 160.

Charlie

$l$ takes values in $l\in[0,n-1]$ as you say

Let me blow the dust off my copy of Griffiths

ACuriousMind

@satan29 $l$ cannot be half-integer. Angular momentum in general can be, but orbital angular momentum cannot.

RewCie

15:21

$m \in \{|l| \leq m\}$

I forgot actually

bolbteppa

If you look at the derivation using separation of variables, you'll note at one stage one has to assume something like $\hat{l}_z \psi = l_z \psi$ so that $\psi(r,\theta,\phi) = \psi(r,\theta) e^{i l_z \phi}$ and for single-valued-ness $l_z$ must be an integer, denoted $m$, this excludes the possibility of half-integer values

ACuriousMind

see physics.stackexchange.com/q/153369/50583

RewCie

I've a question

does anyone know what is a discrete convolution operator?

often denoted with *

Charlie

$m_l\in[-l,...,+l]$, I don't know what you've written above

satan 29

RewCie

15:23

actually I messed up, I'd have to pick griffiths to recall it again

satan 29

@ACuriousMind @Charlie @bolbteppa @RewCie

ACuriousMind

@satan29 see the question I linked

RewCie

No idea... I forgot that.... sorry...

ACuriousMind

the passage you photographed shows that, in general, operators that have the commutation relations of $L_i$ have eigenvalues of integers or half-integers

RewCie

Maybe, I'll pick those parts of QM later on...

bolbteppa

15:25

The resolution is that the separation of variables solution assumes a representation of the angular momentum algebra, the differential operators $\hat{l}_z = - i \frac{\partial}{\partial \phi}$ unavoidably forces one to work only with integer 'spin' representations

ACuriousMind

but that doesn't mean that you have to have all of these possible eigenvalues in every situation

and for the orbital angular momentum in 3d, it's the case that only the integer ones occur

there's no contradiction

bolbteppa

See this answer in the post linked above

satan 29

hmm so its like the commutation thing just solves an abstract mathematical problem

bolbteppa

The point is, the algebra follows from general principles, wave functions live in a representation of this algebra, choosing differential operators in the Schrodinger equation to act on scalar wave functions is like a leap of logic, you skipped the part where you specify the representation more carefully, i.e. everybody does this until faced with the reality of spin forcing one to re-examine it

RewCie

Being in engineering has it's own perks, you forget all the pure physics you studied...

Tho, those are helpful sometimes...

I remember using Maxwell's equations to cheat in Electrical engineering stuff...

satan 29

15:30

@bolbteppa how exactly does spin play a role here?

Charlie

Why would you appeal to complicated things like representation theory when you could make vague, confusing, physical arguments to avoid it :p

RewCie

haha :-)

ACuriousMind

@satan29 spin is the name for the kind of angular momentum that can be half-integer

satan 29

right

Wait

bolbteppa

If the problem admits spherical symmetry, the wave functions live in a representation of the rotation group, the algebra of that group admits irreducible representations classified by 'spin', if you work with the algebra abstractly you find all allowable values for spin, if you work with the differential operator representation of the angular momentum algebra, you've implicitly restricted the set of allowable values of spin to be integers, as the argument I made above shows

satan 29

15:33

the paragraph I clicked was dealing with angular momentum and not orbital angular momentum

bolbteppa

Sure, but you asked for a resolution between why we get integers in one case, but are allowed to also get half-integers in another case

If you know nothing of representation theory, this will force you to face it at least on a basic level

satan 29

hmm

okay hang on though. The orbital angular momentum L is $$\sqrt{l(l+1)}\hbar$$, and the spin AM S, is $$\sqrt{s(s+1)} \hbar$$.

is the total AM = $\sqrt{L^2 + S^2}$?

Charlie

Angular momentum operator

In quantum mechanics, the angular momentum operator is one of several related operators analogous to classical angular momentum. The angular momentum operator plays a central role in the theory of atomic and molecular physics and other quantum problems involving rotational symmetry. Such an operator is applied to a mathematical representation of the physical state of a system and yields an angular momentum value if the state has a definite value for it. In both classical and quantum mechanical systems, angular momentum (together with linear momentum and energy) is one of the three fundamental...

satan 29

if it is, then the commutation method shouldve given solutions of this kind, no? Since the quantity we are workring is the total angular momentum rxP...

Charlie

"$\vec J=\vec L+\vec S$"

satan 29

15:45

@Charlie hmm wikipedia claims $J^2= j(j+1)\hbar$ which explains everything.

Slereah

Gary Coleman is about $\sqrt{2}\ \mathrm{m}$

satan 29

(j is 0,1/2..etc)

But can i find a derivation that derives J^2 from J=L+S ?

nvm that

Another question

the commutation method uses l=rxP to begin its calculation

RewCie

I just wanted to know, the Convolution Neural Nets use filter banks to apply 2D/1D space transformations and to exaggerate certain features using discrete convolution operator, then, with input and output vectors, $x_i, y_i$, the convolutions, $$ y_i = \sum_j k_{ij} \text o \, x_i $$ might mean $y_i = \sum_j {i (n-j) \times x_i}$ right? assuming $k$ is the total rows of a feature map. So, since, $k_ij$ feature maps are arbitrary, doesn't that mean that we can write it straightforward as

$$y_i = \sum_j k_{ij}\times x_i$$?

satan 29

which is how you define angular momentum classically

RewCie

Here, is used $\text o$ to represent discrete convolution operator

satan 29

15:51

however, from what I know electrons dont revolve around the nucleus, so why isnt this blasphemy?

RewCie

@satan29 Yes, that's correct. It is bound to be same as classical analogue

refer 4th chapter of L&L

Charlie

Quantisation basically just involves putting operator hats on classical quantities

satan 29

@RewCie why?

RewCie

I was too surprised with that when I read that... actually...

@satan29 sorry, I don't remember it

sorry.... It has been ages since QM for me...

Charlie

The classical formula for angular momentum is $L_i=\epsilon_{ijk}x_jp_k$, so the quantum version is the same formula but with the position and momentum functions as operators

satan 29

15:54

actually wait

RewCie

Actually, I again messed with my question typeset... sorry

satan 29

I dont suppose you need stuff to necessarily rotate, in order to define a quantity rxP

Charlie

It's a mathematical expression, you don't technically need anything physical for it, it's just a definition

satan 29

yes, exactly

I just realised this

you can compute the "angular momentum" about some origin for perfect , straight line motion too. Its just rxp

Charlie

Sure, it's just a function of the position and momentum

even if it happens to evaluate to zero

satan 29

15:57

@Charlie hmm yeah

But does angular momentum have significance of some kind apart from being just some conserved quantity?

Charlie

What kind of significance are you looking for?

satan 29

because in classical mechanics for instance

if your particle is not rotating. say its following s atraight path.

then you can compute the "angular momentum" but it wont be of significannce, will it?

Charlie

And you're asking if there's some similar interpretation in QM?

satan 29

well yes

in classical mechanics, whenever we used L, we were trying to analyse rotations of some sort.

Charlie

I don't think there is any such similar interpretation. Since we don't think of electrons as orbiting the nucleus you can't interpret the AM eigenvalues as representing something rotating around a point like you can in classical mechanics

satan 29

16:01

However, in QM it is clear that "angular momentum" has no relevance to rotation. Its just being treated as some mathematical quantity that is quantized.

ACuriousMind

@satan29 what do you mean, "it won't be of significance"?

it's still a conserved quantity, it's usually just not a very interesting one for linear motion because linear momentum is a much simpler conserved quantity there

satan 29

" it's usually just not a very interesting one for linear motion" Thats what I wanted to say

ACuriousMind

likewise, it's also not a very interesting thing to look at for a free quantum particle

satan 29

but it seems as if AM is some quantity of utmist importance in QM..

ACuriousMind

why?

it's just an observable

satan 29

16:05

as in, most authors introduce it like that

ACuriousMind

well, it leads you to spin, which is a very non-classical thing

people tend to focus on the bits of QM they find non-intuitive

satan 29

hmm

ACuriousMind

it's not that it is any more important in QM than in CM to know how your stuff transforms under rotation, it's just a bit weirder in QM :P

RewCie

stern garlach experiment

satan 29

ok. So the AM is just some quantised quantity , and an observable. Looks like a good summary?

bolbteppa

16:08

The Angular momentum differential operator is derived as an operator by considering how a wave function changes under infinitesimal rotations

satan 29

oh wow

so there is a link to rotations, but not to electron rotations about the nucleus, i guess

ACuriousMind

also not a quantum thing - angular momentum is the generator of rotations, both classically and quantumly

Nihar Karve

@RewCie yes, there's really no need to "run the indices backwards". Some people call this "cross-correlation" but there's no real difference practically

satan 29

@ACuriousMind what exactly do you mean by "generator of rotations"

ACuriousMind

this is why I say people need to learn Hamiltonian mechanics before quantum mechanics so they can separate the Hamiltonian stuff from the truly quantum stuff :P

satan 29

16:10

XD

bolbteppa

If $d \mathbf{r} = d \vec{\varphi} \times \mathbf{r}$ is an infinitesimal displacement due to a rotation then $\psi(\mathbf{r} + d \mathbf{r}) = \psi(r) + d \mathbf{r} \cdot \nabla \psi = \psi(\mathbf{r}) + ...$ leads to $\mathbf{L} = - i \hbar \mathbf{r} \times \nabla$

RewCie

Yes, that's what I was about to ask... for arbitary filter maps, running indices backwards makes same sense as forwards.

@NiharKarve Thanks, I just found that the way that mathematical expression is written because of historical significance.... We can drop that convolution operator.

ACuriousMind

@satan29 an infinitesimal rotation is $\{L_i,-\}$ classically (acting on the phase space) and $[L_i,-]$ quantumly (Acting on the operators on the Hilbert space

where $L_i = \epsilon_{ijk}x_jp_k$ in both cases, and factors of $\mathrm{i}$ might be missing because I'm lazy

the reason angular momentum generates rotations in QM is because it does in CM and "quantization" in practice is often really just replacing the Poisson bracket by the commutator

satan 29

I guess Ill need to learn HM to understand what a poisson bracket does..

ACuriousMind

it's just like a commutator ;)

RewCie

16:19

What you all think about an AI that codes for you?

I mean an AI that can code and make algorithms like normal human being?

satan 29

okay, another question

RewCie

just ask

satan 29

what about the treatment of magnetic moment , in QM

ACuriousMind

you'll have to be a bit more precise to make that into a question ;P

satan 29

magnetic moment of a current loop is $IA$

Nihar Karve

16:22

@RewCie ah, if you're talking along the lines of neural architecture search, evolutionary programming etc. then I find that really interesting

you should take a look at Google's AutoML-Zero

RewCie

@NiharKarve Actually I was about to read DeepCoder, just check the paper

satan 29

you can define "classically" the magnetic moment of an electron, rotating about a nucleus

RewCie

Just yesterday, I posted a paper on Twitter on Pix2Code - An AI Model that just writes HTML/CSS Codes using just a given UI Image...

satan 29

since it acts as a current loop with area pir^2 and current e/T= ew/2pi

but again, we know this doesnt happen in QM. How do you even define "magnetic moment of an electron"

if it doesnt revolve around the nucleus, i.e doesnt act as a current loop?

ACuriousMind

@satan29 that's actually a pretty subtle question

satan 29

16:32

hmm

ACuriousMind

a working definition might be that the "magnetic moment" is just the gradient of the Hamiltonian (i.e. energy) w.r.t. magnetic field

this works for QM, but it turns out you have to put in strange g-factors to relate this moment to angular momentum

satan 29

I see

@ACuriousMind yeah. classicaly this was 1. but for spin this comes out to be 2

ACuriousMind

it's not exactly 2 :)

satan 29

btw, wikipedia defines it as an "intrinsic property" of an electron like charge. Seems comforting ig?

ACuriousMind

this is one of the best tests of quantum electrodynamics - the spin g-factor for the electron is not 2, but slightly more than 2, and QED can compute this

satan 29

16:34

@ACuriousMind yeah i read there is some correction in QED... 2+ a/pi +...

ACuriousMind

@satan29 more carefully Id say that spin and charge are the intrinsics - the magnetic moment is derived from that

satan 29

well yes, but then that does mean that mm is also an intrinsic, right?

ACuriousMind

depends on your notion of "intrinsic" :P

if you give me charge, mass and spin and the QED Lagrangian, I can derive the magnetic moment from that

satan 29

hmm

ACuriousMind

that still doesn't mean anything is rotating in the classical sense here, but that's not specific to the magnetic moment, that's just the weirdness of spin

Vincent Thacker

16:48

Hey, anyone knows a good resource for classical field theory?

Charlie

Depends how much you want to learn, enough to start QFT or are you looking for a complete book on classical field theory?

Vincent Thacker

@Charlie I would say the latter. I'm looking primarily at Lagrangian and/or Hamiltonian formalism

RewCie

Just heard that Windows is coming up with a new OS?

Vincent Thacker

@RewCie Yeah

Charlie

There's a thread here: physics.stackexchange.com/questions/45941/…

Vincent Thacker

16:51

They promised that Windows 10 will be the last version

6 years later

@Charlie Thanks, I'll look into it

bolbteppa

@VincentThacker e.g. these notes with the book in that link

Vincent Thacker

@bolbteppa You mean Landau and Lifshitz?

bolbteppa

Yeah

Vincent Thacker

@bolbteppa Alright thanks

DanielSank

17:37

@ACuriousMind I only socialize with nice operators.

2

ACuriousMind

life's too short for pathological operators, eh?

napstablook

18:21

18

Q: What is the force between two charged objects when the space between them is partially filled by a dielectric medium?

I am given two charged particles of same charge at a distance of $r$. They initially apply force $F$. Now an infinite dielectric (of dielectric constant $4$) of width $\frac{r}{2}$ is introduced between the particles. What will be the new force? I find this problem confusing because I have onl...

can someone please check if my answer is correct? or am I overcomplicating things and John's answer is correct?

Also is it possible to find a solution to this problem without any assumptions?(assuming I am correct.)

More Anonymous

19:24

Random math question: let's say I give you a line element: $ds^2 = dx^2 + dy^2$ how would you go from this to the inner product of the displacement vector: $\vec s \cdot \vec s = ? $ ?

ACuriousMind

what's a "displacement vector" in this context?

the $\mathrm{d}s^2$ is just notation - it is not the square of the derivative of a function $s$

More Anonymous

@ACuriousMind I know ...

ACuriousMind

so what is your $\vec s$ there?

More Anonymous

@ACuriousMind So lets say this is the line element of a particle. The the displacement vector is the displacement between you and the particle. For example $\vec s = - c e_t + x e_x + y e_y + z e_z $

ACuriousMind

I don't know what "line element of a particle" means

$\mathrm{d}s^2$ is the physics notation for the metric tensor on a manifold

More Anonymous

19:30

So lets say the displacement vector of a particle is given by :

ACuriousMind

you've written down a 2d metric tensor, but you haven't told us what manifold we're on yet

More Anonymous

One moment

ACuriousMind

if you're talking about a displacement vector, you can only talk about that in $\mathbb{R}^n$ - there is no notion of vectors "between" two different points on an arbitrary manifold, and so line elements have nothing to do with displacement vectors in the ordinary sense

More Anonymous

Yes we specify the manifold

So lets say I have a position vector

ACuriousMind

there's no position vector if we're doing diff.geo on a manifold

there is only position as a point on the manifold

More Anonymous

19:36

Cant I have a origin?

ACuriousMind

no

manifolds don't have origins, that is something very specific to $\mathbb{R}^n$/vector spaces

More Anonymous

I'm super confused

ACuriousMind

the parts of physics where we talk about position vectors are not the parts of physics where we have things like line elements

More Anonymous

So vectors only exist in flat manifolds?

ACuriousMind

either you're in a non-relativistic setting where space is Euclidean and you can treat 3d space as just $\mathbb{R}^3$ with the usual Euclidean inner product or you're in a relativistic/cosmological setting where you have to care about the shape of spacetime and variations in the metric and so you need to think about spacetime as a manifold with a metric tensor

@MoreAnonymous no, in the context of manifolds, vectors exist only as tangent vectors

at each point of a manifold, you can talk about the tangent vectors at that point

physically, such tangent vectors are directions/velocities, not distances

More Anonymous

19:41

The book was saying this^

ACuriousMind

sure, Minkowski space is $\mathbb{R}^4$

often written $\mathbb{R}^{3,1}$ to emphasize the choice of Lorentzian inner product

you can talk about position vectors in Minkowski space, but that won't generalize to any other spacetime

More Anonymous

Wait I feel Im entering a mathematical technicality here but $d \vec r \cdot d \vec r r $ can I go to $ \vec r \cdot \vec r $?

@ACuriousMind The author seems to do it successfully

?

Book: Differential Forms and Geometry of General Relativity

ACuriousMind

so you need to think about what $\vec r$ is there

it's a vector field that at each point $\vec r$ of Minkowski space assigns the tangent vector $\vec r$ to that point - since Minkowski space is flat, this is actually well-defined

More Anonymous

@ACuriousMind I thought it was the position of the particle all this time

ACuriousMind

it's not, it's just a vector field

More Anonymous

19:47

Okay ... So how do you go from $d \vec r \cdot d \vec r r $ to $ \vec r \cdot \vec r $ usually?

ACuriousMind

you don't, this is just confusing physics notation

More Anonymous

@ACuriousMind Are you saying $\vec r \cdot \vec r$ is not a mathematical object?

ACuriousMind

$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2$ means that at each point of the manifold, the inner product of the tangent space at that point is $\mathrm{diag}(1,1)$ in the coordinate basis $\partial_x, \partial_y$

so you have that the inner product at any point is $\vec v \cdot \vec w = v_x w_x + v_y w_y$

More Anonymous

Your basically talking about this

right?

ACuriousMind

more generally, you have $\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu \mathrm{d}x^\nu$ and then the inner product is $v\cdot w = g^{\mu\nu}v_\mu w_\nu$

@MoreAnonymous I wasn't talking about Killing vectors, no

More Anonymous

19:51

@ACuriousMind yea I realised the killing part wasn't relevant

But whatever you do with your differential geometry notation I can do with my vector notation imo

I was under the impression they both were kind of saying the same thing

ACuriousMind

they are, but the non-vector notation generalizes directly to manifolds that are not $\mathbb{R}^n$

and doing diff. geo on $\mathbb{R}^n$ is kinda pointless, there you can indeed just think about everything as vectors

More Anonymous

The book I mentioned uses this kinda notation throughout and seems to be successful

ACuriousMind

as you can tell, I would probably not be a large fan of this book :P

that physicists can muddle through with unclear notation and abuse of terminology doesn't mean they should :P

More Anonymous

For example ^

I feel in love with this book!!!

ACuriousMind

that's atrocious

More Anonymous

19:58

@ACuriousMind I thought it was nice and added so much clarity :P

ACuriousMind

like, it probably gets the right result in the end, but that's extremely non-standard notation to the point that I can't tell whether it's a misunderstanding of the underlying math or just a very unusual way to write it

few who haven't read this will understand what you mean by $\mathrm{d}\vec r$ without a lot of explanation

More Anonymous

@ACuriousMind I thought it was a very natural extension of vectors

ACuriousMind

as I haven't read the book I can't say whether that's true, but that's definitely not how the mainstream does differential geometry :P

More Anonymous

@ACuriousMind Okay is it possible to translate this to $\vec r$ to mainstream differential geometry?

(assume it is someone who hasnt read the book)

ACuriousMind

it just looks like a coordinate-dependent mess

the only way to make this into a proper object would be to think of this just as the square root of the metric tensor

More Anonymous

20:03

@ACuriousMind You can do that O_O

ACuriousMind

i.e. you literally take the tensor field $g$ and do $\sqrt{g}$ by taking the square root at every point

that's not quite it, you'd get imaginary stuff, you also need to get rid of the minus in the time component first

More Anonymous

@ACuriousMind Soon your gonna introduce matrices as well aren't you

?

:P

ACuriousMind

Hm? $g$ is a matrix, and so is your $\mathrm{d}\vec r$

More Anonymous

@ACuriousMind I was thinking along the lines of Pauli matrices etc

ACuriousMind

$\mathrm{d}t$ is a covector and $\hat{t}$ would more conventionally be written $\partial_t$ and there is an implicit tensor product between them here, i.e. $\mathrm{d}\vec r$ is a very weird way to write down a (1,1)-tensor

More Anonymous

20:07

So can I integrate the covector ?

ACuriousMind

really, I don't think this formulation works so well - what happens to this $\mathrm{d}\vec r$ in coordinates where the metric $\mathrm{d}s^2$ is not diagonal?

this is a computational trick of some sort, but I don't think it really describes a meaningful coordinate-free object

More Anonymous

@ACuriousMind Lemme see what he does

Okay so far I cant find any non diagonal metric

welp!

:P

But I'm still gonna finish that book since Im enjoying it despite its uniqueness :P

ACuriousMind

yeah, that's what I expected - this thing only makes sense in a particular coordinate system, it's not a genuine geometric object

bolbteppa

20:39

Think of $s$ as standing for 'sector' in the old geometry formula $s = r \theta$, i.e. $ds = r d \theta$. The arc length formula $s = \int ds = \int \sqrt{d \mathbf{r} \cdot d \mathbf{r}}$ is just a generalization beyond circles

ACuriousMind

@bolbteppa crucially, in that context the $\mathrm{d}\mathbf{r}$ is the tangent vector along the arc you're integrating over

that's well-defined and not some free-floating thing as it is used in the snippets above

bolbteppa

A space where light satisfies $c^2 dt^2 = d \mathbf{r}^2$ is a space of points $(t,x,y,z)$ whose distance function is now $ds^2 = - c^2 dt^2 + d \mathbf{r}^2$ so that $ds^2 = 0$ for two points collected by a light-beam, we can write this as $ds^2 = dx \cdot dx = - c^2 dt^2 + (dx^2 + dy^2 + dz^2)$ if the first basis element satisfies $\hat{e}_t \cdot \hat{e}_t = - 1$, it's that simple

Physics Meta

21:34

0

Q: Quick to comment, but never a reply

Why do people comment, with a derogatory statement, almost immediately then fail to reply or interact with me, the author, thus degrading the question entirely? Why do that to someone else. I have a strong feeling that the same person also down voted. That makes me so angry. Virtual Photon Laser

discussion

Justintimeforfun

@bolbteppa cCould you help me put of the last comment Virtual Photon Laser? Weird how your comment timing is somewhat related to my question.

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