The h Bar

Silly Goose

01:33

I am curious who started the trend of referring to things as “being invariant” when it is not the case

qwerty

01:46

@SillyGoose what are you referring to?

More Anonymous

02:20

Does anyone have suggestions for books to learn topology with applications in phsyics?

SnoopyKid

02:51

@naturallyInconsistent ok so the long bone with compressive force to break it at 2 tonnes will support a sudden load at 1 tonne?

naturallyInconsistent

03:02

@SnoopyKid your question is meaningless!

PM 2Ring

@SillyGoose You may enjoy reading this answer about unitary evolution & "quantum leaps".

112

A: Do electrons really perform instantaneous quantum leaps?

Do electrons change orbitals as per QM instantaneously? In every reasonable interpretation of this question, the answer is no. But there are historical and sociological reasons why a lot of people say the answer is yes. Consider an electron in a hydrogen atom which falls from the $2p$ state...

naturallyInconsistent

@Relativisticcucumber its not right. its rite

@Relativisticcucumber you have bands even in completely free quantum electron in a lattice. A good treatment of Bloch's theorem ought to also carry the band index.

but the easiest way to think of the band index is that it carries the information of the principle quantum number as in H atom.

123

05:29

Hello Everyone...

123

05:42

What if the force of gravity proportional to acceleration (not proportional to mass of an object). Do we have different objects different weight?

Ryder Rude

06:21

hi

123

Hi @RyderRude

Ryder Rude

06:35

@123 hi

@123 i don't understand the question...

123

NP i have figured out the problem. : )

Ryder Rude

ok :)

qwerty

07:12

@123 you really should take on the advice that was given to you yesterday. when you encounter something you don't understand, challenge yourself to solve it for yourself. wait at least 1 week before asking someone else to help. this will definitely help you.

123

Hi @qwerty

yes you are right. I always did everything (reading different books, threads, google etc...) before asking or confirming here.

qwerty

no, I didn't say read or google. I said to think. you figured it out within 1hr of asking, by yourself!

2

123

: ) sure, i will...

Ryder Rude

anyone wanna friendly debate on any philosophical topic?

Ryder Rude

07:58

"tensor product in enveloping lie algebra becomes operator composition in vector space representation"

does category theory say anything deep about this result

Slereah

08:18

@SillyGoose Parmenides

123

Newton says gravity is a real force, during free fall objects float in local frame because of fictitious force opposite to gravity act on object.

If there are two equal and opposite forces it produce stress within the object.

If we take example of water droplet free fall from tap. just before fall a water droplet fall from tap it has a shape like egg. If newton was right why during free fall water droplet take a shape of spherical? It should maintain the same shape like egg because of equal and opposite force, and shows stress within the water droplet during free fall.

During free fall water droplet has a shape of spherical as in zero force (isolated) , clearly a indication of gravity is not a force as Einstein says.

Why we still use newton's idea of gravity is a real force?

PM 2Ring

08:41

@123 The equal & opposite forces aren't on the one object. When you drop a rock, the rock and the Earth attract each other with equal & opposite forces.

But the Earth has a huge mass so its acceleration is very small.

123

@PM2Ring Yes but rock has strong interaction (strong electromagnetic force between molecule), so equal and opposite force within the rock do not produce effective stress to change its shape. That's why i took an example of water droplet.

PM 2Ring

When a water drop falls, it becomes (almost) spherical because all the water molecules feel the same acceleration. It's not a perfect sphere, though, because of air friction. And vibrations.

4

Q: Why does the falling of water droplets look the same as an expanded balloon released?

Today I was observing water droplets and I noticed that the drop falling out almost looks like the same if I blow into a balloon with my mouth and let it go. I mostly understand the mechanism of why the drop let stays and falls: due to surface tension initially a hemispherical type surface is fo...

123

Pls ignore air friction and vibration, then it be perfect spherical in shape.

PM 2Ring

@123 Yes, then it would be a perfect sphere.

123

And my question specifically about newtons idea of gravity is real force, which posses problem then water droplet should maintain a shape like egg as it has shape just before drop from the tap.

PM 2Ring

08:47

@123 When something is falling, there's virtually no stress caused by gravity, because all parts of the body are experiencing equal acceleration.

123

If gravity is not a real force. Then no force act on the water droplet there is no stress within the droplet. It should be spherical.

PM 2Ring

But if the falling body is huge, then there are stresses, because the acceleration is proportional to $1/r^2$

123

@PM2Ring Pls think the moment when water just at the tip of tap, it that time droplet has a shape like cone.

@PM2Ring Pls consider small object and near the surface of the Earth. Where g is constant

qwerty

you need to consider surface tension. read the answer PM 2ring linked

123

qwerty i have read about surface tension. I want to conclude is gravity real force or not.

PM 2Ring

08:52

@123 This "gravity is not a real force" stuff is confusing you. In Newtonian mechanics, gravity is a real force.

123

@PM2Ring Yes this is what i am asking. If gravity is not a real force. Why we do not abandoned idea of newton at which gravity is a real force?

PM 2Ring

@123 Because Newton's mechanics is a very good approximation. And the calculations are easier than General Relativity.

123

You can take any example of lighter object where electromagnetic interaction between particles is weak. During free fall it shows the behavior as no stress due to gravity and opposing fictitous force as per newton.

@PM2Ring Oookay... Thanks... : )

PM 2Ring

Even when we do need the high precision of GR, we normally start with a Newton approximation, and then adjust it for HR effects.

123

Oookay... : )

PM 2Ring

08:57

We only need pure GR calculations for stuff like orbits around neutron stars & black holes.

123

I see...

Thanks for the contribution and answer.

PM 2Ring

In Newtonian mechanics, gravity is a bit strange because it operates without direct contact. And because it gives equal acceleration to everything. Other forces don't do that.

123

@PM2Ring Yes i know this problem... you are right

PM 2Ring

But gravity gives equal acceleration because the strength of the gravitational force is proportional to the mass. As was discussed here a few days ago...

123

@PM2Ring That's why i have posted the todays 1st question. Because we feel force by our senses not mass. If gravity act like other forces (e.g. applied force which is constant in magnitude). Then we feel same degree of heaviness of all types of objects. Am i correct?

PM 2Ring

09:04

What is force, really? You can observe acceleration, and velocity & displacement. But how do you observe force? Answer: You can't observe force! But you can calculate it. And when you calculate it, you find that it behaves the way Newton says.

123

@PM2Ring Yes i know this problem also. We don't actually know what the force is. But we can imagine force as a muscular effort.

Force is just the way of interacting objects each other.

PM 2Ring

When Newton was developing his theories, he did lots of mathematics & geometry. But he also did lots of experiments. He made pendulums out of different materials to see how that affected their period. He collided stuff to see how it bounced. Etc.

Ryder Rude

@123 we had discussed this stuff before about gravity being a real force in Newton's mech and not a real force in Einstein

this is exactly what people tell u when they tell u to ask progressing questions

2

123

@RyderRude : ) this was exactly helped me a lot in thinking further.

@RyderRude No i asked different question today. I argued newton's idea of gravity is a real force is wrong. and i gave an example.

Ryder Rude

oh

then it is a progressing discussing. Sorry. U r discussing why we do not abandon wrong theories

123

09:10

@RyderRude Yes exactly... : ) NP my friend

Ryder Rude

like @PM2Ring said, theories are right in some domain of application and wrong in others. so we can still continue to use Newton's idea of gravity in the domain where it is right

123

@RyderRude After discussing newton's law of motion with you. I have learned a lot of lot of things. And got new ideas. I will make youtube video on it in my language to share this information with others. Also i will share it with you, because it is written in english. So you can understand that.

But there is still few questions remain, first i want to clear these in my head.

Ryder Rude

sure.. we will try to help you

123

🫡

Discussion about newton's law of motion with you. I can clearly see the difference in weight and force. and mass as a intrinsic property and its different features and factors in different situations.

Ryder Rude

@PM2Ring i also think Newton had imagined force as a primitive quantity, as in, F=ma was not defining a force but describing the effect of a force according to Newton

maybe Newton thought that forces existed metaphysically

123

09:20

@RyderRude I have read in one book. Newton also have an idea of force as an muscular effort. also as well as primitive term

Ryder Rude

maybe he thought that F=Gm1m2/r2 was describing the force that exists metaphysically and F=ma was describing the effect of the force

@123 oh

@123 Hooke also thought that a force was something a spring can measure

but I feel these ideas r somewhat outdated. Forces arent used all that much outside of Newtonian mech

123

Also we can define force as compression/extension of spring. stretching of rubber band. It will give us idea about force beside primitive term.

@RyderRude Exactly... I gave example above

Ryder Rude

yes. but i prefer defining force the way that that book does it

the one u gave screenshots of

Simon, I think

123

Yes.. you are right. But to teach students at first hand. We need to give them some initial idea about new term. Then we extend these idea to primitive notion.

Ryder Rude

yes. Some books will write things like "a force is a push or pull"

123

09:25

Exactly... So new learner can have some idea. momentum as a quantity of motion.

Because we are very much rely on our senses. Similarly our body also sensitive to temperature .

Ryder Rude

10:55

Anyone see Looper?

it is a good time travel movie, but with some paradoxes

More Anonymous

11:23

@RyderRude I have

Ryder Rude

@MoreAnonymous did u think it is consistent

i think maybe past re-writes future in this movie

like, changes in past re-write future

also, can this kind of time travel be allowed in physics

GR def doesn't allow it. and QM also doesn't allow it as far as we understand QM

think_meaning_builds

11:48

Thanks to you @RyderRude l rewatched 12 Angry Men (1997) and thoroughly enjoyed it :-)

Ryder Rude

@think_meaning_builds i have only seen 50s one

@think_meaning_builds glad that u enjoyed it :)

More Anonymous

12:09

@RyderRude nope

Ryder Rude

12:52

Beethoven - Für Elise | Piano & Orchestra

►

@MoreAnonymous same

@Mr.Feynman hi. how is Hilbert space symmetrisation obtained in string theory? is there a spin statistics theorem

Silly Goose

13:22

@qwerty an example is qft texts calling the “lagrangian invariant with respect to lorentz transformations”

Do you all have a favorite academic resource which you lije specifically because it is written well?

qwerty

@SillyGoose I hadn't come across that yet, what's the correct thing to say?

Relativisticcucumber

13:43

@naturallyInconsistent LOL oh no thanks for letting me know XD

naturallyInconsistent

mewth

@SillyGoose what is wrong with that? The $d^4x$ is Lorentz invariant, and so the Lagrangian density is also usually chosen to be Lorentz invariant.

Ryder Rude

@SillyGoose i like Shankar's QM because it is written well

imbAF

14:11

Can someone explain to me how does one begin to solve the Dirac equation for a free particle at rest, so that the solutions are spinors?

Like for example a solution would be $\psi_1=e^{-i \frac{mc^2}{\hbar}t}(1,0,0,0)^T$

But how do we get the element (1,0,0,0)?

ACuriousMind

@imbAF What do you mean "how do we get it?" - you've just written it down, what is there to get?

imbAF

I have a hard time wording it out

Let's say you have the equation

So, you propose a solution to it, to $\psi$

For example, in the SE you propose as a solution something like: $\psi=Ce^{\alpha t}$ and then you find the value for \alpha

Now, for the first solution to the DE of the case considered I would propose

that

$\psi$=Ce^{\alpha t}(1,0,0,0)$

Why not $\psi$=Ce^{\alpha t}(1,0,1,0)$

ACuriousMind

@naturallyInconsistent I suspect this is a nitpick about scalar functions not actually being "invariant" under the transformation since you don't have $L(x) = L(\Lambda x)$; it's not that the whole function is invariant, but that it takes value in a space whose elements are invariant (in contrast to a vector field, etc)

imbAF

Or anything else for that matter

ACuriousMind

@imbAF why would you choose a basis that makes your life harder than necessary?

imbAF

14:18

Basis?

ACuriousMind

you have four components, so you need to propose four independent solutions. The easiest way to do so is to do this in terms of (1,0,0,0), (0,1,0,0), etc.

@imbAF ...the vectors like (1,0,1,0) are relative to a basis, no?

you can't write down numbers as components of vectors without having chosen a basis

imbAF

One second, by saying basis, you are implying that the solution is a vector, multiplyied with something else

but the spinor is not a vector right

which is the solution to the DE

ACuriousMind

of course a spinor is a vector (in the mathematical sense)

it's an element of $\mathbb{C}^4$

imbAF

As a structure, yes

ACuriousMind

$\mathbb{C}^4$ is a vector space

and if you don't think that the (1,0,0,0) you wrote down there is a vector I have no idea what you think it is

imbAF

14:20

Ofc I think it's a vector

but in the lecture it was said

that the spinor is not a vector

And if a spinor contains a vector expression, but it is not

ACuriousMind

it's not a vector in the sense that it would transform like a vector under Lorentz transformations

it is a vector in the proper mathematical sense, all linear algebra applies here

Silly Goose

@naturallyInconsistent ACM's response is indeed what I had in mind

imbAF

so it's not a four vector?

ACuriousMind

ohysicists tend to use the word "vector" rather confusingly and tie it to the representation of the Lorentz group something transforms under, which is inconsistent with the pure mathematical definition of what vectors and vector spaces are

it's confusing, but we have to get over it

imbAF

Ok

And once again, when you try to solve the D.E and you propose an arbitrary form for the solution

what is your argument ?

For saying that $\psi=Ce^{\alpha t}(1,0,0,0)^T$

ACuriousMind

14:23

you need to find four independent solutions

the most convenient way to attempt this is to just try to find solutions that are non-zero in only one component

that's what the (1,0,0,0) there is - it says we look at an ansatz that is zero in three components

imbAF

which are the components

space and time"?

Or not

ACuriousMind

...(1,0,0,0) has four components, it's zero in three of them. I don't understand the question

imbAF

Nothing

You already said that the solutions are of it's an element of $\mathbb{C}^4$

So, it's the complex four vector space

Silly Goose

@imbAF the four components of the vector $\{1,0,0,0\}$ are not associated to spacetime

imbAF

But what I meant was that for example (1,0,0,0) represents a spin up particle in z direction

Or so I believe

I was reading my notes about the D.E these 2 days and I have a couple of questions. The first was already answered, regarding the solutions for the scenario described

I have another question. In the notes for the spinor, the following symbolic was used $\psi(x^{\mu})$. Is that the same with $\psi(\vec r,t)$ ?

ACuriousMind

14:31

@imbAF Yes. You should have already seen that notation for other relativistic fields.

imbAF

I haven't

I have seen the four vector notation

But not as a variable from which a function depends on, like $\psi(x^{\mu})$

Silly Goose

@naturallyInconsistent more generally to ACM's comment, what is important (seemingly) is that the action is invariant. this does not imply that the lagrangian density itself is lorentz invariant.

imbAF

ACM. For a charged particle in an electromagnetic field, the relation between the linear,kinetic and field momentum is: $p_{kin}=p_{lin}-p_{field}=\vec p -e\vec A$

Do you have an intuitive explanation of the difference between the linear and kinetic momentum ?

I mean what is field momentum anyway ?

naturallyInconsistent

@imbAF how much classical electrodynamics have you done?

imbAF

A lot I would say

I might be rusty, since it has been a while

naturallyInconsistent

14:38

@imbAF then how can you not be aware of field momentum?

imbAF

But I can attempt to try and understand what you will say

Idk, maybe I have forgotten about it

naturallyInconsistent

@SillyGoose "Transforms as a Lorentz scalar, i.e. invariant" is kinda what people usually mean as invariant. I know what you and ACM are saying, but it's soooo pedantic

ACuriousMind

@imbAF I don't know what you mean by "linear", "kinetic" or "field" momentum; the proper terms to use in this context would be kinematic and canonical momentum. There are many questions on these topics, see e.g. physics.stackexchange.com/q/114908/50583, physics.stackexchange.com/a/301351/50583

@naturallyInconsistent I did say this was a nitpick :P

naturallyInconsistent

@ACuriousMind yes, which is why it is not a reply to you lol

Silly Goose

well to me it is like saying my white cup is invariant under transformations which turn my cup blue

invariant literally means left unchanged

naturallyInconsistent

14:41

I mean, a function that is invariant in the way that the silly fowl wants, is pretty useless

and useless things are omitted from usual conversation

Silly Goose

i mean that precise notion of invariant forms the fundamental reason to write down the actions we care about, no?

naturallyInconsistent

But if an action or Lagrangian density is invariant in precisely that way that you argue about, then it is absolutely useless and thus even a listener who is new to the scene can deduce (with some effort, of course) that that would be a misunderstanding

you, of course, can go and invent terminology like domain-invariant and range-invariant if that is what you want

imbAF

Ok another question has to do with the proof of covariance of the D.E. What we need to do is show that $S(\Lambda)$ exists, where: $\psi'(x'^{\mu})=S(\Lambda)\psi(x^{mu})$. In the lecture, it is said:

$\psi'(x'^{\mu})=S(\Lambda)\psi(x^\mu)=S(\Lambda)\psi((\Lambda^{-1})^\mu_{\ \ \nu}x'^\nu)$

$\psi(x^{\mu})=S^{-1}\psi'(x'^\mu)=S^{-1}(\Lambda)\psi'(\Lambda x)$

No intertial frame is excellent therefore:

$\psi(x^\mu)=S(\Lambda^-1)\psi'(x')$.

As a result $S(\Lambda^{-1})=S^{-1}(\Lambda)$

Where O and O' are two reference frames

But I don't understand the part where because no inertial frame of reference is excellent, the following expression holds true:$\psi(x^\mu)=S(\Lambda^-1)\psi'(x')$.

naturallyInconsistent

I mean, I'd completely agree with you if you are saying that the whole "we need to multiply a spinor field with an element of a spinor representation of the Lorentz group when we Lorentz transform the field" thing is extremely confusing as currently presented in QFT, and so terms like domain- and range-invariant is helpful for that. However, for the Lagrangian density being a Lorentz scalar function, that's not something that deserves solving.

@imbAF No inertial frame of reference is excellent ????? Burn that book?

imbAF

It wasn't a book but notes

I translated it to english

ACuriousMind

14:53

It's an obvious mistranslation of German ausgezeichnet; the correct translation is "preferred"

2

imbAF

Slereah

perhaps it is an excellent frame

imbAF

@ACuriousMind But how this argument leads to the expression: \psi(x^\mu)=S(\Lambda^-1)\psi'(x')$

ACuriousMind

I have no idea what the argument is even supposed to be needed for; $S$ should have been constructed as a representation of the Lorentz group and $S(\Lambda^{-1}) = S(\Lambda)^{-1}$ is part of the defining characteristic of a representation

imbAF

Is it perhaps, because we weren't taught anything about representation theory, that the lecturer was trying to derive/show this characteristics of the representation?

naturallyInconsistent

14:57

@ACuriousMind i know. was hyperbole just now

imbAF

But, which in my opinion, he didn't clearly show that

naturallyInconsistent

@imbAF it is typically treated horribly, to be fair to the lecturer

imbAF

Ok, so what do I say, that $S(\Lambda^{-1}) = S(\Lambda)^{-1}$ is a chracteristic of the representation of the group

ACuriousMind

But from your notes it seems to me the argument is clear - one can just ignore the line about "no preferred frame" - you take $\psi(x') = S(\Lambda^{-1})\psi'(x')$ and plug in the expressions from above and then $S(\Lambda^{-1}) = S(\Lambda)^{-1}$ follows

imbAF

and S is one ?

@ACuriousMind Do that, without arguing why

ACuriousMind

15:00

@imbAF I don't know what you mean

imbAF

For $S^{-1}$ there's a clear reason. It is the inverse of $S(\Lambda)$, used when you jump from O' to O

@ACuriousMind Ok, let me explain

Ryder Rude

@imbAF hi. Have u also seen the approach where we decoupled the Dirac eqn into two eqns

imbAF

@RyderRude you mean, where you derive the pauli equation ?

Ryder Rude

sorry it's two coupled equations

i will give the reference

ACuriousMind

it's not made easier by the fact that there's a bunch of typos w.r.t. to primes here if I'm not mistaken; you'd be much better off learning this from any other source but these notes.

Ryder Rude

15:01

@imbAF no

imbAF

@ACuriousMind I think notation, is a way of showing what happens. So:

From O--->O': $\psi'=S(\Lambda)\psi$.
From O'--->O: $\psi=S^{-1}(\Lambda)\psi'$.

So one, initially decides for a convention

And it's the above one

Now suddenly you say: you take $\psi(x') = S(\Lambda^{-1})\psi'(x')$

Why?

Obviously if you take it, then you get $S(\Lambda^{-1})=S^{-1}(\Lambda)$

But before that, should you argue this: $\psi(x') = S(\Lambda^{-1})\psi'(x')$

@ACuriousMind Unless, you already did here

ACuriousMind

@imbAF you can switch the role of $O$ and $O'$ in this (that's what the notes mean by "no preferred frame"!); this switches $\psi$ and $\psi'$ and replaces $\Lambda$ by $\Lambda^{-1}$

then the first line gives you $\psi = S(\Lambda^{-1})\psi'$, and compared with the second line from the original you have to conclude $S^{-1}(\Lambda) = S(\Lambda^{-1})$

imbAF

Ok I see what you mean but I fail to understand

what, what

what roles we switched for this

And, it feels like a convention trick more than anything

I know that there is no preferred frame. But once you start solving something, you establish some conventions/rules

Maybe I am being unclear with what I am saying, but is because of what you just said

Ryder Rude

it is in Shankar section 20.3 but he does not completely solve the Dirac eqn @imbAF

he only solves for the energies

Silly Goose

@naturallyInconsistent hm i am wondering if i am misunderstanding. The actions we consider in textbook qft satisfy $S \xrightarrow{\Lambda} S$, or no?

Ryder Rude

15:10

we solve for the spinors too in the general solution

imbAF

So what would constitute completely solving the D.E

ACuriousMind

@imbAF Maybe it is clearer if we don't write down both directions simultaneously. From $O$ to $O'$ we have $\psi' = S(\Lambda) \psi$. This means that $\psi = S(\Lambda)^{-1} \psi'$ (simply by linear algebra). But at the same time we can consider the transformation from $O'$ to $O$ with $\Lambda^{-1}$ as the matrix for the transformation, which then gives $\psi = S(\Lambda^{-1})\psi'$. Looking at both equations at once results in $S(\Lambda^{-1}) = S(\Lambda)^{-1}$.

imbAF

@RyderRude thx

Ryder Rude

@imbAF it is a general formula for $\psi (x,t)$, sort of like the general solution to free particle QM

naturallyInconsistent

@SillyGoose you'd have to explain your notation

Ryder Rude

15:11

i will get the reference for this...

i read it in a book

@imbAF it is in Schwarz section 11.2 titled "Solving the Dirac equation"

it uses the same technique in which we decouple Dirac eqn into two eqns

@imbAF Ashok Das Lectures on Quantum Field Theory Chapter 2 is titled "Solving the Dirac eqn"

imbAF

Ok, I will read it and hopefully it has a clean derivation

Ryder Rude

This is much better to follow than Schwartz

imbAF

though it also depends from the case considered

Ryder Rude

@imbAF great

imbAF

@ACuriousMind I understand your argument now. Initially I started with O and jumped to O'. Then you start with O' and jump to O, and you name the transformation as $\Lambda^{-1}$, just so you can reach the desired result. It's a small math twick.

@RyderRude any links?

Ryder Rude

15:29

i will delete the link now cuz of the chat's policy

ACuriousMind

@RyderRude Do you really think the point of the policy is that you post the links anyway, then delete them? You don't get to circumvent the rules just by deleting your message afterwards if you posted it in full knowledge of the rules.

Slereah

Get the paddle out

It's time for a spanking

ACuriousMind

They're suspended for a day. No need for violent fantasies :P

Relativisticcucumber

18:23

@Slereah LOL

@naturallyInconsistent i have finally resolved my band index confusion. i see now that if we plug in a bloch form solution to the SE, then we get a set of equations indexed by k but there are also n solutions for each k equation so yippee

indeed it was in a&m all along

Silly Goose

@naturallyInconsistent well say I have $S[\phi, \partial_\mu \phi] = \int d^4 x \mathcal{L}[\phi, \partial_\mu \phi$. Then, $S \xrightarrow{\Lambda} S'[\phi, \partial_\mu \phi] = S[\phi', \partial_\mu \phi']$ where the RHS is the Lorentz transformed action. Then, $S[\phi, \partial_\mu \phi] = S'[\phi, \partial_\mu \phi]$, i.e. the transformed action is literally equal to the untransformed action for the exact same arguments.

Or I mean actions we care about satisfy the constraint that $S = S'$ for the exact same arguments. (To my understanding)

Or another way to clarify my point is that, to my understanding, the class of actions we are interested in are the actions $S$ that satisfy the property $S[\phi, \partial_\mu \phi] = S[\phi', \partial_\mu \phi']$ where $\phi'$ denotes the lorentz transformed field. and where we have specialized to the case of an action depending on a single field (and its spacetime derivatives) for simplicity.

ACuriousMind

If we're already being pedantic, the action is not a function separately of $\phi$ and $\partial_\mu \phi$, it's just a function of the field $\phi$. It's $S[\phi] = \int L(\phi,\partial_\mu \phi)$.

Silly Goose

oh

ACuriousMind

see also the last paragraph of this answer of mine

Silly Goose

The difference I mean to point out is that Schwartz will call a relationship like (the one for scalar fields) $\phi(x) = \phi'(x')$ invariance but then what should one call the relationship (fundamental to describing the actual actions of interest) $S[\phi] = S'[\phi]$? Also invariance but a different invariance? Yes perhaps it is pedantic, but it is nonsensical imo to use the language in this way

Nairit Sahoo

18:38

Is anybody here familiar with the Ranada solution in electromagnetism? They consider solutions where the field lines coincide with level curves of some scalar fields. Isn't this the case with all fields?

ACuriousMind

also I find the insistence on being accurate and then using the horrible notation with the primes inconsistent

Silly Goose

I was trying to put the two expressions in a shared notation :P

ACuriousMind

like, can you tell what the difference between $S'[\phi]$, $S'[\phi']$ and $S[\phi']$ is supposed to be and which of the three should really figure in the invariance statement?

Silly Goose

well $S'[\phi] := S[\phi']$ for me, but this could be an inappropriate definition

i.e., the lorentz transformed action is defined as the action with the fields themselves lorentz transformed

(or this is not my definition, this is what was said in a course i took)

Well perhaps fundamentally one wants the differential equations that pop out of the action principle to have the property that if $\psi$ is a solution, then Lorentz transforms of $\psi$ are also solutions?

Nairit Sahoo

How do you derive the equation 20?

ACuriousMind

18:42

@SillyGoose Not pedantically, since requiring something to be a symmetry of the action is stronger than requiring it to be a symmetry of the equations of motion

Nairit Sahoo

$\nabla \phi$ points towards the orthogonal direction at a point on the curve. But how exactly does the $\nabla \overline{\phi}$ come?

Silly Goose

@ACuriousMind Hm so which is one after fundamentally?

ACuriousMind

I refuse to use the word "fundamentally" :P

Silly Goose

is your position that both notions are useful?

ACuriousMind

I don't understand the question - using Lorentz-invariant action principles has worked pretty well for us

what exactly is the question?

I was merely noting that if we're being this careful about stating things unambiguously, your statement about wanting the equations of motion to be Lorentz-symmetric does not necessarily imply that the action is Lorentz-symmetric

Silly Goose

18:49

@ACuriousMind right, this was my point in starting with "Well perhaps..." because the following notion is not equivalent to the action invariance notion

So I guess my claim should have been (1) Admissible actions in textbook QFT satisfy $S[\phi] = S'[\phi] =: S[\pi(\Lambda)\circ \phi \circ \Lambda^{-1}]$. (2) This condition is literal invariance of the action under Lorentz transformations.

I tried to write the lorentz transformed argument explicitly but that may be wrong :P

naturallyInconsistent

20:09

@Relativisticcucumber IIRC there is more n than before. Still, countably infinite, which you know from Hilbert hotel that everything is fine

@SillyGoose I think it has to be correct but your notation and pedantry is making it so dense that I don't think I should be expending the effort to figure out if it is so.

Allie

21:08

hi guys

Silly Goose

TIL i can render mathjax on my phone browser yippee

Hallo

Allie

when representing an operator $O$ in a different basis as $\Omega = U^\dagger OU$, my first thought is that "U converts to the basis that O is in, then operate O, and then $U^\dagger$ converts back to the original basis". is that the idea?

ACuriousMind

@Allie That's one way to interpret it, yes

Allie

whats the others!! im curious

ACuriousMind

I mean it depends a lot on the context - operators transform that way also under unitary operators when you don't necessarily interpret the $U$ as a basis change

Allie

21:14

i see

ACuriousMind

of course you can always interpret a unitary operator as a basis change, but whether that's a good way to think about it depends on the situation

Allie

ok at least im understanding this correctly

because the current subsection is "change of basis" lol

thanks!!!

imbAF

Hey people! I have a question about what I previously asked, regarding canonical, kinematic and field momentum. I want to check if my understanding is somehow accurate. I am still trying to understand how the kinematic and canonical momentum differ, as in the way that they are manifested, when observed

So if we consider the same system e.g charged particle of some mass, in two scenarios, 1 when the particle is moving in free space and 2 in an electromagnetic field. Will the kinematic momentum be the same and not change, compared to the canonical one who in the first case is equal to the kinematic, but then it changes to the sum of the kinematic and field momentum?

Mr. Feynman

21:37

@RyderRude H-hello. Why do you ask me? I don't even know string theory D:

Nairit Sahoo

Since I posted my question while a conversation was ongoing (i 'm sorry!) and people might have missed it, can anybody have a look at the equations above?

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