The h Bar - 2024-12-09

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imbAF

00:26

In the context of PT in scattering in QFT, the probability of transition is $|\langle f|S|i\rangle|^2$ where $S$ is the scattering matrix. By using the Dyson series it can be written as $|\langle f|T\exp[-i\int H_I(t)dt]|i\rangle|^2$. We then write it as: $|\langle f|T(\phi_1(x_1))\phi_2(x_2)...\phi_n(x_n)|i\rangle|^2$

I searched in peskin how the last jump is made but I didn't find anything on it

How is the jump made?

naturallyInconsistent

00:57

@Relativisticcucumber why else do you think meow meow keeps mentioning Sturm-Liouville?

@SillyGoose The anecdote in Feynman is about one doing auditory counting and the other doing visual counting, so that auxilliary tasks could interfere, but in different ways.

2 hours later…

Relativisticcucumber

02:57

@HerrFeinmann lol what makes u say this?

ive never been suspended -- is that the bar XD

@naturallyInconsistent u r a wise individual

naturallyInconsistent

03:21

aww yiz

Silly Goose

does poynting's theorem apply for radiation in the dipole approximation?

a priori, since we are approximating $E, B$ and so $S$, I do not see why Poynting's theorem should hold.

naturallyInconsistent

03:54

You can treat an ideal dipole, and it would still have to conserve energy-momentum.

5 hours later…

Tobias Fünke

09:13

morning

naturallyInconsistent

mew mew~

Herr Feinmann

09:47

@Relativisticcucumber self projection

naturallyInconsistent

10:12

$$\left|\psi_\text{Feinmann}\right>\left<\psi_\text{Feinmann}\right|$$ ?

Herr Feinmann

10:55

@naturallyInconsistent Indeed, that is my German projector

But I have many identities

2 hours later…

naturallyInconsistent

13:06

What would be the Italian identity?

1 hour later…

DIRAC1930

14:19

@Slereah Hi can you help clear up the mistake I am making

So when people say a 4-vector is a tensor (in SR), what they really mean is that it transforms like a tensor under Lorentz transformations

However under general coordinate transformations, it does not therefore in the strict sense it is not a tensor

Slereah

Does it not

DIRAC1930

Well if you change a position vector from cartesian to polar coordinates, it doesn't transform linearly like that

$\mathrm{d}x^\mu$ does

but not $x^\mu$

Slereah

Ah there's your problem

DIRAC1930

So is what I've said so far correct?

Slereah

$x^\mu$, the position, is not a vector

DIRAC1930

14:26

But $\mathrm{d} x^\mu$ is by defintion right?

Slereah

A dual vector but sure yeah

There is a correspondence between $x^\mu$, a position, and some vector $v$, hence why we can do some blurring between the two, but that does not extend to nonlinear coordinate transformations

DIRAC1930

Okay so when someone checks if a theory is Lorentz invariant, what they mean is that they check whether the action is invariant under Lorentz transformations therefore a Lorentz transformation will be a symmetry of the minimum of the action therefore a solution of the equations of motion in one frame of reference, will also be a solution of the EOM's in another frame of reference

Slereah

Sure

In fact it is that very property that tells you the proper coordinates to use

On a manifold you have a specific subset of coordinates for which the metric transforms properly under lorentz transform

DIRAC1930

So to do this one checks that $S=\int \mathrm{d}s \mathcal{L}(x)$ is invariant under Lorentz transformations which transform both fields $\phi(x) \rightarrow \phi(\Lambda x)$ and the metric $\eta_{\mu \nu} \rightarrow L^{\gamma}{}_{\mu} L^\delta{}_{\nu} \eta_{\gamma \delta} = \eta_{\mu \nu}$

Slereah

Those are the inertial coordinates

Well any field can transform nicely if you transform both the field and the metric, the good part is that this works if you just transform the field

Since $\Lambda^T \eta \Lambda = \eta$

DIRAC1930

14:33

Okay, but when it comes to conformal transformations, since the metric tensor is a tensor, it transforms as a tensor under conformal transformations, but one has to work out how the fields transform since they are only a tensor under Lorentz transformations

Slereah

So if you leave your universe as it is otherwise but take all the matter in the universe and slightly rotate it, it will be the same physics

DIRAC1930

@Slereah But for 4-vectors they only transform inversely to the metric for Lorentz transformations

If I had a general coordinate transformation, it would no longer transform inversely to how the metric transforms

$x^\mu \rightarrow x'{}^{\mu} = \frac{\partial x'{}^\mu}{\partial x^\nu} x^{\nu}$ is only true for $ \frac{\partial x'{}^\mu}{\partial x^\nu} = L^{\mu}{}_{\nu}$

Slereah

Is it?

I mean it doesn't preserve the metric, but it's still a tensor transformation

"tensor" is a broader notion

DIRAC1930

The metric by definition is a tensor so it will transform as a tensor under general coordinate transformations

However $x^\mu$ is only a tensor under Lorentz transformations

Slereah

sure

Well no

DIRAC1930

14:42

Well linear transformations of the coordinates then

The same thing happens in 3 dimensional space $x^i$ is not a tensor but $\mathrm{d}x^i$ is since only the latter will transform like a tensor if I change from cartesian to polar coordinates

Slereah

If you use any coordinate transformation, you have $$\eta_{\mu\nu} x^\mu x^\nu \to J^\mu_{\mu'} J^\nu_{\nu'} \eta_{\mu\nu} J^{\nu'}_{\alpha} J^{\mu'}_{\beta} x^\alpha x^\beta$$

With $J$ the Jacobian

This is invariant

The point is that in the case of a Lorentz invariant theory, this invariance holds if you only transform the $x$ too

DIRAC1930

That is only true for $\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}$

Slereah

I'm assuming some linear transformation here :p

$J$ is some $GL$ matrix

DIRAC1930

Oh okay

So my question is, are the conformal transformations non-linear?

Slereah

No they're perfectly linear

And they make sense in special relativity, they're just when you change the scale you use for your system

But the point is that if you change your scale, you have to change both the vectors and the metric

that's just a change of basis, something that works in any vector space

The lorentz transform is a bit more special in that it's an invariant transformation

DIRAC1930

14:49

So to check if an action is conformally invariant, one does exactly the same thing as checking if an action is invariant under Lorentz transformations

Slereah

Yes

If your action is $S[\varphi]$, on some field $\phi$, test out changing just the field $\varphi$

For instance the EM field

DIRAC1930

So when people talk about having to do an additional Weyl transformation, one also had to do a change of the metric in just the case of Lorentz transformations however it just turned out that the metric is invariant under Lorentz transformations

Is this correct?

Slereah

Yes

DIRAC1930

Secondly, people look at expresions such as $(\partial_{\mu} \phi(x) \partial^{\mu}(x))^n$ and can immediately tell that the action will be Lorentz invariant but why can't they do the same for conformal transformations?

Slereah

People are generally a bit weird with how they handle conformal transformation because they typically still use the metric tensor which is not invariant under conformal transformation, but you can pretend your metric is the conformal metric by just doing it up to rescaling

Well $A^\mu A_\mu$ is just an inner product, and the inner product is always lorentz invariant :p

DIRAC1930

14:59

But then surely if $A^\mu$ transforms like a tensor under any linear transformation, then $A^\mu A_\mu$ should be invariant under any linear coordinate transformation

Slereah

Well no, because raising and lowering indices is not a general tensor notion!

$A^\mu A_\mu = g(A,A)$

This would not be true if you used any other tensor than $g$

If you used like the symplectic form $\omega(A,A)$, that would be a whole other different group

DIRAC1930

Why cant I raise and lower with the metric $\Omega(x) \eta_{\mu \nu}$?

Slereah

Well as I said, the point of "invariance" here is that you're only transforming the vector, not the metric itself

DIRAC1930

What if I just do a passive conformal coordinate transformation in the same way we do a passive Lorentz transformation

So what you are saying is that they start with a theory in Minkowski space that has a metric $\eta_{\mu \nu}$ which is the only tensor that can be used to raise and lower indicies?

@Slereah So is it correct to say $\Omega(x) \eta_{\mu \nu} A^\mu A^\nu$ is invariant under conformal transformations but $\eta_{\mu \nu} A^\mu A^\nu$ is not?

And that one needs to check the later on a case by case basis to see if it is

Slereah

15:18

Well as I said, the metric in general is not invariant under conformal transformation, but what physicists do is that they say that a theory is invariant under conformal transformation if the transformation of the field gets compensated by a similar factor of the metric and measure

That's because physicists don't like to deal with the "conformal metric" because it's a weird object

Anyway the point I am making is that the magical property isn't that $$(\Lambda^{-1} A)^T (\Lambda^T \eta \Lambda) (\Lambda^{-1} A) = A^T \eta A$$

that's just a change of basis and basically says nothing about the product

But $$(\Lambda A)^T \eta (\Lambda A) = A^T \eta A$$

DIRAC1930

But isn't that what we are doing when we do Lorentz transformations

Slereah

That's a "real" transformation

The vector is actually changed

The first one is just the change of basis for a bilinear form : proofwiki.org/wiki/…

That's true for any tensor like that

DIRAC1930

So when we do a passive Lorentz transformation, are we doing the 2nd one, it just turns out that the 1st and 2nd are the same?

Slereah

Well when you use a Lorentz transform, you have the magical identity that $\Lambda^{-1} = \Lambda^T$

and also that $\Lambda^T \eta \Lambda = \eta$

So you can simplify things

As I said, a good way to see this is to consider other bilinear forms

DIRAC1930

When we do a Lorentz transformation, are we not doing $\eta\rightarrow \Lambda^T \eta \Lambda$ and $x \rightarrow \Lambda x$?

And it just happens that $\Lambda^T \eta \Lambda = \eta$

Slereah

15:37

What we do is $x \to \Lambda x$

Therefore $$(\Lambda x)^T \eta (\Lambda x) = x^T (\Lambda^T \eta \Lambda) x$$

I mean it's what we do if we want to check the invariance wrt lorentz transforms

We do the other thing for a coordinate transform, since it's the general version of a basis change

DIRAC1930

I thought a lorentz transformation is a coordinate transformation to a different frame of reference

Slereah

It's just that those coordinate changes are also isometries for the Lorentz group, ie they leave the metric invariant

It can be several thing :p

It's just a group

DIRAC1930

An active transformation is where we stay in our same frame and transform the object and a passive one is when we move ourselves (change coordinates) but keep the object fixed

@Slereah Yes but aren't we also $\eta \rightarrow \eta'$ it just so turns out that by definition, we only consider transformations whose coordinates satisfy $\mathrm{d}s'{}^2=\mathrm{d}t'{}^2- \mathrm{d}x'{}^2 \dots$ which means that we enforce $\eta' = \eta$?

Slereah

It depends on what you're doing!

DIRAC1930

Normal physics lol

Slereah

15:46

There are several ways you can apply the transformation and learn different informations from it

DIRAC1930

Is doing a coordinate transformation the same as doing a change in frame of reference from the normal SR point of view?

Slereah

It is a specific type of coordinate transformation in SR yes

If you do some linear coordinate transform on Minkowski space this corresponds to some change of basis

[that's because the exponential map is bijective on Minkowski space and therefore the vector space and manifold are equivalent in some sense]

DIRAC1930

So we can think of SR as being a normal coordinate transformation with some extra constraints?

So under any coordinate transformation $x \rightarrow x'$ the metric will change as $$\eta_{mn}\rightarrow \eta'_{mn}=S^i{}_n S^j{}_m \eta_{ij}$$

In SR, we just take the transformations such that $\eta' = \eta$

Right?

Slereah

Yes

The isometry group of the metric

DIRAC1930

So generalising this to a conformal transformation one must transform both the coordinates $x\rightarrow x'$ and the metric $\eta\rightarrow \eta'$

So if $A^\mu A_\mu$ is invariant under any linear coordinate transformation, shouldn't it be under conformal transformations also

Slereah

16:00

It's not invariant 😵

DIRAC1930

But why not?

Slereah

I mean what do you mean specifically

DIRAC1930

$(x')^T g' x' = x^T g x$

Slereah

If you mean by changing A then no, if you mean just by changing coordinates then sure, but that's a boring statement

Remember that you can express anything in whatever coordinate system you want if you perform the appropriate transformations

DIRAC1930

But all SR is is coordinate transformations with the constraint that the metric stays invariant

Slereah

16:06

part of the point of relativity

yes, that is a more specific claim

But if you go by that then it's not true for the conformal transformation

DIRAC1930

Well it should be that the laws of nature are invariant under frame of reference

Thats the claim of SR

Slereah

$\Lambda^T \eta \Lambda = \eta$ but $(c \Lambda)^T \eta (c \Lambda) = c^2 \eta$

idk what that sentence means

DIRAC1930

Yes but $x \rightarrow \frac{1}{c} x$

Slereah

yeah sure, but as I said if you go by that logic then that's true of every transformation

DIRAC1930

The question is whether $\phi \rightarrow \phi'$ satisfies the untransformed EOM's

Slereah

16:10

You have general transformations in a theory, and you have special transformations

DIRAC1930

which will not generally be true for any linear trasnformation

Slereah

Those are slightly different

DIRAC1930

16:24

Also how is a special conformal transformation linear en.wikipedia.org/wiki/Special_conformal_transformation

It doesn't look very linear

Slereah

16:36

It's not !

It is a second order symmetry

DIRAC1930

16:58

Ok so then the $4-$vectors of SR are not manifestly tensors of conformal transformations

So $s'=s$ is only true for Lorentz transformations $\mathrm{d}s' = \mathrm{d}s$ is true for all coordinate transformations

Slereah

idk what a tensor of conformal transformation means

4-vectors are tensors, full stop

Try not to make up terminology or it may get confusing

DIRAC1930

Only $\mathrm{d}x^\mu$ is a tensor since it transforms correctly under general coordinate transformations

$x^\mu$ does not, but it does under Lorentz transformations

One can see this if one were to transform from cartesian to polar coordinates

So $\mathrm{d}s'=\mathrm{d}s$ is a trivial statement that is true for all general coordinate transformations, but $s=s'$ is only true for Lorentz transformations

It may well be that you have a theory for which $A^\mu A_\mu$ is also invariant under conformal transformations, but you have to check this for each theory

By $A^\mu$ here I mean some $A$ that is a function of the fields i.e. $A^\mu(\phi)$

@Slereah What are your thoughts

ACuriousMind

17:37

@DIRAC1930 You're running into the problem that physicists say "coordinate transformation" but they don't actually mean it.

The challenge is to develop a language in which it is possible to precisely state the meaning of statements like: 1. The value of an integral is independent of the choice of coordinates. 2. The Lagrangian $F\wedge{\star}F$ of electrodynamics is invariant under Lorentz transformations but not general diffeomorphisms. 3. The Einstein-Hilbert action is invariant under diffeomorphisms.

DIRAC1930

What do physicists mean when they say coordinate transformation?

ACuriousMind

This language can - in my opinion - only be the language of differential geometry. Relevant answers of mine are physics.stackexchange.com/a/759904/50583 (difference between "integrals are independent of choice of coordinates" and invariance under symmetries) and physics.stackexchange.com/a/706483/50583 (GR and EM as gauge theories with different symmetry groups)

DIRAC1930

So when physicists write $x^\mu \rightarrow x^{'\mu}=\Lambda^{\mu}{}_\nu x^\nu$, what exactly do they mean?

I'm thinking that this means that I keep reality the same but just describe it from a different frame of reference with coordinates $x'$

ACuriousMind

did you read the first answer I linked, and how it connects the "coordinate transformations" to the formal notion of a family of diffeomorphisms?

DIRAC1930

Yes but it is too complicated for me

ACuriousMind

17:45

I'm afraid I cannot make it simpler

at least not in a few chat messages

DIRAC1930

But my question should be simple to answer since every single book on SR written in tensor notation is predicated on it

ACuriousMind

I'm afraid my opinion is that it's not - the reason the physics language is so confusing is because it does not spend time building the abstract concepts that would be necessary to make the relevant distinctions; this saving in time is paid for by not being able to answer such structural questions unambiguously

2

Slereah

Fundamentally you're asking about what it means to transform a vector but there is a million situations where you do that in physics

And they all mean something different

ACuriousMind

see also this answer of mine for another very similar deceptively simple phrase - "the generators of translation transform as vectors" that's almost impossible to explain within the usual physics language of indices without the abstract language of representations

DIRAC1930

Ok, so it seems like from my point of view when I say transform like a vector, I mean that the components transform in a certain way keeping the vector the same

So my definition of a vector are coordinates such that $x^i \rightarrow R x^i$

Is this the normal physics way?

ACuriousMind

18:00

@DIRAC1930 You might find texts that say that, but even in physics the careful ones will mind not to confuse vectors and coordinates

coordinates are tuples of numbers that identify a point in a manifold, vectors are tuples of numbers that identify a vector in a tangent space at a point - these two notions are not the same, even if they are both tuples of numbers of the same size

but again, a few messages in chat will not clear this confusion, since you are lacking the entire mathematical framework to state clearly what currently remains nebulous

Slereah

Physics also isn't that good at differentiating vector spaces and affine spaces

qwerty

@DIRAC1930 you should read the start of gravitation by misner Thorne and wheeler. they go into painstaking detail that a position is NOT a vector.

@ACuriousMind if a textbook told me that I wouldn't trust anything else it said

DIRAC1930

18:18

@qwerty Literally every textbook in the first half of the last century talk like that

I think there is some utility in thinking like this

qwerty

it only sort of works if you only ever work in Cartesian coordinates in $\mathbb{R}^n$. I don't know what what you mean by "talk like this" otherwise...

DIRAC1930

Well most physicists will use for example Maxwell's equations to calculate something regarding some experiement they see in front of them

qwerty

whether a position is a vector has nothing to do with experiment though

DIRAC1930

@qwerty I only work in Cartesian coordinates lol

qwerty

@DIRAC1930 ^ clearly you don't which is the reason for confusion :p

DIRAC1930

18:59

I think part of the issue is statements like "These transformations are just changes of variables, every theory is invariant under them" here physics.stackexchange.com/a/447193/310229 which is just not true. The $4-$vectors in theories $A^\mu$ only transform like that under Lorentz transformations. If one had $\mathrm{d}A^\mu$ or something then, yes under general coordinate transformations it will transform as written so it is not a trivial statment to check if an action is invariant

under conformal transformations that are general coordinate transformations

There's a reason noone tries to transform from cartesian to polar coordinates linearly though $x' = A x$

However $\mathrm{d}x' = A \mathrm{d}x$ is a different story

2 hours later…

ArduinoEnthusiast

20:32

Hi all. I am struggling with a HW question. Can someone give me some hints if I am on the right track or not. So this is problem I am trying to solve:

Diatomic molecules have energy levels $$ ^{2S+1} \Lambda _{\xi} $$ where $ \mathrm{S}$ is the total electron spin, $\Lambda = (0, 1, ...) \hbar $ is the total electron orbital angular momentum (similar to $ \mathrm{L} $ number of an atom) projected on the bond axis (i.e. \textit{internuclear axis }connecting centers of two nuclei), and $ \xi $ is the total electron angular momentum (similar to $ \mathrm{J = S + L} $ number of an atom) proje…

DIRAC1930

20:54

And secondly since the conformal transformation are non-linear, it becomes more non-trivial

Silly Goose

21:28

@ACuriousMind vectors are not even tuples of numbers

Silly Goose

22:01

does anyone have suggestions on how to get a deep understanding of field theory?

DIRAC1930

22:23

It seems to define an object $x^\mu$ that transforms like a 4 vector does for Lorentz transformations but for conformal transformations, one needs a 6 dimensional space and then project out. I think this is what Dirac did in 1936

Or something along those lines

ACuriousMind

22:55

@SillyGoose you might have noticed by the rest of the statement ("vectors are tuples...that identify a vector") that I was not making a mathematical definition here :P again, the common language of physics is not well-suited to making the distinctions required to be precise here

qwerty

23:07

although i agree it's fine in general to call the components of a vector in a basis "a vector", given the context of dirac1930s confusion I think silly goose has a point (pun intended)

ACuriousMind

@qwerty yes but you can in fact define vectors via their "components" directly; while I agree that I find the abstract viewpoint of a thing with components more elegant, it is possible to define it as tuples with prescribed transformation behaviour under coordinate changes - the other main options are via equivalence classes of curves and via abstract constructions of the tangent bundle

see e.g. this answer of mine for a discussion of the "component definition" as the cocycle construction of bundles

DIRAC1930

Theres no confusion
People wrongly think that a $4$vector is a tensor

ACuriousMind

@DIRAC1930 I don't think you're in a position to call that wrong unless you supply a proper definition of the terms involved; there are at least two technical meanings of "4-vector", either as an element of Minkowski space or as a (1,0)-tensor on a Lorentzian spacetime. The confusion is precisely that usual usage in physics does not distinguish carefully enough between these notions

DIRAC1930

Lets just go by this for now phys.libretexts.org/Bookshelves/Relativity/…

One will quickly find even in the 3dimensional case, that that is not the how the position 3-vector transforms under transformations from cartesian coordinates to polar coodinates

It is however how $\mathrm{d}x^\mu$ transforms therefore this is a tensor

If one limits the general coordinate transformations to just the Lorentz transformations, then $x^\mu$ does follow that transformation law

Therefore $x^\mu$ is a $4$vector

Which is defined to be an object that transforms as $x\rightarrow L x$ under Lorentz transformations

ACuriousMind

23:25

again, you are not being careful enough about the difference between a point on a manifold, a (tangent) vector and a co(tangent)-vector; if you continue to insist on oversimplifying the problem I have nothing to contribute

the whole issue (not only your questions specifically) has piled up enough that I'm considering trying to write up something that's below a book but longer than my usual answers, but this will be a project over the holidays, if anything

DIRAC1930

That's not important to this though

Then the 2nd issue is that people wrongly think that a conformal transformation is something different to a coordinate transformation because they think that any Lorentz invariant action is trivially invariant under any general coordinate transformation just because $\partial_\mu \phi$ looks like a tensor, when in fact it is a $4$vector

Which is the whole reason why Dirac in 1936 introduced a $6-$vector that transforms as a tensor but only under conformal coordinate transformations

imbAF

23:47

@ACuriousMind ACM what was the book you suggested about gauge transformations ?

qwerty

@DIRAC1930 I don't know how you can claim that there is no confusion when you continue to call a position a vector after multiple people have told you it's not the case. when the basics aren't even correct how can you expect to meaningfully discuss anything else further

ACuriousMind

@imbAF Quantization of Gauge Systems by Hennaux and Teitelboim (now Bunster), but it is probably far more abstract and technical than the kind of thing you're looking for. It is, however, the reference for classical and quantum gauge theories in the Hamiltonian formalism in the vein of Dirac-Bergmann

it's one of the few physics books I have a hard copy of :P

DIRAC1930

Because it doesn't matter. The vectors stay the same in all of this, just the components transform

And in SR, it is obvious what the basis vectors are

ACuriousMind

¯\_(ツ)_/¯

imbAF

@ACuriousMind Do you think I should read something else before I read this book? To make things easier for me?

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