The h Bar

Ryder Rude

18:00

In QM, one takes the adjoint

Nakahara's book has a formula for the adjoint which involves the metric

it is in index notation

lemme check

section 2.2.4 inner product and adjoint

it says when $g_{ij}=\delta _{ij}$ and $G_{\alpha \beta}=\delta {\alpha \beta}$, the adjoint reduces to the transpose

bolbteppa

The Thesis that Killed Academia?

►

Casually bashing the idea of society funding people doing a PhD on something because 'nobody will read it', and whining about not wanting to pay taxes for this kind of thing, basically anti-science at its finest

4

Ryder Rude

I think Nakahara's book is defining an extremely general idea of adjoint. it is the notion of adjoint of functions $f: V \to W$ where $V$ and $W$ can be different vector spaces

in QM, V and W are the same

bolbteppa

15

Q: Definition of adjoint of a linear map

I am having a tough time understanding adjoint of a linear map. Consider a linear map between two vector spaces $\, f:V\rightarrow W,$ let us denote $f^*$ to denote its adjoint. Accroding to this video https://www.youtube.com/watch?v=SjCs_HyYtSo (around time 5:50) the author explains that adjoi...

It's not difficult it's done in an abstract linear algebra course, basically it's just exploiting the relationship between a linear functional and an inner product, you can probably use the Riesz Representation theorem to justify it in a general manner on Hilbert spaces

Ryder Rude

18:16

@bolbteppa oh

@bolbteppa i think any map $v^a_b$ from $V$ to $W$ is simultaneously a map from $W*$ to $V*$

@bolbteppa Sabine is extremely critical

Sabine is critical of physicists but supports superdeterminism ideas

Tobias Fünke

@RyderRude well, "transpose" of a linear map is basis independent as well

@bolbteppa I could not stand this video either...

Ryder Rude

@TobiasFünke i am taking transpose to mean interchanging rows and columns

the result of this does not transform like a tensor

Tobias Fünke

I don't know what this means. I just know that one can define the transpose basis independent, and if you go in a basis, then this coincides what people do when they take the transpose of a matrix

bolbteppa

It's just that a linear map f : V -> W 'induces' a linear map f* : W* -> V* through the fact that for f(v) in W to any linear functional L : W -> R on W we can re-interpret L[f(v)] as the result of a linear functional f*[L] : V -> R on V instead via L[f(v)] := f*[L](v)

Tobias Fünke

"does not transform like a tensor" ...I don't like this phrase, but I think I know what it means in the language of diffgeo...

Ryder Rude

18:28

@bolbteppa yes. this is what I meant. it is apparent in the index notation

@TobiasFünke i think one can define it in the abstract, but one would be defining the adjoint

Tobias Fünke

for the sake of simplicity, let us consider two finite-dimensional complex vector spaces $V,W$, and a linear map $A:V\to W$. Then its transpose $A^t: W^*\to V^*$ is linear and well defined by $A^t(\phi):=\phi\circ A$ for $\phi\in W^*$, i.e. $\phi: W\to \mathbb C$.

no, you don't need the "adjoint", which anyway requires an inner product (if you mean adjoint notion of linear operators on Hilbert spaces)

ACuriousMind

@TobiasFünke I don't like it either, but I discuss a possible formalization of the phrase in this answer of mine

Tobias Fünke

@ACuriousMind thanks, I'll have a look. I thought it just means that a given tensor (field) expressed in a basis, the coefficients transform in a certain (well-defined) way, no?

@bolbteppa you can use MathJax here too, btw. makes it easier to read for everyone

Ryder Rude

@TobiasFünke i will have to think about it

Tobias Fünke

Transpose of a linear map

In linear algebra, the transpose of a linear map between two vector spaces, defined over the same field, is an induced map between the dual spaces of the two vector spaces. The transpose or algebraic adjoint of a linear map is often used to study the original linear map. This concept is generalised by adjoint functors. == Definition == Let X # {\displaystyle X^{\#}} denote the algebraic dual space of a vector space X . {\displaystyle X.} Let...

it is basically just this

bolbteppa

18:34

en.wikipedia.org/w/…

ACuriousMind

@TobiasFünke Sure - if you start with the abstract objects, it's just a statement about the transformation of the coefficients in a basis. The answer is about explaining in terms of the math formalism what we're doing when we claim that just specifying a bunch of numbers that "transform like a tensor" constitutes a definition of a tensor field (rather than already having the abstract object and just expanding it)

Tobias Fünke

@ACuriousMind I see. Thanks. I'll read it later, I am too tired now :d

bolbteppa

The wiki of the transpose in the Hermitian adjoint section explains the subtle difference which is probably what you really meant earlier about QM books vs math books being slightly different

Silly Goose

@ACuriousMind the example I am thinking of is $\Lambda_{\nu}{}^{\alpha} \Lambda_{\mu}{}^{\beta}\eta_{\alpha \beta} \rightsquigarrow [\Lambda] [\eta] [\Lambda^T]$

Tobias Fünke

@RyderRude I mean, it might be that various concepts have similar or even the same maps. But for me it makes the most sense as I've laid out. The adjoint, in my terminology, is a map on the Hilbert space itself (and not on its dual)

Ryder Rude

18:36

@TobiasFünke but this uses the inner product

Tobias Fünke

@RyderRude no

where have I used an inner product in my definition?

Ryder Rude

@TobiasFünke sorry.. i meant the wiki

Tobias Fünke

@RyderRude no, I don't think the brackets mean "inner product"

ACuriousMind

@SillyGoose I suspected that, and that's exactly the special caase of tensors with 2 input/indices - there's no notion of "transpose of a tensor" in general, you tried to abstract the situation too much as so often ;)

Ryder Rude

@TobiasFünke i went inside the link titled "natural pairing"

it uses a non degenerate bilinear form

Silly Goose

18:37

classic silly goose move

Ryder Rude

@TobiasFünke i will have to think about ur definition. thanks for sharing it

ACuriousMind

A (1,1)-tensor at a point is an element of $V\otimes V^\ast$ (where $V$ is the tangent space), so it is equivalently a map $\Lambda : V\to V$. The completely ordinary linear algebra definition of the transpose applies: This has a transpose $\Lambda^T : V^\ast \to V^\ast$, which is equivalently an element of $V^\ast \otimes V$.

bolbteppa

x.com/ThePhDPlace/status/1867380461959569589

Silly Goose

Let $[\Lambda] \rightsquigarrow \Lambda^\mu{}_\nu$. Okay so the identification $[\Lambda^T] \rightsquigarrow (\Lambda^T)_\mu{}^\nu$ reflects that fact that we've dualed the domain: $V \otimes V^* \mapsto V^* \otimes V$ because dualing the domain is precisely reflected by raising/lowering a lowered/raised index?

ACuriousMind

If you're not comfortable with how this relates to the indices, I can understand that since I think one just has to do this kind of symbolic manipulations a lot to become comfortable with them, but there isn't even really anything specific to "tensors" here, it's just linear algebra and switching between the abstract descriptions and the index expressions

Silly Goose

18:42

Since heuristically I can read $v_\mu$ as a "vector index" which eats a covector.

ACuriousMind

@SillyGoose well you have to have a convention for what index position corresponds to an element of $V$. If you would write a vector in $V$ as $v^\mu$, then the elements of the dual $V^\ast$ have the lower position $v_\mu$, and so elements of $V\otimes V^\ast$ have position ${\Lambda^\mu}_\nu$ and elements of its dual $V^\ast \otimes V$ then ${\Lambda_\mu}^\nu$.

Silly Goose

okay i see. is the usual convention $x^\mu$ is a vector?

ACuriousMind

not if $x$ is a coordinate :P but yes, the usual convention is that the upper index is for components of vectors

Tobias Fünke

@RyderRude math.stackexchange.com/questions/484844/… this might help.

Ryder Rude

@TobiasFünke the map $v^a_b$ can be interpreted as a map $A:V-->W$ and as a map $A^T: W*-->V*$. Is this the same as ur idea?

Tobias Fünke

18:47

what is $v^a_b$? but yes, in my definition the "transpose" is exactly like this

Silly Goose

okay i see. so naturally, we should have $\Lambda^\mu{}_\nu \sim \Lambda \in V \otimes V^*$. Then, $\Lambda_\mu{}^\nu \sim \Lambda^T \in V \otimes V^*$. Is it also a convention to choose the tensor factor order?

Ryder Rude

$v^{a}_{b}$ is a (1,1) tensor

Silly Goose

I think $V \otimes V^*$ is most natural to make contact with matrices, but it seems otherwise arbitrary.

ACuriousMind

@SillyGoose I think you messed up the order :P

Ryder Rude

i mean that this tensor defines two maps which u r calling transposes of each other. Is this correct? @TobiasFünke

Tobias Fünke

18:48

@SillyGoose what do you mean?

ACuriousMind

also, sure, you might call it a "convention" to have the indices on the tensor in the order of the factors, but frankly I would find any other convention just wrong :P

Ryder Rude

$v^a_b$ defines a map $P: B-->A$ and another map $Q: A*-->B*$. Then P and Q are transposes of each other @TobiasFünke

ACuriousMind

if you define $T_{\mu\nu}$ such that for an element of $V\otimes V$ the $\mu$ index tracks the second factor you're just deranged

Tobias Fünke

sorry, I cannot quite follow, I am a bit tired already :s

Silly Goose

@ACuriousMind Actually I am confused by your first sentence here. If $\Lambda \in V \otimes V^*$, then I can see how to induce a map $\Lambda': V \to V^*$ or $\Lambda'': V^* \to V$, but I am uncertain what you mean by the map $\Lambda: V \to V$

oh wait

ACuriousMind

18:51

@SillyGoose It's a standard result of linear algebra that $\mathrm{Hom}(V,W)\cong V\otimes W^\ast$ for finite-dimensional $V,W$.

Tobias Fünke

1

Q: Thinking of a linear operator as a (1,1) tensor

I am reading that a linear operator $A$ can be thought of as a (1,1) tensor [where $(r,s)$ corresponds to $r$ vectors and $s$ dual vectors]. This can be done by saying $$A(v,f) \equiv f(Av)$$ where $v$ is a vector and $f$ is a dual vector. I assume the $Av$ in the argument on the rhs corresponds ...

operators tensor-calculus mathematics

that is just the application of the canonical isomorphism in finite dimensions

i.e. the fact that in finite-dimensions it holds that $V\cong V^{\ast \ast}$

Silly Goose

Hm is $\Lambda: V \to V$ not dependent on some $v^* \in V^*$?

I.e. there exists an induced map $\Lambda_{v^*}: V \to V$ for all $v^* \in V^*$

ACuriousMind

@TobiasFünke caution: Hilbert spaces are isomorphic to their duals (and hence to their double duals), but infinite-dimensional Hilbert spaces do not have that $\mathrm{Hom}(H,H)$ is the same as $H\otimes H^\ast$ (the latter are the Hilbert-Schmidt operators, which are a subset of all operators).

Tobias Fünke

well, I said finite-dimensional ;)

and it holds again for all vector spaces, you don't need an inner product

but yes, what you say is correct and important!

Silly Goose

@ACuriousMind oh i did mess up the order in the second statement. I meant $\Lambda^T \in V^* \otimes V$

ACuriousMind

18:57

@SillyGoose I don't know what you mean. For a simple tensor $v\otimes f\in V\otimes V^\ast$, you define the corresponding map $V\to V, w\mapsto f(w) v$. By linearity, this extends to all tensors, defining the isomorphism between the tensors and the maps.

Silly Goose

I was thinking (wrongly) of the induced map for every $v \otimes f \in V \otimes V^*$, define $w \in V \mapsto v \otimes (fw) = (fw)v \in V$.

Which is just a generalization i guess of the actually wanted map

oh wait

ACuriousMind

if you unpack the entire tower of definitions, in the end in index notation this turns out to just be the "contract the vector's upper index with the tensor's lower index"

Silly Goose

this map is precisely what you just said

i guess i was getting confused with the domains. i see now that the map is really some $T: (V \otimes V^*) \times V \to V$. Then we can fix a $f \in V \otimes V^*$ and just get a map $T_f: V \to V$.

@TobiasFünke I think I was saying that there is a way to consistently use $V^* \otimes V$ as the "undual space" and $V \otimes V^*$ as the "dual space" if you make appropriate redefinitions of your "undual space vectors" and etc.

ACuriousMind

that's the curried version of the $V\otimes V^\ast \to \mathrm{Hom}(V,V)$, sure

Silly Goose

@ACuriousMind oh this is a technical term

Slereah

19:03

I thought Germans only cared about currying wursts

ACuriousMind

@SillyGoose Currying is, yes - it means that you can express a function $A\times B\to C$ equivalently as a function $A\to \mathrm{Hom}(B,C)$

Tobias Fünke

@SillyGoose oO I am more confused than before haha. But nevermind, as I said: I am tired

Ryder Rude

@TobiasFünke i think this definition is good too, but i associate a transpose as something which takes a $v^a _b$ tensor to a $v^b _a$ tensor. this idea of a transpose is defined as an adjoint in the abstract

but ur idea of a transpose is different

Tobias Fünke

@Slereah oh my haha

Silly Goose

This is the same problem that I had when in classical EM one introduces the quadrupole "tensor". Or any time in classical physics when a "tensor" is introduced. They say that the order of index do not matter, that whether it is a (1,1) a (2,0) or a (0,2) tensor does not matter. But I think it does matter because one needs to consistently use the standard convention, and I have never had the standard convention spelled out.

Slereah

19:05

it does matter yes

Silly Goose

I asked a question about this on the main site recently (about the quadrupole tensor)

Tobias Fünke

it does matter, no? I mean these are different mathematical objects in a strict sense

Silly Goose

I think it would resolve my confusion completely to just know precisely what the standard convention is.

ACuriousMind

convention for what?

Silly Goose

I agree that it matters. but i think some physicists maybe don't really know what they are talking about when they speak of "tensors"

Tobias Fünke

19:06

@SillyGoose probably lol

ACuriousMind

@SillyGoose now that I won't disagree with :P

Silly Goose

@ACuriousMind For what object the stress-energy tensor should be, the quadrupole tensor should be, the Lorentz transformations in the fundamental representation should be, etc. I mean these should all really be fixed by defining the convention for vectors I imagine.

Tobias Fünke

physics always has the problem that most courses do not establish the mathematics well, and from my experience I can say that it can be quite a mess to find out what precisely is meant

ACuriousMind

@SillyGoose but that all just follows from the definition, no? A Lorentz transformation in its most basic form is just a transformation that maps vectors to vectors and so is equivalently a (1,1)-tensor, etc.

that's not a "convention"

Silly Goose

Concretely, the quadrupole tensor is called a (2,0) tensor a (0,2) tensor or a (1,1) tensor interchangably. But obviously which object you use changes how you do computations concretely with them because you need to appropriately "process" the inputs according to how you're writing down the quadrupole tensor.

ACuriousMind

19:08

I think the problem you are encountering is just that people may use different (but equivalent or at least equivalently useable) definitions, for which there is no cure :P

Silly Goose

@ACuriousMind well this I see. once you fix $v^\mu$ to be a vector, then $\Lambda^\mu{}_\nu$.

ACuriousMind

translating between different conventions for index position or signs or whatever is just a ~~curse~~ fact of life when you're reading more than one source on something :P

Tobias Fünke

:d yep

Silly Goose

well i wish physicists would just stop trying to use the word tensor when they just end up always writing down a matrix anyways!

why call the moment of inertia a tensor if you do not use any of the tensorial properties (at least this was the case in my classical mechanics course)!

ACuriousMind

it sounds cool

Silly Goose

19:11

i mean who started this. physicists always emphasize when something is a tensor and not a matrix. then they proceed to only use a tensor in the form of a matrix. so what are they trying to say :P

ACuriousMind

(wait until you realize that "tensor" to a physicist may mean both "tensor" in the linear algebra sense and "tensor field" in the diff. geo. sense and they never even seem to clearly state that those are distinct notions :P)

Silly Goose

i will (attempt to) never succumb to the physicist usage of vector and tensor (in the transformation sense).

Tobias Fünke

@RyderRude do you mean by that that if you have a matrix $M$, then the transpose $M^T$ is just the transpose "we know"? if yes, then this is precisely what my definition does, I think

ACuriousMind

I share your outrage at this state of affairs but there is no secret text that resolves this - the only winning move is not to play (i.e. become a mathematician) ;P

Tobias Fünke

@SillyGoose I was so confused when starting my physics degree. but I think ACM nailed it: it sounds cool lol

Ryder Rude

19:14

@TobiasFünke i think both definitions have different uses...

Tobias Fünke

@RyderRude after choosing suitable bases*

Silly Goose

blebolus. i mean if they just taught the basics, it would even be much easier to compute the quadrupole tensor. if you treat it as a $(1,1)$ tensor, then all you are doing is taking the outer product (or tensor product of a covector and vector) to produce an appropriate matrix. which is much easier than applying some clunky (but equivalent) explicit formula for the quadrupole moment...

Tobias Fünke

Okay, then so it may be :)

@SillyGoose for that you have to explain what a covector and tensor product is

you just shift the things you have to explain, no?

Ryder Rude

it does not matter what one chooses to call "the transpose" , but both of these have different uses

Silly Goose

well I think all physics students should take an abstract linear algebra early i think :P (not those silly courses that focus on computing row-echelon forms or whatever)

Tobias Fünke

19:15

sure

Ryder Rude

also note that one can't apply ur definition to the object $v^a _b$, as it returns the exact same object @TobiasFünke

but the adjoint can be applied to this object

Silly Goose

well i think for the physics I have done so far it is okay to talk about the tensor product in operational terms of the kronecker product i guess.

no need to do category theory or whatever

@TobiasFünke by the way, are you based in germany as well (as ACM, not as well as in I am also based in germany)?

Tobias Fünke

@SillyGoose yes

Ryder Rude

ur definition takes a map $V->W$ to a map $W*-->V*$. but $v^a_b$ is simultaneously encodes both of those maps.

Silly Goose

(Perhaps this is too off-topic or sensitive) But is germany okay :P? An american news outlet describes germany as currently going through a difficult economic and political time.

Tobias Fünke

19:19

xD

well. ACM, what do you think? :P

Ryder Rude

while the adjoint takes a map $V-->W$ to a map $W-->V$. so this is a non trivial operation on $v^a_b$ which gives $adj(v) ^b_a$

Silly Goose

I suspect that the news outlet has a not totally correct presentation of reality.

Tobias Fünke

I mean, in news it always sounds worse than it is, no? I'd say we are still fine, but globally of course there are many crises; in Europe, for example, many governments are not stable...

and yes, politically it is not easy... we have new elections in 2 months

ACuriousMind

let's say I wouldn't call it worse than what's happening in the rest of the Western world on average, but it's not great :P

Tobias Fünke

if one is optimistic, then one could say that the current situation leads to many opportunities for improvement

@ACuriousMind exactly

Silly Goose

19:26

I see

the new york times seems to love to describe things as collapsing :P

Relativisticcucumber

19:47

@SillyGoose yeah what is up w this btw xD i mean i even took the math majors linear course and the entire time we just did various decompositions by hand

but i just recalled a funny story -- so in my apple notes it tries to autocorrect schrodinger to stronger. when i was a child and i'd right click the red underlining and saw the option "add to dictionary" i would never use that thing bc i thought it was a worldwide dictionary and i was afraid of adding smth wrong to the global dictionary, therefore creating a typo pandemic.

i was like "wow i cant believe i have the power to influence the worldwide dictionary i better use this power wisely"

ACuriousMind

lol

but very responsible of you

Relativisticcucumber

a day that acm lols is a good day

Silly Goose

@Relativisticcucumber each day i grow more pessimistic about the knowledge of others and their will to unbenight (except for the people here ;D)

Relativisticcucumber

@SillyGoose on the contrary, we have the perfect example of this behavior in this chat

nuff said

Herr Feinmann

20:19

@Relativisticcucumber It would take me a long while to decide to save words that are not in the default dictionary :P

And I should decide which words to use and probably change my usage of words to avoid unpleasant choices

Yes, ACM, it's about time you realize I'm batshit insane

Relativisticcucumber

21:04

@HerrFeinmann what xD

ACuriousMind

part of the insanity seems to be that he talks to me when I'm not even there :P

Relativisticcucumber

LOL

Herr Feinmann

21:19

@ACuriousMind in my headcanon you're always reading the chat when you're in the room

qwerty

21:43

@ACuriousMind do you mean the whole physicists not distinguishing a thing evaluated at a point versus on a space slight of hand? or that a tensor field (notion 2) cannot be thought of as a field of tensors (notion 1) straightforwardly?

@SillyGoose it's a watermelon /s chat.stackexchange.com/transcript/71?m=66625578#66625578

Tobias Fünke

ACM, do you by chance know the textbook from M. Schottenloher: "Geometry and Symmetry in Physics"?

ACuriousMind

@qwerty it's some mixture of the two: Of course you can consider a tensor field as taking the value of a tensor at each point, but it's a tensor on a different space at each point (the tangent space at that point)

@TobiasFünke no, but I know (and like) Schottenloher's CFT book

Tobias Fünke

can I ask you a specific question to the book I refer to? I have a quick question which I think you can immediately answer. But of course, only if you have time

ACuriousMind

don't ask about asking, just ask :P

Tobias Fünke

mathematik.uni-muenchen.de/~schotten/bas/… here is the PDF link from the author's uni site. p. 53-54, regarding the definition of the Galilean space-time

qwerty

21:51

@ACuriousMind wait, I was recently reading about tangent bundles - is that not a single space defined using all the tangent spaces?

ACuriousMind

@qwerty The tangent bundle is a space, yes. I don't quite see the relation to what I said.

Tobias Fünke

As far as I understand, he defined $\Delta:V\to\mathbb R$ to mean the time difference between any two events, which I so far understand (hopefully correctly). He then proceeds and defines $V_t:=\Delta{-1}(t)$, which I interpret as the pre-image.

qwerty

@ACuriousMind oh - I thought you were emphasising that the tensors are defined "on a different space at each point"

ACuriousMind

@qwerty they are - at each point $p\in M$, the tensor field corresponds to an element of some tensor power of $T_p M$

Tobias Fünke

but then I am lost. He says that $V_t$ describe in $V$ and thus in $M$ "slices of equal times"...???

ACuriousMind

21:55

$T_p M$ is a vector space, so the usual linear algebra concept of tensor applies and makes this a proper definition. The total bundle space $TM$ is not a vector space, there is no such thing as a "tensor on $TM$" if that's the confusion

qwerty

ahh I see your point

ACuriousMind

@TobiasFünke No, you misunderstood $\Delta$. $\Delta$ is just a map that assigns to any element of $V$ its time (it's the projection $\mathbb{R}^{4}\to\mathbb{R}$ onto the time component of $(t,x,y,z)$ in the standard case).

Tobias Fünke

@ACuriousMind well, yes I see this. But given two events $a,b\in M$, s.t. $a+v=b$, then $\Delta(v)=t$ means that the time difference between $a$ and $b$ is $t$, no?

which is also what the claims "objektive Zeitunterschied", I think

ACuriousMind

@TobiasFünke yes

Tobias Fünke

so my problem is the misunderstanding of $V_t$, I believe. from the mathematical definition it seems something like "all vectors which, when added to an event, yields an event with time difference $t$" (?). But as far as I understand the text, it is more like they "describe" slices of equal times. Also the picture on p.54 is kind of confusing...

ACuriousMind

22:06

@TobiasFünke Ah, yes, the section is formulated confusingly: I think what's meant here is that of course $M$ and $V$ are not really different - the affine space $M$ is just $V$ without an origin chosen. So you can fix some arbitrary origin in $M$ and then the foliation of $V$ by the $V_t$ turns into a foliation of $M$ by some $M_t$.

the picture contains the same conflation of $M$ and $V$

Tobias Fünke

yes, okay, that was also what I've thought, especially with the next paragraph, where he chooses $M=\mathbb R^4$ and $V$ the same (but with vector space structure)...

but choosing an origin feels...not good

haha

anyway, thank you! much appreciated. so far I liked the book, though

ACuriousMind

@TobiasFünke all the choice of origin does to the foliation of $M$ is shift the $t$

Tobias Fünke

@ACuriousMind okay. I will think about it, thank you

ACuriousMind

if you think about it the foliation of $M$ itself remains the same, just the $t$ you label the $M_t$ with shifts by adding a constant (the time difference between the previous and the new origin)

Tobias Fünke

okay. I'll try to write down everything properly as soon as I have time, I am way to tired today ^^

good night everyone, see you

ACuriousMind

22:20

you've been telling us you're tired for 4 hours, go to sleep!

Tobias Fünke

:d yes yes

DIRAC1930

22:35

@SillyGoose One has to be very careful. Most objects encountered in SR are not tensors hence the name $4-$vector. They transform like a tensor but only under Lorentz transformations. If one does a general coordinate transform, one will find that the object they are considering no longer transforms like a tensor. In SR, quantities such as $\mathrm{d}x^\mu$ are tensors since they transform under the correct law for general coordinate transformations, but objects such as $A^\mu$ are not

Since they only follow the tensor law of transformation for Lorentz transformations

You can see they are not tensors if you try to convert a quantity like $A^\mu$ from cartesian to spherical coordinates

However it will work for $\mathrm{d}x^\mu$

Herr Feinmann

22:50

I wonder why there is a "English language learners" SE site and also a "English language" site

For other languages there is just the "~ language" site and people ask question about language learning :P

DIRAC1930

@HerrFeinmann What are you currently working on?

Herr Feinmann

I'm a student, I don't do research :P

DIRAC1930

What are you studying in that case lol

Herr Feinmann

I will start working on superconductivity in the next few days oof

Then I can get back to HEP... Forever

ACuriousMind

hep hep hurray

Herr Feinmann

23:04

Good morning

DIRAC1930

I want to just study electromagnetism and gravity and forget about everything else lol

There is something special about those two forces

I hope one day, determining the mass to charge ratio of the electron from first principles will be solved

Silly Goose

23:22

I want to study applications of topology and geometry to quantum theory :P. I can finally get back to it during winter break…

@qwerty what the watermelon XD

DIRAC1930

What subject area do you mostly work on?

Silly Goose

not anything specifically right now. i am just taking courses. i am hoping to move into some condensed matter theory that involves topology/geometry :P

DIRAC1930

Sounds interesting

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