« first day (4781 days earlier)      last day (173 days later) » 
00:00 - 20:0020:00 - 00:00

8:16 PM
@ACuriousMind can you please delete it, it let me delete it twice then randomly changed it's mind and wont let me any more
 
cya
 
@bolbteppa You could delete it before because Qmechanic's answer was at a net negative score, it is intentional that you cannot delete posts with answers with at least one upvote because it would be unfair to answerers to get their effort deleted just because OP wants to get rid of the question
see "When can't I delete my own posts?" at meta
 
What is your mind most curious about in Physics? @ACuriousMind
 
oh no i have become lost again @ACuriousMind so when you said if we start with a relativistic action -- what exactly does it mean for you to be a relativistic action
oh no that sounds like you are personally a relativistic action but i think you get the question XD
@naturallyInconsistent yeah i thought i got it but i realize now i was wrong. big sad lol
 
@Relativisticcucumber personal identity crisis aside, a relativistic action is an action on field configurations on Minkowksi space
 
8:31 PM
@ACuriousMind so it is not true that a relativistic lagrangian is one which is invariant under poincare transformations, right? or i mean this is true but like that is not the fundamental definition but merely a consequence?
 
@Relativisticcucumber since the action has to be a scalar, by the argument above this more or less immediately means the Lorentz group will show up as a symmetry
 
oh yes
 
so iiuc, we established that being lorentz invariant does not mean a theory is relativistic. i am on board with this. but @SillyGoose asked me why making a theory lorentz invariant does not mean its special relativistic if being lorentz invariant is what it means for the action to be invariant and i became lost again
because these two things seem to be in contradiction
@ACuriousMind lmao
 
note that my argument about "relativistic" implying Lorentz invariance only works if there are no "background terms" in the action
 
8:34 PM
what do you mean by background terms?
 
@ACuriousMind er okay modern is not the right word. But to say the subset of relativistic theories is tautological as you say. I guess what I mean to say is: poincare invariance is present in every physical theory that takes place in minkowski space as its spacetime, or something to this effect
 
if there is some non-dynamical background, e.g. a vector field $V^\mu$ which is not a dynamical variable, it's just there (like when we ask in classical EM what the field configuration corresponding to some given current $j^\mu$ is), then terms like $\partial_\mu \phi V^\mu$ can break Lorentz invariance - just like the "general" rotation invariance of Galilean physics can be broken if I put something like a constant electric field in all of space
 
hm i think the discussion with silly goose above is actually not my issue but i might be conflating what im misunderstanding. i see what you are saying with the background terms though
 
I feel both of you are just reading too much into this whole affair: This is intro to relativistic QFT. The typical situation we want to describe with this theory is one with a Poincaré invariant action. There's no further claim here, no one claims you can magically derive out of nothing that all theories have to be of this form (because they don't have to!), but that this is the set of theories this particular field of physics is concerned with
 
@ACuriousMind my issue is solely with how we make qft relativistic. you said via the action, not by making sure its lorentz invariant, right? but then what do we do to the action?
 
8:41 PM
i would just like the starting point :P or at least some finite list of assumptions that leads one to consider these theories.
 
@Relativisticcucumber We're not "making QFT Lorentz invariant"
Very concretely, the first goal of QFT is to develop a quantum theory of electromagnetism
The Lagrangian of EM, $F_{\mu\nu}F^{\mu\mu} + A^\mu j_\mu$, is Poincaré invariant
this is, both historically and motivationally, the starting point of QFT
it will turn out that this framework is suitable to describe the quantum theory of all elementary particles and can be setup in a way that it applies to all Poincaré invariant actions of fields
 
@ACuriousMind i dont think i meant to say that :0
 
no one is claiming QFT only applies to Poincaré invariant actions (again, cond-mat is full of non-relativistic QFTs), no one is claiming that absolutely every pathological action that technically has "Poincaré invariance" would be called "relativistic", no one is claiming that absolutely every theory we'd called "relativistic" has to have a Lorentz invariance action
none of the strong claims you're asking about are actually being made
 
i think what im actually saying is being misunderstood. actually i think my confusion has nothing to do w qft
i think it's a relativity question.
 
separately, strictly speaking is the target space of a quantum field not a vector space?
 
8:49 PM
@Relativisticcucumber I agree you're probably being misunderstood - I don't understand what the problem is supposed to be
@SillyGoose why would it not be?
 
oh wait nvm lol
 
Let me try to rephrase omitting all QFT terminology. So, 1) in special relativity, we show that the poincare group elements contain all isometries of mink space time. 2) making a theory "special relativistic" is about ensuring the action is invariant under the isometries of space time. 3) an action that is special relativistic is automatically lorentz invariant. 4) an action that is lorentz invariant is not necessarily special relativistic. // are all of these statements true? @ACuriousMind
 
Poincaré invariance in QFT has replaced spinors
 
@Relativisticcucumber 1. Yes. 2. & 3. Yes in the absense of backgrounds (as explained above) and modulo the caveat that there may be pathological exceptions to this no one cares about 4. Yes - my counterexample was a 2+0 conformal theory
@Mr.Feynman and I don't even understand why that happened
But I still want to caution that we're not really "making" special-relativistic theories
we're noticing that our best theories - like electromagnetism - are special relativistic theories
no one sat down and said "let's make EM special relativistic" - it's the other way around, we discovered special relativity to make sense of EM
 
when SR was discovered it made sense of EM
 
9:03 PM
ok ok i get it !!
thank god
 
I can't tell if those are happy or exasperated '!!' ;)
 
why can it only be one
im jk, its a happy !!
@SillyGoose so are u also good now?
 
@ACuriousMind me neither but maybe that's because I have a simplistic view of the issue and my explanation would be a one-liner
 
we can put this to rest once and for all? XD
"but wHaT iS tHe HaMilToNiAn"
 
I'm a bystander
 
9:05 PM
well
 
honk
 
my question is what if i do not want to privilege the action hehe
 
And I'm walking through the streets looking for cats
 
cat!
 
@Mr.Feynman HFDJSLFDSLJFDLSJFDSKFJ
yes
 
9:06 PM
honk
 
@Relativisticcucumber The black one fled
 
@SillyGoose but what acm said is not about action?
its about the mink metric?
 
The orange one sit and watched me, until I took a pic and in that very moment it moved
 
@SillyGoose in this house we put the action on a pedestal
3
 
but still feel the clarifying point is that we dont need to make anything SR, it just is and that gives us LI for freeeeeeee so we only need posit a theory is SR
sr being special relativistic
or not even posit
observe~~~~~
@Mr.Feynman i really love black cats
 
9:08 PM
Come on, I should have stayed home and learn SUSY
 
this is a separate question: when are representations of isomorphic lie algebras equivalent?
 
@Relativisticcucumber acab (all cats are beautiful)
 
By equivalent do you mean isomorphic?
 
@Mr.Feynman i was at manilla bay this summer and i sat down and 5 cats came and surrounded me and just sat next to me. it was the best time of my life
 
@Relativisticcucumber I'm jealous
 
9:10 PM
I thought I understood the logic for passing to the universal cover, but I am missing this step. I know that given connected lie group $G$ and its unique universal cover $\tilde{G}$ we have by definition $\mathfrak{g} \cong \tilde{\mathfrak{g}}$. but it is not clear how represetations of $\tilde{\mathfrak{g}}$ tell us about representations of $\mathfrak{g}$. Let's include details about complexification and all if that is what is done in a physics context.
@Mr.Feynman yes
 
@SillyGoose they're isomorphic, every representation of one is the same representation of the other
 
Two reps are isomorphic is there an intertwiner which is a vector space isomorphism
 
if you have a rep $\rho : \mathfrak{g}_1\to \mathrm{GL}(V_1)$ and a homomorphism $f : \mathfrak{g}_2 \to \mathfrak{g}_1$, you get a representation of $\mathfrak{g}_2$ by $\rho\circ f$
 
hm okay
 
Oh maybe I missed the question
@ACuriousMind have you ever read Tong's notes on SUSY?
 
9:13 PM
I mean I'm not sure I got the question either :P
@Mr.Feynman no
 
I'm going through now
That course is longer than the ST one
 
@ACuriousMind Fine, I added a proper explanation, if you want to remove your comments and leave it as it is
May be the last time I use the main site lets see, I feel dirty not being able to control something as simple as removing a question
 
I miss getting an answer by Qmechanic and being called OP :(
That's an idea for a username
OP Feynman
But without context it would sound like I'm calling myself overpowered
 
I have a question about the product metric of the cartesian product of (1+1) minkowski with itself. say the metrics are $g=dudv$ and $h=drdw$. I think I have to do $g \oplus h$ to obtain the product metric. Am I on the right track?
 
@JohnZimmerman yes
 
9:23 PM
If you want the product metric that is how you do it yes
It is guaranteed to exist
Though the product of 1+1 Minkowski with itself would have signature (--++)
 
Now if I Cauchy foliate both (1+1) spaces with the same foliation, and take the corresponding product foliation I should get a foliation of that (2+2) space equipped with the product metric $g \oplus h$, is that correct?
 
I suppose yes?
 
Then I suppose I can take the foliation of the (2+2) space equipped with $g\oplus h$ and project it onto a subspace to obtain 1 dim. of time and 2 of space. But I don't know how $g \oplus h$ reacts to this projection
 
It's a projection, it works like a projection
Just do the induced metric on that surface
 
okay i think this is the logic...
 
 
2 hours later…
11:43 PM
the r/physics thread on Kerr’s paper is quite enjoyable
 
00:00 - 20:0020:00 - 00:00

« first day (4781 days earlier)      last day (173 days later) »