« first day (3873 days earlier)      last day (55 days later) » 

2:37 AM
Quick question, reading wikipedia I came to vacuum bubles where it is written that $Z= \int e^{-S} D\phi = e^{-HT}$ shouldn't it be $Z= \int e^{-S} D\phi = \langle\phi|e^{-HT}|\phi\rangle$?
 
 
6 hours later…
8:57 AM
Maybe they mean $H$ as the eigenvalue of the vacuum?
 
 
2 hours later…
EVO
11:11 AM
@ACuriousMind Thanks
 
 
2 hours later…
1:31 PM
Hey all this is a mathsy question
Any thoughts?
 
what does $P(x,y)=P(r,\theta)$ even mean
 
Oh $P$ is just a plynomial
*polynomial
So if you apply the coordinate transformations
then you should get the same polnomial
yea should have written $P_1$ and $P_2$
@NiharKarve better?
 
How can you compare polynomials in different variables? Are you saying something like $x^2 y + y$ gets mapped to $r^2\theta + \theta$?
 
@NiharKarve I'm saying $x^2 + y^2 = r^2$
Which is true even in the coordinate transformation used to go from cartesian to polar coordinates
The mapping must be between the metric and the polynomial so maybe $dx^2 + dy^2 \to x^2 + y^2$
and $dr^2 + r^2 d \theta \to r^2$
 
2:01 PM
What do you mean polynomials that are invariant under coordinate transformations?
 
@Charlie I mean whatever procedure I use to map $dx^2 + dy^2 \to x^2 + y^2$ should also map $dr^2 + r^2 d \theta^2 \to r^2$ as well
 
There probably isn't a way to do so if you're only given the metric, the map $r\mapsto\sqrt{x^2+y^2}$ is just the coordinate transformation of $r$ between polar and cartesian coordinates. In other words the relationship is between two coordinate systems, and doesn't really involve the metric
I can't think of a well-defined map that takes you from $ds^2$ in cartesian coordinates to $x^2+y^2$ while also knowing that the corresponding map in polar coordinates is $ds^2\mapsto r^2$
 
the whole point of tensors/differential forms is that they are not functions or polynomials and do not transform like them
if you could map a tensor (the metric) to a simple function, we'd do it all the time
 
Then again perhaps if your coordinate transformation is a diffeomorphism there is a bijection between the polynomials in one coordinate system and the other?
although cartesian to polar coordinates is in particular not a (global) diffeomorphism
 
@Charlie I thought of one but it doesnt work for all metrics
 
2:17 PM
What is your idea?
 
This
0
Q: Directly going from the line element to a coordinate invariant polynomial?

More AnonymousBackground and Question I came up with an algorithm which enables me to go from $ds^2 \to K$ where $K$ is a coordinate invariant quantity. Is the below conjectured algorithm always right? If so, why does it work? What is $K$ geometrically Algorithm with Example Consider the following line element...

P.S I dont know why it works but I dont know any counter examples either
 
You can do it by hand on (presumably) every metric, sure. The step in your question that has to be done by hand is "Replace all the differentials of transformed variables with $dx \to x $ and $dy \to y $ if a variable does not take part in the transformation then replace it with $0$."
 
@Charlie You can only do it on scalable metrics
 
In the first example you're just computing $g(v,v)$ with $v = x\partial_x + y\partial_y$, in the second with $v = t\partial_t$
 
I don't know what a scalable metric is
 
2:22 PM
the result is a proper function on the manifold, so of course it's "coordinate-invariannt"
 
One for which there exists a transformation such that $g_{\mu\nu}\rightarrow \lambda g_{\mu\nu}$?
 
what remains entirely unclear is why you're doing this or what this has to do with the scale transformation
 
@Charlie yes
 
My next question was also going to be why you even want to do this too :P
it doesn't seem especially useful
 
@Charlie I sometimes dabble in problems that don't seem to have any particular use :P
@ACuriousMind You can try the algorithm for another metric
Or different coordinate transformation?
 
2:25 PM
you haven't actually described an algorithm
you've just given two examples
I have no idea how this is supposed to work just after reading your post
 
Oh
Anything which participates in the scaling replace its differential with $dx_1 \to 0$
 
what scaling?
 
those which do not participate replace with $0$
scaling of the metric from $ds^2 \to \lambda ds^2$
 
there are potentially many coordinate transformations that might scale the metric, how do you know which one to pick?
 
Can you give an example of such a metric? Where there is more than one transformation
?
 
2:29 PM
Transformations that rescale the metric are conformal transformations
 
@MoreAnonymous take $\mathrm{d}s^2 = \mathrm{d}x\mathrm{d}y$. You can do $x\mapsto \lambda_1x$ and $y\mapsto \lambda_2 y$ for any $\lambda_1,\lambda_2$ as long as $\lambda_1\lambda_2 = \lambda$.
in particular you can choose one of the $\lambda_i = 1$ making it "not participate" in your unconventional language
 
@ACuriousMind Fair ... In that case you must do all that can participate
 
why?
 
@ACuriousMind I just noticed the conjecture I dont know why its working
If you can spot a counter example let me know?
 
I don't know what "its working" means in this context
 
2:37 PM
@ACuriousMind The procedure yields a polynomial irrespective of what coordinate transformation of it you put into it
 
any prescription of replacing the $\mathrm{d}x^i$ by variables will produce a polynomial
 
@ACuriousMind Yes but $r^2 \theta^2 + r^2 \neq x^2 + y^2\$
It depends on the coordinate system
 
in the case where there is only a single transformation that does this, "the vector field that generates the scale transformation" is a coordinate-independent identification of a vector field
if the transformation is such that the coordinates themselves are scaled by a factor, then the field that generates this transformation is $x^i\partial_{x^i}$ (for the coordinates that participate)
 
ACM's visible confusion at half the things said in this chat is great entertainment :P
5
 
now, $\mathrm{d}x^i(x^k\partial_{x^k})\mathrm{d}x^j(x^k\partial_{x^k}) = x^k\delta^{ik} x^k\delta^{jk}$, so taking $g(v,v,)$ for this vector field gives your polynomial, you're doing exactly what I said at the start - it's just the inner product of a particular vector field
@Charlie lol
 
2:48 PM
@Charlie lol
 
note that this will therefore stop working when you choose coordinates where the scale transformation is not a simple scaling of the coordinates (choose some stuff like $x' = \mathrm{sin}(x), y' = \mathrm{sin}(y)$ on a region where this is bijective and it will fail in those coordinates)
 
@ACuriousMind Can you give an example?
Like of a metric where it fails after coordinate transforming?
 
3:05 PM
@MoreAnonymous I just gave one - we have $\mathrm{d}x = \mathrm{d}\mathrm{arcsin}(x') = \frac{1}{\sqrt{1 - x'^2}}\mathrm{d}x'$ and likewise for $y,y'$ so $\mathrm{d}s^2 = \frac{1}{1 - x'^2}\mathrm{d}x' + \frac{1}{1 - y'^2}\mathrm{d}y'$. The scale transformation is now something ugly that sends $x' = \mathrm{sin}(x)$ to $\mathrm{sin}(\sqrt{\lambda}x)$ and likewise for $y'$
alas, $\mathrm{arcsin}(x')^2 + \mathrm{arcsin}(y')^2$ is definitely not the same as what you get by replacing $\mathrm{d}x'$ by $x'$ and likewise for $y'$ in the metric.
(we're starting from $\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2$ in case that wasn't clear)
 
3:19 PM
In Jackson's "Classical Electrodynamics" (and in other physics texts), it says that we can assume that when solving a PDE, we can assume that the solution can be expanded as $\phi(x,y,z)=X(x)Y(y)Z(z)$ when solving Laplace's equation. I don't understand why this is true. Are there any solutions to Laplace's equation which do not have this property?
Any references on proving why this is true? Furthermore, is it true that, in general, such a mapping from the n-dimensional smooth functions to a set of 1D functions is isomorphic? Is it perhaps only the case if we consider solutions to elliptic PDEs?
 
In mathematics, separation of variables (also known as the Fourier method) is any of several methods for solving ordinary and partial differential equations, in which algebra allows one to rewrite an equation so that each of two variables occurs on a different side of the equation. == Ordinary differential equations (ODE) == Suppose a differential equation can be written in the form d d x f ( x ) = g ( x...
 
1. When you write $x' = sin x$ they don't have same domain so the equivalence of the metrics will only be for a certain coordinate patch (I think)

2. When you do the coordinate transformation $x' \to \lambda x'$ will have a different range than the original. For example if I limit my metric to $ds^2 = dx^2 + dy^2 $ to $(-1,1)$ and transform the metric to $\lamda x$ and $\lambda y$ then my range has contracted to $(-1/lambda,1/lambda)$ in this case the metric has not rescaled since the domain is different
 
but that's how coordinate transformations work
if you say you only want coordinates that cover all of $\mathbb{R}^n$, then you're not really doing stuff in arbitrary coordinates
if you don't like the $\mathrm{sin}$ because you can only do it in a small domain, take the $\mathrm{tan}$
that's a bijection between $\mathbb{R}$ and $(-\pi/2,\pi/2)$
 
and is there a bijection between $C^\infty(\mathbb{R}^d)$ and $$C^\infty(\mathbb{R})^{\times n}$$?
It feels like by doing the separation of variables, we may or may not be restricting to the set of solutions that can be expanded this way
 
3:36 PM
Separation of variables is not always valid
 
I'm wondering if there are some resources on why it isn't always valid and on what lets us safely do this separation in physical contexts
or rather, since we're solving the PDE for the set of functions where this expansion is possible, why do we usually ignore in physics those solutions where this expansion is not possible
 
@JohnK there is a bijection between $L^2(\mathbb{R}^n)$ and $L^2(\mathbb{R})^{\otimes n}$
 
"The method of separation of variables is used when the partial differential equation and the boundary conditions are linear and homogeneous"
 
i.e. you can write any square-integrable function in $n$ variables as the sum of products of functions of the individual variables
so it's always valid to try to write your function as such a product, it's just not always a useful ansatz
 
3:40 PM
Oh right so putting 1 and 1 together, can we say that for all linear and homogeneous boundary conditions, the solution will be square integrable?
and since we require physical functions to be square integrable, that means we have to impose linear and homogeneous boundary conditions?
 
Wavefunctions needs to be square integrable, I don't know about all situations having that constraint
 
I imagine it's the same for the electrostatic potential, since the energy density has an $|E^2|$ expression in it. Is this true?
 
4:09 PM
If you're talking for instance about an infinite charged rod I wouldn't expect $\phi$ to be square integrable on $\Bbb R^3$ for instance
 
123
4:53 PM
Hi All...
 
5:11 PM
From Wikipedia "In magnets, the original rotational symmetry (present in the absence of an external magnetic field) is spontaneously broken such that the magnetization points into a specific direction. The Goldstone bosons then are the magnons, i.e., spin waves in which the local magnetization direction oscillates."
I don't understand this statement. A crystal does not have full rotational symmetry (continuous symmetry), only discrete symmetries.
 
@B.Brekke the symmetries of the crystal are not necessarily the symmetries of the theory
Case in point: The Heisenberg model that describes the theory with magnons has full "rotational" symmetry if you just rotate the $S_i$ and leave the positions alone
 
@ACuriousMind Are you sure you can leave the positions alone? All rotations must leave one point invariant, but you have two points in the Heisenberg model. Both can't be invariant
 
I'm just saying there's an $\mathrm{SO}(3)$ symmetry there where you just rotate the $S_i$ and do nothing on the positions
it's not really relevant whether this is a "real" rotation, but this is certainly a transformation we can do and it is the symmetry that is broken by the ground states of the model
 
Hmm, so this $\mathrm{SO}(3)$ symmetry is not necessarily the obvious spatial rotation.
 
well, the model doesn't really know that it's describing something spatial - the grid positions are just labeled by indices
perhaps a less ambiguous phrasing would be that the symmetry that's broken is the symmetry that rotates the spins
 
5:26 PM
Okay, this is very helpful. At one point I thought that I had to wave goodbye to Goldstone modes for all solid state systems
 
 
1 hour later…
6:53 PM
$ds^2 = \frac{1}{ax^4+by^4} dx^2 + \frac{1}{cx^4+dy^4}dy^2$ does not result in a 'polynomial' by this 'algorithm'
 
7:05 PM
@bolbteppa Sorry I should have just used the word function
 
What do you mean 'invariant', $x^2+y^2$ looks different under a coordinate transformation, it's $r^2$ in polar coordinates
 
@bolbteppa So the goal was to map the metric to a function via some procedure. Such that if I map $ds^2(x,y) \to f(x,y)$ and $ds^2(r,\theta) \to g(r,\theta)$ then f(x,y) = g(r,\theta)
 
@bolbteppa what they were observing was that their "algorithm" in some cases was compatible with coordinate transformation, i.e. doing the substituitons first and then coordinate-transforming yields the same result as the other way around
i.e. the function it produced looked to be a proper function on the manifold, in other words
but as I said, it works that way only for special coordinates in which the scale transformation is linear
and in the end there's nothing special about the function produced, it's just the norm of the vector field generating the scale transformation
 
@ACuriousMind Why am I referred to as they? :p
 
@MoreAnonymous what would you like to be referred as?
I default to 'they' unless gender is obvious
 
7:20 PM
@ACuriousMind Oh that's pretty woke
I prefer "SoGeniusHeRemainedAnonymous"
But you can also refer to me as "MoreAnonymous" :P
I was watching an interesting video
P.S I am politically neither left nor right
But politically ignorant
 
@MoreAnonymous well I'm not gonna type that every time but the 'he' in there at least answers the question :P
 
@ACuriousMind Whoa! thats not gender neutral! Whats your thoughts on the new gender theories?
 
By "new gender theories" are you just referring to the existence people who identify as non-binary?
 
@Charlie Yea
 
@MoreAnonymous Huh? I'm just saying since you referred to yourself as 'he' there, I guess that's how you want to be referred to
 
7:26 PM
@ACuriousMind I was joking ... I am a male and prefer to be called "he"
 
I think my personal thoughts on larger philosophical issues with gender are rather irrelevant to the basic politeness of not talking to people in a way that upsets them needlessly
 
@ACuriousMind I think theres something in western culture that pisses them off ... It was only after india was colonized that we started having a problem with them.
 
this is the internet, you could be a dog, an AI or an alien for all I know and care, so just as I'll call you by your display name, I'll call you by the pronouns you prefer
 
Who's "them"?
 
@Charlie transgenders for example
 
7:29 PM
Actually you used "them" twice, which is referring to transgender people?
Unless it's both
 
Its referring to them twice
*transgender people twice
 
I guess I don't see how the colonization of India is related to issues surrounding transgenderism :P
 
@Charlie Our cultural identity was destoryed
 
Are you Indian?
 
7:32 PM
I still don't see how the two are related though lol
 
We inherited western thought
And our own culture was destroyed
In fact a law was passed by the british
to make homosexuality illegal
 
@Charlie I think the thesis here is that Indian culture was much more tolerant with non-binary gender conceptions and non-heterosexuality before colonization?
 
@ACuriousMind Yup
Man my communication skills are terrible
 
Oh right sorry of course I keep forgetting this is through the lens of someone who is from india
 
As I don't know anything about Indian culture before colonization, I have nothing to contribute to that discussion
 
7:36 PM
So you're talking specifically about transgenderism in india and I was trying to see how the colonization of india influenced the issue in my country lol
 
@Charlie lol
 
$x^2+y^2$ is not a coordinate invariant quantity
 
@bolbteppa Yes but the algorithm gives you $r^2$ in polar coordinates
So it seems to give you something which independent of the coordinate system you put into it
 
What is the 'second' version of the line element for the second example in your post
 
I don't know much about the colonization of india either, but since it's not something we in the UK were proudly taught in history class it presumably wasn't great
 
7:38 PM
@bolbteppa in 2d it is - it's the norm of the vector field $x\partial_x + y\partial_y$, which is a scalar function
 
$x^2+y^2$ is $uv$ in coordinates where $u = x + iy, v = x - iy$
 
@Charlie I want my Kohinoor back
 
I think that's the name of a diamond right, I only know that from Dr Who
 
hmm
 
@Charlie Yup
I find it a joke how its displayed in Buckingham Palace
 
7:40 PM
@bolbteppa he really means "the procedure gives the same scalar function regardless of order of substitution/coordinate transformation", let's not get into another confusing discussion about the meaning of "invariance" ;P
 
And this is how we looted country x
and this is what we looted from y
and we proudly display ourselves as woke moral citizens
 
honestly, many Western historical museums are like that :P
 
Right but $x^2+y^2$ is different from $r^2$, this really doesn't make much sense as it stands
 
Unfortunately because it's not actively taught in schools (unless that's changed in the last 5 years since I was at school) the reason we have those things is probably not largely remembered
 
@bolbteppa it's the same function - the 2d function $x^2 + y^2$ in $(r,\theta)$ coordinates is $r^2$
 
7:43 PM
@Charlie I wont be surprised that the Tories didn't want to include many cruelties of your civilization
 
Nationalism is a hell of a drug
 
fraid not
 
If I knew how the world was gonna go downhill I might have tried to be something more charitable like a welfare economist, a well educated social activist
2
,etc
 
 
2 hours later…
9:30 PM
@JohnRennie Thank you :)
 

« first day (3873 days earlier)      last day (55 days later) »