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2:10 AM
@Ten I suspect part of the cause is that particle physics got a hear-start on almost everyone else in using the internet—even in pre-web days, then invented the web, and kinda stayed on the bleading edge.
So we've had Spire (now known as InSpire) with it's comprehensive job listing for particle physics since the mid 1990s.
But having one of the most internet active (and large in terms of raw numbers) sub-disciplines set up their own private arranegment makes it that much harder to get a broad initiative off the ground.
I got all three of my post-docs by searching [In]Spitre listings. I also got all my (many; oh so many!) faculty rejections from there.
@SirCumference We'd have to involve the community mods for that. The power of mere site mods only extends to 365 days. If you see someone with a longer suspension then they either (a) gave the team reason to believe they were younger than 13 by more than a year or (b) really annoyed people in high places.
2:35 AM
@dmckee Welp, in JD's case he's been banned on 5 or something SE sites in the last year
And this happens every year
How liberal is your use of "really annoying" here
@SirCumference My observation is that most people who've gotten mega bans have made work for the the team. Either by making a lot of fuss on the mother meta or by engaging in cross-site games.
Perhaps someone who is getting quietly stomped by a lot of site mods isn't making work for the CMs?
I'm just saying that they'd probably take action if the issue is mentioned to them. I doubt they'd find it acceptable after hearing the problem
If no one says anything though we'll remain in the status quo
@ACuriousMind LaTeX 3 is the Duke Nukem 3 of the serious software world. We hear rave reviews every few years and nothing every happens.
@SirCumference The CMs are usually aware of major events on the sites. Or at least some CM is usually aware of the major events on physics. I assume they divide up the network somehow or another.
I don't suppose that has to mean that anyone in particular is necessarily aware of the extent of any particular case when it cross many sites.
2 hours later…
4:40 AM
@PM2Ring I've attempted a proper answer to this.
1 hour later…
5:44 AM
2 hours later…
7:29 AM
@Slereah got some time?
I'm struggling to understand partial derivatives
morgen Jungen <3
in the first example, the constants simply dissapear, but in the second one it's even more confusing/
I understand that the "with respect" value we simply treat as what we usually treat but all other variables we treat as constants.
In the second example, (where it's with respect to x), the y and z, simply stay there, but in the first one where it's with respect to x again, the y dissapears.
I think I should practise normal derivatives a bit more first :c
Consider the function $g(x,y,z) = 3yz$
Then take $g(x + dx, y, z)$
This is still $g(x + dx, y, z) = 3yz$, since there's no dependence on $x$.
Therefore, $dg = g(x + dx, y, z) - g(x,y,z) = 0$
So $dg/dx = 0$
$g(x+dx,y,z)$ what's that?
mathsisfun.com/calculus/derivatives-rules.html this says that constants are turned into 0, so what's the problem?
I mean, in the video, all "not with respect to" variables should be just turned into 0
7:45 AM
do you remember what the definition of the derivative is
$$\frac{\partial f(x,y,z)}{\partial x} = \lim_{h \to 0} \frac{f(x + h, y, z) - f(x,y,z)}{h}$$
What still confuses you
https://www.mathsisfun.com/calculus/derivatives-rules.html this says that constants are turned into 0, so what's the problem?
I mean, in the video, all "not with respect to" variables should be just turned into 0
7:49 AM
What problem are you refering to
Seems like they are indeed "turned into zero" in the video
not in the second example f(x, y, z)
in the fx part of the second example, y and z stay there
even though they should be treated as constants and turned into 0
When constants are multiplied by a variable, the constants stay untouched?
Do you mean $f(x,y,z) = x y^2 z^3 + 3 y z$, $$\frac{\partial f}{\partial x} = y^2 z^3 + 0$$
This is entirely correct
y and z are supposed to be treated as constants and therefore turned into 0
The first term isn't a constant, it's linear in $x$
7:53 AM
Therefore $y^2 z^3$ is a factor of $x$
On the other hand, it is now a constant of $f_{,x}$
So that the second derivative will be zero
(ax)' = a?
is this what's happening?
got it finally
You don't even need to think about it, really, you could just apply the definition
7:57 AM
f(x + dx, y, z) - f(x,y,z) &=& (x + dx) y^2 z^3 - x y^2 z^3\\
&=& dx y^2 z^3
Therefore $$f_{,x} = y^2 z^3$$
Got it. That's the (ax)' = a rule but you showed me how to get to it from the definition?
For the linear case it's fairly easy to show
@Slereah Thanks!
Just use the distributive property
hey balarka
Hi @Slereah
8:03 AM
you're on the NASA website
use latex you swine
<table width="730" cellspacing="0" cellpadding="10" border="0" bgcolor="#FFFFFF" align="center">
Oh I see the problem
It's in HTML 2
This page was probably written 20 years ago
8:21 AM
@Slereah Thanks again! I did some partial derivatives exercises and everything works. I simply had to review the derivative rules a bit better :D
Are partial derivatives multivariable calculus :P?
The product rule applies to logartihms but not trigonometry functions?
why would it
Trig functions have their own identities to deal with such things
last example in the video
he uses product rule and chain rule on a ln
Wait what do you mean by "the product rule"
if you mean $(ab)' = a'b + b'a$ then yes, it applies to all functions
8:33 AM
He uses the product rule between the r and the ln in the last example
I ain't watching a 7 minutes video of basic calculus at work
Ok, don't watch it. I'm just telling you he is using the product rule on $r$ (variable, the "with respect to" one) and $ln$.
@JohnRennie Thanks. It looks good to me, although I expect that safesphere & JD will reject it as non-physical.
8:52 AM
What do the partial derivatives tell me anyways? For example I have $f(x, y) = x^2 + y^2$ with $\frac{\partial f}{\partial x} = 2x$ and $\frac{\partial f}{\partial y} = 2y$ and so? I mean, I've graphed the 3D structure of the function and how am I supposed to find the slope (as what partial derivatives are for) of a point such as $x = 2$ and $y = 3$?
The partial derivatives tell you the slope of the tangent plane at that point
Ok, the function gives me the z coordinate when calculated right?
I thought the slope is supposed to be a single number? How am I supposed to get a single number from the partial derivatives of (with respect to) x and y?
I mean, for $x = 2$ and $y = 3$, the slope at that point is $(4, 6)$?
The first number $4$ shows how steep it is in the x axis and the $6$ shows how steep it is in the y axis?
Am I correct?
And the plain function $f(x, y)$ gives you the z coordinate, right?
9:44 AM
@NovaliumCompany suppose you're standing on a hill. You can walk downhill in which case the slope is negative. Of you can walk uphill in which case the slope is positive. Or you can walk along a contour line in which case the slope is zero.
So the gradient of a hill is not simply a number because it depends on the direction.
In fact the gradient is an object called a one form. When you combine a one form with a vector it gives you a number i.e. a scalar, and in this case that vector is the direction.
If you have a height function $z(x,y)$ then the gradient one form is given by $(\partial z/\partial x, \partial z/\partial y)$. Note that it has two components.
10:15 AM
For constraints to work, the second derivatives of the Lagrangian have to be invertible on the submanifold of the configuration space, is that it?
10:28 AM
@JohnRennie Got it. So my explanation is somewhat correct? That $\partial z/\partial x$ shows the "rotation" of the tangent line in the x axis and $\partial z/\partial y$ in the y axis?
@NovaliumCompany I wouldn't say rotation.
$\partial z/\partial x$ gives the gradient of the slope in the $xz$ plane.
Should I google what is a gradient?
Wait, is gradient just another fancy way to say slope?
@NovaliumCompany do you need to Google it. Isn't it pretty simple?
Maybe it is simple, but I don't know what it is to tell you so
The word slope doesn't have a precise meaning in physics.
10:36 AM
The word gradient does have a precise meaning because it is the object returned by the grad operator.
In vector calculus, the gradient is a multi-variable generalization of the derivative. Whereas the ordinary derivative of a function of a single variable is a scalar-valued function, the gradient of a function of several variables is a vector-valued function. Specifically, the gradient of a differentiable function f {\displaystyle f} of several variables, at a point P {\displaystyle P} , is the vector whose components are the partial derivatives of f {\displaystyle f}...
:O ... :c
More things for me to get confused by
I know what gradient is in css
background: linear-gradient(..., ...);
So in the simplest way you can phrase it, when I get $(\partial z/\partial x, \partial z/\partial y)$ for the rate of change of the 3D graph, what exactly does it mean? What exactly does it describe?
I learned about slopes in single variable derivatives that they describe the rate of change, the rise over run... and now you are telling that I can't convert that into 3 dimensions? :P
Suppose you start at some point $(x,y)$ and move a small distance $dx$ in the $\hat x$ direction. Then $\partial z/\partial x \times dx$ gives you the change in $z$.
Likewise, if you you move a small direction $dy$ in the $\hat y$ direction then $\partial z/\partial y \times dy$ gives you the change in $z$.
10:53 AM
It's a bit confusing how y in the 2D graph plays the up/down role, and in the 3D graph, the z plays the up/down role :P
@JohnRennie Confused me even more :ccc
@NovaliumCompany , whats the problem? what are you trying to do?
11:08 AM
@Gyromagnetic In the function $z(x, y) = x^2 + y^2$ understand what $(\partial z/\partial x, \partial z/\partial y)$ mean. I thought they represented the slope's rotation or something, but no
@NovaliumCompany mhmm, learning about them partial derivatives today?
@Gyromagnetic yup
@NovaliumCompany for funsies?
@Gyromagnetic wot?
@NovaliumCompany , just for fun?
11:13 AM
@Gyromagnetic yup, curiosity
@NovaliumCompany I recommend just doing a bunch of exercises and eventually you'll get it.
@Gyromagnetic Yeah. I'll watch a few khan academy vids and I believe things will click.
@JohnRennie Nvm, just watched this khanacademy.org/math/multivariable-calculus/… Everything clicked.
I'll just watch this whole thing khanacademy.org/math/multivariable-calculus
@NovaliumCompany Cool :-)
11:30 AM
@JohnRennie hi m8, do you miss the good old days as a researcher?
and whats your favorite colloid science effect ;)?
@Gyromagnetic I was only in academic research for the three years of my PhD. And by the end of that I'd had enough of the topic I was working on (photodissolution).
Then I worked as an industrial scientist, which is completely different to academic life.
And after twelve years of that I'd had enough again :-)
I'm actually quite content as a semi-retired computer nerd and I don't miss working in research.
Help I am stuck in hell
i wish for a rigorous demonstration of constrained surfaces of phase space but either they don't use the submanifold or they don't have examples
bloody mathematicians
too important to use actual examples
11:47 AM
@JohnRennie you never tried teaching?
or any form of tutoring
@skullpetrol I wouldn't be a good teacher. Teachers have to be able to deal with pupils who aren't motivated, and I wouldn't have the patience or that. The great thing about answering questions on this site is that the people here are all highly motivated and interested in physics.
yeah, you can lead a horse to water...
@Slereah what do you mean about a rigorous demonstration of constrained surfaces
12:01 PM
@bolbteppa I am a bit unclear about first class constraints and would like a proper demonstration for a specific example
So far the idea I get from all of this is
1) Find some submanifold of the phase space with dimension equal to the rank of $L_{,\dot{x}\dot{x}}$
2) Define the Hamiltonian upon it
Is that roughly the idea
Submanifold such that this matrix is invertible upon it, 'course
I get the overall feeling that a lot of sources on this topic are a bit vague of when the Hamiltonian is defined and such because they tend to mix $q$, $\dot q$ and $p$, even though $p(\dot{q})$ isn't invertible in general
Like you can't define the Hamiltonian before you have the constraint surface, no?
Also is there a general process to build the constraint surface?
I know they are not unique
Right, so if the Hamiltonian is zero then there are no dynamics at all
12:19 PM
I guess there's some Thing that's like
A map between the configuration space and the phase space
and we need the derivatives of the Lagrangian to be invertible on one to define the Hamiltonian on the other?
I'm not sure
I s'ppose the map would be the Momenta
f : C &{}\to& P\\
(q, \dot{q}) &{}\mapsto& (q, p(\dot{q}))
Something like that
configuration space is the manifold $M$ and the phase space is $T^* M$
The map you want is the zero section $M \to T^* M$
I thought the configuration space was the tangent bundle
and the phase space the cotangent bundle
there's a map $TM \to T^*M$ as well
I have seen it said
just what the symplectic form does
its a nondeg 2-form
12:27 PM
So do I look for a submanifold of the configuration space on which $\partial^2 L / \partial \dot{x}^a \partial \dot{x}^b$ is invertible everywhere, map this to the phase space, and define the Hamiltonian there?
using the fancy $f^{-1}$ which is defined on this restriction
does that sound like a slice of fried gold
That's the point
Ah yes
It didn't come across clearly in the text I've looked into
Is there a constructive way of choosing the submanifolds, or do you just do it on gut feeling?
I guess in the case of the Dirac parametrized non-relativistic point particle it's just gonna be every hypersurface defined by $t(\tau) = f(\tau)$, for every diffeomorphism $f$
Which I think basically means that the submanifold is always $\mathbb{R}^3$
Which makes sense since that's the space for the unconstrained version
12:43 PM
Basically, given a Lagrangian formulation of a theory, by expanding the EL equations
0 &= \frac{d}{dt} \frac{\partial L}{\partial \dot{q}^a}(t,q^a,\dot{q}^a) - \frac{\partial L}{\partial q^a} \\
&= \frac{\partial}{ \partial t} \frac{\partial L}{\partial \dot{q}^a} + \frac{\partial^2 L}{\partial q^b \partial \dot{q}^a} \dot{q}^b + \frac{\partial^2 L}{\partial \dot{q}^b \partial \dot{q}^a} \ddot{q}^b - \frac{\partial L}{\partial q^a} \\
&= \left( \frac{\partial^2 L}{\partial \dot{q}^b \partial \dot{q}^a} \right) \ddot{q}^b + \left( \frac{\partial^2 L}{\partial q^b \partial \dot{
yeah that much I got
A system is singular when $\det M_{ba} = 0$, for example
L &= \frac{1}{2}(\dot{x}^2 + \dot{y}^2), \ \ \ \to \ \ M_{ba} = \begin{bmatrix}
1 & 0 \\ 0 & 1 \end{bmatrix} \ \ \to \ \ \det(M_{ba}) = 1 \neq 0, \\
L &= \frac{1}{2}(\dot{x} - y)^2 \ \ \ \ \ \to \ \ M_{ab} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \ \ \to \ \ \det(M_{ba}) = 0.
it is once you go to the Hamiltonian that things started to be a bit unclear
The usual physicist problem of not expressing clearly what variables we're dealing with
They just mix $p$'s and $\dot{q}$'s like nobody's business
If we define $p_a = \frac{\partial L}{\partial \dot{q}^a}$ then from $M_{ba} = \frac{\partial^2 L}{\partial \dot{q}^b \partial \dot{q}^a} = \frac{\partial}{\partial \dot{q}^b} \frac{\partial L}{\partial \dot{q}^a} = \frac{\partial}{\partial \dot{q}^b} p_a$ we see that if we treated $p_a = p_a(q^b,\dot{q}^b)$ then in it's Taylor expansion we see $M_{ba}$ is the Jacobian of the transformation and so you can't use the implicit/inverse function theorem to invert this in terms of $q^b,\dot{q}^b$
In other words, there will exist momenta which cannot be expressed in terms of the velocities, and so some of the momenta will at most be functions of the coordinates and other momenta alone - these are called primary constraints, expressions relating the coordinates and momenta that follow directly from the Lagrangian. For example $p^2 + m^2 = 0$ for the relativistic point particle.
Before even going to Hamiltonians, ask what happens to constrained Lagrangian systems
Consider the Lagrangian
L = \frac{1}{2}(\dot{x}^2 + \dot{y}^2).
We know the equations of motion giving the time evolution of the $x$ and $y$ coordinate space variables are
\ddot{x} &= 0, \\
\ddot{y} &= 0,
and we can solve these explicitly as
x &= x_0 + v_0^x t, \\
y &= y_0 + v_0^y t.
Thus, the equations of motion coupled with boundary conditions which fix the integration constants completely determine the motion of the system. As you see above $\det M_{ba} \neq 0$.
We now re-analyse this system from a Hamiltonian perspective. The canonical momenta are
p_x &= \frac{\partial L}{\partial \dot{x}} = \dot{x}, \\
p_y &= \frac{\partial L}{\partial \dot{y}} = \dot{y},
and the canonical Hamiltonian is
H_c &= p_x \dot{x} + p_y \dot{y} - L \\
&= p_x p_x + p_y p_y - \frac{1}{2}(\dot{x}^2 + \dot{y}^2) \\
&= p_x^2 + p_y^2 - \frac{1}{2}(p_x^2 + p_y^2) \\
&= \frac{1}{2}(p_x^2 + p_y^2).
The possibility of constructing $H$ clearly relies on being able to solve for $\dot{x}$ and $\dot{y}$ as functions of $p_x$ and $p_y$.
We can now compute the time evolution from the Poisson brackets, which satisfy $\{x,p_x \} = 1$,
\dot{x} &= \{x,H_c \} = \{x,\frac{1}{2}(p_x^2 + p_y^2) \} = p_x, \\
\dot{y} &= \{y,H_c \} = \{y,\frac{1}{2}(p_x^2 + p_y^2) \} = p_y, \\
\dot{p}_x &= \{p_x,H_c \} = 0 \ \ \ \leftrightarrow \ \ \ \ddot{x} = 0 \\
\dot{p}_y &= \{p_y,H_c \} = 0 \ \ \ \leftrightarrow \ \ \ \ddot{y} = 0.
This shows that if we can move freely between the Lagrangian and Hamiltonian descriptions, we can reproduce the equations of motion in either formalism. An important lesson from this is that t
12:56 PM
Consider the Lagrangian
L = \frac{1}{2}(\dot{x} - y)^2
The equations of motion are
\ddot{x} - \dot{y} &= 0, \\
\dot{x} - y &= 0.
This system has two degrees of freedom, $x$ and $y$, and two equations of motion, however the two equations of motion are not independent - the first equation is simply the time derivative of the second, in other words it is a consequence of the second equation of motion. But the second equation has no acceleration, this it is not an equation of motion, and since the equation with acceleration is simply a consequ
If we consider the solution $x(t)$ as a Taylor expansion
x(t) = \alpha + \beta t + \frac{\gamma}{2} t^2 + \phi(t)
then the constraint $\dot{x} = y$ implies
y(y) = \beta + \gamma t + \dot{\phi}(t).
The boundary conditions on this system are
\phi(0) &= \dot{\phi}(0) = \ddot{\phi}(0) = 0, \\
x(0) &= \alpha, \dot{x}(0) = \beta, \\
y(0) &= \beta, \dot{y}(0) = \gamma.
The function $\phi(t)$ is otherwise completely arbitrary, a particular choice of boundary conditions can fix $\alpha, \beta, \gamma$ but it can't fix $\ph
Now let us consider the Hamiltonian formulation of this constrained system. The canonical momenta are
p_x &= \frac{\partial L}{\partial \dot{x}} = \dot{x} - y, \\
p_y &= \frac{\partial L}{\partial \dot{y}} = 0.
Here we see that $\dot{x}$ can be inverted (in this case for $p_x$), however $\dot{y}$ cannot be inverted (not for $p_y$, nor in general for $p_x$). Another problem is that this system is already incompatible with the basic Poisson algebra for dynamical systems, which should satisfy $\{y,p_y \} = 1$ however here we see $\{y,p_y \} = 0$, again indicating that
How much of this did you have around
These are basically my notes from those videos
I did see the video
I did find the whole \begin{align} x(t) = \alpha + \beta t + \frac{\gamma}{2} t^2 + \phi(t) \end{align}
A bit weird
I mean you could just write $x(t) = \phi(t)$
From the Hamiltonian perspective, the problem with the $L = \frac{1}{2}(\dot{x}-y)^2$ example is clearly the existence of relations between the coordinates and momenta which do not also involve the velocities, in this case we found the relation $p_y = 0$ which does not involve a velocity and so causes problems, whereas $p_x = \dot{x} - y$ is perfectly fine.
If $\phi$ is completely arbitrary, it could also include that polynomial term, no?
I mean I can see why it's there, semantically
but it's a bit weird
1:01 PM
It's basically just that this Lagrangian ends up producing the EOM $\ddot{x} = \ddot{x}$ and the BC's amount to just specifying the first and second derivative, but the higher order Taylor expansion on $x$ has no conditions
Ah yes
I see
i.e. it's a convoluted way of specifying the first and second derivative and nothing more
Who knew solving free particles would be so hard :p
1:34 PM
I'm gonna have to do GHOST FIELDS for non-relativistic classical particles
who you gonna call?
$L = p_{\mu} \dot{x}^{\mu} - \frac{1}{2}(p^2 + m^2) - i c \partial_{\tau} b$ is the Ghost Lagrangian for a relativistic point particle, is it's non-rel reduction the non-rel ghost Lagrangian :p
@SirCumference That's an old ass one
@bolbteppa The non-relativistic one is slightly fancier I think
Because the Lagrangian is $$L = \frac{\dot{x}^2}{\dot{t}}$$
The inverse function might play tricks on those ghosties
The theorem is variously known as the:

Kronecker–Capelli theorem in Poland, Romania and Russia;
Rouché–Capelli theorem in Italy;
Rouché–Fontené theorem in France;
Rouché–Frobenius theorem in Spain and many countries in Latin America;
Frobenius theorem in the Czech Republic and in Slovakia.
I don't even get why
Kronecker is German
and so's Frobinius
at least name the theorem to pretend you invented it
1:53 PM
Gauss-Ostrogradskii etc
Holy moly @bolbteppa you surely are not lazy to write all of these mathematical essays
@bolbteppa Do you mean the Newton-Leibniz-Gauss-Ostrogradski-Green-Stokes-Poincare theorem
I think you forgot Euler and Fontaine math.toronto.edu/mgualt/wiki/samelson_forms_history.pdf
It's the name given to it in Arnold
2:10 PM
What is the weak equality supposed to be, rigorously, anyway
Equal on the constraint surface?
If so how do you define the Hamiltonian not on the constraint surface, since the inversion of the momentum isn't defined?
Do you define it by extension?
It's an equality that holds after you apply the Poisson brackets, e.g. $\{y,p_y \} = 1$ should hold however if $p_y = 0$ then $\{y,p_y \} = 0$ clearly holds contradicting $\{y,p_y \} = 1$ so that you only assume $p_y = 0$ after applying PB's e.g. $\{y,p_y^2\} = 2 \{y,p_y \} p_y = 2 p_y$ should hold after which you can then invoke $p_y = 0$. Any functions which are weakly equal differ at most by a linear combination of constraints
Is there any one-dimensional example of a gauge Lagrangian
would be nice to be able to actually draw the phase space
I guess not any good one because that doesn't leave a lot of constraints to do
Just a zero dimensional path
Actually know what might be a decent example to do?
Newtonian mechanics with electric fields
2:39 PM
What’s a gauge lagrangian @Slereah
A lagrangian such that the action of which is invariant under a gauge transformation
The relativistic point particle action $S = - m\int \sqrt{-\dot{x}^2} ds$ under coordinate transformations?
Well if it's relativistic it's gonna be at least two degrees of freedom
So phase space in 4D
hard to draw
Oh right
What is supposed to be one dimensional here
2:43 PM
To have only one degree of freedom
So that the action is $S[x(t), \dot{x}(t)]$
But I don't think there's any interesting case for that
I guess $S = \int \dot{x}$ would be reparametrization invariant, but it is also trivial
Henneaux actually has some graphical examples
I should work that out
Can't wait for the holidays rly
Two weeks 'til I get some time off
Then I can rest up, gather my sciences, take some cocaine and work all that out
Some Garry gum
3:11 PM
Looking back at the constrained system $L = \frac{1}{2}(\dot{x}-y)^2$ with it's canonical Hamiltonian $H = \frac{1}{2} p_x^2 + p_x y$ which leads to inconsistencies, we note the inconsistencies lie in the fact that $\{y,p_y\} = 0$ due to $p_y = 0$ when we should have $\{y,p_y \} = 1$. To make this system consistent we use Lagrange multipliers to form the constrained Hamiltonian
$$\tilde{H} = \frac{1}{2} p_x^2 + p_x y + \lambda p_y $$
and then impose the constraints 'weakly', i.e. $p_y = 0$ is to hold after applying the PB's, and even more the constraint $p_y = 0$ should not change with tim
It may be for some given Lagrangian that the primary constraint $\phi_a$ is not consistent i.e. $\dot{\phi}_a = \{\phi_a,\tilde{H} \} $ is not $\approx 0$ i.e. not weakly zero, the result it gives is then called a 'secondary constraint' which should then be added to the original Hamiltonian as a constraint with Lagrange multipliers and the process repeated until you find a Hamiltonian such that the time derivatives of the constraints are weakly zero
This process is how you find the correct 'manifold'
@NovaliumCompany Again, I advise staying away from multivariable calc for the time being :P
In EM it turns out Gauss Law is a secondary constraint (ch.2)
I'll give it a looksy
@SirCumference who is the time being
is it him
@Slereah what did you even search to find this lol
In trying to satisfy $\dot{\phi}_a = \{\phi_a,\tilde{H} \} = \{ \phi_a,H_c \} + \lambda^b \{\phi_a,\phi_b\} \approx 0$ you're led to different cases and the notion of 'first and second class constraint's'
3:24 PM
@SirCumference That is the Clock King
a Batman villain
you may know him better under this form
No amount of tangent bundles are going to make what's going on here make more sense :p
If you're not too young
I'm 21...
@SirCumference Might be too young then
@bolbteppa I do wonder if the whole Hamiltonian-as-a-bundle thing does constraints
It probably uses principal bundles
the horror
Well here we're defining manifolds via systems of equations so yeah it would for sure
3:27 PM
@SirCumference lol guess answers NC-17s question about "any harmful effects of math" o_O
@bolbteppa alternative pov, dont succumb to the FUD of a Big TOE :P
@NovaliumCompany heres a factoid to take (national) pride in :) Bulgaria’s Bitcoin Holdings Surpass Their Gold Reserves trustnodes.com/2019/07/21/…
@bolbteppa old insider term originating with microsoft during dotcom period. Fear Uncertainty Dread
@Slereah I thought you'd scrubbed the deviantart barrel for a fan character :P
He's basically a clock in pajamas
This constraint stuff probably explains why GR can't be unified with QM
Pretty sure in LQG they use this
3:36 PM
@bolbteppa what constraint stuff
Discussion above
@bolbteppa i mean point particles are also diffeomorphism invariant
And rhey quantize fine
As you know QM is just 0+1 dimensional QFT :p
hmm yeah
I should try to get around to writing up the first chapter of that Dirac book on this stuff
4:21 PM
The gravity quantization issue is the usual problems I think rly
All the time slicing and renormalization shenanigans
2 hours later…
5:58 PM
any GRists around?
say, @JohnRennie?
@EmilioPisanty, "a basic assumption on the side of GR, namely, that spacetime is locally Minkowski". This is not an accurate description of GR. GR does not preclude a solution of zero metric $g_{\mu\nu}=0$, which is by no means "locally Minkowski". Locally Minkowskian metric is an accidental fact. — MadMax 1 hour ago
is that true?
that sounds like BS to me, but I'm not a specialist
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