The h Bar

Slereah

7:40 AM

morgen

John Rennie

@EmilioPisanty it looks as if there might be hope after all. Bojo the clown just got a bloody nose in parliament!

Slereah

$$S[\Phi(\vec{x}(t))] = \int_{\alpha(t_a)}^{\alpha(t_b)} (\frac{ds}{dt} \frac{d}{dt}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s)) \cdot (\frac{ds}{dt} \frac{d}{ds}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s))) \frac{dt}{ds} ds$$

plz kill me

I am in a hell of my own making

Now to show that this contains the Galilean group

gonna be a bit tough

With just $f$ or just $\alpha$ it's alright, but both at the same times

Who knows what kind of weird transformations could occur that I have to rule out

All the FORBIDDEN SYMMETRIES

Forbidden by the Coleman-Mandula theorem

Hm

I have to deal with something like $$\int \left[ f(x) g(x) + h(x) \right] dx = \int f(x) dx$$

If this is true for all $f$, does this imply $h(x) = 0$?

Not quite the fundamental lemma of the calculus of variation

@ACuriousMind @bolbteppa halp

I'm 99% sure this is true but as usual

Never trust an obvious statement

Slereah

8:35 AM

Oh wait

Obviously not for $h$

Can just be a function that integrate to $0$ on that range

BUT IS IT TRUE IF THE EQUALITY IS TRUE FOR EVERY DOMAIN OF INTEGRATION

bolbteppa

9:25 AM

Can be rewritten as $\int f(x)[g(x) - 1]dx = - \int h(x) dx$

Slereah

Does this help

bolbteppa

Fundamental lemma of calculus of variations

In mathematics, specifically in the calculus of variations, a variation δf of a function f can be concentrated on an arbitrarily small interval, but not a single point. Accordingly, the necessary condition of extremum (functional derivative equal zero) appears in a weak formulation (variational form) integrated with an arbitrary function δf. The fundamental lemma of the calculus of variations is typically used to transform this weak formulation into the strong formulation (differential equation), free of the integration with arbitrary function. The proof usually exploits the possibility to choose...

Version for 2 functions

Maybe you can write it in this way

Slereah

Second part doesn't depend on $f$ tho

bolbteppa

$\int \{ f(x)[g(x) - 1] + h(x)g'(x) \} dx = 0$

hmm

Slereah

I need to work out the full integral of this rly

it is an unpleasant one but it shouldn't be too hard

Emilio Pisanty

9:34 AM

@JohnRennie counting chickens?

I mean, it's good and all

But not enough on its own, yet

John Rennie

@EmilioPisanty yes, it could still go in lots of different ways. BoJo wants to call a general election, but that needs to be agreed by parliament and the opposition have said they'll vote against it.

Plus as I understand it the motion was to ask for another three month extension in the event of no deal being approved, but the EU may not agree to another extension on the grounds nothing will change in three months.

But it's still good to see that mendacious scumbag get a kicking :-)

Slereah

9:50 AM

BoJo's bizarre adventures

John Rennie

Bizarre is exactly the word I'd use. Possibly also incredible in the literal meaning of the word.

Slereah

If I'm not fucking up, the final transformed action is something along the lines of $$S[\Phi(x)] = \int_{\Phi(t_a)}^{\Phi(t_b)} \dot{\alpha}(\alpha^{-1}(s)) \left[ (f'(x, s))^2 \dot{x}^2 + 2 \dot{x} f'(x, s) \dot{f}(x, s) + \dot{f}^2(x,s) \right] ds$$

And the end result should be $$\int_{\Phi(t_a)}^{\Phi(t_b)} \dot{\alpha}(\alpha^{-1}(s)) \left[ (f'(x, s))^2 \dot{x}^2 + 2 \dot{x} f'(x, s) \dot{f}(x, s) + \dot{f}^2(x,s) \right] ds = \int_{\Phi(t_a)}^{\Phi(t_b)} \dot{x}^2 ds $$

Obviously this works all fine for $f'$ a Galilean transformation and $\alpha$ a time translation

But showing that these are the unique transformations seems tricky

Wait is it even correct for a Galilean transformation

\begin{eqnarray}
f(x(s), s) &=& \pm x(s) + vs + a\\
\dot{f}(x(s), s) &=& v\\
f'(x(s), s) &=& \pm 1
\end{eqnarray}

Hm

I guess the terms don't cancel out but they are total derivatives?

and therefore bla bla bla

Emilio Pisanty

10:52 AM

@JohnRennie so the opposition's plan is a vote of no confidence but only if it doesn't lead to an election?

That's bonkers

@JohnRennie it's three months? Yeah, good luck with that =/

John Rennie

@EmilioPisanty It's because the fear is that BoJo will drag the election out to leave not enough time to sort out what to do before the Oct 31st deadline.

Emilio Pisanty

It's still not enough time

John Rennie

An election could easily mean Parliament was dissolved for a month.

Emilio Pisanty

And the result could well be a strong mandate for a second referendum

@Slereah I think I like de Pfeffel better than BoJo

Student404Mus

I wonder where on earth the negative sign in the third step comes from?!. The anticommnutation relations implies a positive sign since we have pair anticommutations

$\begin{aligned} C \overline{\psi} \psi C &=\left(-i \gamma^{0} \gamma^{2} \psi\right)^{T}\left(-i \overline{\psi} \gamma^{0} \gamma^{2}\right)^{T}=-\gamma_{a b}^{0} \gamma_{b c}^{2} \psi_{c} \overline{\psi}_{d} \gamma_{d e}^{0} \gamma_{e a}^{2} \\ &=+\overline{\psi}_{d} \gamma_{d e}^{0} \gamma_{e a}^{2} \gamma_{a b}^{0} \gamma_{b c}^{2} \psi_{c}=-\overline{\psi} \gamma^{2} \gamma^{0} \gamma^{0} \gamma^{2} \psi \\ &=+\overline{\psi} \psi \end{aligned}$

Emilio Pisanty

10:58 AM

Brexit: raising the bar on knuckle whiteness since 2016

John Rennie

@EmilioPisanty the thing is that there is simply no acceptable deal. The land border between Northern Ireland and Ireland makes that impossible.

The EU will not consider any deal that doesn't allow free traffic across the Irish border. End of story.

And that's unacceptable to the Ulster Unionists because then there would have to be a traffic control between Northern Ireland and the mainland.

In effect it would be a step (a large step!) towards the reunification of Ireland.

Ultradark

11:43 AM

Is space times time info entropy?

PM 2Ring

12:30 PM

Is Nick Bostrom's simulation hypothesis on topic, or do we VTC as non-mainstream? physics.stackexchange.com/questions/500176/…

JMac

I'm tempted to VTC that one as unclear or something. It's not clear to me what types of differences they expect

Slereah

Whether mainstream or not doesn't matter much

it's not a physical hypothesis

more of a philosophy stack exchange question

PM 2Ring

@JMac Sure, but there's a possibility of the OP clarifying that. Whereas if it is off-topic, then no amount of clarification or other editing will help.

@Slereah I tend to agree. OTOH, you could say the same about most questions asking about the various interpretations of QM. ;)

Emilio Pisanty

0

Q: Robust ways to find zeros of the Tricomi confluent hypergeometric function as a function of its parameters

I'm solving a quantum mechanical problem, and the quantization condition requires me to solve the equation $$ U\left(\frac12(\ell+1-E), \ell+1, r^2\right) = 0, $$ where $U(a,b,z)$ is the confluent hypergeometric function of the second kind, $r>0$ is a fixed positive real number, and $\ell\geq0$ i...

@Semiclassical any insights appreciated =)

JMac

@PM2Ring Right, but if I can't even really understand if question relates to physics, it's hard to actually say it is off topic, and not just a poorly worded physics question.

Emilio Pisanty

12:42 PM

The last time I was here I had some funky Bessel function combinations and this package saved me. I'm afraid that with this one I won't be quite as lucky.

Student404Mus

This is the step I missed,

$\begin{aligned} C \overline{\psi} \psi C &=\left(-i \gamma^{0} \gamma^{2} \psi\right)^{T}_{ab}\left(-i \overline{\psi} \gamma^{0} \gamma^{2}\right)^{T}_{bc} \\ &= \left(-i \gamma^{0} \gamma^{2} \psi\right)_{ba}\left(-i \overline{\psi} \gamma^{0} \gamma^{2}\right)_{cb} \end{aligned}$

and the fact that only gamma's with "same index" anticommutes

PM 2Ring

@EmilioPisanty This is way out of my league, but I suspect that if Mathematica can't handle it, things may be grim. Does it help if you use more precision? In Python, I use mpmath for arbitrary precision maths. It's got some good root solvers, and it supplies a bunch of hypergeometric functions. mpmath.org/doc/current/functions/hypergeometric.html

Slereah

@PM2Ring Get rid of them too

I don't mind

At least interpretations of QM usually come with some physical theory behind it

This is has no possible prediction

Transcript for