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Qmechanic

02:59

@Slereah Andreas is active on Math and Mathoverflow: mathoverflow.net/users/64141/andreas-cap

Slereah

05:45

Good to know if I have questions for him I guess!

imbAF

09:33

@ACuriousMind I have one question, which emerged as I was continuing with my studies of SR. If, for the rod in it's rest frame I'd calculate the spacetime interval, which as we concluded it's the proper length, well I have a sign problem here. the spacetime interval is $\Delta s^2=c\Delta t -\Delta x$, which for spacelike events $\Delta t=0$ therefore $\Delta s^2<0$ and than this means that proper length (square root of the spacetime interval) is imaginary. What am I confusing here?

ACuriousMind

@imbAF The sign in the spacetime interval is just a convention; both proper time and proper distance have to be defined such that they don't end up imaginary (i.e. by taking the absolute value of the r.h.s. before taking the root)

imbAF

so $l=\sqrt{-\Delta s^2}$ for proper length?

ACuriousMind

in your sign convention, yes

imbAF

oh ok

Thanks

ACuriousMind

fair warning: the sign really is arbitrary, and there is no universally preferred choice: You will see both $\Delta t^2 - \Delta x^2$ and $\Delta x^2 - \Delta t^2$ so don't start despairing if you see different texts where the signs of a lot of stuff are different

imbAF

09:47

Yes, I did notice that exact thing

35

Q: "Reality" of length contraction in SR

I was in argument with someone who claims that length contraction is not "real" but only "apparent", that the measurement of a solid rod in its rest reference frame is the "real length" of the rod and all other measurements are somehow just "artificial" and "apparent". Seemed to me like a bad con...

imbAF

10:36

That's why I asked you. Because the spacetime interval is, the scalar product of the 4 position vector, so it has to be c\Delta t^2..

But I understand

Ryder Rude

10:49

hello

Padé Approximants

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please see this for a better version of Taylor series

when u truncate this series, it still tries to stay close to the original function when the truncated Taylor series may blow up

ACuriousMind

What a silly thing to say. The purpose of Taylor series is usually not to be the "best" approximation to a function. There's a whole zoo of other approximations that are "better" for the variety of use cases one might have.

Ryder Rude

yes. i meant that this is better for certain applications

it is for when the truncated Taylor series blows up but the actual function does not blow up

the truncated taylor series blows up for $e^{-x}$

Ryder Rude

11:09

Les Champs-Elysées | Joe Dassin | Pomplamoose ft. John Schroeder

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This is a very fun song @PM2Ring

bolbteppa

John's answer in that SR question is really great

@RyderRude that looks great

Ryder Rude

yes

bolbteppa

They show up in these lectures

6

Q: Free lecture notes to Carl Bender's Mathematical Physics video lecture course?

I am just watching Carl Bender's Mathematical Physics video lecture course (about asymptotics and its application in physics) http://www.perimeterscholars.org/328.html which is great! Are there any lecture notes for this great course uploaded somewhere? After watching the videos, I'd like to r...

reference-request asymptotics mathematical-physics

Think they show up in this one:

Mathematical Physics 07 - Carl Bender

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Ryder Rude

11:43

@bolbteppa thanks

Ryder Rude

12:36

youtube.com/shorts/jtZqwSXAAGg?si=d8m7E2-Zt4_Jx610 This problem took 270 years to solve

i guess they were looking for a closed form solution. infinite series solutions may have existed

because u can write the area function Area=F(L) and write a Taylor expansion of $F^{-1}(Area)$

Ryder Rude

14:39

in Special Relativity, mass shows up in two equations. the first is the length of the energy momentum four vector $E^2-p^2=m^2$ and the other in Newton's second law: $m\frac{d^2x^{\mu}}{d\tau ^2}= F^{\mu}$

in principle, could these constants be different?

is it an experimental fact that they r equal?

Ryder Rude

14:53

it seems like in principle u could have $p^{\mu}(\tau)$ such that $|p^{\mu}(0)|=m_1$ and $a\frac{dp^{\mu}}{d\tau}=F^{\mu}$. if we choose $a \neq 1$, then the two masses will be different

bolbteppa

15:24

@RyderRude Absolutely not, on varying $S = - \int [mc \sqrt{\dot{x}^2} + F x ] d \tau$ (check signs/constants) you should get your F = ma with the acceleration term having a square root, and from the same action you derive the energy-momentum relation

Does the form of $E^2 - p^2 = m^2$ change form with that extra term in the action, I don't want to check

(e.g. the way $P^{\mu} = (E_{part} + e \phi, \mathbf{p} + e \mathbf{A})$ does)

Ryder Rude

@bolbteppa yes. this is correct. thanks

@bolbteppa if we use the Maxwell Lagrangian plus the interaction term in place of $Fx$ in the Lagrangian, we would get E=$\sqrt{p^2+m^2}+ electromagnetic energy$

so these two constants are the same

Ryder Rude

15:52

we should note that both $|p^{\mu}|=m$ and $m\frac{dp^{\mu}}{d\tau}=F^{\mu}$ are written for the kinetic momentum $p_{kinetic}^{\mu}$. for the canonical momentum $p_{canonical}^{\mu}$, the expression of energy need not be $\sqrt{p^2+m^2}$

for the kinetic momentum $E=\sqrt{p^2+m^2}+fieldenergy+interaction energy$ should be true

for the canonical momentum, it's more like $E=\sqrt{(p-A)^2+m^2}+q\phi+fieldenergy$

PM 2Ring

16:29

@RyderRude Here's a humorous one in French from Pomplamousse, from a couple of years ago. Assedic youtu.be/klIwRQ8kxvU?si=ciX1Ledx6KujPSQk That song's a cover, but they're currently writing songs for a new French album, eg Tu peux pas savoir youtu.be/RASImzptZ3c?si=1uCnmIPk3GsN897J IMHO, it's pretty cool that their French songs are so popular. They originally just did an EP of French songs, but it went so well that they expanded it to an album.

Here's a fun song from the force of Nature that is Lawrence. Gracie has the remarkable ability to be simultaneously silly and intelligent. :) I guess I should give a language warning...

Lawrence - i’m confident that i’m insecure (acoustic-ish)

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PM 2Ring

16:50

@RyderRude Right. Truncated Taylor series often blow up. You can often get a much better approximation using a Chebyshev polynomial. Padé approximants usually give better approximations (if the denominator is well-behaved) and generally converge over a wider range than the truncated Taylor series they're derived from. Unfortunately, you can't easily create a Padé-like thing from a Chebyshev poly.

A Chebyshev poly is close to a minimax poly, which minimizes the maximum error. You can get the true minimax using the Remez algorithm en.wikipedia.org/wiki/Remez_algorithm but sometimes that algorithm just won't work if there are vertical asymptotes too close to the region of interest.

PM 2Ring

17:15

A simple way to compute Padé coefficients is to use Wynn's Epsilon Method. Oddly, it isn't given on Wikipedia or Mathworld. And the notation in Wynn's original paper is a bit messy. I guess it's designed for pen & paper calculation, rather than implementing in code.

Actually, I should clarify that. ;)

Wynn's Epsilon Method is a sequence accelerator. But if your coding language lets you do symbolic calculations (not just numeric ones) you can easilt use Wynn's algo to convert a Taylor sequence in symbolic form into a Padé.

Here's a demo in Sage of Wynn' epsilon accelerating the classic Leibniz series for pi, which is just the Taylor series of $4\arctan(1)$. That converges very slowly. But the accelerator gets over 60 bits of ptecision in 25 terms.

sagecell.sagemath.org/…

PM 2Ring

18:04

Here's a (rather terse) demo of using Wynn's to make Padés for arctan. The plot shows the error. I restrict the domain to $\pm\tan(\pi/8)$ to improve the convergence. It's easy to reduce any tan into that interval.

sagecell.sagemath.org/…

Oops. Sorry about the typos. They can be hard to notice in this tiny window on a phone...

imbAF

18:24

Hello, I want to know if the following expression is accurate:
$g^{\mu\nu}=\frac{\partial x_\mu}{\partial x_\nu}$ ?
I am considering a covariant four-vector

Slereah

19:11

Accurate to what

bolbteppa

$\frac{\partial x_\mu}{\partial x_\nu}= \partial^{\nu} x_{\mu} = \delta^{\nu}_{\mu}$

imbAF

19:50

@Slereah the left side giving the right side of the eq.

Slereah

Is it supposed to be a metric tensor?

imbAF

Now that I think about it

it must be the identity matrix

Slereah

I mean yes

That's a Jacobian and that's the Jacobian of the identity transformation

which is the identity matrix

imbAF

who is the jacobian?

Slereah

This is the Jacobian

The Jacobian is the derivatives given by the change between two coordinate systems

imbAF

19:55

would Physics change totally, if we were to find a magnetic monopole ?

bolbteppa

You're probably thinking of $\partial_{\mu} x_{\nu} = \partial_{\mu} \eta_{\nu \rho} x^{\rho} = \eta_{\nu \rho} \partial_{\mu} x^{\rho} = \eta_{\nu \rho} \delta_{\mu}^{\rho} = \eta_{\mu \nu}$

Slereah

at first I thought it was some attempt at the first fundamental form, but that don't work either

bolbteppa

That's in Minkowski space, in curved space we have $\partial_{\mu} x_{\nu} = \partial_{\mu} [ g_{\nu \rho}(x) x^{\rho}] \neq g_{\nu \rho}(x) \partial_{\mu} x^{\rho}$

imbAF

I am at the minkowski space

thanks for the equation

bolbteppa

A magnetic monopole would basically mean adding magnetic terms to all of Maxwell's equations and complicating absolutely everything

Slereah

20:00

If there are any I don't think we'd have that much complication since that means they are pretty rare

and the number of monopole is invariant iirc

imbAF

how can you conclude that the nr. of monopoles would be invariant?

Slereah

Depends on the theory I guess, but usually it stems from the topology of the bundle involved in EM

imbAF

I have a question about the four vector notation. If we have $A_\mu^{\nu}$ and the subscript $\mu$ is in front or comes before the superscript $\nu$, would it represent a different thing compared to when the superscript $\nu$ would be before the subscript $/mu$. I am describing it, cuz I can't do it on latex, to add blank space

Slereah

You can, ie ${A^\mu}_\nu$ and ${A_\nu}^\mu$

imbAF

ah yeah

do they represent different things?

Slereah

20:07

It is technically different since one will be a tensor $TM \otimes T^*M$ while the other will be $T^* M \otimes TM$

But I think they're equivalent IIRC?

So usually people don't care that much about ordering in that aspect

Tensor product

In mathematics, the tensor product V ⊗ W {\displaystyle V\otimes W} of two vector spaces V and W (over the same field) is a vector space to which is associated a bilinear map V × W → V ⊗ W {\displaystyle V\times W\to V\otimes W} that maps a pair ( v , w ) , v ∈ V , w ∈ W {\displaystyle (v,w),\ v\in V,w\in...

imbAF

But hold on

Can you perform this: $B^\nu {A^\mu}_\nu$ ? Shouldn't the contracted indices be one after the other i.e $B^\nu{A_\nu}^\mu$

or it doesn't matter?

Slereah

It doesn't matter no

imbAF

ahaa, I thought it did

Slereah

It does matter for the non-commutative case, IIRC, though

Like if you're using spinors with indices, you have to watch out

imbAF

Haven't reached the spinors yet, but I know that they are super weird

I mean, do they have a mathematical entity that represents them?

They aren't vectors right?

Slereah

20:13

They are vectors in the broad sense of the word

imbAF

how so?

they require a 720 degree rotation

Slereah

You know, you can add them together and multiply them by scalars

imbAF

to go in their initial place?

ACuriousMind

21:07

Spinors are, mathematically, vectors. They just transform in a non-standard (projective) representation of the rotation group

the 720° thing is misleading to some degree

imbAF

But they have nothing to do with spin as a concept ?

ACuriousMind

oh, they're very closely related to the notion of spin - they're spin-1/2 objects

imbAF

ahaa

ACuriousMind

but instead of making mystical statements about 720° rotations, I prefer to think of them in terms of the proper (projective) representation theory, see physics.stackexchange.com/a/167470/50583, physics.stackexchange.com/q/203944/50583

imbAF

I will have to admit that things like symmetry groups, representation theory are foreign to me. Unfortunately we haven't touched upon these in my Uni

ACuriousMind

21:15

I mean that's fine (for introductory pedagogy), but I personally don't think you can really understand these things from a mathematical perspective without those concepts

without them, you just have to accept spinors are kinda weird :P

imbAF

yeah, but in like 2 3 days I will revise a lecture on Spinors, which will probably suck the life out of me for the upcoming week, as I try to understand it

ACuriousMind

y'know, I think that's one of these things that's fine not to understand fully the first time around?

imbAF

But you use it, if I am not mistaken in the Dirac equation right?

If I remember correctly it has 4 components

which are matrices?

ACuriousMind

sounds like you should wait for the lecture :P

imbAF

We did took the lecture a while ago, I have the notes, and it's like 6 pages max. I don't think they dwelled deep enough for a fundamental understanding

ACuriousMind

21:25

like, the "matrix" bit can be right or very confused depending on what you mean by it :P

imbAF

well, it looks like a vector with 4 components, but each component is not a scalar but a matrix, a matrix were /gamma 's are used. I believe

I don't have a clear memory of it

ACuriousMind

yeah, okay, you're confused

there's $\gamma$-matrices and there's 4-component spinors

the $\gamma$-matrices act on the spinors

it's not that they are the spinors or their components

imbAF

I see

Yeah, I don't have a clear memory of it

One thing I remember though

The Dirac eq. emerged because , said simply, the Klein Gordan eq. wasn't positive definite, right?

maybe other reasons too but

One of those was the fact that it wasn't positive definite?

ACuriousMind

you really need to be more precise with the language

an equation can't be "positive-definite"

imbAF

the

charge density

ACuriousMind

21:30

I know what you probably mean but that's not how one says it

imbAF

that can be derivated from it

Right?

ACuriousMind

why would that be a problem for a charge density?

charges can be negative

(also: what do you mean by "derivated"?)

imbAF

probability density *

xD

Klein-Gordon equation does not have a positive definate probability density

ACuriousMind

you mean the solutions to the KG equation cannot be interpreted as probability densities because they are not guaranteed to always be non-negative.

imbAF

And by derived I mean, starting with the KG Equation and having an expression that can be considered as a continuity equation, and the expression which would be the probability density, can be negative

But anyway, since the KG equation has this problem, why in many cases it is said that it's the solution to massive particles without spin?

Is the probability density affected as to whether the particles in consideration have or have not spin?

ACuriousMind

21:42

@imbAF [citation needed]

anyone careful about their statements should claim that the KG equation needs to be interpreted as a field equation in QFT in order to fully cure all the problem with it

imbAF

The solutions of the Klein-Gordon equation describe massive, spin-0 particles.

ACuriousMind

@imbAF no they don't

imbAF

In wikipedia

Klein–Gordon equation

The Klein–Gordon equation (Klein–Fock–Gordon equation or sometimes Klein–Gordon–Fock equation) is a relativistic wave equation, related to the Schrödinger equation. It is second-order in space and time and manifestly Lorentz-covariant. It is a quantized version of the relativistic energy–momentum relation E 2 = ( p c ) 2 + ( m...

ACuriousMind

or, rather, rQM can get away with pretending they do but ultimately it is not consistent

imbAF

Its solutions include a quantum scalar or pseudoscalar field, a field whose quanta are spinless particles.

ACuriousMind

21:44

note that "a field whose quanta are spinless particles" is not the same as claiming the solution to the KG equation is a probability amplitude for such a particle

this claim is correct: in order to fully cure the pathologies, we need to do QFT, not QM

but, uh, don't worry about this now

if you're only barely learning about spinors you're not yet doing QFT

imbAF

Most definitely

I am not as of right now

I believe that a mathematician is a better physicist than an actual physicist

If you know the math, all you have to do is apply it

ACuriousMind

it doesn't really work like that

a lot of derivations that look very simple in the typical "physicist" fashion where we use physical intuition and handwavy math are much more complicated to do correctly within rigorous mathematics

and we don't even have rigorous QFT at all (for practical applications)

every physicist has to find the correct amount of rigor vs. handwaving that works for them

imbAF

Perhaps, but if we can manage with handwavy, we certainly can do it with rigorous math

So it looks to me

ACuriousMind

no we can't

sometimes the handwaving is just wrong, sometimes it's correct but it took decades for something to figure out how to do it rigorously, sometimes it's still unknown what the heck is going on

it's an open problem how to do QFT fully rigorously yet QFT has been an amazingly successful theory; if we had waited for someone to come up with proper mathematics for it we'd still be waiting

imbAF

I have yet to learn QFT, but I find it hard that we are operating in a way that doesn't comply with math

Sanjana

22:10

I was wondering, which equation counts as a gauge choice...
I can set $\partial_\mu A^\mu=0 $ or $\nabla \cdot A=0$ or $A_0=0$
but certainly I can't choose $A_\mu=0$ which is not a gauge choice (unless it's pure gauge). Also it will kill all the dof at once...

So what kind of equations are gauge choices and what are not?...I guess it has to be a "scalar equation" but any other restriction other than that? Maybe this isn't the best way to see it...idk

ACuriousMind

It has to be some equation $F(A) = 0$ so that, for any given $A_\mu$, you can solve $F(A') = 0$ for $A'_\mu = A_\mu + \partial_\mu\alpha$ for $\alpha$

i.e. you have to show that it is always possible to fulfill the gauge condition by performing a gauge transformation

Sanjana

What's the problem with $A_\mu=0$? It holds for constant $\alpha$.
If you say that it holds for a specific function $\alpha$ which is the constant function here then doesn't Lorenz Gauge also hold only when functions satisfying the wave equation are chosen?

ACuriousMind

@Sanjana no it doesn't

you have to solve $A_\mu + \partial_\mu \alpha = 0$ for arbitrary $A$

Physics Meta

0

Q: Mainstream Physics: Unconfirmed Particles

Asking regarding potential questions I might pose soon. While asking a well formulated question about tachyons in special relativity is allowed (as I did last week), I’m wondering how far that goes. I’ll be specific: I want to ask about more obscure hypotheticals. Specifically graviphotons and gr...

support scope questions non-mainstream-physics

ACuriousMind

$\alpha = 0$ (or constant) does not solve that :P

Sanjana

22:18

Oh I misread your first reply...umm is this supposed to be trivial...I mean I have not seen it written anywhere explicitly?

ACuriousMind

Contrast this to the condition $A_0 = 0$ ("temporal gauge"): We get $A_0 + \partial_0\alpha = 0$, which yields with $\partial_0\alpha = A_0$ a differential equation that has solutions for any $A$ under mild niceness assumptions

Sanjana

oh wow nice...makes sense now.thanks

ACuriousMind

@Sanjana it's "trivial" if you state carefully what a gauge choice even is but you're right that many texts are bad at discussing this :P

The core of gauge theory is that $A_\mu$ and $A_\mu + \partial_\mu \alpha$ are, for physical purposes, equivalent for arbitrary $\alpha$. A "gauge choice" is some constraint $F(A) = 0$ we want to impose on our $A_\mu$ using this freedom. So what we're saying when we impose $F(A) = 0$ is "if we start with some $A$ that does not fulfill this equation, solve the equation $F(A + \mathrm{d}\alpha) = 0$ to transform it into a physically equivalent $A$ that does fulfill it"

Charlie

23:19

@imbAF QFT doesn't "obviously" involve non-rigorous steps. Unless you're planning on learning it rigorously from the start - which you won't - there are no steps where you have to do something clearly illogical and "look the other way" in order to get things to work

QFT just sort of looks like a messier version of quantum mechanics, not a visibly incomprehensible offense to mathematics

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